Answer:
The velocity of the star is 0.532 c.
Explanation:
Given that,
Wavelength of observer = 525 nm
Wave length of source = 950 nm
We need to calculate the velocity
If the direction is from observer to star.
From Doppler effect

Put the value into the formula







Negative sign shows the star is moving toward the observer.
Hence, The velocity of the star is 0.532 c.
No...........................................................................Work for them urself
Explanation:
Given:
v₀ = 0 m/s
a = 2.50 m/s²
t = 4 s
Find: v
v = at + v₀
v = (2.50 m/s²) (4 s) + 0 m/s
v = 10 m/s
It can help determine substances that appear similar but react differently under the same circumstances.
Answer:
The speed of the ball is 42.5 m/s
Explanation:
The initial kinetic energy of the ball is:
= 85.75 J
The speed of the ball after leaving the bat is:

V=47.92 m/s
Using kinematic equation we can find the speed of the ball after being 25 m above the point of collision:




