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Anika [276]
3 years ago
14

An astronaut is out in space with an extremely precise timing device measuring the speed of various fast‑moving objects. A laser

gun is mounted on a spaceship and aimed backward. As the spaceship flies away from the astronaut at a speed of 200 million m/s, the laser shines a beam of light toward him. Determine the speed that would be measured in each of the cases listed. The speed of the laser light as measured in the astronaut's reference frame. The speed of the laser light measured in the spaceship's reference frame. The speed of the laser gun as measured in the astronaut's reference frame. The speed of the laser gun measured in the spaceship's reference frame.
Physics
1 answer:
Leviafan [203]3 years ago
5 0

Answer and Explanation:

with reference to Einstein's theory of special relativity, the speed of an electromagnetic radiation, here, laser will not change in any inertial frame or remains same irrespective of any change in inertial frame.

Therefore, the speed of light measured in both the cases, i.e., in astronaut's reference frame and spaceship's reference frame will be equal to the speed of light in vacuum, i.e., 3\times 10^{8} m/s.

The laser gun's speed in astronaut's reference frame is the same as the speed of the spaceship as it mounted on it, i.e., the speed of the laser gun is 200 million m/s.

The laser gun's speed measured in spaceship's reference frame will be zero, as it is mounted on the spaceship and is stationary in the spaceship's reference frame.

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The formula is x = 1/2 at^2 and I have managed to fill in the variables as this. d = 1/2 9.81 m/s^2 1^2
Artyom0805 [142]

Right, as you mentioned in the comments, you find d by plugging in the different values of t.

For t=1\,\mathrm s, we have

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)(1\,\mathrm s)^2

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(1\,\mathrm s^2\right)

d=4.905\,\mathrm m

Similarly, for t=2\,\mathrm s, you get

d=\dfrac12\left(9.81\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(2\,\mathrm s\right)

d=\left(4.905\,\dfrac{\mathrm m}{\mathrm s^2}\right)\left(4\,\mathrm s^2\right)

d=19.62\,\mathrm m

8 0
3 years ago
I just need help with these seven!
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21. light changes its direction when travelling through a new medium because denser mediums have a higher angle of refraction.
5 0
3 years ago
A bicycle is ridden along a horizontal road with a driving force of 400, n,400n. its speed is constant at 12, m, slash, s,12m/s.
mr_godi [17]

The magnitude of the sum of the frictional forces acting on the bike and its rider is 400N.

<h3>What is friction force?</h3>

The friction force is the opposing force which acts on the object which is in relative motion.

The driving force is equal and opposite to the friction force acting between road and bicycle.

Friction force = 400N

The friction force between rider and bike is zero.

So the magnitude of sum of friction force = 400N +0 = 400N

Thus, the magnitude of the sum of the frictional forces acting on the bike and its rider.

Learn more about friction force.

brainly.com/question/1714663

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4 0
1 year ago
A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the
Novay_Z [31]

Answer:

It will be cut in half

Explanation:

The diffraction of a slit is given by the formula

a sin θ = m where

a = width of the slit,

λ = wavelength and

m = integer that determines the order of diffraction.

Next we divide both sides by a, we have

sin θ = m λ / a

Also, recall that

a’ = 2 a

Then we substitute in the previous equation

2asin θ' = m λ, if divide by 2a, we have

sin θ' = (m λ / 2a).

Now again, from the first equation, we said that sin θ = m λ / a, so we substitute

sin θ ’= sin θ / 2

Then we use trigonometry to find the width, we say

tan θ = y / L

Since the angle is small, we then have

tan θ = sin θ / cos θ

tan θ = sin θ, this then means that

sin θ = y / L

we will then substitute

y’ / L = y/L 1/2

y' = y / 2

this means that when the slit width is doubled the pattern width will then be halved

4 0
3 years ago
What happens to the charge on the conductive sphere when it is connected to a source of charge such as the electrostatic voltage
hichkok12 [17]

Answer and Explanation:

The charge on the conductive sphere spreads out non-uniformly over the surface of the sphere.

Normally, the charge on such spherical surface stay on this surface uniformly, but the presence of a voltage source tampers with that dynamic.

8 0
2 years ago
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