Answer:
a) a = + 1,758 10¹¹ m / s , b) = √ (2 1,758 10¹¹ E x)
Explanation:
a) A charged particle is an electric field undergoing force given by the expression
F = qE
where q is the charge of the paticle and E electric field.
In this case we are told that the particle is positron
q = + 1.6 10⁻¹⁹ C
let's calculate the force
F = + 1.6 10⁻¹⁹ E
we write the positive sign, to show that the particle accelerates in the same direction of the electric field
let's write Newton's second law to find the acceleration
F = ma
a = F / m
a = + 1.6 10-19 / 9.1 10-31 E
a = + 1,758 10¹¹ m / s
b) the velocity of the particle starting from rest
v² = v₀² + 2 a x
v = √ (2 1,758 10¹¹ E x)
A point charge of -4.28 pC is fixed on the y-axis, 2.79 mm from the origin. What is the electric field produced by this charge at point P, which is on the x-axis, 9.83 mm from the origin
Answer:
E = (-3.61^i+1.02^j) N/C
magnitude E = 3.75N/C
Explanation:
In order to calculate the electric field at the point P, you use the following formula, which takes into account the components of the electric field vector:
[tex]\vec{E}=-k\frac{q}{r^2}cos\theta\ \hat{i}+k\frac{q}{r^2}sin\theta\ \hat{j}\\\\\vec{E}=k\frac{q^2}{r}[-cos\theta\ \hat{i}+sin\theta\ \hat{j}][/tex] (1)
Where the minus sign means that the electric field point to the charge.
k: Coulomb's constant = 8.98*10^9Nm^2/C^2
q = -4.28 pC = -4.28*10^-12C
r: distance to the charge from the point P
The point P is at the point (0,9.83mm)
θ: angle between the electric field vector and the x-axis
The angle is calculated as follow:
[tex]\theta=tan^{-1}(\frac{2.79mm}{9.83mm})=74.15\°[/tex]
The distance r is:
[tex]r=\sqrt{(2.79mm)^2+(9.83mm)^2}=10.21mm=10.21*10^{-3}m[/tex]
You replace the values of all parameters in the equation (1):
[tex]\vec{E}=(8.98*10^9Nm^2/C^2)\frac{4.28*10^{-12}C}{(10.21*10^{-3}m)}[-cos(15.84\°)\hat{i}+sin(15.84\°)\hat{j}]\\\\\vec{E}=(-3.61\hat{i}+1.02\hat{j})\frac{N}{C}\\\\|\vec{E}|=\sqrt{(3.61)^2+(1.02)^2}\frac{N}{C}=3.75\frac{N}{C}[/tex]
The electric field is E = (-3.61^i+1.02^j) N/C with a a magnitude of 3.75N/C
A small object with mass 3.80 kg moves counterclockwise with constant speed 1.65 rad/s in a circle of radius 2.70 m centered at the origin. It starts at the point with position vector 2.70 m. Then it undergoes an angular displacement of 8.70 rad.
(a) What is its new position vector?
in meters
(b) In what quadrant is the particle located and what angle does its position vector make with the positive x-axis?
(c) What is its velocity?
in m/s
(d) In what direction is it moving?
_____° from the +x direction.
(e) What is its acceleration?
in m/s2
(f) What total force is exerted on the object?
in N
Answer:
Explanation:
angular velocity
ω = 1.65 rad /s
radius R = 2.70 m
angular displacement = 8.70 rad
a )
New position vector in vector form
= R cos8.7 i + R sin8.7 j
= 2.7 cos8.7 i + 2.7 sin8.7 j
= 2.7 x .748 i + 2.7 x .663 j
= 2.01 i + 1.79 j
b )
8.7 radian = 180/π x 8.7 degree
= 498.72 degree
= 498.72 - 360
= 138.72 degree
It will be in second quadrant .
angle made with positive x - axis
= 138.72 degree .
c )
velocity
v = ω R
= 1.65 x 2.7
= 4.455 m /s
d )
It is moving in a direction making 138.72° with positive x direction .
e )
acceleration will be centripetal acceleration
= v²/ R
= 4.455² / 2.7
= 7.35 m /s²
f ) force = mass x acceleration
= 3.8 x 7.35
= 27.93 N .
A small truck has a mass of 2065 kg. How much work is required to decrease the speed of the vehicle from 23.0 m/s to 12.0 m/s on a level road
Answer:
397.51 kJ
Explanation:
Since the change in velocity is done on a level road, there is no change in the potential energy. The workdone is the workdone on reducing the kinetic energy.
Workdone W = change in kinetic energy ∆K.E
W = ∆K.E
K.E = 0.5mv^2
∆K.E = 0.5m(m1^2 - m2^2)
Given;
Mass m = 2065 kg
Initial velocity v1 = 23.0 m/s
Final velocity v2 = 12.0 m/s
W = ∆K.E = 0.5m(m1^2 - m2^2)
Substituting the given values;
W = ∆K.E = 0.5×2065(23^2 - 12^2)
W = 397512.5J
W = 397.51 kJ
Consider two coils, with the first coil having twice as many loops as the second. Given the flux Φ though each loop of the first coil due to current in the second coil, what can be said about the flux through each loop of the second coil due to an equal current in the first coil?
Answer:
[tex]$ \phi_{21} = \frac{\phi_{12}}{2} $[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
Explanation:
The flux through each loop of the first coil due to current in the second coil is,
[tex]\phi_{12} = \phi[/tex]
The number of loops in the first coil is
no. of loops = 2N
Total flux passing through the first coil is
[tex]\phi_{12} = 2N\phi[/tex]
The flux through each loop of the second coil due to current in the first coil is,
[tex]\phi_{21} = \phi[/tex]
The number of loops in the second coil is
no. of loops = N
Total flux passing through the second coil is
[tex]\phi_{21} = N\phi[/tex]
Comparing both
[tex]\phi_{12} = \phi_{21} \\\\ 2N\phi = N\phi\\\\\phi_{21} = \frac{\phi_{12}}{2}[/tex]
Which means that the flux through each loop of the second coil is half as much as flux through each loop of first coil.
A chain lying on the ground is 10 m long and its mass is 70 kg. How much work (in J) is required to raise one end of the chain to a height of 6 m
Answer:
The work done in moving the chain is 4116 J
Explanation:
Given;
mass of the chain, m = 70 kg
length of the chain, l = 10 m
vertical height through which one end of the chain was raised, h = 6 m
Work done in raising this chain to this height is equal to potential energy due to this vertical height
W = PE
W = mgh
where;
m is mass of the chain
g is acceleration due to gravity
h is the vertical height through which the chain was raised
W = 70 x 9.8 x 6
W = 4116 J
Therefore, the work done in moving the chain is 4116 J
A body of mass 10 kg moving on a circular path of radius 1 m completes a rotation in 2 pi sec the centripetal acceleration is A 1m/s 2 B 2m/s 2 C 4m/s 2 D none
Answer:
1 m/s²
Explanation:
a = v² / r
a = ω² r
a = (2π rad / 2π s)² (1 m)
a = 1 m/s²
A piece of iron rests on top of a piece of wood floating in a bathtub. If the iron is removed from the wood, and kept out of the water, what happens to the water level in the tub?
a. It goes up.
b. It does not change.
c. It goes down.
d. It is impossible to determine from the information given.
Answer:
It goes down.
The water level remain the same.
Explanation:
This can be explained using Archimedes principle which states that a body fully or partially submerged in a fluid is acted by an upward bouyant force which is equal to the weight of the fluid the body displaced.
The wood will only sink if the weight of the wood is greater than the weight of the fluid the wood displaced, but the weight of the wood is equal to the weight of fluid displaced, therefore the wood will float.
Therefore, the weight of the wood is the same as the weight of the fluid displaced, so the wood will be at the same level as the water.
If the iron is removed, the level of the water goes down because iron weight is bigger than the water displaced and it tends to increase the water level but since it is removed, the water level will decrease.
4. The capacitance of a capacitor is increased by a factor of 1.5 when it is completely filled
with a certain dielectric material. Find the dielectric constant of the material and its
electric susceptibility
Answer:
a. Dielectric constant, ε = 1.5 b. Electric susceptibility, χ = 0.5
Explanation:
a. Dielectric constant
Since the capacitance of the capacitor is increased by a factor of 1.5. Let its initial capacitance be C and its final capacitance after adding the material be C'.
Since C' = εC where ε = relative permittivity,
Also, C' = 1.5C
Comparing both equations for C', ε = 1.5.
Since ε = relative permittivity = dielectric constant,
dielectric constant = 1.5
So, the dielectric constant = 1.5
b. Electric susceptibility
The electric susceptibility χ is given by
χ = ε - 1 where ε = dielectric constant
Since ε = 1.5,
χ = ε - 1
χ = 1.5 - 1
χ = 0.5
So the electric susceptibility χ = 0.5
A spring with k = 35.5 N/m has a mass of 5.50 kg attached to it. An external force F = (4.40 N)sin[(6.80 s−1)t] drives the spring mass system so that it oscillates without any resistive forces. What is the amplitude of the oscillatory motion of the spring-mass system?
Answer:
A = 0.02 m
Explanation:
The spring constant, k = 35.5 N/m
The attached mass, m = 5.50 kg
The expression for the external force, F = (4.40 N)sin[(6.80 s⁻¹)t].....(1)
The general expression for the external force, F = F₀ sin (wt).............(2)
Comparing equations (1) and (2):
The forced frequency, [tex]\omega = 6.80 rad/s[/tex]
F₀ = 4.40 N
The natural frequency can be calculated using the formula:
[tex]\omega_0 = \sqrt{\frac{k}{m} } \\\\\omega_0 = \sqrt{\frac{35.5}{5.5} } \\\\\omega_0 = 2.54 rad/s[/tex]
The amplitude of oscillation of a spring-mass system in the steady state:
[tex]A = \frac{F_0}{m(\omega^2 - \omega_o^2)} \\\\A = \frac{4.4}{5.5(6.8^2 - 2.54^2)} \\\\A = 0.02 m[/tex]
A brass rod with a mass of 0.300 kg slides on parallel horizontal iron rails, 0.440 m apart, and carries a current of 15.0 A. The coefficient of friction between the rod and rails is 0.300. What vertical, uniform magnetic field is needed to keep the rod moving at a constant speed
Answer:
The magnitude of the magnetic field is [tex]B = 0.0890 \ T[/tex]
Explanation:
From the question we are told that
The mass of the rod is [tex]m =0.300 \ kg[/tex]
The distance of separation is [tex]d = 0.440 \ m[/tex]
The current is [tex]I = 15.0 \ A[/tex]
The coefficient of friction is [tex]\mu = 0.300[/tex]
Generally for the rod the rod to continue moving at a constant speed
The frictional force must equal to the magnetic field force so
[tex]F_m = F_f[/tex]
Where [tex]F_m = B* I * d[/tex]
and [tex]F_f = \mu * m * g[/tex]
[tex]B*I *d = \mu * m * g[/tex]
=> [tex]B = \frac{\mu * m * g }{I * d }[/tex]
substituting values
[tex]B = \frac{0.2 * 0.300 * 9.8 }{ 15 * 0.440 }[/tex]
[tex]B = 0.0890 \ T[/tex]
An evacuated tube uses an accelerating voltage of 50 kV to accelerate electrons to hit a copper plate and produce X-rays. Non-relativistically, what would be the maximum speed (in m/s) of these electrons
Answer:
Explanation:
Accelerating voltage = 50 x 10³ V
energy created = q V where q is charge on electron and V is accelerating potential
So qV = 1/2 m v² , m is mass of electron and v is velocity
Putting values
1.6 x 10⁻¹⁹ x 50 x 10³ = 1/2 x 9.1 x 10⁻³¹ x v²
v² = 175.8 x 10¹⁴
v = 13.25 x 10⁷ m /s
An object of mass 4kg is moving along a horizontal plane. If the coefficient of kinetic friction is 0.2 find the friction force acting on the object.
Answer:
The friction force acting on the object is 7.84 N
Explanation:
Given;
mass of object, m = 4 kg
coefficient of kinetic friction, μk = 0.2
The friction force acting on the object is calculated as;
F = μkN
F = μkmg
where;
F is the frictional force
m is the mass of the object
g is the acceleration due to gravity
F = 0.2 x 4 x 9.8
F = 7.84 N
Therefore, the friction force acting on the object is 7.84 N
The net torque acting on an object is zero. What can you say about the angular momentum (about the same axis) of the object?
Answer:
The angular momentum (about the same axis) of the object is constant
Explanation:
Torque is the time time derivative of angular momentum.
τ = dL / dt
where;
dL is the change in angular momentum
τ is torque acting on the object
dt is change in time
when net torque acting on an object is zero, then
dL / dt = 0
Change in angular momentum (ΔL) is zero, therefore we can say that the angular momentum (L) is constant.
Thus, the angular momentum (about the same axis) of the object is constant.
Each charge is equidistant from the origin. In which direction is the net electric field at the point P on the y-axis?
Answer:
"Upwards and towards the left" is the right answer.
Explanation:
The magnitude of the field will be:
⇒ [tex]E=\frac{kq}{r^2}[/tex]
And direction -> for negative charges, to positive charges, except charges.
Charging across the y-axis. It would be up to the aggregate field. Because the x-axis needs to charge. Total production is to the west.Thus the net field is upwards as well as to the left.
The bases of developing convective cumulus clouds will be relatively higher at a location with a relatively ______ difference between the surface temperature and surface dew point temperature.
Answer:
large
Explanation:
Cumulus clouds is a term in metrology that defines the type of clouds which are characterized by its low altitude, puffy appearance, and fair-weather nature. They are generally considered as low-level clouds, with less than than 2,000m in altitude except they are the more vertical cumulus congestus form.
Thus, it can be noted that, the difference between the surface dew point temperature and the surface temperature is related to relative humidity. Hence, in a situation when there is a LARGE difference between the surface temperature and the surface dewpoint temperature, then the relative humidity is very low (e.g., 10%).
Therefore, the bases of developing convective cumulus clouds will be relatively higher at a location with a relatively LARGE difference between the surface temperature and surface dew point temperature.
Using the image below, identify the numbered parts of the wave
Answer:
1) Wavelength
2) Peak of Wave
3) Trough of Wave
4) Amplitude of Wave
(1) Wavelength
(2) Peak of Wave
(3) Trough of Wave
(4) Amplitude of Wave
What are the terms shown in the image of wave?Wavelength -The distance between two successive crests or troughs of the wave is called as wavelength
Amplitude- Maximum distance between highest point to the axix of the wave is amplitude.
Crest - Positive amplitude of the wave is crest
Trough- Negative amplitude of the wave is called trough.
Hence
(1) Wavelength
(2) Peak of Wave
(3) Trough of Wave
(4) Amplitude of Wave
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An eagle is flying horizontally at a speed of 4.6 m/s when the fish in her talons wiggles loose and falls into the lake 4.2 m below. Calculate the vertical component of the velocity of the fish relative to the water when it hits the water.
Answer:
v = 9.07 m/s
the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s
Explanation:
Given;
An eagle is flying horizontally at a speed of 4.6 m/s
Initial horizontal velocity uh = 4.6 m/s
Initial vertical velocity uy = 0
Height to fall d = 4.2 m
Acceleration due to gravity g = 9.8 m/s^2
The final vertical velocity of the fish when it hits the water can be calculated using the equation of motion;
v^2 = u^2 + 2as
v^2 = uy^2 + 2gd
uy = 0
v^2 = 2gd
v = √(2gd)
Substituting the given values;
v = √(2×9.8×4.2)
v = 9.073036977771 m/s
v = 9.07 m/s
the vertical component of the velocity of the fish relative to the water when it hits the water is 9.07 m/s
79. An example of an electrical insulator is
a. graphite.
b. glass.
c. aluminum
d. tungsten
Answer:
B. Glass
Explanation:
An electrical insulator is a substance that does not conduct electricity.
Glass has tightly bounded electrons, that is why it is an insulator of electricity.
Hey there! The answer to your question is below.
The correct answer would be B.GLASS
Glass is a insulator and rubber too
Glass, wood, plastic and more are all insulators
So the answer would be B. Glass
Hope this helps!
By: xBrainly
Wind gusts create ripples on the ocean that have a wavelength of 3.03 cm and propagate at 3.37 m/s. What is their frequency (in Hz)?
Answer:
Their frequency is 111.22 Hz
Explanation:
Wavelength is the minimum distance between two successive points on the wave that are in the same state of vibration and is expressed in units of length (m).
Frequency is the number of vibrations that occur in a unit of time. Its unit is s⁻¹ or hertz (Hz).
The propagation speed of a wave is the quantity that measures the speed at which the wave's disturbance propagates throughout its displacement. The speed at which the wave propagates depends on both the type of wave and the medium through which it propagates. Relate wavelength (λ) and frequency (f) inversely proportional using the following equation:
v = f * λ.
Then the frequency can be calculated as: f=v÷λ
In this case:
λ=3.03 cm=0.0303 m (1m=100 cm)v= 3.37 m/sReplacing:
[tex]f=\frac{3.37 \frac{m}{s} }{0.0303 m}[/tex]
Solving:
f=111.22 Hz
Their frequency is 111.22 Hz
A woman is listening to her radio, which is 174 m from the radio station transmitter. (a) How many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz
Explanation:
It is given that,
The distance between the radio and the radio station is 174 m
We need to find how many wavelengths of the radio waves are there between the transmitter and radio receiver if the woman is listening to an AM radio station broadcasting at 1540 kHz.
f = 1540 kHz
Wavelength,
[tex]\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1540\times 10^3}\\\\\lambda=194.8\ m[/tex]
Let there are n wavelengths of the radio waves. So,
[tex]n=\dfrac{d}{\lambda}\\\\n=\dfrac{174}{194.8}\\\\n=0.89\ \text{wavelengths}[/tex]
There are 0.89 wavelengths.
How did the horizontal velocity vector component change during the flight of the cannonball in the simulation
Answer:
The horizontal velocity vector of the canonball does not change at all, but is constant throughout the flight.
Explanation:
First, I'll assume this is a projectile simulation, since no simulation is shown here. That been the case, in a projectile flight, there is only a vertical component force (gravity) acting on the body, and no horizontal component force on the body. The effect of this on the canonball is that the vertical velocity component on the canonball goes from maximum to zero at a deceleration of 9.81 m/s^2, in the first half of the flight. And then zero to maximum at an acceleration of 9.81 m/s^2 for the second half of the flight before hitting the ground. Since there is no force acting on the horizontal velocity vector of the canonball, there will be no acceleration or deceleration of the horizontal velocity component of the canonball. This means that the horizontal velocity component of the canonball is constant throughout the flight
A point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. The dipole is initially oriented so that the charge Q is located in the plane that bisects the dipole. Assume that r>>s
Immediately after the dipole is released:
a. What is the magnitude of the force on the dipole?
b. What is the magnitude of the torque on the dipole?
Answer:
a) the magnitude of the force is
F= Q([tex]\frac{kqs}{r^3}[/tex]) and where k = 1/4πε₀
F = Qqs/4πε₀r³
b) the magnitude of the torque on the dipole
τ = Qqs/4πε₀r²
Explanation:
from coulomb's law
E = [tex]\frac{kq}{r^{2} }[/tex]
where k = 1/4πε₀
the expression of the electric field due to dipole at a distance r is
E(r) = [tex]\frac{kp}{r^{3} }[/tex] , where p = q × s
E(r) = [tex]\frac{kqs}{r^{3} }[/tex] where r>>s
a) find the magnitude of force due to the dipole
F=QE
F= Q([tex]\frac{kqs}{r^3}[/tex])
where k = 1/4πε₀
F = Qqs/4πε₀r³
b) b) magnitude of the torque(τ) on the dipole is dependent on the perpendicular forces
τ = F sinθ × s
θ = 90°
note: sin90° = 1
τ = F × r
recall F = Qqs/4πε₀r³
∴ τ = (Qqs/4πε₀r³) × r
τ = Qqs/4πε₀r²
Part A: The expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].
Part B: The expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].
How do you calculate the force and torque on the dipole?Given that a point charge Q is held at a distance r from the center of a dipole that consists of two charges ±q separated by a distance s. Also, r>>s.
Part A
The electric field due to dipole at a distance r is given below.
[tex]E_r = \dfrac {qs}{4\pi \epsilon_0 r^3}[/tex]
The magnitude of the force can be given as below.
[tex]F = QE_r[/tex]
[tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex]
Hence the expression of the force on the dipole is [tex]F = \dfrac {Qqs}{4 \pi \epsilon_0 r^3}[/tex].
Part B
The torque on the dipole will dependent on the perpendicular forces on the dipole. The expression of the torque is given below.
[tex]\tau = F \times r \times sin \theta[/tex]
For the perpendicular forces, θ = 90°. Hence the torque is given below.
[tex]\tau = F\times r[/tex]
[tex]\tau = \dfrac {Qqs}{4 \pi \epsilon_0 r^3} \times r[/tex]
[tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex]
Hence the expression of the torque on the dipole is [tex]\tau = \dfrac {Qqs}{4\pi\epsilon_0 r^2}[/tex].
To know more about force and torque, follow the link given below.
https://brainly.com/question/18992494.
The study of charges in motion and their
interaction with magnetic fields is known as....?
(A) electrostatics
(B) electrodynamics
(C) electromotive
(D) electrons
(E) none of the above
2. Any material with an unequal number of
electrons and or protons could generally be
termed as .....?
Answer:
The study of charges in motion and their interaction with magnetic fields is known as electromagnetism.
Any material with an unequal number of electrons and or protons could generally be termed as ion.
Explanation:
i hope this will help you :)
Which one of the following is the shortest length?
A)
100 meters
C)
104 millimeters
E)
10 nanometers
B)
10² centimeters
D)
105 micrometers
Answer:
Option E (10 nanometers) is the shortest lengthExplanation:
From,
1cm = [tex]10^{-2}m[/tex]
1mm = [tex]10^{-3}m[/tex]
1nanometer = [tex]10^{-9[/tex]
1micrometer = [tex]10^{-6[/tex]
Therefore,
A) [tex]10^0[/tex] meters = 1meter
B) [tex]10^2[/tex] cm = [tex]10^2 * 10^{-2} = 1meter[/tex]
C) [tex]10^4[/tex] mm = [tex]10^4 * 10^{-3} = 10meter[/tex]
D) [tex]10^5[/tex] micrometer = [tex]10^5 * 10^{-6} = 0.1meter[/tex]
E) [tex]10[/tex] nanometer = [tex]10 * 10^-9 = 1*10^{-8}[/tex]
Therefore 10nanometers is the shortest length
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A grandfather clock (with a pendulum) keeps perfect time on Earth. If you were to transport this clock to the Moon, would its period of oscillation increase, decrease, or stay the same? Would its frequency of oscillation increase, decrease, or stay the same? Explain.
Answer:
Frequency of oscillation will increase
Explanation:
Because The period of a pendulum is inversely proportional to the square root of the gravitational acceleration, so as. Gravity increases on the moon period of oscillation decreases whereas period of oscillating pendulum is inversely proportional to frequency thus frequency is directly proportional to gravity so as gravity increases on the moon the frequency will increase
Answer:
a) Therefore the period of oscillation in the moon will increase.
[tex]T_{E}<T_{M}[/tex]
b) The frequency in the moon will decrease
[tex]\omega_{E}>\omega_{M}[/tex]
Explanation:
The period of the pendulum is given by this equation:
[tex]T=2\pi\sqrt{\frac{L}{g}}[/tex] (1)
As we can see, T is proportional to L and inversely proportional to the gravity vector. We know that g in the earth is grader than g in the moon.
[tex]g_{E}>g_{M}[/tex] (2)
Therefore the period of oscillation in the moon will increase.
[tex]T_{E}<T_{M}[/tex] (3)
Now, the frequency of oscillation is:
[tex]\omega=\sqrt{\frac{g}{L}}[/tex]
In this case, ω is proportional to g, then using (1) we can conclude that:
[tex]\omega_{E}>\omega_{M}[/tex]
Therefore the frequency in the moon will decrease.
I hope it helps you!
A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential of the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.
Answer:
The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]
Explanation:
From the question we are told that
The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]
The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]
The radius of the sphere is
[tex]r = \frac{d}{2}[/tex]
substituting values
[tex]r = \frac{0.45}{2}[/tex]
[tex]r = 0.225 \ m[/tex]
The potential on the surface is mathematically represented as
[tex]V = \frac{k * Q }{r }[/tex]
Where k is coulomb's constant with value [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
given from the question that there is no other charge the Q is the excess charge
Thus
[tex]Q = \frac{V* r}{ k}[/tex]
substituting values
[tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]
[tex]Q = 3.5 *10^{-7} \ C[/tex]
A wheel 2.40 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.40 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
A. What is the tangential speed?
B. Total acceleration
C. Angular position of point P.
Answer:
Explanation:
Radius of wheel = 1.2 m
A )
To know angular speed after t sec , we use the formula
ω = ω₀ + α t , where ω₀ is initial velocity , α is angular acceleration
ω = 0 + 4.4 x 2
= 8.8 rad / s
v= ωR , v is tangential speed , ω is angular speed , R is radius of wheel .
= 8.8 x 1.2 = 10.56 m /s
B )
radial acceleration
Ar = v² / R
= 10.56² / 1.2
= 92.93 m /s²
Tangential acceleration
At = angular acceleration x radius
= 4.4 x 1.2 = 5.28 m /s²
Total acceleration
= √ ( At² + Ar² )
=√ (5.28² +92.93²)
= 93 m /s²
C )
θ = ωt + 1/2 α t² where θ is angular position after time t .
= 0 + .5 x 4.4 x 2²
= 8.8 rad
= 180x 8.8/ 3.14 = 504.45 degree
initial position = 57.3°
final position = 504 .45 + 57.3
= 561.75 °
= 561.75 - 360
= 201.75 ° .
Position of radius vector of point P will be at angle of 201.75 from horizontal axis .
If the diameter of a radar dish is doubled, what happens to its resolving power, assuming that all other factors remain unchanged? Its resolving power
a. is reduced to one-half its original value.
b. Quadruples.
c. is reduced to one-quartet its original value.
d. Halves.
e. Doubles.
Answer:
e. Doubles.
Explanation:
Resolving power is given by the formula as follows :
[tex]\dfrac{1}{d\theta}=\dfrac{D}{1.22\lambda}[/tex]
Here, [tex]d\theta[/tex] is the angle subtended by two distant objects
D is diameter of the telescope
Here, the diameter of a radar dish is doubled, assuming all other factors remain unchanged, then the resolving power gets doubled. Hence, the correct option is (e).
To water the yard, you use a hose with a diameter of 3.0 cm. Water flows from the hose with a speed of 2.2 m/s. If you partially block the end of the hose so the effective diameter is now 0.50 cm, with what speed does water spray from the hose
Answer:
v₂ = 79.69 m/s
Explanation:
The initial diameter of the hose, d₁ = 3.0 cm = 0.03 m
Initial Cross Sectional Area, A₁ = πd₁²/4
A₁ = (π* 0.03²)/4
A₁ = 0.00071 m²
The initial speed of water from the hose, v₁ = 2.2 m/s
The diameter of the hose after blocking the end, d₂ = 0.50 cm = 0.005 m
Cross Sectional Area of the hose after blocking the end, A₂ = πd₂²/4
A₂ = (π* 0.005²)/4
A₂ = 0.0000196 m²
To get the speed, v₂, at which the water spray from the hose after blocking the end, we will use the continuity equation:
A₁v₁ = A₂v₂
0.00071 * 2.2 = 0.0000196 v₂
0.001562 = 0.0000196 v₂
v₂ = 0.001562/0.0000196
v₂ = 79.69 m/s
Total energy of a particle executing S.H.M of amplitude A is proportional to:
(a)A?
(b) A-2
(c) A
(d) A-
Answer:
If x = A sin w t where w is the angular frequency
then v = w A cos w t
Since KE = 1/2 m Vmax^2 and Vmax = w A maximum KE
the total energy is proportional to A^2
Also, since the maximum potential energy is
PEmax = 1/2 K A^2 where the KE is zero (maximum amplitude)
one can again see that the total energy is proportional to A^2