<span>Ans : Initial E = KE = ½mv² = ½ * 1.2kg * (2.2m/s)² = 2.9 J
max spring compression where both velocities are the same: conserve momentum:
1.2kg * 2.2m/s = (1.2 + 3.2)kg * v → v = 0.6 m/s
which means the combined KE = ½ * (1.2 + 3.2)kg * (0.6m/s)² = 0.79 J
The remaining energy went into the spring:
U = (2.9 - 0.79) J = 2.1 J = ½kx² = ½ * 554N/m * x²
x = 0.0076 m ↠(a)</span>
Explanation:
u=166m/s, v=0(at it's highest point final velocity is zero), a=9.8m/s², t=8.6s
by the formula, S=ut+½at².
S=[166×8.6+½.×9.8×(8.6)²]. ...by calculation
S = 1427.6+362.404
S=1790.004m
hope this helps you.
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:

Explanation:
As we know that the wave equation is given as

now we have


so we have



also we have

so we have



now we know that at t = 0 and x = 0 wave is at y = 0.19 m
so we have

so we have

Answer:
According to the Conservation of Momentum,
Momentum of the gun = momentum of the bullet
M(gun)×V(gun)=m(bullet)×v(bullet)
4kg × V = 0.3kg × 600m/s²
V = (0.3 × 600)/4 = 45 m/s
The recoil velocity on the gun is <em><u>45 m/s</u></em>
<h3><u>45 m/s</u> is the right answer.</h3>