NB: The diagram of the pulley system is not shown but the information provided is sufficient to answer the question
Answer:
Power = 2702.56 W
Explanation:
Let the power consumed be P
Energy expended = E = mgh
height, h = 5 m
E = 80 * 9.8 * 5
E = 3920 J

To calculate the time, t
From F = ma
F = 900 N
900 = 80 a
a = 900/80
a = 11.25 m/s²
From the equation of motion, 
The drill head starts from rest, u = 0 m/s

Power, P = E/t
P = 3920/0.0.943
P = 4157.79 W
But Efficiency, E = 0.65
P = 0.65 * 4157.79
Power = 2702.56 W
<h3><u>Answer</u>;</h3>
≈ 5 Kgm²/sec
<h3><u>Explanation</u>;</h3>
Angular momentum is given by the formula
L = Iω, where I is the moment of inertia and ω is the angular speed.
I = mr², where m is the mass and r is the radius
= 0.65 × 0.7²
= 0.3185
Angular speed, ω = v/r
= (2 × 3.142 × r × 2.5) r
= 15.71
Therefore;
Angular momentum = Iω
= 0.3185 × 15.71
= 5.003635
<u>≈ 5 Kgm²/sec</u>
Answer:
you'll see rocks that have been there for years
Answer:

Explanation:
We are given that







We have to find the exit temperature.
By steady energy flow equation



Substitute the values




By definition we know that the force is the vector product of the vector of the current by the length with the magnetic field vector. The current in this case goes in a positive "Y" direction. If we assume that the magnetic field goes in the positive "K" direction, then the result will be in the positive "X" direction. Attached solution.