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Ivanshal [37]
3 years ago
14

Suppose that a person riding on the top of a freight car shines a searchlight beam in the direction in which the train is travel

ing.
How does the speed of the light beam relative to the ground compare to the speed of beam when the train is at rest?
Physics
1 answer:
Snezhnost [94]3 years ago
6 0

Answer:

same

Explanation:

Acc. to Einstien's postulate of special theory of

Relativity , Velocity of the light beam is same in all frames of references

(a) If the freight car is at rest

The frame we can assumed as Non - inertial frame  of reference s

In the inertial frame of reference , velocity  of the light beam  has its own value as : 3 x 10^8 m/s

(b) If the freight car is moving , the frame we can assumed as  Non -inertial frame of reference    

In thus case also , The velocity of the light beam  will also have  the same value as ; 3 x 108 m/s

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Energy flows from the sun to _______ to consumers and eventually to _______
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3 years ago
1.- Un barco recorre la distancia que separa Gran Canaria de Tenerife (90 km) en 6 horas. ¿Cuál es
bezimeni [28]

Answer:

The speed is 15 km/h or 4.16 m/s.

Explanation:

A boat travels the distance that separates Gran Canaria from Tenerife (90 km) in 6 hours. Which  the speed of the boat in km / h? And in m / s?

Given that,

Distance, d = 90 km = 90000 m

Time, t = 6 hours = 21600 s

Speed = distance/time

v=\dfrac{90\ km}{6\ h}\\\\=15\ km/h

or

v=\dfrac{90000\ m}{21600\ s}\\\\=4.16\ m/s

So, the required speed is 15 km/h or 4.16 m/s.

7 0
2 years ago
An object of mass m is hung from a spring and set into oscillation. The period of the oscillation is measured and recorded as T.
sergejj [24]

Answer:\sqrt{2}T

Explanation:

Given

object of mass m is suspended from spring and set in oscillation with time Period T

We know Time period of a mass in oscillation is given by

T=2\pi \sqrt{\frac{m}{k}}

where k=spring constant

When mass m is replaced by a mass of 2 m time period is given by

T'=2\pi \sqrt{\frac{2m}{k}}

T'=\sqrt{2}\times 2\pi \sqrt{\frac{m}{k}}

T'=\sqrt{2}T

i.e. New time period becomes \sqrt{2} times of previous one

                         

7 0
2 years ago
4. How much force is required to stop a 60 kg person traveling at 30 m/s during a time of a)
11111nata11111 [884]

Explanation:

F = ma, and a = Δv / Δt.

F = m Δv / Δt

Given: m = 60 kg and Δv = -30 m/s.

a) Δt = 5.0 s

F = (60 kg) (-30 m/s) / (5.0 s)

F = -360 N

b) Δt = 0.50 s

F = (60 kg) (-30 m/s) / (0.50 s)

F = -3600 N

c) Δt = 0.05 s

F = (60 kg) (-30 m/s) / (0.05 s)

F = -36000 N

3 0
3 years ago
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