The impulse imparted to the shells equals the change in the momentum:
Fav*(Delta t)= Delta m*v.
The mass change is
Delta m= n*m= (89.9shells)*(88.7g)=7.97Kg
So the average force is
F=((v)*(Delta m))/t= ((929)*(7.97))/4.84=1529.78 N
Since the velocity of the shells is much greater than the velocity of the helicopter, there is no need to use relative velocity.
Answer:
7.75 s
Explanation:
Newton's second law:
∑F = ma
35 N = (70 kg) a
a = 0.5 m/s²
Given v₀ = 0 m/s and Δx = 15 m:
Δx = v₀ t + ½ at²
(15 m) = (0 m/s) t + ½ (0.5 m/s²) t²
t = 7.75 s
"Energy and Momentum" is always conserved in an inelastic condition
Hope this helps!
Answer:
F = −10093.41 N
Explanation:
Given that,
Mass of a baseball, m = 143 g = 0.143 kg
Initial speed of the baseball, u = +38.8 m/s
The hitter's bat is in contact with the ball for 1.20 ms and then travels straight back to the pitcher's mound at a speed of 45.9 m/s, v = -45.9 m/s
We need to find the average force exerted on the ball by the bat. So, Force is given by :

a is acceleration

So, the average force exerted on the ball by the bat has a magnitude of 10093.41 N.