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[supernoy]

According to an observer on Earth, two spaceships A and B are

moving towards each other. Spaceship A is moving at 0.80c to the

observer's left while spaceship B is moving at 0.65c to her right. An

astronaut in spaceship A turns a laser on for an experiment.

Relative to an astronaut in spaceship B, how fast are the photons

coming from the laser in spaceship A travelling? Express the speed in

terms of c.

0.25c

0.95c

I.ooc

1.45c

 

[🤖]

To determine the speed of the photons coming from the laser in spaceship A as observed by an astronaut in spaceship B, we need to use the relativistic velocity addition formula:

v = (u + v)/(1 + uv/c^2)

where v is the relative velocity we want to find, u is the velocity of the laser photons relative to spaceship A (which we know is c), and v is the velocity of spaceship A relative to spaceship B.

We can first find v by using the velocity addition formula again, but with spaceship A and B as the reference frames:

v = (vA + vB)/(1 + vAvB/c^2)

where vA is the velocity of spaceship A relative to the observer on Earth (which is -0.8c), and vB is the velocity of spaceship B relative to the observer on Earth (which is 0.65c). Plugging in the values, we get:

v = (-0.8c + 0.65c)/(1 - 0.8c(0.65c)/c^2) = -0.15c/0.48 = -0.3125c

Note that the negative sign indicates that spaceship A and B are moving towards each other.

Now, we can use this value of v to find the velocity of the laser photons relative to spaceship B:

v = (c + v)/(1 + cv/c^2) = (c - 0.3125c)/(1 - 0.3125c^2/c^2) = 0.688c

Therefore, the speed of the photons coming from the laser in spaceship A as observed by an astronaut in spaceship B is 0.688c.

 

[supernoy]

To determine the speed of the photons coming from the laser in spaceship A as observed by an astronaut in spaceship B, we need to use the relativistic velocity addition formula:

v = (u + v)/(1 + uv/c^2)

where v is the relative velocity we want to find, u is the velocity of the laser photons relative to spaceship A (which we know is c), and v is the velocity of spaceship A relative to spaceship B.

We can first find v by using the velocity addition formula again, but with spaceship A and B as the reference frames:

v = (vA + vB)/(1 + vAvB/c^2)

where vA is the velocity of spaceship A relative to the observer on Earth (which is -0.8c), and vB is the velocity of spaceship B relative to the observer on Earth (which is 0.65c). Plugging in the values, we get:

v = (-0.8c + 0.65c)/(1 - 0.8c(0.65c)/c^2) = -0.15c/0.48 = -0.3125c

Note that the negative sign indicates that spaceship A and B are moving towards each other.

Now, we can use this value of v to find the velocity of the laser photons relative to spaceship B:

v = (c + v)/(1 + cv/c^2) = (c - 0.3125c)/(1 - 0.3125c^2/c^2) = 0.688c

Therefore, the speed of the photons coming from the laser in spaceship A as observed by an astronaut in spaceship B is 0.688c.

4. An ambulance is traveling at 50 km/hr along a highway with heavy

traffic. A stationary car with an audio frequency counter is in front of

the ambulance. The ambulance turns on its siren of 700-Hz frequency

to inform cars in front. Assume that the speed of sound is 343 m/s.

What frequency will the counter read?

700 Hz

730 Hz

750 Hz

770 Hz

4. An ambulance is traveling at 50 km/hr along a highway with heavy

traffic. A stationary car with an audio frequency counter is in front of

the ambulance. The ambulance turns on its siren of 700-Hz frequency

to inform cars in front. Assume that the speed of sound is 343 m/s.

What frequency will the counter read?

700 Hz

730 Hz

750 Hz

770 Hz

1. A rectangular wooden barge (p = 750 kg/m3) has dimensions of

2.5m x 1.5m x 15 cm.

If it is placed on seawater (p = 1027 kg/m3), how high above the

surface of the seawater will the barge's upper surface be?

4.0 cm

7.8 cm

11 cm

None of these. The barge will sink in seawater.

 

[🤖]

4. An ambulance is traveling at 50 km/hr along a highway with heavy

traffic. A stationary car with an audio frequency counter is in front of

the ambulance. The ambulance turns on its siren of 700-Hz frequency

to inform cars in front. Assume that the speed of sound is 343 m/s.

What frequency will the counter read?

700 Hz

730 Hz

750 Hz

770 Hz

The frequency heard by the stationary car with the audio frequency counter will be different from the actual frequency of the siren due to the Doppler effect.

The Doppler effect is given by the formula:

f' = (v ± u)/(v ± s) * f

where,

f' is the frequency heard by the observer

f is the actual frequency of the source

v is the speed of sound in the medium

u is the velocity of the observer

s is the velocity of the source

Here, the ambulance is moving towards the stationary car, so the relative velocity of the ambulance with respect to the car is u = 50 km/hr = 50/3.6 m/s. The speed of sound in air is given as v = 343 m/s, and the frequency of the siren is f = 700 Hz. The velocity of the source s is 0, since the siren is fixed to the ambulance.

Substituting these values, we get:

f' = (v ± u)/(v ± s) * f

f' = ((343 m/s) + (50/3.6 m/s))/(343 m/s) * 700 Hz

f' = 770 Hz

Therefore, the frequency read by the audio frequency counter in the stationary car will be 770 Hz.

 

[supernoy]

The frequency heard by the stationary car with the audio frequency counter will be different from the actual frequency of the siren due to the Doppler effect.

The Doppler effect is given by the formula:

f' = (v ± u)/(v ± s) * f

where,

f' is the frequency heard by the observer

f is the actual frequency of the source

v is the speed of sound in the medium

u is the velocity of the observer

s is the velocity of the source

Here, the ambulance is moving towards the stationary car, so the relative velocity of the ambulance with respect to the car is u = 50 km/hr = 50/3.6 m/s. The speed of sound in air is given as v = 343 m/s, and the frequency of the siren is f = 700 Hz. The velocity of the source s is 0, since the siren is fixed to the ambulance.

Substituting these values, we get:

f' = (v ± u)/(v ± s) * f

f' = ((343 m/s) + (50/3.6 m/s))/(343 m/s) * 700 Hz

f' = 770 Hz

Therefore, the frequency read by the audio frequency counter in the stationary car will be 770 Hz.

A vertical cylindrical conducting shell has a charge —q distributed

evenly at either bases with radius r and height h.

Which gaussian surface has an electric field equal to zero?

Vertical cylinder with R > r and H <= h

Vertical cylinder with R =< r and H > h

Vertical cylinder with R > r and H >= h

Vertical cylinder with R =< r and H < h

 

[🤖]

A vertical cylindrical conducting shell has a charge —q distributed

evenly at either bases with radius r and height h.

Which gaussian surface has an electric field equal to zero?

Vertical cylinder with R > r and H <= h

Vertical cylinder with R =< r and H > h

Vertical cylinder with R > r and H >= h

Vertical cylinder with R =< r and H < h

The electric field due to a uniformly charged cylinder along its axis is given by:

E = (λ / 2πε0) * [R2 / (R2 + L2)^(3/2)]

where λ is the linear charge density (charge per unit length) of the cylinder, R is the distance from the center of the cylinder to the point where the electric field is to be calculated, L is the length of the cylinder, and ε0 is the permittivity of free space.

In this case, the cylinder has a finite height h, and the charge is distributed evenly at the two bases. Therefore, the linear charge density λ = -q / h.

To find the Gaussian surface where the electric field is zero, we need to imagine a cylinder of radius R and height H around the original cylinder. The electric field at any point inside this Gaussian surface will be zero if the enclosed charge is zero.

Let's consider the four options given:

1. Vertical cylinder with R > r and H <= h:
In this case, the Gaussian surface does not enclose any charge. Therefore, the electric field is zero.

2. Vertical cylinder with R =< r and H > h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.

3. Vertical cylinder with R > r and H >= h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.

4. Vertical cylinder with R =< r and H < h:
In this case, the Gaussian surface encloses only a portion of the charge -q. Therefore, the enclosed charge is not zero, and the electric field is not zero.

Therefore, the correct answer is option 1: Vertical cylinder with R > r and H <= h.

 

[supernoy]

According to an observer on Earth, two spaceships A and B are

moving towards each other. Spaceship A is moving at 0.80c to the

observer's left while spaceship B is moving at 0.65c to her right. An

astronaut in spaceship A turns a laser on for an experiment.

Relative to an astronaut in spaceship B, how fast are the photons

coming from the laser in spaceship A travelling? Express the speed in

terms of c.

The electric field due to a uniformly charged cylinder along its axis is given by:

E = (λ / 2πε0) * [R2 / (R2 + L2)^(3/2)]

where λ is the linear charge density (charge per unit length) of the cylinder, R is the distance from the center of the cylinder to the point where the electric field is to be calculated, L is the length of the cylinder, and ε0 is the permittivity of free space.

In this case, the cylinder has a finite height h, and the charge is distributed evenly at the two bases. Therefore, the linear charge density λ = -q / h.

To find the Gaussian surface where the electric field is zero, we need to imagine a cylinder of radius R and height H around the original cylinder. The electric field at any point inside this Gaussian surface will be zero if the enclosed charge is zero.

Let's consider the four options given:

1. Vertical cylinder with R > r and H <= h:
In this case, the Gaussian surface does not enclose any charge. Therefore, the electric field is zero.

2. Vertical cylinder with R =< r and H > h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.

3. Vertical cylinder with R > r and H >= h:
In this case, the Gaussian surface encloses the entire charge -q. Therefore, the electric field is not zero.

4. Vertical cylinder with R =< r and H < h:
In this case, the Gaussian surface encloses only a portion of the charge -q. Therefore, the enclosed charge is not zero, and the electric field is not zero.

Therefore, the correct answer is option 1: Vertical cylinder with R > r and H <= h.

According to an observer on Earth, two spaceships A and B are
moving towards each other. Spaceship A is moving at 0.80c to the
observer's left while spaceship B is moving at 0.65c to her right. An
astronaut in spaceship A turns a laser on for an experiment.
Relative to an astronaut in spaceship B, how fast are the photons
coming from the laser in spaceship A travelling? Express the speed in
terms of c.

 

[supernoy]

Three vectors are define as

A=2i-3j
B=-i+5j
C=-4k

then what is the value of expression
A * (C x B)

 

[🤖]

First, we need to find the cross product of vectors C and B:

C x B = (0-(-4)5)i - (0-(-4)(-1))j + ((-4)*(-1)-0)k
C x B = -20i + 4j + 4k

Now we can find the dot product of vector A with the result of the cross product of vectors C and B:

A * (C x B) = (2i - 3j) * (-20i + 4j + 4k)
A * (C x B) = (2*-20) + (-3*4) + (0)
A * (C x B) = -40 - 12
A * (C x B) = -52

Therefore, the value of expression A * (C x B) is -52.

 

[supernoy]

First, we need to find the cross product of vectors C and B:

C x B = (0-(-4)5)i - (0-(-4)(-1))j + ((-4)*(-1)-0)k
C x B = -20i + 4j + 4k

Now we can find the dot product of vector A with the result of the cross product of vectors C and B:

A * (C x B) = (2i - 3j) * (-20i + 4j + 4k)
A * (C x B) = (2*-20) + (-3*4) + (0)
A * (C x B) = -40 - 12
A * (C x B) = -52

Therefore, the value of expression A * (C x B) is -52.

According to an observer on Earth, two spaceships A and B are moving towards each other. Spaceship A is moving at 0.80c to the observer's left while

spaceship B is moving at 0.65c to her right. An astronaut in spaceship A turns a laser on for an experiment.

Relative to an astronaut in spaceship B, how fast are the photons coming from the laser in spaceship A travelling? Express the speed in terms of c.

 

[🤖]

According to an observer on Earth, two spaceships A and B are moving towards each other. Spaceship A is moving at 0.80c to the observer's left while

spaceship B is moving at 0.65c to her right. An astronaut in spaceship A turns a laser on for an experiment.

Relative to an astronaut in spaceship B, how fast are the photons coming from the laser in spaceship A travelling? Express the speed in terms of c.

From the observer's point of view, the velocity of spaceship A is -0.8c (to the left) and the velocity of spaceship B is +0.65c (to the right). Let's use the relativistic velocity addition formula to calculate the velocity of the photons as seen by an astronaut in spaceship B.

v = (u + v) / (1 + uv/c^2)

where
u = velocity of spaceship A as seen by the observer on Earth = -0.8c
v = velocity of spaceship B as seen by the observer on Earth = +0.65c
c = speed of light

Plugging in the values, we get:

v = (-0.8c + 0.65c) / (1 + (-0.8c)(0.65c)/c^2)
v = -0.15c / (1 - 0.52)
v = -0.15c / 0.48
v = -0.3125c

Therefore, the velocity of the photons coming from the laser in spaceship A as seen by an astronaut in spaceship B is -0.3125c (to the left).