Answer:
Explanation:
Given
Length of rope 
Weight of rope 
weight density
Work done to lift rope 33 m


![W=73.45\left [ \left ( \frac{h^2}{2}\right )\right ]^{33}_0](https://tex.z-dn.net/?f=W%3D73.45%5Cleft%20%5B%20%5Cleft%20%28%20%5Cfrac%7Bh%5E2%7D%7B2%7D%5Cright%20%29%5Cright%20%5D%5E%7B33%7D_0)
v^2 = v0^2 +2ad
v^2 = 22^2 + 2*3.78*45 = 824.2
v= √824.2 = 28.7 m/s
Answer:
a = g = 9.81[m/s^2]
Explanation:
This problem can be solve using the second law of Newton.
We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.
m*g = m*a
where:
g = gravity = 9.81[m/s^2]
a = acceleration [m/s^2]
Note: If the skydiver will be under air resistance forces his acceleration will be different.
Answer:
v = 54 m/s
Explanation:
Given,
The maximum height of the flight of golf ball, h = 150 m
The velocity at height h, u = 0
The velocity of the golf ball right before it hits the ground, v = ?
Using the III equations of motion
<em> v² = u² + 2gh</em>
Substituting the given values in the above equation,
v² = 0 + 2 x 9.8 x 150 m
= 2940
v = 54 m/s
Hence, the speed of the golf ball right before it hits the ground, v = 54 m/s
The statements that are held true with regards to the static equilibrium of bodies are:
<span>The net torque acting on the object must equal zero
</span><span>The net torque on the object does not have to be zero if the net force on the object is zero
Furthermore, when a body is in a state of static equilibrium, the summation of all forces, either vertically or horizontally, must be equal to zero. </span>