Answer:
The time it can operate between chargins in minutes is

Explanation:
Given:
,
, 
a). The rotational kinetic energy






b). The power average 0.8kW un range time can be find

Solve to t'



Answer:
The net force acting on the body is 10N directed to the left.
Explanation:
Magnitude of force to the right = 5N
Magnitude of force to the left = 15N
Net force acting on the object and in what direction;
Solution:
It is the vector sum of all forces acting on a body. This net force is the single force that will replace the forces acting on a body;
For the problem;
Net force = Force to the left + Force to the right
Let us take left to be negative and right to be positive;
Force to the left = -15N
Net force = -15N + 5N = -10N
The net force acting on the body is 10N directed to the left.
The problem you would encounter is measuring the height of two different people, a tall one and a short one, and getting the same answer for both of them.
No matter WHAT we're hearing out of the White House these days, you CAN'T bend and stretch your standard measuring devices, or any other 'facts', to make them fit the thing that you're measuring. This does not work. You're always entitled to your own opinions, but you're not entitled to your own facts.
Answer:
![125\sqrt[4]{8}](https://tex.z-dn.net/?f=125%5Csqrt%5B4%5D%7B8%7D)
Explanation:
A number of the form

can be re-written in the radical form as follows:
![\sqrt[n]{a^m}](https://tex.z-dn.net/?f=%5Csqrt%5Bn%5D%7Ba%5Em%7D)
In this problem, we have:
a = 1,250
m = 3
n = 4
So, if we apply the formula, we get
![1,250^{\frac{3}{4}}=\sqrt[4]{(1,250)^3}](https://tex.z-dn.net/?f=1%2C250%5E%7B%5Cfrac%7B3%7D%7B4%7D%7D%3D%5Csqrt%5B4%5D%7B%281%2C250%29%5E3%7D)
Then, we can rewrite 1250 as

So we can rewrite the expression as
![=\sqrt[4]{(2\cdot 5^4)^3}=5^3 \sqrt[4]{2^3}=125\sqrt[4]{8}](https://tex.z-dn.net/?f=%3D%5Csqrt%5B4%5D%7B%282%5Ccdot%205%5E4%29%5E3%7D%3D5%5E3%20%5Csqrt%5B4%5D%7B2%5E3%7D%3D125%5Csqrt%5B4%5D%7B8%7D)
FREQUENCY is the number of complete waves that pass a given point in a certain amount of time.
Good luck :)