If the spaceship's Physicist happens to be hanging out of one side
of the ship, and he measures the speed of the photons as they pass
him and leave the ship, he'll see them passing him at 'c' ... the speed
of light.
When those photons pass somebody who happens to be in their
path, and he decides to measure their speed, he'll see them move
past him at 'c' ... the speed of light.
It doesn't matter whether the observer who measures them is
moving, or at what speed.
And it doesn't matter what source the photons come from, or
whether the source is moving, or at what speed.
And it doesn't matter what the photons' wavelength/frequency is ...
anything from radio to gamma rays.
The photons pass everybody at 'c' ... the speed of light.
Yes, I hear you. That can't be true. It's crazy.
Maybe it's crazy, but it's true.
Answer:
3257806.62409 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of Sun = 
r = Radius of Star = 20 km
u = Initial velocity = 0
v = Final velocity
s = Displacement = 16 m
a = Acceleration
Gravitational acceleration is given by
The gravitational acceleration at the surface of such a star is
The velocity of the object would be 3257806.62409 m/s
Answer:
100watts
Explanation:
Given parameters:
Workdone = 500Nm
Time taken = 5s
Unknown:
Power in watts = ?
Solution:
Power is the rate at which work is done;
Power =
Input the parameters and solve;
Power = 
= 100watts
Answer:
Explanation:
for vertical movement , time to reach the top = time to reach the hand = 2.5 s
v = u - gt
At the top , v = 0 , time t = 2.5 s
0 = u - g x 2.5
u = 2.5 x 9.8 = 24.5 m /s
velocity of throw = 24.5 m /s
So , when it passes the hand on its way down , it will have velocity equal to 24.5 m /s and it will accelerate downwards . Let its velocity down by 22 m be v
v² = u² + 2 g s
= 24.5² + 2 x 9.8 x 22
= 600.25 + 431.2
= 1031.45
v = 32.11 m /s .
