Answer:
T = approximately 24 hs.
Explanation:
In order to keep the satellite over a fixed point on the equator, as the earth rotates, the satellite must have the same angular velocity that Earth has, which means that it must have a period equal to the time used by Earth to complete an entire rotation on itself, which is almost exactly 24 Hs.
Mathematically, this can be obtained taking into account that the force that keeps the satellite in orbit is the centripetal force, which is actually the gravitational force exerted by Earth, so we can write the following equality:
Fg = Fc ⇒ G*ms*me / (re +rsat)² = ms*ω²*(re +rsat)
By definition, ω =ΔФ / Δt
For a complete revolution, ΔФ = 2*π, and Δt = T (period of the rotation),
so we can replace ω by (2*π/T), solving then for T:
T= 86,313 sec. (24 hs are exactly 86,400 sec, so the value is actually very close to the theorical one).
The period of a satellite in a geosynchronous orbit is 24 hours, matching the Earth's rotation. The satellite follows a circular orbit keeping it in the same place relative to Earth's surface. Satellites in this orbit provide various services due to their continuous presence above a particular location.
Explanation:The period of a satellite in a geosynchronous orbit is 24 hours. It's called geosynchronous because it matches the Earth's rotation period which is also 24 hours, this is how they are able to stay over the same point on the Earth's surface. The satellites move in elliptical orbits, but those in geosynchronous orbit follow a circular path so they maintain a constant altitude and remain in constant position relative to the Earth's surface.
As there are many satellites launched each year, being in a geosynchronous orbit helps these satellites to provide continuous services like weather tracking, communications, global positioning systems, etc. These satellites also avoid being captured by Earth's gravity and pulled into the atmosphere because they are at a distance of approximately 36000km from the earth's surface, where gravity is much weaker.
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The Sun orbits the Milky Way galaxy once each 2.60 x 10^8 y, with a roughly circular orbit averaging 3.00 x 10^4 light years in radius. (A light year is the distance traveled by light in 1 y.) Calculate the centripetal acceleration of the Sun in its galactic orbit..
Final answer:
To calculate the centripetal acceleration and average speed of the Sun in its galactic orbit, we utilize relevant formulas and astronomical data. The concept of centripetal acceleration is fundamental in understanding circular motion in celestial bodies like the Sun as it orbits the Milky Way galaxy.
Explanation:
Centripetal acceleration is the acceleration directed toward the center of a circular path. To calculate the centripetal acceleration of the Sun in its galactic orbit, we use the formula: a = v^2 / r, where v is the speed of the Sun and r is the radius of its orbit. Given that the Sun's orbit radius is 3.00 x 10^4 light years and it takes 2.60 x 10^8 years to orbit the Milky Way galaxy, we can also calculate the average speed of the Sun in its galactic orbit. To determine if a nearly inertial frame of reference can be located at the Sun, we need to consider the motion relative to the galactic center.
a cat is on a merry-go-round that completes 1 full rotation in 6 seconds. The cat sits at a radius of 8.4 metres from the centre.
Find the minimum coefficient of friction to prevent the cat from sliding off.
To solve this problem we will use the concepts related to the uniform circular movement from where we will obtain the speed of the object. From there we will go to the equilibrium equations so that the friction force must be equal to the centripetal force. We will clear the value of the coefficient of friction sought.
The velocity from the uniform circular motion can be described as
[tex]v = \frac{2 \pi r}{T}[/tex]
Here,
r = Radius
T = Period
Replacing,
[tex]v = \frac{2\pi (8.4)}{6}[/tex]
[tex]v =8.7964 m/s[/tex]
From equilibrium to stay in the circle the friction force must be equivalent to the centripetal force, therefore
[tex]F_f = F_c[/tex]
[tex]\mu N = \frac{mv^2}{r}[/tex]
Here,
[tex]\mu =[/tex] Coefficient of friction
N = Normal Force
m = mass
v = Velocity
r = Radius
The value of the Normal force is equal to the Weight, then
[tex]\mu(mg) = \frac{mv^2}{r}[/tex]
Rearranging to find the coefficient of friction
[tex]\mu = \frac{v^2}{gr}[/tex]
Replacing,
[tex]\mu = \frac{(8.7964)^2}{(9.8)(8.4)}[/tex]
[tex]\mu =0.9399[/tex]
Therefore the minimum coefficient of friction to prevent the cat from sliding off is 0.9399
A(n) _________ is a line joining the points of equal magnetic declination. Contour line isohyet isotherm isomag isogonic line
Answer:
Isogonic line is a line joining the points of equal magnetic declination.
Explanation:
Isogonal line is a line that joins the places of equal declination. Also isogonal line is known as the line which connects the point having the same magnetic declination.
An isogonic line joins points of equal magnetic declination on a map. This is important for navigation, as it reflects the angle difference between true north and magnetic north. Isotherms, isohyets, and isobars are other types of isolines used in geography. Thus option 5. isogonic line is correct.
A isogonic line is a line joining points of equal magnetic declination. Magnetic declination is the angle between magnetic north (the direction the north end of a compass needle points) and true north. These lines are important for navigational purposes and are often shown on special maps known as isogonic charts.
Other types of isolines include:
Isotherm: connects points of equal temperature.Isohyet: connects points of equal precipitation.Isobar: connects points of equal atmospheric pressure.Understanding these different lines helps in various geographical and meteorological analyses, making it easier to interpret maps and forecasts.
Complete question.
A(n) _________ is a line joining the points of equal magnetic declination.
Contour lineisohyet line isotherm lineisomag lineisogonic lineThe displacement z of a particle of rest mass m0, resulting from a constant force m0g along the z-axis is including relativistic effect. Find the displacement z as a power series in time t. Compare with the classical result,
Final answer:
The displacement z of a particle with rest mass m0 subject to a constant force m0g along the z-axis, accounting for relativistic effects, involves integrating the work-energy theorem's relativistic expression for kinetic energy and comparing it with the classical displacement z = 1/2 gt².
Explanation:
The problem given involves finding the displacement z of a particle with rest mass m0 under the influence of a constant force along the z-axis, considering relativistic effects. The comparison with the classical result implies a consideration of the relativistic factor y in kinetic energy expressions.
We recall that the relativistic kinetic energy is more complex than the classical expression K = 1/2 mu², because as the velocity u of the particle approaches the speed of light c, the kinetic energy diverges due to the increase in y.
The task involves using the work-energy theorem in a relativistic context, where the work done by the force m0g produces a change in the relativistic kinetic energy. As such, relativistic mechanics dictates that energy and momentum are dependent on the relativistic factor y, which is a function of velocity.
Therefore, to expand the displacement z as a power series in time t, one would start by expressing the relativistic momentum and kinetic energy, then integrate to find the time-dependent expressions for velocity and displacement.
Comparing with the classical result, where z is simply obtained by z = 1/2 gt² (ignoring air resistance), shows how relativistic effects modify the simple parabolic trajectory of a particle in a gravitational field. The relativistic displacement would involve a series with higher-order terms in time, accounting for the varying mass with velocity.
If y(t)y(t) describes the position with time, what is the proper formula for velocity with time? (Recall velocity is related to the derivative of position with time.)
Answer:
Although this question is not complete, I would give a general solution to this kind of problems.
If y(t) describes the position of a body with time such that
y(t) = at^(n) + bt^(m) + C
Then
V(t) = dy(t)/dt = ant^(n-1) + bmt^(m-1)
Explanation:
As an example supplies the position of a particle is given by
y(t) = 4t³- 3t² + 9
V(t) = 4x3t²- 3x2t¹
V(t) = d(t)/dt = 12t² - 6t.
Another example,
If y(t) = 15t³ - 2t² + 30t -80
V(t) = d(t)/dt = 15x3t² - 4t +30 = 45t² + 4t + 30.
Basically, in the equations above the powers of t reduces by one when computing the velocity function from y(t) by differentiation (calculating the derivative of y(t)). The constant term C (9 and 80 in the functions of y(t) in examples 1and 2 above) reduces to zero because the derivative of a constant (and ordinary number without the t attached to it) is always zero.
One last example,
y(t) = 2t^6 -3t²
V(t) = d(t)/dt = 12t^5 - 6t
ou plan to excite electrons in a material by exposing it laser radiation. If you want to jump electrons from the 2 shell to the 6 shell, what wavelength of laser should you use
Answer:
[tex]410.2 nm[/tex]
Explanation:
We are given that
[tex]n_1=2,n_2=6[/tex]
We have to find the wavelength of laser should you used.
We know that
[tex]\frac{1}{\lambda}=R(\frac{1}{n^2_1}-\frac{1}{n^2_2})[/tex]
Where [tex]R=1.097\times 10^7/m[/tex]=Rydberg constant
[tex]\lambda[/tex]=Wavelength
Using the formula
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{2^2}-\frac{1}{6^2})[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{1}{4}-\frac{1}{36})[/tex]
[tex]\frac{1}{\lambda}=1.097\times 10^7(\frac{9-1}{36}=1.097\times 10^7\times \frac{8}{36}[/tex]
[tex]\frac{1}{\lambda}=\frac{1.097\times 10^7}{4}[/tex]
Using identity:[tex]\frac{1}{a^x}=a^{-x}[/tex]
[tex]\lambda=\frac{4}{1.097}\times 10^{-7}[/tex]=[tex]4.102\times 10^{-7} m[/tex]
1 nm=[tex]10^{-9} m[/tex]
[tex]\lambda=4.102\times 100 \times 10^{-9}=410.2\times 10^{-9} [/tex] m=410.2 nm
Hence, the wavelength of laser=[tex]410.2 nm[/tex]
Which of the following quantities are vectors? Select all that apply. Which of the following quantities are vectors? Select all that apply. 27 J/s 1980 kg m/s due south 35 m/s 8.2 m/s2 north-west 3.2 mi straight up 2.9 m/s2 293 K 200 s
Vector quantities have magnitude and direction. The quantities that are vectors from the given list are:
1980 kg m/s due south,
8.2 m/s^2 north-west,
3.2 mi straight up.
Explanation:Quantities that have a magnitude and direction are called vectors. Based on this definition, the following quantities from the list are vectors:
1980 kg m/s due south: This quantity has both magnitude (1980 kg m/s) and direction (south).8.2 m/s2 north-west: This quantity represents acceleration with both magnitude (8.2 m/s2) and direction (north-west).3.2 mi straight up: This quantity has a magnitude (3.2 mi) and a specific direction (upward).Therefore, these three quantities are vectors.
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A long string is stretched and its left end is oscillated upward and downward. Two points on the string are labeled A and B.At the instant shown, orient v(A) and v(B) to correctly represent the direction of the wave velocity at points A and B.At each of the points A and B, rotate the given vector to indicate the direction of the wave velocity.Figure:A string is attached and moving up and down
In an oscillating string, the wave velocity at a point on the string is oriented opposite to the direction of the string's movement at that point. Thus, if point A is moving up, v(A) is pointed down, and if point B is moving down, v(B) is pointed up.
Explanation:
The orientation of the velocity of the wave at points A and B is determined by the direction of the wave's movement. For an upward and downward oscillating string, the wave moves perpendicular to the string. When a segment of the string is moving upwards, the wave velocity at that point is downward, and vice versa.
So, if point A is currently moving upwards, then v(A) should be pointed downwards to represent the direction of the wave velocity. Similarly, if point B is moving downwards, then v(B) should be pointed upwards. The velocity vectors v(A) and v(B) are always opposite to the direction of string's movement at those points.
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An ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant, What?
1. potential energy
2. velocity
3. pressure
4. total energy
5. kinetic energy
Answer:
4. total energy
Explanation:
According to Bernoulli's principle at any two points along a streamline flow The total energy that is sum of pressure energy , Kinetic energy and potential energy of the liquid all taken in per unit volume remains constant. Therefore,
for ideal fluid flows through a pipe of variable cross section without any friction. The fluid completely fills the pipe. At any given point in the pipe, the fluid has a constant Total Energy.
A boat is traveling at 3.2 m/s in the same direction as an ocean wave of wavelength 30 m and speed 6.8 m/s. If the boat is on the crest of a wave, how much time will elapse until the boat is next on a crest?
To solve this problem we apply the kinematic equations of linear motion. For which the speed is described as the distance traveled in a time interval. This would be,
[tex]v = \frac{x}{t} \rightarrow t = \frac{x}{v}[/tex]
Our values are given as,
[tex]\text{The speed of the boat} = v_b = 3.8m/s[/tex]
[tex]\text{The speed of the ocean} = V = 6.8m/s[/tex]
[tex]\text{The wave length of the wave is the same distance traveled by boat} = d = \lambda = 30m[/tex]
[tex]\text{The relative speed of the boat} = v_r = -3.8 +6.8 = 3m/s[/tex]
Replacing we have,
[tex]t = \frac{d}{v_r}[/tex]
[tex]t = \frac{30m}{3m/s}[/tex]
[tex]t = 10s[/tex]
Therefore will take until the boat is next on a crest around to 10s
A circular test track for cars has a circumference of 3.5 kmkm . A car travels around the track from the southernmost point to the northernmost point.
-What distance does the car travel? (In km)
-What is the car's displacement from its original position? (In km)
Answer:
(A) Distance will be equal to 1.75 km
(B) Displacement will be equal to 1.114 km
Explanation:
We have given circumference of the circular track = 3.5 km
Circumference is given by [tex]2\pi r=3.5[/tex]
r = 0.557 km
(a) It is given that car travels from southernmost point to the northernmost point.
For this car have to travel the distance equal to semi perimeter of the circular track
So distance will be equal to [tex]=\frac{3.5}{2}=1.75km[/tex]
(b) If car go along the diameter of the circular track then it will also go from southernmost point to the northernmost point. and it will be equal to diameter of the track
So displacement will be equal to d = 2×0.557 = 1.114 m
When two point charges are 2.0 cm apart, each one experiences a 1.0-N electric force due to the
other charge. If they are moved to a new separation of 8.0 cm, the electric force on each of them
is closest to
A) 1.0 N.
B) 4.0 N.
C) 16 N.
D) 0.25 N.
E) 0.063 N.
Answer:
option (E)
Explanation:
F = 1 N
r = 2 cm
r' = 8 cm
F' = ?
According to the Coulomb's law
[tex]F \alpha \frac{1}{r^{2}}[/tex]
So, [tex]\frac{F'}{F}=\left (\frac{r}{r'} \right )^{2}[/tex]
[tex]\frac{F'}{1}=\left (\frac{2}{8} \right )^{2}[/tex]
F' = 0.063 N
When they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
Given:
The point charges are initially at a distance of [tex]d=2[/tex] cm.
The initial force experienced by them is [tex]f=1[/tex] N.
The electrostatic force experienced by two charges q and Q is defined as,
[tex]f=K\dfrac{qQ}{r^2}[/tex]
where K is the constant and r is the distance between the charges.
The initial force experienced by them will be,
[tex]f=K\dfrac{qQ}{d^2}=1\rm\; N[/tex]
Now, the final distance between the charges is changed to 8 cm which is equal to 4d.
So, the new force on the charges will be,
[tex]F=K\dfrac{qQ}{(4d)^2}\\F=K\dfrac{qQ}{d^2}\times \dfrac{1}{16}\\F=f\times \dfrac{1}{16}\\F=0.0625\approx0.063\rm\;N[/tex]
Therefore, when they are moved to a new separation of 8.0 cm, the electric force on each of them will be closest to 0.063 N. Option E is correct.
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(II) At t =0, an 885-9 mass at rest on the end of a horizontal Spring (K: 184 N/m) is struck by a hammer which gives it aninitial speed of 226 m/s. Determine (a) the period and frequency ofthe motion, (b) the amplitude, (c) the maximumacceleration, (d) the total energy, and (e) the kinetic energy when x =O.4O A where A is the amplitude.
Answer:
A. T = 0.4358s and f = 2.29hz
B. A = 15.67m
C. amax = 3258.71m/s
D. amax = 22601J
E. Ek = 3616.16J
Explanation:
A. The period of the motion, T = 2pi*(sqrt(m/k))
Where m is the mass of the body in motion = 885g = 0.885kg
k = the spring constant = 184N/m2
T = 2pi*(sqrt(0.885/184))
= 0.4358s
Frequency of the motion, f = 1/T
T = 0.4358s
f = 2.2949hz
B. Maximum speed, Vmax = A*(sqrt(k/m))
Where A = amplitude of the motion
Making amplitude subject of formula,
A = Vmax(sqrt(m/k))
= 226*(sqrt(0.885/184))
= 15.6739m
C. Maximum acceleration, amax = A*(k/m)
= 15.6739*(184/0.885)
= 3258.71m/s
D. Total energy, Etotal = 1/2*(m * Vmax)2
= 1/2 * 0.885 * (226)2
= 22601J
E. Kinetic energy, Ek = Etotal - mechanical energy
Ek = 1/2*(k*A2) - 1/2*(k*x2)
Where x = 0.40A
Ek = 1/2*((k*A2) - (k*0.40A)2)
= 1/2*k*A2*(1 - 0.16)
= 1/2*k*A2*0.16
But 1/2*k*A2 = 22601J
Therefore, Ek = 22601*0.16
= 3616.16J
What is the electric potential energy of a system that consists of two protons 2.1×10−15 m apart-as might occur inside a typical nucleus? Express your answer using two significant figures.
The electric potential energy of a system comprising two protons 2.1 x 10^-15 m apart can be calculated using Coulomb's law, which gives us a result of approximately 0.68 MeV.
Explanation:The electric potential energy of a system depends upon the charge of the components and the distance between them. In this case, we can calculate the electric potential energy using Coulomb's law which states that the electric potential energy 'V' between two charges is given by the equation V = k*q1*q2/r where 'k' is Coulomb's constant (8.99 × 10^9 N m^2/C^2), 'q1' and 'q2' are the two charges, and 'r' is the distance between the charges. Given that the charges are two protons, they both have the same charge (1.6 × 10^-19 Coulombs). The distance 'r' is given as 2.1 × 10^-15 m. Substituting these values in, we get: V = (8.99 × 10^9 N m^2/C^2)* (1.6 × 10^-19 C) * (1.6 × 10^-19 C) / (2.1 × 10^-15 m) which results in an electric potential energy of approximately 0.68 MeV (mega electron volts).
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A 1.50 kg rock whose density is 4700 kg/m3 is suspended by a string such that half of the rock's volume is under water.
What is the tension in the string?(In N)
Answer:
Tension T = 13.14N
Explanation:
Given:
Mass of rock m = 1.50kg
Density of rock p = 4700kg/m^3
Volume of rock V = mass/density = m/p
V = 1.50kg/4700kg/m3 = 3.19×10^-4m3
Taking the summation of forces acting on the rock;
T-W+Fb = 0
T = W - Fb .....1
T = tension
W = weight of rock
Fb = buoyant force
Fb = pw(0.5V)g = density of water × Volume under water×™ acceleration due to gravity
g = 9.8m/s^2
T = mg - pw(0.5V)g
T = 1.50×9.8 - 1000kg/m^3 ×0.5(3.19 × 10^-4) × 9.8
T = 13.14N
Final answer:
To find the tension in the string, we can use the concept of buoyancy. By equating the buoyant force and weight of the rock, we can solve for the tension. The tension will be equal to the weight of the rock minus the buoyant force.
Explanation:
In order to find the tension in the string, we can use the concept of buoyancy. Since half of the rock's volume is under water, it experiences an upward buoyant force equal to the weight of the water displaced by that volume. The buoyant force can be found using the following equation:
Buoyant force = density of water * volume of water displaced * acceleration due to gravity
The weight of the rock is equal to its mass multiplied by the acceleration due to gravity.
By equating the buoyant force and weight of the rock, we can solve for the tension in the string. The tension will be equal to the weight of the rock minus the buoyant force.
When the ambulance passes the person (switching from moving towards him to moving away from him), the perceived frequency of the sound he receives decreases by 9.27 percent. How fast is the ambulance driving?
To develop this problem we will apply the considerations made through the concept of Doppler effect. The Doppler effect is the change in the perceived frequency of any wave movement when the emitter, or focus of waves, and the receiver, or observer, move relative to each other. At first the source is moving towards the observer. Than the perceived frequency at first
[tex]F_1 = F \frac{{343}}{(343-V)}[/tex]
Where F is the actual frequency and v is the velocity of the ambulance
Now the source is moving away from the observer.
[tex]F_2 = F\frac{343}{(343+V)}[/tex]
We are also so told the perceived frequency decreases by 11.9%
[tex]F_2 = F_1 - 9.27\% \text{ of } F_1[/tex]
[tex]F_2 = F_1-0.0927F_1[/tex]
[tex]F_2 = 0.9073F_1[/tex]
Equating,
[tex]F\frac{343}{(343+V)}= 0.9073(F\frac{343}{(343-V)})[/tex]
[tex]\frac{1}{(343+V)}= 0.9073\frac{1}{(343-V)}[/tex]
[tex]0.9073(343+V) = 343-V[/tex]
[tex](0.9073)(343)+(0.9073)V = 343-V[/tex]
[tex]V+0.9073V = 343-(0.9073)(343)[/tex]
Solving for V,
[tex]V = 16.67 m/s[/tex]
This is a computer program, why is there an error between the computer values and the nominal values of R?
Answer:
the difference is due to resistance tolerance
Explanation:
In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.
This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values and the values measured with the instruments, multimeters.
Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost
In summary, the difference is due to resistance tolerance.
In the hydrogen atom, what is the total energy of an electron that is in an orbit that has a radius of 8.784 × 10^(-10) m?
The total energy of an electron in a hydrogen atom with a given orbit radius can be determined using the formula E = -13.6 eV / n². By calculating the principal quantum number n from the given radius, we can find the total energy of the electron. In this case, the total energy is -3.4 eV.
Explanation:The total energy of an electron in the hydrogen atom can be determined using the formula:
E = -13.6 eV / n²
where E is the energy, n is the principal quantum number, and eV represents electron volts.
Given that the radius of the electron's orbit is 8.784 × 10⁽⁻¹⁰⁾ m, we can find the value of n using the formula for the radius of the nth orbit:
r = 0.529 × n² / Z Å
where r is the radius, n is the principal quantum number, and Z is the atomic number. Assuming the atomic number for hydrogen is 1, solving for n gives us:
n = √(r * Z / 0.529)
Substituting the given values, we find that n = 2.
Finally, plugging in n into the energy formula, we find:
E = -13.6 eV / (2²) = -3.4 eV
Water from a 100 ft x 200 ft lawn converges into a gully. During a 1-hour storm, discharge into the gully is 0.37 cfs (ft3/s). The interception by grass is 0.02 acre-ft. Overall, 70% of the rain is infiltrated. What is the amount of rainfall?
Answer:
7344 ft^3
Explanation:
rain flown = 0.37 ft^3/s * 3600 s/hr = 1332 ft^3
rain intercepted by grass = 0.02 acre-ft = 0.02 * 43559.9 = 871.2 ft^3
AS 1 acre-ft = 43559.9 ft^3
LET THE TOTAL RAINFALL BE 'X'
Thus the equations become
x = 0.7x + 871.2 +1332 (as 70 % of rain in infiltrated)
x = 7344 ft^3
A positively charged particle initially at rest on the ground accelerates upward to 160 m/s in 2.10 s. The particle has a charge-to-mass ratio of 0.100 C/kg and the electric field in this region is constant and uniform.What are the magnitude and direction of the electric field? Express your answer to two significant digits and include the appropriate units.
Answer:
The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]
Explanation:
Given that,
Time t = 2.10 s
Speed = 160 m/s
Specific charge =Ratio of charge to mass = 0.100 C/kg
We need to calculate the acceleration
Using equation of motion
[tex]a=\dfrac{v-u}{t}[/tex]
Put the value into the formula
[tex]a=\dfrac{160-0}{2.10}[/tex]
[tex]a=76.19\ m/s^2[/tex]
We need to calculate the magnitude of the electric field
Using formula of electric field
[tex]E=\dfrac{F}{q}[/tex]
[tex]E=\dfrac{ma}{q}[/tex]
[tex]E=\dfrac{a+g}{\dfrac{q}{m}}[/tex]
Put the value into the formula
[tex]E=\dfrac{76.19+9.8}{0.100}[/tex]
[tex]E=8.6\times10^{2}\ N/C[/tex]
The direction is upward.
Hence, The magnitude of the electric field is [tex]8.6\times10^{2}\ N/C[/tex]
Answer:
E = 8.6 x 10² N/C
Explanation:
given,
initial speed of charge,u = 0 m/s
final speed of charge,v = 160 m/s
time,t = 2.1 s
charge-to-mass ratio = 0.100 C/kg
Electric field of the region = ?
Acceleration of the charge
[tex]a = \dfrac{v-u}{t}[/tex]
[tex]a = \dfrac{160 - 0}{2.1}[/tex]
a = 76.19 m/s²
specific charge = [tex]\dfrac{q}{m} = 0.1[/tex]
now,
Electric field,
[tex]E = \dfrac{F}{q}[/tex]
charge is moving upwards so,
[tex]E = \dfrac{(a + g)}{\dfrac{q}{m}}[/tex]
[tex]E = \dfrac{(76.19+9.8)}{0.1}[/tex]
E = 860 N/C
electric field , E = 8.6 x 10² N/C
hence, the magnitude of electric field is equal to E = 8.6 x 10² N/C
Consider a steam power plant that operates on the ideal reheat Rankine cycle. The plant maintains the boiler at 17.5 MPa, the reheater at 2 MPa, and the condenser at 50 kPa. The temperature is 550°C at the entrance of the high-pressure turbine, and 300°C at the entrance of the low-pressure turbine. Determine the thermal efficiency of this system.
The ideal thermal efficiency for the steam power plant operating on the reheat Rankine cycle is approximately 57.0%.
To determine the thermal efficiency of the steam power plant operating on the ideal reheat Rankine cycle, we will follow the steps in the Problem-Solving Strategies for Thermodynamics:
Identify the Temperatures and Pressures
Boiler Pressure (P1): 17.5 MPa (High-pressure turbine inlet)
Reheater Pressure (P2): 2 MPa (Low-pressure turbine inlet)
Condenser Pressure (P3): 50 kPa
High-Pressure Turbine Inlet Temperature (T1): 550°C = 823 K
Low-Pressure Turbine Inlet Temperature (T2): 300°C = 573 K
Calculate Maximum Efficiency (Carnot Efficiency)
The maximum theoretical efficiency (Carnot efficiency) for a heat engine operating between two temperatures is given by the formula:
[tex]\eta_{max} = 1 - \frac{T_c}{T_h}[/tex]
where:
[tex]T_h[/tex] = Highest temperature (at the boiler, T1) = 823 K
[tex]T_c[/tex] = Lowest temperature (at the condenser) = Convert 50 kPa to temperature
Using steam tables or Mollier diagrams, the saturation temperature corresponding to 50 kPa is approximately 81°C = 354 K.
Now substituting:
[tex]\eta_{max} = 1 - \frac{354}{823} \approx 0.570 \text{ or } 57.0\%[/tex]
You are driving down the highway at 65 m p h, which is 29 m/s. Your tires have a radius of 0.30 m. a. How many times per second does each tire rotate? b. What is the speed of a point at the top of a tire, relative to the ground?
To solve this problem we will apply the concepts related to linear velocity and angular velocity to perform the respective conversion with the given values. To find the velocity in the upper part of the tire we will use the mathematical relation that expresses that it is twice the linear velocity. Let's start
PART A)
[tex]\omega = \frac{v}{r}[/tex]
[tex]\omega = \frac{29}{0.3}[/tex]
[tex]\omega = 96.66 rad/s[/tex]
Now we now that [tex]2\pi rad = 1 rev[/tex], then
[tex]\omega = 96.66rad/s (\frac{1 rev}{2\pi rad})[/tex]
[tex]\omega = 15.38rev/s[/tex]
PART B)
[tex]v = 2v_0[/tex]
[tex]v = 2(29)[/tex]
[tex]v = 58m/s[/tex]
A horse canters away from its trainer in a straight line, moving 126 m away in 13.0 s . It then turns abruptly and gallops halfway back in 5.0 s . Calculate the (a) its average speed and (b) itsaverage velocity for the entire trip, using "away from the trainer"as the positive direction.
Final answer:
The average speed of the horse for the entire trip is 10.5 m/s, and the average velocity is 3.5 m/s, calculated based on the total distance covered and total displacement over the entire duration of the trip.
Explanation:
To calculate the average speed and average velocity of the horse for the entire trip, we consider both phases of the journey: when the horse canters away and when it gallops back.
Part (a) - Average Speed
Average speed is total distance covered divided by total time taken. The horse travels 126 m away and then comes back halfway, which is 63 m, resulting in a total distance of 126 m + 63 m = 189 m. The total time taken for the trip is 13.0 s + 5.0 s = 18.0 s. Therefore, the average speed is:
average speed = total distance / total time = 189 m / 18.0 s = 10.5 m/s
Part (b) - Average Velocity
Average velocity is the total displacement divided by the total time taken. Since the horse ends up halfway back, the displacement is 126 m - 63 m = 63 m, in the initially defined positive direction (away from the trainer). Using a total time of 18.0 s:
average velocity = total displacement / total time = 63 m / 18.0 s = 3.5 m/s
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V.
Use ϵ0 = 8.85×10⁻¹² C²/N⋅m².
The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
Answer:
U_eq = 1.99 * 10^(-10) J
Explanation:
Given:
Plate Area = 10 cm^2
d = 0.01 m
k_dielectric = 3
k_air = 1
V = 15 V
e_o = 8.85 * 10 ^-12 C^2 / N .m
Equations used:
U = 0.5 C*V^2 .... Eq 1
C = e_o * k*A /d .... Eq 2
U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3
For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:
U_electric = 0.5 e_o * k_1*A*V^2 /2*d
U_air = 0.5 e_o * k_2*A*V^2 /2*d
The total Energy is:
U_eq = U_electric + U_air
U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d
U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d
Plug the given values:
U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01
U_eq = 1.99 * 10^(-10) J
The energy U2 of the partially dielectric-filled capacitor is found by calculating the total capacitance Ct when the capacitor is half-filled with the dielectric and then using the formula U2 = ½ CtV2 with the given voltage.
Explanation:To calculate the energy U2 of the partially dielectric-filled capacitor, we start by finding the capacitance when the capacitor is half-filled with the dielectric. With a dielectric constant k of 3.00 and a vacuum permittivity ε0 of 8.85×10−12 C2/N·m2, the initial capacitance Ci entirely filled with the dielectric is:
Ci = kε0A/d
When the capacitor is half-filled, the capacitance can be represented as two capacitors in parallel: one with dielectric (Cd) and one without (C0). These can be described by:
Cd = ε0kA/(2d) and C0 = ε0A/(2d)
The total capacitance Ct when the capacitor is half-filled is then:
Ct = Cd + C0
Finally, the energy stored in the capacitor U2 is given by:
U2 = ½ CtV2
This formula allows us to plug in the calculated capacitance Ct and the constant voltage V to find the desired energy U2.
A force of 35 N acts on a ball for 0.2 s. If the ball is initially at rest:
a. What is the impulse on the ball?
b. What is the final momentum of the ball?
To solve this problem we will apply the concepts related to momentum and momentum on a body. Both are equivalent values but can be found through different expressions. The impulse is the product of the Force for time while the momentum is the product between the mass and the velocity. The result of these operations yields equivalent units.
PART A ) The Impulse can be calculcated as follows
[tex]L= F\Delta t[/tex]
Where,
F = Force
[tex]\Delta t =[/tex]Change in time
Replacing,
[tex]L = (35N)(0.2s)[/tex]
[tex]L= 7N\cdot s[/tex]
PART B) At the same time the momentum follows the conservation of momentum where:
Initial momentum= Final momentum
And the change in momentum is equal to the Impulse, then
[tex]\Delta p = L[/tex]
And
[tex]\Delta p = p_f - p_i[/tex]
There is not initial momentum then
[tex]\Delta p = p_f[/tex]
[tex]L = p_f[/tex]
[tex]p_f = 7N\cdot s = 7kg\cdot m/s[/tex]
Final answer:
The impulse on the ball, which is the product of the force and the time interval, is 7 N·s. Since the ball is initially at rest, the final momentum of the ball is also 7 N·s.
Explanation:
To solve the student's question, we need to apply the concepts of impulse and momentum in physics.
Impulse and Momentum
Impulse is defined as the product of the force acting on an object and the time interval over which it acts. From Newton's second law, the impulse on an object is equal to the change in its momentum.
a) The impulse on the ball is calculated using the formula Impulse = Force × Time. Given that the force is 35 N and the time is 0.2 s, the impulse can be calculated as:
Impulse = 35 N × 0.2 s = 7 N·s
b) The final momentum of the ball is equal to the impulse, since the ball was initially at rest (momentum = 0). Therefore, the final momentum of the ball is:
Final momentum = Impulse = 7 N·s
Suppose a balloons was laying by the window at night. The next day, when the sun came up, it warmed the gas (air) that was in the balloon. What would be true about the density of the air in the balloon?
As the gas (air) in the balloon warms, it expands due to thermal expansion, causing its density to decrease. This lower density compared to the surrounding air leads to buoyancy, making the balloon rise.
Explanation:When the sun warms the gas (air) inside a balloon, the air expands due to an increase in temperature. This process, known as thermal expansion, causes the molecules in the air to move faster and spread out more, occupying a larger volume. As a result, the density of the air inside the balloon decreases because density is defined as mass per unit volume, and while the mass of the air remains constant, its volume increases. In the context of a hot air balloon, or any closed system where air is heated, this decrease in density compared to the cooler surrounding air leads to buoyancy. Buoyancy is the force that makes things float, which in the case of the balloon, causes it to rise since the hot air inside it is less dense than the cooler external environment.
Two runners start simultaneously from the same point on a circular 200-m track and run in the same direction. One runs at a constant speed of 6.20 m/s and the other runs at a constant speed of 5.50 m/s. g
Answer
given,
speed of the fastest runner = 6.2 m/s
speed of the slower runner = 5.5 m/s
Assume, we need to calculate time when both the runner meet for the first time.
distance cover by the fast runner = distance cover by the slow runner
6.2 t = 5.5 t + 200
0.7 t = 200
t = 285.71 s
time after which both the runner meet.
Distance they covered after starting
D = s x t
D = 6.2 x 285.71
D = 1771.43 m.
now, calculating the second time both will meet.
they will take double time to meet again
t' = 2 x 285.71
t '= 571.42 s
The sound source of a ship’s sonar system operates at a frequency of 22.0 kHz . The speed of sound in water (assumed to be at a uniform 20∘C) is 1482 m/s . What is the difference in frequency between the directly radiated waves and the waves reflected from a whale traveling straight toward the ship at 4.95 m/s ? Assume that the ship is at rest in the water.
Your target variable is Δf, the magnitude of the difference in frequency between the waves emitted from the sonar device and the waves received by the device after reflecting off the whale. Write an expression for Δf in terms of the relevant frequencies using the subscript notation introduced above.
Express your answer in terms of some or all of the variables fLe, fLr, fSe, and fSr.
Δf =
Answer:
[tex]\Delta f=f_{Lr}-f_{Se}[/tex]
147.45 Hz
Explanation:
v = Speed of sound in water = 1482 m/s
[tex]v_w[/tex] = Speed of whale = 4.95 m/s
Frequency of the wave in stationary condition
[tex]f_{Lr}=f\dfrac{v+v_w}{v-v_w}[/tex]
Ship's frequency which is reflected back
[tex]f_{Se}=f\dfrac{v}{v-v_w}[/tex]
The difference in frequency is given by
[tex]\Delta f=f\dfrac{v+v_w}{v-v_w}-f\dfrac{v}{v-v_w}\\\Rightarrow \Delta f=f_{Lr}-f_{Se}[/tex]
[tex]\mathbf{\Delta f=f_{Lr}-f_{Se}}[/tex]
[tex]f_{Lr}=22\times \dfrac{1482+4.95}{1482-4.95}\\\Rightarrow f_{Lr}=22.14745\ kHz[/tex]
[tex]f_{Se}=22\ kHz[/tex]
[tex]\Delta f=f_{Lr}-f_{Se}\\\Rightarrow \Delta f=22.14745-22\\\Rightarrow \Delta f=0.14745\ kHz\\\Rightarrow \Delta f=147.45\ Hz[/tex]
The difference in wavelength is 147.45 Hz
Assume this field is generated by a point charge of Q = 5 × 10-9 C. How far away is this charge located? Give your answer in meters.
Final answer:
The distance from the charge is 30 meters.
Explanation:
The distance from the point charge can be calculated using the equation E = kQ/r^2, where E is the electric field strength, k is the electrostatic constant (9 x 10^9 Nm^2/C^2), Q is the charge, and r is the distance. In this case, the electric field strength is the same at any point 5.00 mm away from the charge. Therefore, we can set up the equation as follows:
E = kQ/r^2
5.00 x 10^-3 m = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/r^2
By rearranging the equation and solving for r, we can find the distance:
r^2 = (9 x 10^9 Nm^2/C^2)(5 x 10^-9 C)/(5.00 x 10^-3 m)
r^2 = 900 m^2
r = sqrt(900 m^2) = 30 m
Therefore, the charge is located 30 meters away.
In a recent airline disaster, an airliner ying at 30,000ft,550mi/h, lost power and fell to Earth. The mass of the aircraft was 255,000 lb. If the magnitude of the work done by drag force on the plane during the fall was 2.96 × 106 Btu, estimate the velocity of the aircraft at the time of impact, in mi/h. Let g = 32.08 ft/s2.
To develop this problem we will apply the energy conservation theorem and the principle of work. Basically we will have that the kinetic, potential and work energy must be equivalent to the kinetic energy just before the impact. We will start converting the units given to the British system and then proceed with the Calculations.
Remember that according to the energy balance in this case it would be balanced like this
[tex]T_1 +\sum U_{1-2} = T_2[/tex]
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Here
m = mass
[tex]v_{1,2}[/tex]= Velocity at each moment
[tex]W_{drag}[/tex]= Work by drag
h = Height
g = Acceleration due to gravity
Mass
[tex]m =255000 lb (\frac{1slug}{32.174lb})[/tex]
[tex]m = 7925.654slug[/tex]
Initial Velocity
[tex]v_1 = 550 \frac{mi}{h} (\frac{5280ft}{1mi})(\frac{1hr}{3600s})[/tex]
[tex]v_1 = 806.667ft/s[/tex]
Work by drag
[tex]W_{drag} = (2.96*10^6BTU)(\frac{778.169lb\cdot ft}{1BTU})[/tex]
[tex]W_{drag} = 2303380240lb\cdot ft[/tex]
By work energy principle
[tex]\frac{1}{2} mv_1^2+mgh -W_{drag} = \frac{1}{2} mv_2^2[/tex]
Replacing,
[tex]\frac{1}{2} (7925.654slug)(806.667ft/s)^2 +(7925.654slug)(32.08ft/s^2)(30000ft) -2303380240lb\cdot ft = \frac{1}{2} (7925.654slug)v_2^2[/tex]
Solving for [tex]v_2[/tex], we have that
[tex]v_2 = 1412.2 ft/s[/tex]
Converting this value,
[tex]v_2 = 1412.2 ft/s (\frac{1mi}{5280ft})(\frac{3600s}{1h})[/tex]
[tex]v_2 = 962.85mi/h[/tex]
Therefore the velocity of the aircraft at the time of impact is 962.85mi/h
To estimate the velocity of the aircraft at the time of impact, use the work-energy principle. The estimated velocity is 256.15 mi/h.
Explanation:To estimate the velocity of the aircraft at the time of impact, we can use the work-energy principle. The work done by the drag force is equal to the change in kinetic energy of the aircraft. The work done by the drag force is given as 2.96x10^6 Btu. We need to convert this energy into foot-pounds, and then use the kinetic energy equation to find the velocity. The equation is:
K = (1/2) mv^2
Where K is the kinetic energy, m is the mass of the aircraft, and v is the velocity. Rearranging the equation, we have:
v = √((2K) / m)
Plugging in the given values, we get:
v = √((2(2.96x10^6 Btu) * (3.968x10^8 ft-lbf/Btu)) / (255,000 lb * 32.08 ft/s^2)) = 256.15 mi/h
Therefore, the estimated velocity of the aircraft at the time of impact is 256.15 mi/h.