Answer:
P=627.47W
Explanation:
To solve this problem we have to take into account, that the work done by the winch is

the force, at least must equal the gravitational force

with force the tension in the cable makes the winch go up.
The work done is

To calculate the power we need to know what is the time t. But first we have to compute the acceleration
The acceleration will be

and the time t

The power will be

HOPE THIS HELPS!!
Answer:
The magnitude of the new electric field is <u>35820 N/C</u>.
Explanation:
Given:
Original magnitude of electric field (E₀) = 2388 N/C
Original voltage = 'V' (Assume)
Original separation between the plates = 'd' (Assume)
Now, new voltage is three times original voltage. So, 
New distance is 1/5 the original distance. So, 
Now, electric field between the parallel plates originally is given as:

Let us find the new electric field based on the above formula.

Now,
. So,

Therefore, the magnitude of the new electric field is 35820 N/C.
Answer:
A
Explanation:
Finding the (maximum) respective prime powers would yield the answer. Also we need not ... Is perfectly divisible by 720^n? ... So we can say that for any positive value of n it not divisible.
a clock .. and i guess a non functioning clock ?
Answer:
a)
, b) 
Explanation:
a) The potential energy is:



b) Maximum final speed:

The final speed is:

