Answer:
The time he can wait to pull the cord is 41.3 s
Explanation:
The equation for the height of the skydiver at a time "t" is as follows:
y = y0 + v0 · t + 1/2 · g · t²
Where:
y = height at time "t".
y0 = initial height.
v0 = initial velocity.
t = time.
g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).
First, let´s calculate how much time will it take for the skydiver to hit the ground if he doesn´t activate the parachute.
When he reaches the ground, the height will be 0 (placing the origin of the frame of reference on the ground). Then:
y = y0 + v0 · t + 1/2 · g · t²
0 m = 15000 m + 0 m/s · t - 1/2 · 9.8 m/s² · t²
0 m = 15000 m - 4.9 m/s² · t²
-15000 m / -4.9 m/s² = t²
t = 55.3 s
Then, if it takes 4.0 s for the parachute to be fully deployed and the parachute has to be fully deployed 10.0 s before reaching the ground, the skydiver has to pull the cord 14.0 s before reaching the ground. Then, the time he can wait before pulling the cord is (55.3 s - 14.0 s) 41.3 s.
Explanation:
Momentum is mass times speed.
p = mv
a) p = (1500 kg) (25.0 m/s) = 37,500 kg m/s
b) p = (40,000 kg) (1.00 m/s) = 40,000 kg m/s
The truck has more linear momentum.
Momentum in the y direction:
pᵧ = (1500 kg) (25.0 m/s) = 37,500 kg m/s
Momentum in the x direction:
pₓ = (1500 kg) (15.0 m/s) = 22,500 kg m/s
Total linear momentum:
p² = pₓ² + pᵧ²
p² = (22,500 kg m/s)² + (37,500 kg m/s)²
p = 43,700 kg m/s
Explanation:
a) The height of the ball h with respect to the reference line is
so its initial gravitational potential energy 
is
b) To find the speed of the ball at the reference point, let's use the conservation law of energy:
We know that the initial kinetic energy 
as well as its final gravitational potential energy 
are zero so we can write the conservation law as
Note that the mass gets cancelled out and then we solve for the velocity v as
Answer:
1. Torque → F. Study of forces
2. C.O.G → D. Point of action of weight.
3. Plumb line → A. Line of C.O.G
