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user100 [1]
2 years ago
6

Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf ball at 100 mph from the deck of the

yacht. You see the ball move away from you at 100mph, while a person standing on a near by beach would observe your golf ball traveling at 120 mph (20 mph + 100 mph). Now imagine you are aboard the Hermes spacecraft traveling at 0.1c (1/10 the speed of light) past Mars and shine a laser from the front of the ship. You would see the light traveling at c (the speed of light) away from your ship. According to Einstein’s special relativity, how fast will a person on Mars observe the light to be traveling?
Physics
2 answers:
Novosadov [1.4K]2 years ago
8 0
A naive guess would be 1.1c, according to classical mechanics. But Special relativity posited a new special law of adding velocities; so when one is travelling at 5m/s on a platform moving 5m/s, the result is very very close to 10m/s, but it is not exact. This deviation is more visible in higher speeds. There is a specific formula that gives us the speed of an object when it moves in a frame of reference, but in this case the answer is simple. The speed of light is an absolute barrier to the speed of any object and it is preserved in all frames of reference. Thus, a person will also measure a velocity of c for the light.
Whitepunk [10]2 years ago
6 0

Before solving this question, first we have to understand the special theory of relative.

As per classical mechanics, the velocity of light will be different in different frame of reference. The light moves in the ether medium which exists every where in the entire universe.

Let us consider a body which moves with a velocity v. Let light is coming along the direction of the body. As per classical mechanics,the velocity of light with respect to the body will be [ c-v].

Let us consider that light is coming from opposite direction. Hence, the velocity of light with respect to the observer will be c+v.

From above we see that velocity of light is different in both the cases which is wrong.

As per Einstein's special theory of relativity, the velocity of light will be same in every frame of reference i.e c=300000 km/s.

As per the question ,the space craft is moving with a velocity 0.1 c.

We are asked to calculate the velocity of the light with respect to an observer present in Mars.

Considering Einstein's theory of relativity, the velocity of light will be c [300000 km/s] with respect to the person in Mars.

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azamat

Answer:

87.5 m/s

Explanation:

The speed of a wave is given by

v=\lambda f

where

v is the wave speed

\lambda is the wavelength

f is the frequency

In this problem, we have

f=250 Hz is the frequency

\lambda=0.35 m is the wavelength

Substituting into the equation, we find

v=(0.35 m)(250 Hz)=87.5 m/s

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Hans Langseth's beard measured 5.33 m in 1927. Consider two charges, q1 = 2.42 nC and an unspecified charge, q2, are separated 5
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Answer:

-7.89 * 10^(-9) C

Explanation:

Parameters given:

q1 = 2.42 nC = 2.42 * 10^(-9) C

Distance between q1 and q2 = 5.33 m

q3 = 1.0 nC = 1 * 10^(-9) C

Distance between q1 and q3 = 1.9 m

Distance between q2 and q3 = 5.33 - 1.9 = 3.43 m

The net force acting on q3 is:

F = F(q1, q3) + F(q2, q3)

F = (k*q1*q3)/1.9² + (k*q2*q3)/3.43²

F = (9 * 10^(9) * 2.42 * 10^(-9) * 1 * 10^(-9))/3.61 + (9 * 10^(9) * q2 * 1 * 10^(-9))/11.7649

F = 6.033 * 10^(-9) + 0.765*q2

If the net force is zero:

0 = 6.033 * 10^(-9) + 0.765*q2

-0.765*q2 = 6.033 * 10^(-9)

=> q2 = -[6.033 * 10^(-9)]/0.765

q2 = -7.89 * 10^(-9) C

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lesantik [10]

Answer:

a)\lambda=6.63\times10^{-31}m

b)\lambda=6.63\times10^{-39}m

c)\lambda=9.97\times10^{-11}m

d)\lambda=4.03\times10^{-36}m

e)λ=∞

Explanation:

De Broglie discovered that an electron or other mass particles can have a wavelength associated, and that wavelength (λ) is:

\lambda=\frac{h}{P}=\frac{h}{mv}

with h the Plank's constant (6.63\times10^{-34}\frac{m^{2}kg}{s}) and P the momentum of the object that is mass (m) times velocity (v).

a)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}kg*1.0)}

\lambda=6.63\times10^{-31}m

b)\lambda=\frac{6.63\times10^{-34}}{(1.0\times10^{-3}*(1.00\times10^{8}))}

\lambda=6.63\times10^{-39}m

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\lambda=9.97\times10^{-11}m

d)\lambda=\frac{6.63\times10^{-34}}{(74*2.22)}

\lambda=4.03\times10^{-36}m

e) \lambda=\frac{6.63\times10^{-34}}{(74*0)}

λ=∞

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