It's a projectile near the earth under the influence of gravity only.
Answer:
Vf = 4.77 m/s
Explanation:
During the downward motion we can easily find the final velocity or the velocity with which the ball hits the ground, by using third equation of motion. The third equation of motion is given as follows:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height = 1.16 m
Vf = Final Velocity of Ball = ?
Vi = Initial Velocity of Ball = 0 m/s (Since, ball was initially at rest)
Therefore, using these values in the equation, we get:
(2)(9.8 m/s²)(1.16 m) = Vf² - (0 m/s)²
Vf = √(22.736 m²/s²)
<u>Vf = 4.77 m/s</u>
Answer:
3.30 x 10^-7 Pascal
Explanation:
distance r = 1.5 m
power P = 700 W
the radiation pressure is given as
Pr = P/A*c
where
area of the surface A = 4πr^2
calculate for A
speed of light is c = 3×10^8 m/s
plugging above values in equation above gives
Pr = 3.30 x 10^-7 Pascal
Answer:
The velocity of the gun relative to the ground is 19.66 m/s
Explanation:
Given data,
The mass of the gun, M = 15.0 kg
The mass of the bullet, m = 50 g
The velocity of the train, v = 75 km/h
= 20.83 m/s
The velocity of bullet relative to train, V' = 350 m/s
The velocity of bullet relative to ground, V = 350 + 20
= 370 m/s
According to the law of conservation of momentum,
Mv' + mV' = 0


= -1.17 m/s
Therefore, the velocity of the gun with,
v₀ = V + v'
= 20.83 - 1.17
= 19.66 m/s
Hence, the velocity of the gun relative to the ground is 19.66 m/s