Answer:
The rate clock is about
F = 8 GHz
Explanation:
f₁ = 4 G Hz , t₁ = 10 s , t₂ = 6s , f₂ = 1.2 f₁
Can organize to find the rate clock the designer build to the target so
X / 4 Ghz = 10 s , 1.2 X / Y = 6 s
X * Y = 10 s ⇒ F = 10 s
1.2 * 4 G Hz = 6 s
F = 10 * ( 1.2 * 4 G Hz ) / 6
F = 10 * ( 1.2 * 4 x 10 ⁹ Hz ) / 6
F = 8 x 10 ⁹ Hz
F = 8 GHz
Final answer:
To run a program in 6 seconds on Machine B, which needs 1.2 times as many clock cycles as Machine A, the target clock rate should be 8 GHz.
Explanation:
The student asked for the target clock rate needed for computer B to run a program in 6 seconds, given that computer A runs it in 10 seconds with a 4 GHz clock and that machine B needs 1.2 times as many clock cycles as machine A for the same program. To solve this, we know that time (T) is equal to the number of cycles (N) divided by the clock rate (C), or T = N / C. For machine A, TA = NA / CA and for machine B, TB = (1.2 * NA) / CB. If machine A completes the program in 10 seconds, the number of cycles it uses is CA * TA, which is 4 GHz * 10 s, yielding 40 billion cycles.
Machine B needs to run these 40 billion cycles in 6 seconds. Also, machine B requires 1.2 times the number of cycles of machine A; thus we have (1.2 * 40 billion) / 6 s to find CB, the clock rate for machine B. This simplifies to 8 GHz. Thus, for machine B to run the program in 6 seconds, the target clock rate should be 8 GHz.
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to:
Complete Question:
A machinist turns the power on to a grinding wheel, which is at rest at time t = 0.00 s. The wheel accelerates uniformly for 10 s and reaches the operating angular velocity of 25 rad/s. The wheel is run at that angular velocity for 37 s and then power is shut off. The wheel decelerates uniformly at 1.5 rads/s2 until the wheel stops. In this situation, the time interval of angular deceleration (slowing down) is closest to
Answer:
t= 16.7 sec.
Explanation:
As we are told that the wheel is accelerating uniformly, we can apply the definition of angular acceleration to its value:
γ = (ωf -ω₀) / t
If the wheel was at rest at t-= 0.00 s, the angular acceleration is given by the following equation:
γ = ωf / t = 25 rad/sec / 10 sec = 2.5 rad/sec².
When the power is shut off, as the deceleration is uniform, we can apply the same equation as above, with ωf = 0, and ω₀ = 25 rad/sec, and γ = -1.5 rad/sec, as follows:
γ= (ωf-ω₀) /Δt⇒Δt = (0-25 rad/sec) / (-1.5 rad/sec²) = 16.7 sec
Which of the following statements about revitalization movements is incorrect ?
a. The purpose of revitalization movements is to reform society.
b. Revitalization movements always fail because they require too much change to be tolerated.
c. All known major religions, including Judaism, Christianity, and Islam, began as revitalization movements
d. Revitalization movements may be completely unrealistic.
e. Revitalization movements may be adaptive and give rise to long-lasting religions.
Give an example of a revitalization movement.
Answer:
Correct answer is (b) Revitalization movements always fail because they require too much change to be tolerated.
Example of a revitalization movement is the Ghost Dance that swept through western Native American cultures from 1870-1890.
Explanation:
Another example of Revitalization movements are mostly associated with religion. They often occur in disorganized societies due to warfare, revolution but not necessarily an animistic societies.
The incorrect statement is that 'Revitalization movements always fail because they require too much change to be tolerated.' Revitalization movements have led to significant societal changes and have even given rise to major religions like Christianity, Judaism, and Islam. The correct option is b.
Explanation:The statement 'Revitalization movements always fail because they require too much change to be tolerated' is incorrect. Revitalization movements are deliberate, organized, conscious efforts by members of a society to construct a more satisfying culture. While it's true that they can face intense resistance because they often push for significant changes, it's not true that they always fail. Many have had long-lasting impacts. For example, the Protestant Reformation was a revitalization movement that sought to reform the Roman Catholic Church and led to the creation of Protestant churches. It instigated enormous change in society, and it certainly didn't fail.
All major religions, including Judaism, Christianity, and Islam, did indeed begin as revitalization movements. These movements were adapted to the changing needs of society, and they have resulted in lasting religions that are still prevalent today. Hence revitalization movements can have profound and lasting impacts on society, institutions and cultures, far from always being impractical or bound to fail.
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A horizontal conductor in a power line carries a current of 5500 A from south to north. Earth's magnetic field (60.0 μT) is directed toward the north and is inclined downward at 67.0° to the horizontal. Find the magnitude and direction of the magnetic force on 160 m of the conductor due to Earth's field.
Answer:
[tex]|\vec{F}| =48.60\ N[/tex]
Explanation:
given,
Current in the power line = I = 5500 A
earth's magnetic field = 60.0 μT
inclination downward = 67°
Length = 160 m
magnetic force = ?
[tex]\vec{F} = I (\vec{L}\times \vec{B})[/tex]
[tex]|\vec{F}| = I |(\vec{L}\times \vec{B})|[/tex]
[tex]|\vec{F}| =I LB sin \theta [/tex]
[tex]|\vec{F}| = 5500 \times 160 \times 60 \times 10^{-6}\times sin 67^0[/tex]
[tex]|\vec{F}| =48.60\ N[/tex]
According to the right hand rule the direction of the force is perpendicular to the plane of the length and the magnetic field so, it is to west.
Which of the following sets of driver's actions are multi-task performances necessary for the safe vehicle operation?a. Signaling; Changing Gears ;Checking mirrorsb. Signaling; Brakes ;Checking mirrorsc. Signaling; Changing Gears ; Door opening
The sets of driver's actions “Signalling; Changing Gears; Checking mirrors” are multi-task performances necessary for the safe vehicle operation.
Answer: Option A
Explanation:
Signaling is what drivers plan to do is important for security because other drivers can only know if he tells them. Here are some general rules about Signalling or turns.
Signal early: Signal well before turning back is the best way to prevent others from trying to get through you. Continuous signal: To turn safely, the driver must have both hands on the steering wheel. He should not stop the signal until he completed the turn. Cancel the alarm: The driver should remember to turn off signaling after completing his turn (if you have no attenuation signals).The correct gear change is important. If he can't put the vehicle in the right gear while driving, he has less control. And also, he must check the exterior mirrors on both sides regularly. He needs to check his vehicle mirrors to make sure nobody stands or walks past him.
The nicotine content in a single cigarette of a particular brand has a distribution with mean 0.5 mg and standard deviation 0.1 mg. If 100 of these cigarettes are analyzed, what is the probability that the resulting sample mean nicotine content will be less than 0.49? (Round your answers to four decimal places.)
Answer:
0.1587
Explanation:
Given data
μ = 0.5 mg
standard deviation σ =0.1 mg
n=100
we know that
[tex]P(\overline X<0.49)[/tex]
[tex]Z= (\frac{\overline X-\mu}{\sigma/\sqrt{n} })[/tex]
putting the values we get
[tex]Z= (\frac{0.49-0.5}{0.1/\sqrt{100} })[/tex]
Z=-1
Area under the curve for z =-1 is 0.1587 (from z score table)
P(X<0.49) = 0.1587
P(X<0.37) = 0.0013
Some amount of ideal gas with internal energy U and initial temperature 1000C was compressed to half of volume meanwhile absolute pressure inside of a container increased twice. We can say that internal energy of this gas after compression in terms of U is (20.2, 20.1, 19.4, 19.5) Group of answer choices
Answer:
U. With no variation.
Explanation:
Note- since temperature remains constant when pressure becomes twice and volume becomes half, and internal energy of ideal gas is function of only temperature so it remains constant. The internal energy is independent of the variables stated in the exercise.
Which of the following is the best explanation of why it is difficult to discriminate the color of an object at night?
A. Cones come in three types, each sensitive to different wavelengths of light.
B. The foveae are densely packed with cones.
C. At night, the amount of light entering the eye is insufficient to stimulate the cone cells but is sufficient to stimulate the rod cells.
D. As many as 100 cones may converge on one ganglion cell.
Answer:C
Explanation:
There are two types of cells in the eyes rod and cone cells. Rod cells Provide vision during the night or dim light also called of scotopic vision whereas cone cells provide vision during day time or at bright light also called photopic vision.
Rod cells do not support the color vision that is why it is difficult to differentiate between colors.
Also, the amount of light entering the eyes is low as result cone cells are unable to stimulate.
It is hard to see color at night because cones, which are responsible for color vision, need more light to be stimulated than rods, which allow us to see in low light but only in grayscale.
Explanation:The best explanation for why it is difficult to discriminate the color of an object at night is option C. At night, the amount of light entering the eye is insufficient to stimulate the cone cells but is sufficient to stimulate the rod cells. The rods are highly sensitive to light, allowing us to see in low light conditions but do not provide color information. Cones require brighter light to function and are responsible for our color vision. Since cones do not react to low-intensity light, our vision at night is predominantly in shades of gray, serviced by the activity of rods.
Two cylinders with the same mass density rhoC = 713 kg / m3 are floating in a container of water (with mass density rhoW = 1025 kg / m3). Cylinder #1 has a length of L1 = 20 cm and radius r1 = 5 cm. Cylinder #2 has a length of L2 = 10 cm and radius r2 = 10 cm. If h1 and h2 are the heights that these cylinders stick out above the water, what is the ratio of the height of Cylinder #2 above the water to the height of Cylinder #1 above the water (h2 / h1)? h2 / h1 =
Answer:
Explanation:
Given
density of cylinder is [tex]\rho _c=713 kg/m^3[/tex]
Length of first cylinder is [tex]L_1=20 cm[/tex]
radius [tex]r_1=5 cm[/tex]
For cylinder 2 [tex]L_2=10 cm [/tex]
[tex]r_2=10 cm[/tex]
[tex]h_1[/tex] and [tex]h_2[/tex] are the height above water
E
as object is floating so its weight must be balanced with buoyant force
[tex]\rho _c\frac{\pi }{4}d_1^2L_1g=\rho _w\frac{\pi }{4}d_1^2(L_1-h_1)g----1[/tex]
For 2nd cylinder
[tex]\rho _c\frac{\pi }{4}d_2^2L_2g=\rho _w\frac{\pi }{4}d_2^2(L_2-h_2)g----2[/tex]
Dividing 1 and 2 we get
[tex]\frac{L_1}{L_2}=\frac{L_1-h_1}{L_2-h_2}[/tex]
[tex]\frac{20}{10}=\frac{20-h_1}{10-h_2}[/tex]
[tex]2h_2=h_1[/tex]
[tex]\\\Rightarrow\frac{h_2}{h_1}=\frac{1}{2}[/tex]
The plane of a conducting loop is oriented parallel to the x-y plane. A magnetic field is directed in the -z direction.
Which one of the following actions will not change the magnetic flux through the loop?
A) Decrease the area of the loop.
B) Decrease the strength of the magnetic field.
C) Increase the strength of the magnetic field.
D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop.
E) Rotate the loop about an axis that is directed in the y direction and that passes through the center of the loop.
Answer:
D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop
Explanation:
The magnetic flux is defined as the total magnetic field times the area normal to the magnetic field lines.
Mathematically:
[tex]\phi_B=\vec{B}.\vec{A}[/tex]
where:
[tex]\vec{A}=[/tex] area vector directed normal to the surface
[tex]\vec{B}=[/tex] magnetic field vector
Now as the area of the loop changes there will be a change in magnetic flux.Change in the magnetic field strength will also change the flux accordingly.Since the loop lies in the x-y plane we will get a different area of normal projection on the plane whenever the inclination of the loop changes in xy-plane.Since the area of the loop all remains in the magnetic field while it rotates about the z axis to its center hence this will not affect area subjected to the magnetic field.Final answer:
Rotating the conducting loop about an axis in the z direction will not change the magnetic flux through the loop because the angle between the magnetic field and the normal to the plane of the loop remains constant. so the correct option is D
Explanation:
The question pertains to the change in magnetic flux through a conducting loop when subjected to different actions. According to Faraday's Law of Electromagnetic Induction, the magnetic flux through a loop is given by the product of the magnetic field strength, the area of the loop, and the cosine of the angle between the magnetic field direction and the normal to the loop. In options A, B, and C, changes to the area or the strength of the magnetic field alter the magnetic flux since these factors directly affect the calculation of flux. Option D involves rotating the loop about an axis in the z direction; such a rotation does not change the angle between the magnetic field and the normal to the loop's plane, therefore would not change the flux. In contrast, option E, where the loop is rotated about an axis in the y direction, changes this angle and thus the flux.
Therefore, the action that will not change the magnetic flux through the loop is:
D) Rotate the loop about an axis that is directed in the z direction and that passes through the center of the loop.
A ladder of length L is placed against a smooth wall such that it forms a angle θ with the wall and the friction force between the ladder and the floor is fs .
a. On your paper, derive an expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.
b. Try out your derivation. If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?
Answer:
a. [tex]m=\frac{2f_{s}tan\theta}{g}[/tex]
b. m=5.99kg
Explanation:
a.
In order to solve this problem, we can start by drawing a diagram of the situation. Drawing a diagram is really important since it will help use understand the problem better and analyze it as well. (See attached picture).
In the diagram we can see the forces that are acting on the ladder. We will assume the ladder is static (this is it doesn't have any movement) and analyze the respective forces in the x-direction and the forces in the y-direction, as well as the moments about point B.
So we start with the sum of forces about y, so we get:
[tex]\sum F_{y}=0[/tex]
N-W=0
N=W
N=mg
Next we can do the sum of forces about x, so we get:
[tex]\sum F_{x}=0[/tex]
which yields:
[tex]f_{s}-F_{W}=0[/tex]
so:
[tex]f_{s}=F_{W}[/tex]
Next the torque about point B, so we get:
[tex]\sum M_{B}=0[/tex]
so:
[tex]f_{s} L sin\theta - NLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]
From the sum of forces in the y-direction we know that N=mg (this is because the wall makes no friction over the ladder) so we can directly substitute that into our equation, so we get:
[tex]f_{s} L sin\theta - WLcos\theta + \frac{WL}{2}cos\theta = 0[/tex]
We can now combine like terms, so we get:
[tex]f_{s} L sin\theta -\frac{WL}{2}cos\theta = 0[/tex]
we know that W=mg, so we can substitute that into our equation, so we get:
[tex]f_{s} L sin\theta -\frac{mgL}{2}cos\theta = 0[/tex]
which can now be solved for the mass m:
[tex]\frac{mgL}{2}cos\theta = f_{s} L sin\theta[/tex]
If we divided both sides of the equation into L, we can see that the L's get cancelled, so our equation simplifies to:
[tex]\frac{mg}{2}cos\theta = f_{s}sin\theta[/tex]
we can now divide both sides of the equation into g so we get:
[tex]\frac{m}{2}cos\theta = \frac{f_{s}sin\theta}{g}[/tex]
next we can divide both sides of the equation into cos θ so we get:
[tex]\frac{m}{2}= \frac{f_{s}sin\theta}{g cos\theta}[/tex]
and finally we can multiply both sides of the equation by 2 so we get:
[tex]m=\frac{2f_{s}sin\theta}{g cos\theta}[/tex]
we know that:
[tex]tan \theta=\frac{sin \theta}{cos \theta}[/tex]
so we can simplify the equation a little more, so we get:
[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]
b.
So now we can directly use the equation to find the mass of the ladder with the data indicated by the problem:
θ=32° and [tex]f_{s}=47N[/tex]
we also know that [tex]g=9.8m/s^{2}[/tex]
so we can use our equation now:
[tex]m=\frac{2f_{s}tan \theta}{g}[/tex]
so we get:
[tex]m=\frac{2(47N)tan (32^{o})}{9.8m/s^{2}}[/tex]
which yields:
m=5.99kg
Answer:
a) m = 2fs(tanθ)/g
b) m = 5.99 kg
Explanation:
a) Expression for the ladder's mass, m, in terms of θ, g, fs, and/or L.
Data
m₁ : mass of the lader
g: acceleration due to gravity
L : ladder length
θ : angle that makes the ladder with the floor
µ = 0 : coefficient of friction between the ladder and the wall
fs : friction force between the ladder and the floor
Forces acting on the ladder
W =m*g : Weight of the ladder (vertical downward) , m: mass of the lader
FN :Normal force that the floor exerts on the ladder (vertical upward) (point A)
fs : friction force that the floor exerts on the ladder (horizontal to the left) (point A)
N : Forces that the wall exerts on the ladder (horizontal to the right)
Equilibrium of the forces in X
∑Fx=0
N -fs = 0
N = fs
The equilibrium equation of the moments at the point contact point of the ladder with the floor:
∑MA = 0
MA = F*d
Where:
∑MA : Algebraic sum of moments in the the point (A) (contact point of the ladder with the wall)
MA : moment in the point A ( N*m)
F : Force ( N)
d :Perpendicular distance of the force to the point A ( m )
Calculation of the distances of the forces at the point A
d₁ = (L/2)*cosθ : Distance from W to the point A
d₂ = L*sinθ : Distance from N to the point A
Equilibrium of the moments at the point A
∑MA = 0
N(d₂)-W(d₁) = 0
W( (L/2)*cosθ)= N(L*sinθ )
mg( (L/2)*cosθ)= fs(L*sinθ )
We divided by L both sides of the equation
mg (cosθ/2) =fs(sinθ)
m=2fs(sinθ)/ g( cosθ)
m = 2fs(tanθ)/ g
b) If θ=32° and the friction force on the floor is 47 N, what is the mass of the ladder?
m = 2fs(tanθ)/ g
m = 2(47)(tan32°)/(9,8)
m = 5.99 kg
Communications satellites are placed in orbits so that they always remain above the same point of the earth's surface.A Part completeWhat must be the period of such a satellite?Express your answer in hours to the nearest integer.TT = 24 h Previous Answers CorrectB What is its angular velocity?
Answer:
24 hours
[tex]7\times 10^{-5}\ rad/s[/tex]
Explanation:
If a satellite is in sync with Earth then the period of each satellite is 24 hours.
[tex]T=24\times 60\times 60\ s[/tex]
Angular velocity is given by
[tex]\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{24\times 60\times 60}\\\Rightarrow \omega=7\times 10^{-5}\ rad/s[/tex]
The angular velocity of the satellite is [tex]7\times 10^{-5}\ rad/s[/tex]
A communications satellite in a geostationary orbit must have a period of 24 hours and an angular velocity of approximately 0.262 radians per hour.
Explanation:To remain above the same point on Earth's surface, a communications satellite must be placed in a geostationary orbit. The period of such a satellite must be 24 hours, which is the same as one Earth day. This means that the satellite completes one orbit around the Earth in 24 hours. The angular velocity of the satellite depends on its position in the orbit and can be calculated using the formula:
Angular velocity = 2π / Period
Since the period is 24 hours, the angular velocity of the satellite is approximately 0.262 radians per hour.
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Three people are standing on a horizontally rotating platform in an amusement park. One person is almost at the edge, the second one is (3/5)R from the center, and the third is (1/2)R from the center.
1-If the period of rotation of the first person is T1 = T, find in terms of Tthe period of rotation of the second person.
2- If the period of rotation of the first person is T1 = T, find in terms of T the period of rotation of the third person.
3-If the speed of the first person is v1 = v find in terms of v the speed of the second person
4- If the speed of the first person is v1 = v find in terms of v the speed of the third person.
5- If the radial acceleration of the first person is a1 = a, find in terms of athe radial acceleration of the second person.
6- If the radial acceleration of the first person is a1 = a, find in terms of athe radial acceleration of the third person.
Answer:
1- Period is the time for which one full rotation is completed. Regardless of their positions on the platform, periods of all three are the same. T2 = T.
2- Similarly, T3 = T.
3- The platform is making rotational motion. So, the relation between the angular velocity and the linear velocity is
[tex]v = \omega R[/tex]
For all the people, angular velocity is the same. Their linear velocities are different.
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{3R/5}\\v_2 = \frac{3v}{5}[/tex]
4- Similarly,
[tex]\omega = \frac{v_1}{R} = \frac{v_2}{R/2}\\v_2 = \frac{v}{2}[/tex]
5- Radial acceleration in constant circular motion is
[tex]a_{rad} = \frac{v^2}{R}[/tex]
For the second person:
[tex]a_2 = \frac{v_2^2}{3R/5} = \frac{(3v/5)^2}{3R/5} = \frac{9v^2/25}{3R/5} = \frac{3v^2}{5R} = 3a/5[/tex]
6- Similarly,
[tex]a_3 = \frac{v_3^2}{R/2} = \frac{(v/2)^2}{R/2} = \frac{v^2/4}{R/2} = \frac{v^2}{2R} = a/2[/tex]
Explanation:
As a result, the period is same for every object on the rotating platform, as they all complete their revolutions at the same time. Their speed and radial acceleration is different according to their distance to the center.
The rotational kinematics relations allow to find the results for the questions about the movement of the three people on the turntable are:
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
Rotational kinematics studies the rotational motion of bodies looking for relationships between angular position, angular velocity, and angular acceleration.
In the case where the angular accleration is zero, the expression for the velocity is:
[tex]w = \frac{\Delta \theta }{\Delta t}[/tex]
Where w is the angular velocity and Δw and Δt are the variation in angle t over time.
1 and 2)
Indicates that people are on a turntable, the period is when we have a complete rotation θ = 2π rad in time, therefore the period and the angular velocity are related.
[tex]w= \frac{2\pi }{T} \\T = \frac{2\pi }{w}[/tex]
In the apparatus of parks the angular velocity is constant and we see that it does not depend on the radius, therefore the period for all the people is the same.
T = T₁ = T₂
3) They indicate that the speed of the 1 person who is in the position r=R on the plate is v, let's calculate the speed for the 2 person who is in the position r = [tex]\frac{3}{5} \ R[/tex]
Linear and angular variables are related.
v = w r
Let's substitute for the 1st person.
v = w R
For the 2nd person.
v₂= w ( [tex]\frac{3}{5} R[/tex])
We solve these two equations.
[tex]v_2 = \frac{3}{5} \ v[/tex]
4) We carry out the same calculation for the 3rd person.
v₃ = w ½ R
We solve the two equations.
v₃ = ½ v
5) Ask for radial acceleration.
The relationship between radial and angular acceleration is.
a = α R
We substitute for the 1st person.
a = α R
For the second person.
a₂ = α ( [tex]\frac{3}{5} R[/tex])
We solve the two equations
a₂ = [tex]\frac{3}{5} \ a[/tex]
6) Ask the radial acceleration of the 3rd person.
We substitute.
a₃ = α (½ R)
We solve.
a₃ = ½ a
In conclusion, using the rotational kinematics relations we can find the results for the questions about the movement of the three people on the turntable are
1 and 2) All periods are equal, T₂ = T and T₃ = T.
3) The linear velocity of the 2nd person is: v₂ = [tex]\frac{3}{5} \ v[/tex]
4) The linear velocity of the 3rd person is: v₃ = ½ v
5) The linear acceleration of the 2nd person is: a₂ = [tex]\frac{3}{5} \ a[/tex]
6) The linear acceleration of the 3rd person: a₃ = ½ a
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A person slaps her leg with her hand, which results in her hand coming to rest in a time interval of 2.75 ms from an initial speed of 2.25 m/s . What is the magnitude of the average contact force exerted on the leg, assuming the total mass of the hand and the forearm to be 1.65 kg ?
Answer:
1350N
Explanation:
[tex]2.75 ms = 2.75*10^{-3}s[/tex]
The force exerted on the hand would be the momentum divided by the duration of contact.
As the hand is coming to rest, final velocity would be 0
[tex]F = \frac{\Delta P}{\Delta t} = \frac{m(0 - v)}{\Delta t} = \frac{1.65*(2.25 - 0)}{2.75 * 10^{-3}} = -1350 N[/tex]
The magnitude of the force would be 1350N
5.00 kg of liquid water is heated to 100.0 °C in a closed system. At this temperature, the density of liquid water is 958 kg/m3. The pressure is maintained at atmospheric pressure of 1.01 × 105 Pa. A moveable piston of negligible weight rests on the surface of the water. The water is then converted to steam by adding an additional amount of heat to the system. When all of the water is converted, the final volume of the steam is 8.50 m3. The latent heat of vaporization of water is 2.26 × 106 J/kg. How much heat is added to the system in the isothermal process of converting all of the water into steam?
Answer:
[tex]1.04\times 10^7\ J.[/tex]
Explanation:
In the question given :
Pressure is constant
Therefore, Work done, [tex]W=P\times\Delta V[/tex]
Pressure, P=1.01 × 105 Pa.
Final volume, [tex]V_f=8.50\ m^3.[/tex]
Initial volume, [tex]V_i=\dfrac{Mass}{density}=\dfrac{5}{958}=5.22\times10^-3\ m^3.[/tex]
Therefore, W=8.58\times 10^{5}\ J.
Also, Heat Given, [tex]Q=m\times L=5\times 2.26\times 10^{6}\ J=1.13\times 10^7\ J.[/tex]
Also, according to First law of thermodynamics:
[tex]\Delta U=Q-W=(1.13\times 10^7)-(8.58\times 10^5)=1.04\times 10^7\ J.[/tex]
Hence, this is the required solution.
The energy needed to convert all of the water into steam in this isothermal system is 11.30 MJ. This is calculated using the formula Q = mLv, substituting the given values for mass and latent heat of vaporization.
Explanation:The heat required to convert water from liquid into steam, in an isothermal process, can be found using the formula Q = mLv, where 'm' is the mass of the water, and 'Lv' is the latent heat of vaporization.
Given that we have 5.00 kg of water and the heat of vaporization for water under atmospheric pressure is 2.26 × 106 J/kg, we can substitute these values into our formula:
Q = 5.00 kg × 2.26 × 106 J/kg.
This gives us Q = 11.30 × 106 J, or 11.30 MJ. Therefore, 11.30 MJ of heat is added to the system to convert all the water into steam at atmospheric pressure.
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A particle with a mass of 0.01 kg and a net charge of -0.05 C accelerates from rest through a uniform electric field. If the strength of the field is 2000 V/C, what is the speed of the particle after traveling for 0.5 m?
Answer:
The speed of the particle after travelling for 0.5 m is 100 m/s.
Explanation:
It is given that,
Mass of the particle, m = 0.01 kg
Net charge on the particle, q = -0.05 C
Electric field strength, E = 2000 V/C
Distance travelled by the particle, d = 0.5 m
The work done due to motion of the particle is balanced by the change in kinetic energy as :
[tex]Fd=\dfrac{1}{2}mv^2[/tex]
v is the speed of the particle
F is the electric force
[tex]qEd=\dfrac{1}{2}mv^2[/tex]
[tex]v=\sqrt{\dfrac{2qEd}{m}}[/tex]
[tex]v=\sqrt{\dfrac{2\times 0.05\times 2000\times 0.5}{0.01}}[/tex]
v = 100 m/s
So, the speed of the particle after travelling for 0.5 m is 100 m/s. Hence, this is the required solution.
A binary star system consists of two stars of masses m1 and m2. The stars, which gravitationally attract each other, revolve around the center of mass of the system. The star with mass m1 has a centripetal acceleration of magnitude a1. Note that you do not need to understand universal gravitation to solve this problem.Find a2, the magnitude of the centripetal acceleration of the star with mass m2.
Answer:
a₂ = m₁ / m₂ a₁
Explanation:
For this exercise we note that the attraction between the two stars is an action and reaction force, therefore it has the same magnitude, but it is applied to each of the bodies
Let's apply Newton's second law on the star 1
F₁ = m₁ a₁
Newton's second law in star 2
F₂ = m₂ a₂
| F₁ | = | F₂ |
m₁ a₁ = m₂ a₂
a₂ = m₁ / m₂ a₁
A typical person's eye is 2.5 cm in diameter and has a near point (the closest an object can be and still be seen in focus) of 25 cm, and a far point (the farthest an object can be and still be in focus) of infinity. 1) What is the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye? (Express your answer to two significant figures.)
Answer:
The minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm.
Explanation:
The diameter of person's eye is 2.5 cm. The close point or the near point of the eye is 25 cm and the far point is infinity. We need to determine the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye. Let f is the minimum effective focal length. It can be calculated using lens formula as :
[tex]\dfrac{1}{f}=\dfrac{1}{25}+\dfrac{1}{2.5}[/tex]
f = 2.27 cm
So, the minimum effective focal length of the focusing mechanism (lens plus cornea) of the typical eye is 2.27 cm and it is at the nearest point. Hence, this is the required solution.
An electromagnetic wave in a vacuum traveling in the +x direction generated by a variable source initially has a wavelength λ of 235 μm and a maximum electric field Emax in the +y direction of 7.70×10−3 V/m . If the period of the wave is then increased by a factor of 2.70, what is the equation of the resulting magnetic field component of the wave?
Answer:
If the period of the wave is increased by the factor of 2.70, the wavelength of the wave is also increased by a factor of 2.70. So,
[tex]\lambda_2 = 235\times 2.70 = 634.5 ~nm[/tex]
The magnetic field component can be written as
[tex]\vec{B} = \frac{E_{max}}{c}e^{i(\vec{k}\vec{z}-\omega t)}\^{z}[/tex]
The magnetic field is in the z-direction, because the E-field is directed towards +y and the wave is propagating in the +x-direction. The right-hand rule gives us the direction of the B-field.
[tex]\vec{E} \times \vec{B} = \vec{S}[/tex]
S is the Poynting vector which gives us the propagation of the wave.
We will use the following relationships
[tex]k = 2\pi / \lambda\\f = \omega / 2\pi\\c = \lambda f = \lambda \omega / 2\pi\\\omega = 2\pi c/\lambda[/tex]
[tex]\vec{B} = \frac{7.7\times 10^{-3}}{3\times 10^8}e^{(\frac{2\pi}{3\times 10^8}z - \frac{2\pi\times 3\times 10^8}{634.5})}\\\vec{B} = 2.56\times10^{-11} e^{(2.09\times10^{-8}z - 2.96\times10^{6}t)}\^{z}[/tex]
To find the resulting magnetic field equation of an electromagnetic wave after the period is increased, we use relationships between the wave's wavelength, frequency, period, and electric and magnetic fields, applying the changes to the wave's properties based on Maxwell's theory of electromagnetism.
Explanation:The question is asking for the equation of the resulting magnetic field component of an electromagnetic wave after its period has been increased by a factor of 2.70. Given the initial wavelength λ of 235 μm (micrometers) and the maximum electric field Emax in the +y direction of 7.70×10⁻³ V/m, we can find the initial frequency ν using the relationship ν = c / λ, where c is the speed of light. Since the period T is the reciprocal of the frequency (ν = 1/T), when the period is increased by a factor of 2.70, the frequency decreases by the same factor. The maximum magnetic field Bmax is related to the maximum electric field Emax by Bmax = Emax / c. The resulting magnetic field B can be described by the function B(x, t) = Bmax sin(kx - ωt + φ), where k is the wave number, ω is the angular frequency (2πν), and φ is the phase constant.
A tank of water is in the shape of a cone (assume the ""point"" of the cone is pointing downwards) and is leaking water at a rate of 35 cm3/sec. The base radius of the tank is 1 meter and the height of the tank is 2.5 meters. When the depth of the water is 1.25 meters at what rate is the (a) depth changing and (b) the radius of the top of the water changing?\
Answer:
a) dh/dt = -44.56*10⁻⁴ cm/s
b) dr/dt = -17.82*10⁻⁴ cm/s
Explanation:
Given:
Q = dV/dt = -35 cm³/s
R = 1.00 m
H = 2.50 m
if h = 125 cm
a) dh/dt = ?
b) dr/dt = ?
We know that
V = π*r²*h/3
and
tan ∅ = H/R = 2.5m / 1m = 2.5 ⇒ h/r = 2.5
⇒ h = (5/2)*r
⇒ r = (2/5)*h
If we apply
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = 3*35/π = 105/π ⇒ d(r²*h)/dt = -105/π
a) if r = (2/5)*h
⇒ d(r²*h)/dt = d(((2/5)*h)²*h)/dt = (4/25)*d(h³)/dt = -105/π
⇒ (4/25)(3*h²)(dh/dt) = -105/π
⇒ dh/dt = -875/(4π*h²)
b) if h = (5/2)*r
Q = dV/dt = -35 = d(π*r²*h/3)*dt
⇒ d(r²*h)/dt = d(r²*(5/2)*r)/dt = (5/2)*d(r³)/dt = -105/π
⇒ (5/2)*(3*r²)(dr/dt) = -105/π
⇒ dr/dt = -14/(π*r²)
Now, using h = 125 cm
dh/dt = -875/(4π*h²) = -875/(4π*(125)²)
⇒ dh/dt = -44.56*10⁻⁴ cm/s
then
h = 125 cm ⇒ r = (2/5)*h = (2/5)*(125 cm)
⇒ r = 50 cm
⇒ dr/dt = -14/(π*r²) = - 14/(π*(50)²)
⇒ dr/dt = -17.82*10⁻⁴ cm/s
The rate at which the depth of the water is changing is -0.178 cm/sec, and the rate at which the radius of the top of the water is changing is -0.14 cm/sec.
Explanation:The subject of this question is related to Calculus and the specific topic is related rates. For this problem, in addition to the rate at which water is leaking, we need to consider the geometric relationship within the cone-shaped water tank.
First, we are given that V' = -35 cm^3/sec (we make it negative because the volume is decreasing) and we know that the volume of a cone is V = (1/3)πr²h. Our tank parameters are r = 1m and h = 2.5m, but we need everything in the same units, so we convert our radius to 100 cm.
We know through similar triangles that r/h = R/H, where r and R are the radii at any given time, and h and H are the heights at any given time respectively. Thus, r = (Rh)/H. Substituting r in our volume equation, we get: V = (1/3)π((Rh)^2)/H=h²πR²/H, and hence V=hπR². Differentiating this implicitly with t gives V' = h'πR²+2hπRr'.
Substituting for V', h and r from our given information, we get: -35=h'πR²+2(1.25)πRr'. We can solve for h' and r' separately.
(a) To find h' we set r' = 0, because we only want to know how depth h is changing. Solving for h' we find it to be -0.178 cm/sec.
(b) To find r', we set h' = 0, because we are only interested in how radius r is changing. Solving for r' we get -0.14 cm/sec.
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A hot (70°C) lump of metal has a mass of 250 g and a specific heat of 0.25 cal/g⋅°C. John drops the metal into a 500-g calorimeter containing 75 g of water at 20°C.
The calorimeter is constructed of a material that has a specific heat of 0.10 cal/ g⋅°C.
When equilibrium is reached, what will be the final temperature? cwater = 1.00 cal/g⋅°C.
a. 114°C
b. 72°C
c. 64°C
d. 37°C
Answer:
d. 37°C
Explanation:
When Equilibrium is reached,
Heat lost = heat gain.
Heat lost by the hot metal = c₁m₁(T₁-T₂)............. equation 1
Where c₁ = specific heat capacity of the metal = 0.25 cal/g⋅°C, m₁ = mass of the metal = 250 g, T₁ = initial Temperature of the metal = 70°C, T₂ final temperature of the metal .
Substituting this values into equation 1,
Heat lost by the metal = 62.5(70-T₂)
Also,
Heat gain by the water = c₂m₂(T₂-T₁)................. equation 2
c₂ = 1.00 cal/g⋅°C., m₂ = 75 g, T₁ = 20°C
Substituting this values into equation 2,
Heat gain by water = 1× 75 (T₂ - 20)
Heat gained by water = 75(T₂ - 20)
Also,
Heat gained by the calorimeter = c₃m₃(T₂-T₁)............. equation 3
Where c₃ = 0.10 cal/ g⋅°C, m =500 g, T₁ =20°C.
Substituting this values into equation 3
Heat gained by the calorimeter = 0.1 × 500(T₂ - 20)
Heat gained by the calorimeter = 50(T₂ - 20)
Heat lost by the metal = heat gained by water + heat gained by the calorimeter.
62.5(70-T₂) = 50(T₂ - 20) + 75(T₂ - 20)
4375 - 62.5T₂ = 50T₂ - 1000 + 75T₂ - 1500
Collecting like terms,
-62.5T₂ - 50T₂ - 75T₂ = - 1000 - 1500 - 4375
-187.5T₂ = -6875
Dividing both side by the coefficient of T₂
-187.5T₂ / -187.5 = -6875 /-187.5
T₂ = 36.666°C
T₂ ≈ 37°C
The final temperature = 37°C
One of the hazards facing humans in space is space radiation: high-energy charged particles emitted by the sun. During a solar flare, the intensity of this radiation can reach lethal levels. One proposed method of protection for astronauts on the surface of the moon or Mars is an array of large, electrically charged spheres placed high above areas where people live and work. The spheres would produce a strong electric field \underset{E}{\rightarrow} to deflect the charged particles that make up space radiation. The spheres would be similar in construction to a Mylar balloon, with a thin, electrically conducting layer on the outside surface on which a net positive or negative charge would be placed. A typical sphere might be 5 m in diameter. Suppose that to repel electrons in the radiation from a solar flare, each sphere must produce an electric field \underset{E}{\rightarrow} of magnitude 1 × 106 N/C at 25 m from the center of the sphere.What is the magnitude of E⃗ just outside the surface of such a sphere?a) 0b) 106 N/Cc) 107 N/Cd) 108 N/C
Final answer:
The magnitude of the electric field just outside the surface of a sphere designed to deflect solar radiation on the moon or Mars is 10⁷ N/C, calculated based on the given requirements for the electric field at a distance from the sphere.
Explanation:
The question asks for the magnitude of an electric field E just outside the surface of a sphere that is proposed to protect astronauts on the surface of the moon or Mars by deflecting high-energy charged particles emitted by the sun.
The sphere, assumed to have a uniform surface charge distribution, generates an electric field whose magnitude at a distance r from its center can be found using the formula E = kQ/r², where k is Coulomb's constant (8.99 x 10⁹ N·m²/C²), Q is the charge on the sphere, and r is the distance from the center of the sphere.
Given that the sphere's electric field's magnitude at a distance of 25 m from its center needs to be 1 x 10⁶ N/C, we find the charge Q required to produce this field.
Once Q is determined, we can calculate the electric field's magnitude just outside the sphere's surface (at r = 2.5 m, which is the radius of the 5 m diameter sphere) using the same formula.
The calculation reveals that the magnitude of E just outside the surface of the sphere is of the order 10⁷ N/C, making option c) 10⁷ N/C the correct answer.
The magnitude of the electric field just outside the surface of the sphere is [tex]\(10^7 \, \text{N/C}\),[/tex] so the correct answer is (c) [tex]\(10^7 \, \text{N/C}\).[/tex]
To find the magnitude of the electric field just outside the surface of the sphere, we'll use Gauss's Law, which states:
[tex]\[\oint \vec{E} \cdot d\vec{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}\][/tex]
Where:
- \(\vec{E}\) is the electric field vector
- \(d\vec{A}\) is a differential area vector
- \(Q_{\text{enc}}\) is the enclosed charge
- \(\varepsilon_0\) is the permittivity of free space
Since the electric field is radial, we can express it as [tex]\(E = E(r)\hat{r}\)[/tex], where [tex]\(\hat{r}\)[/tex] is a unit vector pointing radially outward from the center of the sphere.
The magnitude of the electric field \(E\) just outside the surface of the sphere is equal to the magnitude of the electric field produced by a point charge at the center of the sphere, which is given by Coulomb's Law:
[tex]\[E = \frac{k |Q|}{r^2}\][/tex]
Where:
- \(k\) is Coulomb's constant [tex](\(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2\))[/tex]
- \(|Q|\) is the magnitude of the charge enclosed by the Gaussian surface
- \(r\) is the distance from the center of the sphere
Given that [tex]\(E = 1 \times 10^6 \, \text{N/C}\) at \(r = 25 \, \text{m}\), we can solve for \(|Q|\):[/tex]
[tex]\[1 \times 10^6 \, \text{N/C} = \frac{k |Q|}{(25 \, \text{m})^2}\][/tex]
[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (25 \, \text{m})^2}{k}\][/tex]
[tex]\[|Q| = \frac{(1 \times 10^6 \, \text{N/C}) \cdot (625 \, \text{m}^2)}{8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2}\][/tex]
[tex]\[|Q| = \frac{625 \times 10^6}{8.99}\][/tex]
[tex]\[|Q| \approx 69.633 \, \text{C}\][/tex]
Now, let's calculate the electric field just outside the surface of the sphere using this charge:
[tex]\[E = \frac{k |Q|}{r^2}\][/tex]
[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{(5 \, \text{m})^2}\][/tex]
[tex]\[E = \frac{(8.99 \times 10^9 \, \text{N}\cdot\text{m}^2/\text{C}^2) \cdot (69.633 \, \text{C})}{25 \, \text{m}^2}\][/tex]
[tex]\[E = \frac{629.586 \times 10^9}{25}\][/tex]
[tex]\[E \approx 25.183 \times 10^7 \, \text{N/C}\][/tex]
Therefore, the magnitude of the electric field just outside the surface of the sphere is approximately [tex]\(2.5183 \times 10^8 \, \text{N/C}\).[/tex]
The closest option is d) [tex]\(10^8 \, \text{N/C}\).[/tex]
A block with mass m =6.7 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m. While at this equilibrium position, the mass is then given an initial push downward at v = 4.6 m/s.
The block oscillates on the spring without friction.1) What is the spring constant of the spring? ANSWER = (234.5)2) What is the oscillation frequency? ANSWER = (0.942)3) After t = 0.47 s what is the speed of the block?4) What is the magnitude of the maximum acceleration of the block?5)At t = 0.47 s what is the magnitude of the net force on the block?6)Where is the potential energy of the system the greatest?---At the highest point of the oscillation.---At the new equilibrium position of the oscillation.---At the lowest point of the oscillation
The spring constant of the spring is approximately 234.50 N/m.
Explanation:To find the spring constant of the spring, we can use Hooke's Law, which states that the force exerted by a spring is proportional to the displacement of the spring from its equilibrium position. In this case, we know that the mass of the block is 6.7 kg and the displacement of the spring is 0.28 m. Therefore, we can use the equation F = kx, where F is the force, k is the spring constant, and x is the displacement. Plugging in the given values, we get:
F = kx
F = (6.7 kg)(9.8 m/s^2)
F = 65.66 N
65.66 N = k(0.28 m)
k = (65.66 N)/(0.28 m)
k ≈ 234.50 N/m
Therefore, the spring constant of the spring is approximately 234.50 N/m.
Seat belts and air bags save lives by reducing the forces exerted on the driver and passengers in an automobile collision. Cars are designed with a "crumple zone" in the front of the car. In the event of an impact, the passenger compartment decelerates over a distance of about 1 m as the front of the car crumples. An occupant restrained by seat belts and air bags decelerates with the car. By contrast, an unrestrained occupant keeps moving forward with no loss of speed (Newton's first law!) until hitting the dashboard or windshield. These are unyielding surfaces, and the unfortunate occupant then decelerates over a distance of only about 5 mm .
a. A 60 kg person is in a head-on collision. The car's speed at impact is 15 m/s. Estimate the net force on the person if he or she is wearing a seat belt and if the air bag deploys.b. Estimate the net force that ultimately stops the person if he or she is not restrained by a seat belt or air bag.c. How does these two forces compare to the person's weight?
Answer:
a) -6,750 N b) -1.35*10⁶ N c) 11.5 times d) 2,295 times
Explanation:
a) In order to answer the question, we can apply Newton's 2nd Law to the person, as follows:
Fnet = m*a
Assuming that the deceleration is uniform, we can use any of the kinematic equations.
In this particular case, examining the givens that we have (the final speed (which is 0), the initial speed (impact speed), and the distance of deceleration (1 m) the most useful equation is the following:
vf²-v₀² = 2*a*d
Replacing by the givens, and solving for a, we get:
a = (15m/s)² / 2* (1 m) = -112.5 m/s²
With this value of a, we can get the net force F:
F = 60 kg* (-112.5 m/s²) = -6,750 N
b) For this part, same reasoning applies, the only difference being the deceleration distance, which for this case is only 5 mm.
We can apply the same kinematic equation:
vf²-v₀² = 2*a*d
Once again, replacing by the givens, and solving for a, we get:
a = (15m/s)² / 2* (0.005 m) = -22,500 m/s²
With this value of a, we can get the net force F:
F = 60 kg* (-22,500 m/s²) = -1.35*10⁶ N
c) If we compare the forces that we got above with the weight of the person (which is the same to compare the acceleration with g), we have, for the first case (person restrained) a value of approximately 12 g, but for the unrestrained case, we got a value of 2,295 g!
In cloud formation, water vapor turns into water droplets which get bigger and bigger until it rains.
This will cause the temperature of the air in the clouds to:
a. get warmer.
b. get cooler.
c. will not affect the temperature of the air in the clouds.
d. There is no air in clouds.
Answer:
a. get warmer.
Explanation:
When the water vaporous reach the upper layer of the atmosphere they get a cooler air to which they loose their temperature and condense to form clouds as a the temperature of the air increases.
It may be noted that the water looses its high amount of latent heat of vaporization to condense into water this significantly increases the temperature of the air in contact.
If heat is flowing from a table to a block of ice moving across the table, which of the following must be true?
a. The table is rough and there is friction between the table and ice.b. The ice is cooler than the table.c. The ice is changing phase.d. All three are possible, but none is absolutely necessary.
Answer:
The ice is cooler than the table.
Explanation:
The difference in temperature is a necessary condition for the transfer of heat between two bodies.
Here if the heat is being transferred from the table to an ice block this means that the temperature of the table is greater than the ice which drives the heat energy from the table to the ice block and the heat will continue to flow until they both attain a common temperature.
However,
we can tell nothing about the friction between the two bodies from the given information.the ice may or may not be changing its phase based upon the fact that the whole process might be occurring below zero degree Celsius such that the ice is having more cooler temperature than table.Final answer:
Heat flow from a table to a block of ice enables the ice to undergo a phase change from solid to liquid, known as the latent heat of fusion, without a change in temperature .
Explanation:
If heat is flowing from a table to a block of ice moving across the table, which of the following must be true? The correct answer from the choices provided is that the ice is changing phase. Heat flow from the table to the ice suggests that the ice is absorbing energy. When ice absorbs heat, it undergoes a phase change, where the ice melts and becomes liquid water without any change in temperature. This process is known as the latent heat of fusion, where heat is transferred to cause a phase change without altering the temperature of the system. Therefore, while the table may be rough and friction may exist, and the ice might indeed be cooler than the table, the guaranteed phenomenon occurring is a phase change from solid to liquid as the ice absorbs heat.
true or false conceptual physics 2 questions from chapter 25.
1. [ ] The electrostatic potential difference between point A and B, ∆V_AB, can be calculated using a path integral, and ∆V_AB does not depend on the integral path used.
2.[ ] A common reference point for electrostatic potential is V = 0 at [infinity].
3. [ ] The magnitude of work required to bring a charge q from point A to point B is |qV_BA|.
4. [ ] eV is a unit for potential.
5. [ ] When a charged particle is accelerated from rest in an electric field, the amount of kinetic energy increased equals the amount of potential energy decreased, i.e., |1/2 mv^2| = |qΔV|
6. [ ] An electric field line points to the direction of higher electric potential.
7. [ ] An equipotential surface is always parallel to electric field lines.
8. [ ] The electrostatic potential generated by multiple charges, is the vector addition of the potentials generated by individual charges.
9. [ ] Inside a conductor, the electric field and potential are both zero.
10. [ ] The electric fields created by a point charge, a conducting sphere, and an insulator sphere are identical, if they all have the same amount of charge and the field is measured outside the spheres.
Answer:
Explanation:
1) TRUE; potential difference can be calculated using path integral. Since the electric field is a conservative, the potential difference can be calculated using any path.
2) TRUE; since potential due to a charge is inversely dependent on distance, at infinity the potential will be almost zero.
3) TRUE, W = q.VBA.
4) FALSE; eV is a unit for work (or) energy.
5) TRUE; since the electric force is conservative force. There will be no loss in energy, the decreased potential energy will be coverted to kinetic energy.
6) FALSE; in the direction of electric field the potential decreases.
7) FALSE; equipotential surface is perpendicular to the electric field lines.
8) FALSE; electrostatic potential is scalar quantity. It depends only on the charge and distance from it.
9) FALSE; Inside a conductor the electric field is zero but the electric potential is constant at the value that is at the surface of the conductor.
10) TRUE; as long as the field is being measured outiside the body the bodies act as point charges. So electric fields due to all types of bodies charged identically will be equal.
Identical twins Anna and Hannah visit you at the optical clinic. Anna, whose eyes can easily focus on distant objects (her far point), is also able to focus on objects within 20 cm of her eyes (her near point). Assuming the diameter and, hence, the distance between the cornea and retina, of Anna's eye is 20 mm, what is the range (in diopters) of Anna's vision? The limits of this range correspond to the total refractive power of her eyes at their far point and and the refractive power at their near point.
a) from 50 to 50.5 diopters
b) from 50 to 55 diopters
c) from 50 to 60 diopters
d) from 0 to 5 diopters
Answer:
c
Explanation:
from. 50 to 60 diopters
The range of Anna's vision will be between 0-5 diopters.
What will be the range of Anna's eyes?
It is given that
Anna is able to focus on objects within 20 cm of her eyes (her near point)
Now the distance between the cornea and retina, of Anna's eye is 20 mm
The focal length will be f= 20+2= 22cm
Now the power of eyes will be given by
[tex]P= \dfrac{1000}{f(in \ mm)}[/tex]
[tex]P= \dfrac{1000}{ 22}=4.54 \ diopters[/tex]
Thus the range of Anna's vision will be between 0-5 diopters.
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In 1956, Frank Lloyd Wright proposed the construction of a mile-high building in Chicago. Suppose the building had been constructed. Ignoring Earth's rotation, find the change in your weight if you were to ride an elevator from the street level, where you weigh 469 N, to the top of the building.
Answer:
-0.23694 N
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
m = Mass of the Earth = 5.972 × 10²⁴ kg
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
r = Radius of Earth = 6371000 m
dr = Height = 1 mile = 1609.34 m
Acceleration is given by
[tex]a=\dfrac{GM}{r^2}[/tex]
Change in acceleration is given by
[tex]da=-2\dfrac{GM}{r^3}dr[/tex]
[tex]w=ma\\\Rightarrow w=m\dfrac{GM}{r^2}\\\Rightarrow w=469\ N[/tex]
[tex]dw=mda\\\Rightarrow dw=-m2\dfrac{GM}{r^3}dr\\\Rightarrow dw=-2w\dfrac{dr}{r}\\\Rightarrow dw=-2\times 469\times \dfrac{1609.34}{6.371\times 10^{6}}\\\Rightarrow dw=-0.23694\ N[/tex]
The change in weight is -0.23694 N
The change in your weight if you were to ride an elevator from the street level where you weigh 469N to the top of the building is; -0.237 N
The formula for acceleration here is;
a = GM/r²
Where;
G is gravitational constant = 6.67 × 10⁻¹¹ m³/kg.s²
M is mass of earth = 5.972 × 10²⁴ kg
r is distance from center of earth
Since we are trying to find change in weight, let us first find the change in acceleration with respect to r;
da/dr = -2GM/r³
da = -(2GM/r³) dr
Thus, change in weight from top to bottom is;
W_top - W_bottom = m(da)
Now, weight at bottom is gotten from the formula;
W_bottom = GmM/r²
Also, W_bottom = m(da) since we are dealing with change in weight.
Thus;
m(da)= -(2GmM/r³) dr
Recall that GmM/r². Thus;
m(da) = -2W_bottom × dr/r
where;
W_bottom = 469 N
r is radius of earth = 6371000 m
dr = 1 mile = 1609.34 m
Thus;
m(da) = -2 × 469 × 1609.34/6371000
m(da) = -0.237 N
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A Doppler flow meter uses ultrasound waves to measure blood-flow speeds. Suppose the device emits sound at 3.5 MHz, and the speed of sound in human tissue is about 1540 m/s. What is the expected beat frequency observed at the flow meter if blood is flowing in large leg arteries at 3.0 cm/s directly away from the flow meter?
To solve this problem we will use the concept of the Doppler effect applied to the speed of blood, the speed of sound in the blood and the original frequency. This relationship will also be extrapolated to the frequency given by the detector and measured the change in frequencies through the beat frequency. So:
[tex]f_{blood} = f (1-\frac{v_{blood}}{v_{snd}})[/tex]
Where
[tex]f_{blood}[/tex] = Frequency of the blood flow
f = Frequency of the original signal
[tex]v_{blood}[/tex] = Speed of the blood flow
[tex]v_{snd}[/tex] = Speed of sound in blood
[tex]f''_{detector} = \frac{f_{blood}}{(1+\frac{v_{blood}}{v_{snd}})}[/tex]
[tex]f''_{detector} = f (\frac{(1-\frac{v_{blood}}{v_{snd}})}{(1+\frac{v_{blood}}{v_{snd}})})[/tex]
[tex]f''_{detector} = f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]
Now calculating the beat frequency is
[tex]\Delta f = f-f''_{detector}[/tex]
Replacing this latest value we have that,
[tex]\Delta f = f-f \frac{(v_{snd}-v_{blood})}{(v_{snd}+v_{blood})}[/tex]
[tex]\Delta f = f \frac{2v_{blood}}{v_{snd}+v_{blood}}[/tex]
Replacing we have,
[tex]\Delta f = (3.5*10^6)(\frac{2*(3*10^{-2})}{1.54*10^3+3*10^{-2}})[/tex]
[tex]\Delta f = 136.36Hz[/tex]
Therefore the beat frequency is 136.36Hz
Using the beat frequency relation, the expected beat frequency observed at the flow meter would be 136.36 Hz
Given the Parameters :
Frequency of device, F = 3.5 MHz = 3.5 × 10³ Hz Speed of sound in tissue = Vs = 1540 m/s Velocity of blood = Vb = 3 cm/s = (0.01 × 3)m/s = 0.03 m/s = 3 × 10¯² m/s Beat frequency = ΔfThe expected beat frequency observed can be calculated uisng the relation :
[tex] \delta F = F \frac{2V_{b}}{V_{s} + V_{b}}[/tex]Substituting the values into the formula :
[tex] \delta F = 3.5 \times 10^{3} \frac{2 \times 0.03}{(1540 + 0.03}[/tex]
[tex] \delta F = 3.5 \times 10^{3} \frac{0.06}{(1540.03}[/tex]
[tex] \delta F = 136.36 Hz [/tex]
Therefore, the expected beat frequency observed at the flow meter will be 136.36 Hz
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11–8 Consider a heavy car submerged in water in a lake with a flat bottom. The driver’s side door of the car is 1.1 m high and 0.9 m wide, and the top edge of the door is 10 m below the water surface. Determine the net force acting on the door (normal to its surface) and the location of the pressure center if (a) the car is well-sealed and it contains air at atmospheric pressure and (b) the car is filled with water.
Answer:
Explanation:
position of centre of mass of door from surface of water
= 10 + 1.1 / 2
= 10.55 m
Pressure on centre of mass
atmospheric pressure + pressure due to water column
10 ⁵ + hdg
= 10⁵ + 10.55 x 1000 x 9.8
= 2.0339 x 10⁵ Pa
the net force acting on the door (normal to its surface)
= pressure at the centre x area of the door
= .9 x 1.1 x 2.0339 x 10⁵
= 2.01356 x 10⁵ N
pressure centre will be at 10.55 m below the surface.
When the car is filled with air or it is filled with water , in both the cases pressure centre will lie at the centre of the car .
The net force acting on the door (normal to its surface) and the location of the pressure should be [tex]2.01356 \times 10^5 N[/tex]
Calculation of net force & pressure location:The position of center of mass of door from the surface of water should be
[tex]= 10 + 1.1 \div 2[/tex]
= 10.55 m
Now
Pressure on center of mass
= atmospheric pressure + pressure due to water column
[tex]= 10^5 + 10.55 \times 1000 \times 9.8\\\\= 2.0339 \times 10^5 Pa[/tex]
Now
the net force acting on the door (normal to its surface)
[tex]= .9 \times 1.1 \times 2.0339 \times 10^5\\\\= 2.01356 \times 10^5 N[/tex]
So,
pressure centre will be at 10.55 m below the surface.
Therefore,
When the car is filled with air or it is filled with water , in both the cases pressure center should be lie at the centre of the car .
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