The average act on her during the deceleration is 4.47 meters per second.
<u>Explanation</u>:
<u>Given</u>:
youngster mass m = 50.0 kg
She steps off a 1.00 m high platform that is s = 1 meter
She comes to rest in the 10-meter second
<u>To Find</u>:
The average force and momentum
<u>Formulas</u>:
p = m * v
F * Δ t = Δ p
vf^2= vi^2+2as
<u>Solution</u>:
a = 9.8 m/s
vi = 0
vf^2= 0+2(9.8)(1)
vf^2 = 19.6
vf = 4.47 m/s .
Therefore the average force is 4.47 m/s.
Answer:
v = 54.2 m / s
Explanation:
Let's use energy conservation for this problem.
Starting point Higher
Em₀ = U = m g h
Final point. Lower
= K = ½ m v²
Em₀ = Em_{f}
m g h = ½ m v²
v² = 2gh
v = √ 2gh
Let's calculate
v = √ (2 9.8 150)
v = 54.2 m / s
Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,

For 8 shot burst, average recoil force on the gun is :

So, the average recoil force on the gun during that 0.40 s burst is 45 N.
The answer is 4,045.1 meters
.................yes...................