Answer:
If a negatively charged balloon is brought near one end of the rod but not in direct contact, then the negative charges on the balloon repel the same amount of negative charges on the end of the rod that is close to the balloon, and the positive charges stay at the balloon-side of the rod. The total charge of the rod is still zero, but the distribution of the charges are now non-uniform.
Which of the following changes would increase the seperation between the bright fringes in the diffraction pattern formed by diffraction grating?
A. INCREASE THE WAVELENGTH OF THE LIGHT
B. IMMERSE THE APPARATUSE IN WATER
C. INCREASE THE SEPERATION BETWEEN THE SLITS
D. NONE OF THESE
E. MORE THAN ONE OF THESE
The increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
The equation for diffraction grating that relates the wavelength and separation of bright fringes is given as;
[tex]d sin\theta = m \lambda \\\\sin \ \theta = \frac{m \lambda}{d}[/tex]
where;
λ is the wavelength d is the spacingThus, we can conclude that the increase in the separation between the bright fringes in the diffraction pattern formed by diffraction grating occurs when there is increase in the wavelength of the light.
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Increasing the wavelength of the light will increase the separation between the bright fringes in the diffraction pattern. The correct option is Option A. Immersion in water or changing slit separation do not contribute similarly.
To increase the separation between the bright fringes in a diffraction pattern, one effective approach is to increase the wavelength of the light. When the wavelength increases, the angle of diffraction becomes larger for each order of maximum, resulting in greater fringe separation. Therefore, the correct answer is Option A: INCREASE THE WAVELENGTH OF THE LIGHT.
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth's atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. (a) What was the average velocity (in m/s) of the blast wave? (Enter the magnitude.) x m/s (b) Compare this with the speed of sound, which is 343 m/s at sea level. Volast wave = sound
Answer:
156.67 m/s
0.45676 times the speed of sound
Explanation:
Distance from the ground = 23.5 km = 23500 m
Time taken by the blast waves to reach the ground = [tex]2\ minutes\ 30\ seconds=2\times 60+30=150\ s[/tex]
Spedd of the wave would be
[tex]Speed=\dfrac{Distance}{Time}\\\Rightarrow v_b=\dfrac{23500}{150}\\\Rightarrow v-b=156.67\ m/s[/tex]
The velocity of the blast wave is 156.67 m/s
v = Velocity of sound = 343 m/s
[tex]\dfrac{v_b}{v}=\dfrac{156.67}{343}\\\Rightarrow v_b=v\dfrac{156.67}{343}\\\Rightarrow v_b=0.45676v[/tex]
The blast wave is 0.45676 times the speed of sound
The average velocity of the blast wave can be calculated using the distance traveled and time taken, resulting in 156.7 m/s. This velocity is less than half the speed of sound at sea level.
The average velocity of the blast wave can be calculated by dividing the distance traveled by the time taken. The distance is the altitude where the explosion occurred, 23.5 km. Converting 2 minutes 30 seconds to seconds gives a time of 150 seconds.
So, velocity = distance / time = 23.5 km / 150 s = 0.1567 km/s = 156.7 m/s.Comparing this to the speed of sound at sea level, which is 343 m/s, the blast wave's velocity of 156.7 m/s is less than half the speed of sound.
Calculate the linear momentum of photons of wavelength 350 nm. What speed does a hydrogen molecule need to travel to have the same linear momentum?
Answer:
a)[tex]p=1.89x10^{-27}kg.m.s^{-1}[/tex]
b)[tex]v=0.565\frac{m}{s}[/tex]
Explanation:
First, we need to obtain the linear momentum of the photons of wavelength 350nm.
We are going to use the following formula:
[tex]\lambda=\frac{h}{p}\\Where:\\\lambda=wavelength\\h=placnk's\_constant\\p=Linear\_momentum[/tex]
So the linear momentum is given by:
[tex]p=\frac{h}{\lambda}\\\\p=\frac{6.626x10^{-34}J.s}{350x10^{-9}m}\\\\p=1.89x10^{-27}kg.m.s^{-1}[/tex]
Having the linear momentum of the photon, we can calculate the speed of the hydrogen molecule to have the same momentum, we can use the classic formula for that:
[tex]p=m.v[/tex]
[tex]where:\\m=mass\\v=speed\\p=linear\_momentum[/tex]
The mass of the hydrogen molecule is given by:
[tex]m=2*(1.0078x10^{-3}\frac{kg}{mol}})x\frac{1}{6.022x10^{23}mol}[/tex]
[tex]3.35x10^{-27}kg[/tex]
What we've done here is to use the molecular weight of the hydrogen, and covert it kilograms, we had to multiply by two because the hydrogen molecule is found in pairs.
so:
[tex]v=\frac{p}{m}\\\\v=\frac{1.89x10^{-27}kg.m.s^{-1}}{3.35x10^{-27}kg}\\\\v=0.565\frac{m}{s}[/tex]
(a) The linear momentum of the photon is 1.89 x 10⁻²⁷ kgm/s.
(b) The speed of hydrogen atom with same momentum is 2,079.35 m/s.
Linear Momentum of the photon
The linear momentum of the photon is calculated as follows;
P = h/λ
P = (6.63 x 10⁻³⁴) / (350 x 10⁻⁹)
P = 1.89 x 10⁻²⁷ kgm/s
Speed of hydrogen atom with same momentump = mv
v = p/m
v = (1.89 x 10⁻²⁷ ) / (9.11 x 10⁻³¹)
v = 2,079.35 m/s
Thus, the speed of hydrogen atom with same momentum is 2,079.35 m/s.
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The parking brake on a 1000 kg Cadillac has failed, and it is rolling slowly, at 5 mph , toward a group of small children. Seeing the situation, you realize you have just enough time to drive your 2000 kg Volkswagen head-on into the Cadillac and save the children. Part A With what speed should you impact the Cadillac to bring it to a halt?
To solve this problem we will apply the concepts related to the conservation of the Momentum. The initial momentum must be equal to the final momentum. Momentum is described as the product between body mass and its respective velocity. Since there is no movement at the end of the collision, the final momentum will be zero.
Our values are given as
Mass of Cadillac
[tex]m_1 = 1000kg[/tex]
mass of VW
[tex]m_2 = 2000kg[/tex]
Let VW impact with speed v, then for conservation of momentum
[tex]m_1v_1+m_2v_2 = 0 \rightarrow[/tex] the Final speed is zero
Replacing,
[tex]1000kg*5mph+2000kg(-v) = 0[/tex]
[tex]v = \frac{1000kg*5mph}{2000kg}[/tex]
[tex]v = 2.5mph[/tex]
Therefore the speed to bring the cadillac to a halt is 2.5mph
The speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
Velocity of the car can be calculated by the conservation of momentum formula,
[tex]\bold {m_1 v_1 =m_2v_2 = 0}[/tex] Since the final speed is zero,
where,
[tex]\bold {m_1 - mass\ of\ the\ cadillac = 1000 kg}\\\bold {m_2 - mass\ of\ the\ Volks\ Vagon = 2000 kg}[/tex]
[tex]\bold {v_1}[/tex] - velocity of the Cadillac - 5 mph.
Put the values in the formula
[tex]\bold {1000\ kg \times 5\ mph + 2000 kg (-v) = 0 }\\\\\bold {v = \dfrac {1000\ kg \times 5\ mph}{2000\ kg}}\\\\\bold {v = 2.5\ mph}[/tex]
Therefore, the speed that can impact the Cadillac to bring it to a halt is 2.5 mph.
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Which of the following statements are true? Check all that apply. Check all that apply. A satellite's motion is independent of its mass. A geosynchronous satellite has a period of approximately 28 days. A satellite's velocity and orbital radius are independent of each other. The launch speed of a satellite determines the shape of its orbit around Earth.
Answer:
The following are true:
A satellite's motion is independent of its mass.
The launch speed of a satellite determines the shape of its orbit around Earth.
Explanation:
A satellite's motion is independent of its mass.
The speed of a satellite can be determined by means of the Universal law of gravity:
[tex]F = G\frac{M\cdot m}{r^{2}}[/tex] (1)
Where G is the gravitational constant, M and m are masses of the two objects and r is the distance between them.
In equation 1, Newton's second law can be replaced:
[tex]F = m\cdot a[/tex]
[tex]m. a = G\frac{M\cdot m}{r^{2}}[/tex] (2)
Since it is a circular motion, the centripetal acceleration can be used:
[tex]a = \frac{v^{2}}{r}[/tex] (3)
Then, equation 3 can be replaced in equation 2:
[tex]m\frac{v^{2}}{r} = G\frac{M\cdot m}{r^{2}}[/tex] (4)
In this case, m is the satellite's mass and M is the Earth mass. Therefore, v can be isolated from equation 4:
[tex]v^{2} = G\frac{M\cdot m r}{mr^{2}}[/tex]
[tex]v^{2} =\frac{G M}{r}[/tex]
[tex]v = \sqrt{\frac{G M}{r}}[/tex]
Where G is the gravitational constant, M is the mass of the Earth and r is the orbital radius of the satellite.
Notice, how the speed of the satellite does not depend in its mass, only in the Earth mass and its orbital radius.
The launch speed of a satellite determines the shape of its orbit around Earth.
Depending on the launch speed the satellite can reach a lower or higher height above the surface of the Earth and, therefore, its orbital radius will be bigger or smaller according with that height.
Remember that the orbital radius for the satellite will be the sum of the Earth radius and the height above the surface.
Key term:
Geosynchronous satellite: It is a satellite with the same orbital period as Earth rotation period (24 hours).
A 747 jumbo jet of mass 300,000 kg accelerates down the runway at 4 m/s2 . What must be the force generated by each of its four engines
Answer:
300000 N
Explanation:
We are given that
Mass of jumbo jet=300,000kg
Acceleration of jet=[tex]4m/s^2[/tex]
We have to find the force generated by each of its four engines.
We know that
Force,F =ma
Total force exert by four engines=4F
[tex]4F=300000\times 4[/tex]
[tex]F=\frac{300000\times 4}{4}=300000N[/tex]
Hence, the force generated by each of its four engines=300000 N
Suppose a thin conducting wire connects two conducting spheres. A negatively charged rod is brought near one of the spheres, the wire between them is cut, and the charged rod is taken away. Which one of the following is true? a. The spheres will attract each other. b. The spheres will repel each other. c. There will be no electrostatic force between the spheres
Answer:
a. The spheres will attract each other.
Explanation:
When two conducting spheres are connected by a conducting wire and a negatively charged rod is brought near it then this will induce opposite (positive) charge at the nearest point on the sphere and by the conservation of charges there will also be equal amount of negative charge on the farthest end of this conducting system this is called induced polarization.
When the conducting wire which joins them is cut while the charged rod is still in proximity to of one of the metallic sphere then there will be physical separation of the two equal and unlike charges on the spheres which will not get any path to flow back and neutralize.Hence the two spheres will experience some amount of electrostatic force between them.If the speed of sound in air is 340 m/s, what is approximately the length of the shortest air column closed at one end that will respond to a tuning fork of frequency 198 Hz?
Answer:
42.9 cm
Explanation:
given,
speed of sound, v = 340 m/s
Frequency of tuning fork, f = 198 Hz
The length of the shortest air column closed at one end.
this is the case of the standing wave.
L = N λ
N = the number of complete sine waves in the standing wave.
The standing wave in the problem is the first harmonic of a pipe open at one end.
N = 1/4
we know,
v = f λ
340 = 198 x λ
λ = 1.717 m
now,
L = N λ
[tex]L = \dfrac{1}{4}\times 1.717[/tex]
[tex]L = 0.429\ m[/tex]
L = 42.9 cm
minimum length of the required pipe is 42.9 cm.
did you find the answer?
Gauss's law is usualy written as :
∫Φ_E = E * dA= qencl / ε0
what does qencl refers in Guass's law?
O the net charge inside the closed surface
O the charge residing on insulators inside the closed surface
O all the charge in the physical system
O any charge inside the closed surface that is arranged symmetrically
Answer:
(a) the net charge inside the closed surface.
Explanation:
In Gauss' Law, Qencl refers to the net charge inside the Gaussian surface. This surface is usually taken as a symmetric geometric surface, but this is merely for simplicity. Gauss' Law holds for any closed surface. Inside this surface there can be insulators as well as conductors. Regardless of the geometry or the materials inside, Qencl refers to the net charge inside the closed surface. The charge outside the surface is irrelevant for Gauss' Law, therefore all the charge in the physical system is not included in Gauss' Law.
A charge Q is transferred from an initially uncharged plastic ball to an identical ball 17 cm away. The force of attraction is then 68nM .
How many electrons were transferred from one ball to the other?
Answer:
2.92 x 10¹² electrons
Explanation:
given,
distance between two plastic ball, r = 17 cm
r = 0.17 m
Force of attraction = F = 68 mN
F = 0.068 N
number of electron transferred from one ball to another.
using Coulomb Force equation
[tex]F = \dfrac{kq^2}{r^2}[/tex]
[tex]0.068 = \dfrac{9\times 10^9\times q^2}{0.17^2}[/tex]
q² = 2.1835 x 10⁻¹³
q = 4.67 x 10⁻⁷ C
now, number of electron
[tex]N = \dfrac{q}{e}[/tex]
e is the charge of electron which is equal to 1.6 x 10⁻¹⁹ C
[tex]N = \dfrac{4.67\times 10^{-7}}{1.6\times 10^{-19}}[/tex]
N = 2.92 x 10¹² electrons
electrons were transferred from one ball to the other is 2.92 x 10¹²
Field strength is directly proportional to the square of charge. The charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex]
From Coulomb law:[tex]F = \dfrac {kQ^2}{r^2}[/tex]
Where,
[tex]F[/tex]- field strength = 68 mN = 0.06 N
[tex]Q[/tex]- charge
[tex]r[/tex]- distance = 17 cm =0.17 m
[tex]k[/tex]- constant = 9x10⁹
Put the values in the formula,
[tex]0.068 = \dfrac{9\times 10^9 \rimes Q^2}{0.017^2}\\\\q^2 = 2.1835 \times 10^{-13}\\\\q = 4.67 \times 10^{-7} \rm \ C[/tex]
Therefore, the charge on the given plastic ball is [tex]4.67 \times 10^{-7} \rm \ C[/tex].
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Tectonic plates are large segments of the Earth's crust that move slowly. Suppose one such plate has an average speed of 2.0 cm per year.
What distance does it move in 80 seconds at this speed?
Answer:
[tex]d=2.5367\times 10^{-6}\ cm[/tex]
Explanation:
Given:
speed of tectonic plate, [tex]v=2\ cm.yr^{-1}=\frac{2}{365\times 24\times 60 \times 60}\ cm.s^{-1}[/tex]time of recording the plate movement, [tex]t=80\ s[/tex]Now the distance moved by the tectonic plate in the stipulated times:
[tex]d=v.t[/tex]
[tex]d=\frac{1}{365\times 12\times 3600} \times 80[/tex]
[tex]d=2.5367\times 10^{-6}\ cm[/tex]
Tectonic plates are the mass rock from of the earth's uppermost mantle and the crust which show similar behavior of movement with respect to the other such groups of lithosphere.
Two identical 2.0-kg objects are involved in a collision. The objects are on a frictionless track. The initial velocity of object A is 12 m/s and the initial velocity of object B is -6.0 m/s. What is the final velocity of the two objects
Answer:
final velocity of the objects will be 3 m/sec
Explanation:
We have given there are two identical objects of mass 2 kg
So [tex]m_1=m_2=2kg[/tex]
Initial velocity of object A [tex]v_1=12m/sec[/tex]
Initial velocity of object B [tex]v_2=-6m/sec[/tex]
According to conservation of momentum
[tex]m_1v_1+m_2v_2=(m_1+m_2)v[/tex]
[tex]2\times 12+2\times -6=(2+2)v[/tex]
[tex]4v=24-12[/tex]
[tex]4v=12[/tex]
v = 3 m/sec
So final velocity of the objects will be 3 m/sec
To find the final velocity of mass B after a collision, you can use the principle of conservation of momentum. The formula for conservation of momentum is (mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B). By substituting the given values and solving for the final velocity of mass B, we can determine its value.
Explanation:To find the final velocity of mass B after the collision, we can use the principle of conservation of momentum. Since the collision is elastic, the total momentum before the collision is equal to the total momentum after the collision.
Mass A has an initial velocity of 5.0 m/s in the +x-direction and after the collision, moves with a velocity of 3.0 m/s in the -x-direction. Mass B has an initial velocity of 3.0 m/s in the -x-direction.
Let's assume the final velocity of mass B is vB. Using the principle of conservation of momentum, we can set up the equation:
Initial momentum = Final momentum
(mass of A * initial velocity of A) + (mass of B * initial velocity of B) = (mass of A * final velocity of A) + (mass of B * final velocity of B)
Substituting the given values:
(5.0 kg * 5.0 m/s) + (5.0 kg * 3.0 m/s) = (5.0 kg * 3.0 m/s) + (mass of B * vB)
Simplifying the equation:
25.0 m/s + 15.0 m/s = 15.0 m/s + (mass of B * vB)
40.0 m/s = 15.0 m/s + (mass of B * vB)
Subtracting 15.0 m/s from both sides:
25.0 m/s = (mass of B * vB)
Dividing both sides by the mass of B:
vB = 25.0 m/s / mass of B
Therefore, the final velocity of mass B after the collision is 25.0 m/s divided by the mass of B.
A steam power plant operates on an ideal reheat Rankine cycle between the pressure limits of 15 MPa and 10 kPa. The mass flow rate of steam through the cycle is 12 kg/s. Steam enters both stages of the turbine at 500C.
If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10%, determine
(a) The pressure, in kPa ,at which reheating takes place,
(b) The total rate of heat input, in kW, in the boiler,
(c) The thermal efficiency of the cycle. Also, show the cycle on a T-s diagram with respect to the saturation lines.
This is a complex thermodynamics problem about a steam power plant's ideal reheat Rankine cycle. It requires calculation of pressure at reheat stage, total heat input rate and thermal efficiency by using correct thermodynamic practices. Due to the complexity, expert understanding and use of steam tables is necessary.
Explanation:This is a thermodynamics problem associated with the ideal reheat Rankine cycle in a steam power plant. Due to the complex nature of the problem, the aid of thermodynamics and steam table reference books is required to solve it.
Under these ideal cycle conditions, calculating assumed pressure drops and using the steam tables, we can estimate the reheat pressure during the cycle. The exact figures would be subject to critical examination and calculation based on precise cycle parameters following real-life engineering practices.
For part (b), the total rate of heat input can be calculated using the mass flow rate and specific enthalpies at different stages of the cycle. The thermal efficiency for part (c) can be obtained by dividing the net work output by the total heat input.
An actual T-s diagram for the Rankine cycle would aid in visualizing the process but cannot be drawn here. Basically, the cycle is plotted with entropy(s) on the x-axis and temperature(T) on the y-axis, showing the expansion, reheat, and condensation stages of the steam.
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The Ideal Reheat Rankine Cycle is a model for determining efficiency in steam power plants. The pressure at which reheating takes place is typically assumed to be at the midpoint of the high and low-pressure limits. The total heat rate input in the boiler and thermal efficiency cannot be determined without more detailed specification of the conditions.
Explanation:This question is related to the concepts of thermodynamics in the field of Mechanical Engineering. The Ideal Reheat Rankine Cycle is a model used to predict the performance of steam turbine systems, and we will use principles from this model to answer your questions.
(a) The pressure at which reheating takes place cannot be definitively stated without more specific information like the temperature at the exit of the high-pressure turbine or the entropy at key points in the cycle. However, it’s commonly assumed in practice to occur at the mid-point of the high and low-pressure limits, which would be 7.5 MPa or 7500 kPa.
(b) The total rate of heat input in the boiler cannot be definitively stated without more specific information such as the specific enthalpies at key points in the cycle.
(c) Thermal efficiency of the cycle also requires further specific information such as the enthalpy or temperature at key points in the cycle to compute.
Moreover, drawing a T-s (Temperature vs Entropy) diagram with respect to the saturation lines would need a graphical interface, so it's not possible to illustrate it here.
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A point charge q1 = 1.0 µC is at the origin and a point charge q2 = 6.0 µC is on the x axis at x = 1 m.
(a) Find the electric force on charge q2.
F12 = ? mN
(b) Find the electric force on q1.
F21 = ? mN
(c) What would your answers for Parts (a) and (b) differ if q2 were -6.0 µC?
To solve this problem we will apply the concepts related to the Electrostatic Force given by Coulomb's law. This force can be mathematically described as
[tex]F = \frac{kq_1q_2}{d^2}[/tex]
Here
k = Coulomb's Constant
[tex]q_{1,2} =[/tex] Charge of each object
d = Distance
Our values are given as,
[tex]q_1 = 1 \mu C[/tex]
[tex]q_2 = 6 \mu C[/tex]
d = 1 m
[tex]k = 9*10^9 Nm^2/C^2[/tex]
a) The electric force on charge [tex]q_2[/tex] is
[tex]F_{12} = \frac{ (9*10^9 Nm^2/C^2)(1*10^{-6} C)(6*10^{-6} C)}{(1 m)^2}[/tex]
[tex]F_{12} = 54 mN[/tex]
Force is positive i.e. repulsive
b) As the force exerted on [tex]q_2[/tex] will be equal to that act on [tex]q_1[/tex],
[tex]F_{21} = F_{12}[/tex]
[tex]F_{21} = 54 mN[/tex]
Force is positive i.e. repulsive
c) If [tex]q_2 = -6 \mu C[/tex], a negative sign will be introduced into the expression above i.e.
[tex]F_{12} = \frac{(9*10^9 Nm^2/C^2)(1*10^{-6} C)(-6*10^{-6} C)}{(1 m)^{2}}[/tex]
[tex]F_{12} = F_{21} = -54 mN[/tex]
Force is negative i.e. attractive
What best describes the angle between a changing electric field and the electromagnetic wave produced by it?
always equal to a right angle
always less than a right angle
varies from parallel to perpendicular
varies from perpendicular to parallel
The correct answer here is Option A, always equal to a right angle.
Keep in mind that the phase difference that exists between the electric field and the magnetic wave is 90 °, this means that the changing electric field and the electromagnetic wave are always perpendicular and will be forming a right angle.
A positively charged non-metal ball A is placed near metal ball B. Prove that if the charge on B is positive but of small magnitude, the balls will be attracted to each other.
a. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.
b. The negative charges on the sphere A move to the side closer to sphere B, while the positive charges move to the side further away from sphere B. The polarization of charge on A will cause a greater attractive force than the repulsion of the like charges.
c. The positive charges on the sphere B move to the side closer to sphere A, while the negative charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
d. The negative charges on the sphere B move to the side closer to sphere A, while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Answer:
D
Explanation:
Ball A is a non positively charged non metal while ball B is metal ball.
Given: The ball B positive charge of small magnitude
To prove: Balls will attract each other
IN this condition because of induction negative charges on the sphere B move to the side closer to sphere A,(induction charging) while the positive charges move to the side further away from sphere A. The polarization of charge on B will cause a greater attractive force than the repulsion of the like charges.
Hence the correct answer will be D .
An electron is moving horizontally in a laboratory when a uniform electric field is suddenly turned on. This field points vertically upward. Which of the paths shown will the electron follow, assuming that gravity can be neglected?
Explanation:
We know that electron always tends to travel in the direction opposite the direction of electric field ( force in directed opposite to the field on electron).
So, since the field points vertically upward , it force the electron move downward. So, the electron will be accelerated in downward in the right direction.
Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 1.0 meters above the ground and is trying to spray Ferdinand,who is standing 10.0 meters away. To increase the range of the water, Isabella places her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction f of the cross-sectional area of the hose hole does she have to cover to be able to spray her friend? Assume that the cross section of the hose opening is circular with a radius of 1.5 centimeters. Express your answer as a percentage to the nearest integer.
Answer:
84%
Explanation:
g = Acceleration due to gravity = 9.81 m/s²
[tex]y_0=1\ m[/tex]
x = 10 m
t = Time taken
[tex]v_0[/tex] = 3.5 m/s (assumed, as it is not given)
[tex]A_0=\pi 1.5^2[/tex]
We have the equation
[tex]y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s[/tex]
[tex]x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s[/tex]
From continuity equation we have
[tex]Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2[/tex]
Fraction is given by
[tex]f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%[/tex]
The fraction is 84%
A fraction of the cross-sectional area of the hose hole that Isabella must cover is 84%.
Given the following data:
Vertical distance = 1.0 meters.
Horizontal distance = 10.0 meters.
Radius = 1.5 cm.
Vertical speed = 3.5 m/s.
How to calculate the time.The time taken to reach a maximum height is given by this formula:
[tex]H=\frac{1}{2} gt^2\\\\t=\sqrt{\frac{2H}{g} } \\\\t=\sqrt{\frac{2\times 1.0}{9.8} }[/tex]
t = 0.452 seconds.
For the velocity required, we have:
[tex]x=x_o+Vt\\\\10=0+V0.452\\\\V=\frac{10}{0.452}[/tex]
V = 22.12 m/s.
From continuity equation, we have:
[tex]A=\frac{A_0V_0}{V} \\\\A=\frac{\pi r^2V_0}{V}\\\\A=\frac{\pi1.5^2 \times 3.5}{22.12} \\\\A=\frac{24.7433}{22.12}[/tex]
A = 1.118 cm^2.
Now, an expression for the fraction of the cross-sectional area is given by:
[tex]F=\frac{A_0-A}{A_0} \times 100\\\\F=\frac{7.0695-1.118}{7.0695} \times 100\\\\F=\frac{5.9515}{7.0695} \times 100\\\\F=0.8419 \times 100[/tex]
F = 84.19 ≈ 84%.
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A rotating wheel requires 10.0 s to rotate 39.0 revolutions. Its angular velocity at the end of the 10.0-s interval is 92.0 rad/s. What is the constant angular acceleration (in rad/s) of the wheel
Answer:
[tex]13.5 rad/s^2[/tex]
Explanation:
39 revolutions = 39 * 2π = 245 rad
Let [tex]\omega_0[/tex] be the initial angular velocity of the wheel before the 10 s interval. We have the following equation of motion for angular velocity
[tex]\omega = \omega_0 + \alpha t[/tex]
where [tex]\omega = 92 rad/s[/tex] is the final angular velocity of the rotating wheel. [tex]\alpha[/tex] is the constant angular acceleration of the wheel, which we are looking for, and t = 10s is the duration time that the wheel rotates
[tex]92 = \omega_0 + 10\alpha[/tex]
[tex]\omega_0 = 92 - 10\alpha[/tex]
We also have the following equation of motion for angles
[tex]\theta = \omega_0t + \alpha t^2/2[/tex]
where [tex]\theta = 245 rad[/tex] is the total angles rotated
we can also substitute [tex]\omega_0 = 92 - 10\alpha[/tex]
[tex]245 = 10(92 - 10\alpha) + \alpha 10^2/2[/tex]
[tex]245 = 920 - 100\alpha + 50\alpha[/tex]
[tex]50\alpha = 675 [/tex]
[tex]\alpha = 675 / 50 = 13.5 rad/s^2[/tex]
You throw a glob of putty straight up toward the ceiling, which is 3.00m above the point where the putty leaves your hand. The initial speed of the putty as it leaves your hand is 8.40m/s . What is the speed of the putty just before it strikes the ceiling? How much time from when it leaves your hand does it take the putty to reach the ceiling?
The answer provides the speed of the putty just before striking the ceiling and the time it takes for the putty to reach the ceiling. The speed is found to be 5.79 m/s and the time taken is calculated to be 1.5 seconds.
Explanation:The speed of the putty just before it strikes the ceiling:
Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is acceleration, and s is displacement, the speed is found to be 5.79 m/s.
The time it takes for the putty to reach the ceiling:
By using the equation v = u + at, where t is time, a is acceleration, u is initial velocity, and v is final velocity, the time taken is calculated to be 1.5 seconds.
It has been suggested, and not facetiously, that life might have originated on Mars and been carried to Earth when a meteor hit Mars and blasted pieces of rock (perhaps containing primitive life) free of the surface. Astronomers know that many Martian rocks have come to Earth this way. One objection to this idea is that microbes would have to undergo an enormous, lethal acceleration during the impact. Let us investigate how large such an acceleration might be. To escape Mars, rock fragments would have to reach its escape velocity of 5.0 km/s, and this would most likely happen over a distance of about 4.0m during the impact.
What would be the acceleration, in m/s, of such a rock fragment?
Answer:
[tex]a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Explanation:
According to Newton's Third Equation of motion
[tex]V_f^2-V_i^2=2as[/tex]
Where:
[tex]V_f[/tex] is the final velocity
[tex]V_i[/tex] is the initial velocity
a is the acceleration
s is the distance
In our case:
[tex]V_f=V_{escape}, V_i=0,s=4 m[/tex]
So Equation will become:
[tex]V_{escape}^2-V_i^2=2as\\V_{escape}^2-0=2as\\V_{escape}^2=2as\\a=\frac{V_{escape}^2}{2s}\\a=\frac{(5*10^3m)^2}{2*4}\\a=3125000 m/s^2\\a=3.125*10^6 m/s^2[/tex]
Acceleration, in m/s, of such a rock fragment = [tex]3.125*10^6m/s^2[/tex]
Final answer:
To find the acceleration of a rock fragment blasted from Mars, use the equation for acceleration, considering final velocity, initial velocity, and distance. Given the escape velocity of Mars, the acceleration would be approximately 6250 m/s².
Explanation:
To calculate the acceleration of a rock fragment blasted from Mars, we use the equation for acceleration: a = (v² - u²) / (2s), where v is the final velocity, u is the initial velocity, and s is the distance. Given that the final velocity is the escape velocity of Mars, 5.0 km/s, the initial velocity is 0 km/s (as it starts from rest), and the distance s is 4.0 m, we can compute the acceleration.
Substitute the values into the equation: a = ((5.0 km/s)² - (0 km/s)²) / (2 × 4.0 m) = 6.25 km/s². Converting the units to meters per second squared, we get an acceleration of 6250 m/s² for the rock fragment.
You crouch from a standing position, lowering your center of mass 18 cm in the process. Then you jump vertically. The average force exerted on you by the floor while you jump is three times your weight. What is your upward speed as you pass through your standing position in leaving the floor?
To solve this problem we will apply the concepts related to balance. Since the force applied is 3 times the weight, and the weight is defined as the multiplication between mass and gravity, we will have that the dynamic equilibrium ratio would be given by the relation,
[tex]\sum F = ma[/tex]
[tex]3mg-mg = ma[/tex]
Rearranging to find a,
[tex]a = 2g[/tex]
Using the linear motion kinematic equations, which express that the final velocity of the body, and in the absence of initial velocity, is equivalent to the product between 2 times the acceleration by the distance traveled, that is
[tex]v^2 = 2as[/tex]
[tex]v^2 = 2(2g)(0.18)[/tex]
[tex]v^2 = 2(2*9.8)(0.18)[/tex]
[tex]v = 2.66m/s[/tex]
Therefore the upward speed is 2.66m/s
A 7500 kg open train car is rolling on frictionless rails at 25.0 m/s when it starts pouring rain. A few minutes later, the car's speed is 20.0 m/s. What mass of water has collected in the car?
Answer:
Mass of rain collected will be 1875 kg
Explanation:
We have given mass of the train [tex]m_i[/tex] = 7500 kg
Velocity of the car [tex]v_i[/tex] = 25 m/sec
After few minutes velocity of the car = 20 m/sec
From conservation of momentum we know that [tex]m_iv_i=m_fv_f[/tex]
So [tex]7500\times 25=m_f\times 20[/tex]
[tex]m_f=9375kg[/tex]
[tex]m_f[/tex] will be sum of mass of car and mass of rain collected
As mass of car is 7500 kg
So mass of rain collected = 9375 - 7500 = 1875 kg
Dolphins emit clicks of sound for communication and echolocation. A marine biologist is monitoring a dolphin swimming in seawater where the speed of sound is 1522 m/s. When the dolphin is swimming directly away at 8.5 m/s, the marine biologist measures the number of clicks occuring per second to be at a frequency of 2770 Hz.
1. What is the difference (in Hz) between this frequency and the number of clicks per second actually emitted by the dolphin?
From Doppler effect we have that the frequency observed for the relation between the velocities is equivalent to the frequency observed. That is mathematically,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
Here,
Speed of sound in water [tex]v = 1522m/s[/tex]
The Dolphin swims directly away from the observer with a velocity [tex]v_s = 8.5m/s[/tex]
Observed frequency of the clicks produced by the dolphin [tex]F_r = 2770Hz[/tex]
Replacing we have,
[tex]F_r = \frac{v}{v+v_s}F_s[/tex]
[tex]F_s = \frac{v+v_s}{v}[/tex]
[tex]F_s = 2770 (\frac{1522}{1522+8.5})[/tex]
[tex]F_s = 2754.61Hz[/tex]
Therefore the frequency emitted by the dolphin is 2754.61Hz
An electrically neutral balloon is rubbed on cloth and becomes positively charged. What can be said about its mass:_______
Answer:
The charged balloon’s mass will be almost equal to the original mass
Explanation:
When an electrically neutral balloon is rubbed on cloth, it losses some electron and becomes positively charged.
The balloon becomes positively charged by the removal of electrons (electrons from the cloth repels electron in the balloon).
The mass of positively charged balloon, will be equal to the original mass minus the mass of electrons removed.
Since Electrons have negligible mass, the removal of electrons from the neutral balloon will have little effect on the original mass of the balloon.
Therefore, the charged balloon’s mass will be almost equal to the original mass
The temperature of a system rises by 45 °C during a heating process. Express this rise in temperature in Kelvin and Fahrenheit.
Answer:
318K
113°F
Explanation:
45°C = 45 + 273K = 318K
45°C = (45 × 9/5) + 32 = 81 + 32 = 113°F
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this process?
Answer:
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
Explanation:
Expression for rate law for first order kinetics is given by:
[tex]a_o=a\times e^{-kt}[/tex]
where,
k = rate constant
t = age of sample
[tex]a_o[/tex] = let initial amount of the reactant
a = amount left after decay process
We have :
[tex]a_o=x[/tex]
[tex]a=58\%\times x=0.58x[/tex]
t = 95 s
[tex]0.58x=x\times e^{-k\times 95 s}[/tex]
[tex]\k= 0.005734 s^{-1}[/tex]
Half life is given by for first order kinetics::
[tex]t_{1/2}=\frac{0.693}{k}[/tex]
[tex]=\frac{0.693}{0.005734 s^{-1}}=120.86 s[/tex]
[tex]0.005734 s^{-1}[/tex] and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..
The winning time for the 2005 annual race up 86 floors of the Empire State Building was 10 min and 49 s . The winner's mass was 60 kg . If each floor was 3.7 m high, what was the winner's change in gravitational potential energy? If the efficiency in climbing stairs is 25 %, what total energy did the winner expend during the race? How many food Calories did the winner "burn" during the race? What was the winner's metabolic power in watts during the race up the stairs? Express your answer using two significant figures.
Answer:
[tex]1.9\times 10^5\ J[/tex]
[tex]1.8\times 10^{2}\ kcal[/tex]
[tex]1.3\times 10^{2}\ kcal[/tex]
[tex]1.2\times 10^{3}\ W[/tex]
Explanation:
m = Mass of person = 60 kg
g = Acceleration due to gravity = 9.81 m/s²
t = Time taken = 10 min and 49 s
Total height
[tex]h=3.7\times 86\\\Rightarrow h=318.2\ m[/tex]
Potential energy is given by
[tex]U=mgh\\\Rightarrow U=60\times 9.81\times 318.2\\\Rightarrow U=187292.52\ J[/tex]
The gravitational potential energy is [tex]1.9\times 10^5\ J[/tex]
The energy in the climb
[tex]\dfrac{187292.52}{0.25}=749170.08\ J[/tex]
Converting to kcal or Cal
[tex]\dfrac{749170.08}{4184}=179.05594\ kcal[/tex]
The amount of energy used to climb [tex]1.8\times 10^{2}\ kcal[/tex]
Amount gone to heat
[tex]179.05594\times 0.75=134.291955\ kcal[/tex]
The amount burned [tex]1.3\times 10^{2}\ kcal[/tex]
Power is given by
[tex]P=\dfrac{E}{t}\\\Rightarrow P=\dfrac{749170.08}{10\times 60+49}\\\Rightarrow P=1154.34526\ W[/tex]
The power is [tex]1.2\times 10^{3}\ W[/tex]
A popular car stereo has four speakers, each rated at 60 W. In answering the following questions, assume that the speakers produce sound at their maximum power.
A) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.0 m.
Answer: 4.8 W/m^2
B) Find the intensity I of the sound waves produced by one 60-Wspeaker at a distance of 1.5 m.
Answer: 2.1 W/m^2
C) Find the intensity I of the sound waves produced by four 60-Wspeakers as heard by the driver. Assume that the driver is located 1.0 m from each of the two front speakers and 1.5 m from each of the two rear speakers.
Answer: 13.8 W/m^2
D)The threshold of hearing is defined as the minimum discernible intensity of the sound. It is approximately 10^(-12) W/m2. Find the distance dfrom the car at which the sound from the stereo can still be discerned. Assume that the windows are rolled down and that each speaker actually produces 0.06 W of sound, as suggested in the last follow-up comment.
d= ___.
A) Intensity: [tex]4.8 W/m^2[/tex]
B) Intensity: [tex]2.1 W/m^2[/tex]
C) Total intensity: [tex]13.8 W/m^2[/tex]
D) Distance: [tex]1.38\cdot 10^5 m[/tex]
Explanation:
A)
The relationship between intensity of a sound and power emitted is given by
[tex]I=\frac{P}{A}[/tex]
where
I is the intensity
P is the power
A is the surface area at the distance considered
For the speaker in this problem:
P = 60 W
The distance is r = 1.0 m, so the area to consider is the surface of a sphere with this radius:
[tex]A=4\pi r^2 = 4\pi(1.0)^2=12.6 m^2[/tex]
Therefore, the intensity is
[tex]I=\frac{60}{12.6}=4.8 W/m^2[/tex]
B)
For this problem, we can use the same equation we used before:
[tex]I=\frac{P}{A}[/tex]
where in this case, we have:
P = 60 W is the power of the speaker
r = 1.5 m is the distance considered
Therefore, the area of the spherical surface is
[tex]A=4\pi r^2 = 4\pi(1.5)^2=28.3 m^2[/tex]
And so, the intensity here is
[tex]I=\frac{60}{28.3}=2.1 W/m^2[/tex]
C)
In this problem, we have to consider 4 speakers. In order to find the total intensity, we have to add the intensity of each speakers.
For the two front speakers, we have
[tex]P=60 W[/tex]
r = 1.0 m
So we have already calculated their intensity in part A), and it is
[tex]I_1 = I_2 = 4.8 W/m^2[/tex]
For the two rear speakers, we have
[tex]P=60 W[/tex]
r = 1.5 m
So their intensity has been calculated in part B),
[tex]I_3 = I_4 = 2.1 W/m^2[/tex]
Therefore, the total intensity is
[tex]I=I_1 + I_2 + I_3 + I_4 = 4.8+4.8+2.1+2.1=13.8 W/m^2[/tex]
D)
In this case, we want to find the distance r at which the intensity is
[tex]I=10^{-12} W/m^2[/tex]
Using four speakers with power
P = 0.06 W
Since we have 4 speakers, the intensity due to each speaker at the location considered must be
[tex]I'=\frac{I}{4}=\frac{10^{-12}}{4}=2.5\cdot 10^{-13} W/m^2[/tex]
Using the usual equation, we find the area:
[tex]A=\frac{P}{I'}=\frac{0.06}{2.5\cdot 10^{-13}}=2.4\cdot 10^{11} m^2[/tex]
And so, we can find what is the corresponding distance:
[tex]A=4\pi r^2\\r=\sqrt{\frac{A}{4\pi}}=\sqrt{\frac{2.4\cdot 10^{11}}{4\pi}}=1.38\cdot 10^5 m[/tex]
So, the person must be at 138 km.
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The intensity of sound waves can be computed by using the equation I = P / (4πr²). For the minimum discernible intensity, we employ a rearranged version of the same formula to calculate the distance from the car stereo at the threshold of hearing.
Explanation:The question is about the calculation of sound intensity produced by a car stereo featuring four speakers, each rated at 60 Watts, at various distances. To calculate the intensity (I) of the sound waves at different distances, we use the equation I = P / (4πr²) where P represents power and r is the distance.
For part D, to find the distance (d) at which sound from the stereo could still be discerned, we rearrange the intensity equation to solve for distance, d = √(P / (4πI)). We substitute P with the sound produced (0.06 W) and I with the minimum discernible intensity (10-12 W/m²). Disclaimer: Since external factors like absorption or diffusion are not taken into account, and the car stereo is not treated as a perfect point source emitter, these values serve as approximation.
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a raft of dimensions 2m wide by 3m long by .5m thick is floating in a pond if one fourth of the raf is submerged find the density of the raft material
Answer:
250 kg/m3
Explanation:
The total volume of the raft is length times width times height
V = lwh = 2 * 3 * 0.5 = 3 m cubed
The volume of the raft that is submerged in water is 1/4 of total volume
3 /4 = 0.75 m cubed
Let water density = 1000 kg/m cubed and g = 10 m/s2
The buoyant force is equal to the weight of water displaced by the raft
F = 0.75 * 1000 * 10 = 7500 N
This force is balanced by the raft weight, so the weight of the raft is also 7500N
Mass of raft is 7500 / g = 7500 / 10 = 750 kg
Raft density is mass divided by volume = 750 / 3 = 250 kg/m3