Complete Question:
A basketball player tosses a basketball m=1kg straight up with an initial speed of v=7.5 m/s. He releases the ball at shoulder height h= 2.15m. Let gravitational potential energy be zero at ground level
a) Give the total mechanical energy of the ball E in terms of maximum height hn it reaches, the mass m, and the gravitational acceleration g.
b) What is the height, hn in meters?
Answer:
a) Energy = mghₙ
b) Height, hₙ = 5.02 m
Explanation:
a) Total energy in terms of maximum height
Let maximum height be hₙ
At maximum height, velocity, V=0
Total mechanical energy , E = mgh + 1/2 mV^2
Since V=0 at maximum height, the total energy in terms of maximum height becomes
Energy = mghₙ
b) Height, hₙ in meters
mghₙ = mgh + 1/2 mV^2
mghₙ = m(gh + 1/2 V^2)
Divide both sides by mg
hₙ = h + 0.5 (V^2)/g
h = 2.15m
g = 9.8 m/s^2
V = 7.5 m/s
hₙ = 2.15 + 0.5(7.5^2)/9.8
hₙ = 2.15 + 2.87
hₙ = 5.02 m
Answer:
D
Explanation:
The amount of momentum that an object has is dependent upon two variables: how much stuff is moving and how fast the stuff is moving. Momentum depends upon the variables mass and velocity. In terms of an equation, the momentum of an object is equal to the mass of the object times the velocity of the object.
HOPE THIS HELPS :)
:D Anyla... <3
Answer:
The answer is "
".
Explanation:
Its minimum velocity energy is provided whenever the satellite(charge 4 q) becomes 15 m far below the square center generated by the electrode (charge q).

It's ultimate energy capacity whenever the satellite is now in the middle of the electric squares:

Potential energy shifts:


Now that's the energy necessary to lift a satellite of 100 kg to 300 km across the surface of the earth.



This satellite is transmitted by it system at a height of 300 km and not in orbit, any other mechanism is required to bring the satellite into space.
Answer:

Explanation:
Given that :
mass of the SUV is = 2140 kg
moment of inertia about G , i.e
= 875 kg.m²
We know from the conservation of angular momentum that:

![mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=mv_1%20%2A0.765%20%3D%20%5BI%2Bm%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)
![2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2](https://tex.z-dn.net/?f=2140v_1%2A0.765%20%3D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%20%5Comega_2)



From the conservation of energy as well;we have :

^2 -2140(9.81)[\sqrt{0.76^2+0.895^2} -0.765]] =0](https://tex.z-dn.net/?f=%5B%5Cfrac%7B1%7D%7B2%7D%20%5B875%2B2140%280.765%5E2%2B0.895%5E2%29%5D%280.4262%20%5C%20v_1%29%5E2%20-2140%289.81%29%5B%5Csqrt%7B0.76%5E2%2B0.895%5E2%7D%20-0.765%5D%5D%20%3D0)





