That would be a subduction zone.
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:
The maximum error is 
Explanation:
From the question we are told that
The length is 
The radius is 
The pressure is 
The rate is 
The viscosity is 
The error in the viscosity is mathematically represented as

Where 
and 
and 
So
![\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7Br%5E4%7D%7Bv%7D%20%20%7C%20%2A%20%20%5CDelta%20%20P%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%20P%20%2A%20r%5E3%7D%7Bv%7D%20%20%7C%2A%20%20%5CDelta%20%20r%20%2B%20%20%7C-%5Cfrac%7BP%2A%20r%5E4%7D%7Bv%5E2%7D%20%20%7C%2A%20%20%5CDelta%20%20v%5D)
substituting values
![\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7B%280.002%29%5E4%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%20%2A%20%201750%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%204%20%2A10%5E%7B5%7D%20%2A%20%280.002%29%5E3%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%2A%20%200.0002%20%2B%20%20%7C-%5Cfrac%7B%204%2A10%5E%7B5%7D%2A%20%280.002%29%5E4%7D%7B%280.5%2A10%5E%7B-9%7D%29%5E2%7D%20%20%7C%2A%20%200%20%5D)
![\Delta \eta = \frac{\pi}{8} [56 + 5120 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B56%20%20%2B%20%205120%20%5D)


Answer:
absorption and insolation.
Explanation:
Answer:
kinetic energy
Explanation:
a certain amount of energy is transferred by the kick. The ball gains an equal amount of energy, mostly in the form of kinetic energy.