The power of a machine depend on two factors are work and time.
Option C
<u>Explanation:</u>
In science, power defined as the amount of work done in a unit time. i.e. delivering work in a rate of time or energy supply, expressed in input of work or transmitted energy divided by the time interval (t) or W/t.
Example: Some work can be done in the long run with a low-power engine or in a short time with a motor with high performance. The equation for power can be given as


Answer:
The mother has to sit 2.17 ft from the center on the other side of the seesaw.
Explanation:
We are trying to find the sum of torques given by the weights of mother and daughter to be zero.
If the torque of the daughter on one side of the pivoting point is given by:
5.5 ft x 63.5 lb x g = 349.25 g ft lb
we need that the absolute value of the torque exerted by the mom (160.9 lb) to be the same in magnitude (and of course opposite direction). So we assume that "d" is the distance at which the mother locates to make this torque equal in magnitude to her daughter's torque:
d x 160.9 lb x g = 349.25 g ft lb
d = 2.17 ft
Answer:
Current = 8696 A
Fraction of power lost =
= 0.151
Explanation:
Electric power is given by

where I is the current and V is the voltage.

Using values from the question,

The power loss is given by

where R is the resistance of the wire. From the question, the wire has a resistance of
per km. Since resistance is proportional to length, the resistance of the wire is

Hence,

The fraction lost = 
<span>c. What is the magnitude of the tension in the string at the bottom of the circle if you are swinging it at 3.37 m/s?
</span>
Answer: d= 0.57* l
Explanation:
We need to check that before ladder slips the length of ladder the painter can climb.
So we need to satisfy the equilibrium conditions.
So for ∑Fx=0, ∑Fy=0 and ∑M=0
We have,
At the base of ladder, two components N₁ acting vertical and f₁ acting horizontal
At the top of ladder, N₂ acting horizontal
And Between somewhere we have the weight of painter acting downward equal to= mg
So, we have N₁=mg
and also mg*d*cosФ= N₂*l*sin∅
So,
d=
* tan∅
Also, we have f₁=N₂
As f₁= чN₁
So f₁= 0.357 * 69.1 * 9.8
f₁= 241.75
Putting in d equation, we have
d=
* tan 58
d= 0.57* l
So painter can be along the 57% of length before the ladder begins to slip