(a) The work done by the force applied by the tractor is 79,968.47 J.
(b) The work done by the frictional force on the tractor is 55,977.93 J.
(c) The total work done by all the forces is 23,990.54 J.
<h3>
Work done by the applied force</h3>
The work done by the force applied by the tractor is calculated as follows;
W = Fd cosθ
W = (5000 x 20) x cos(36.9)
W = 79,968.47 J
<h3>Work done by frictional force</h3>
W = Ffd cosθ
W = (3500 x 20) x cos(36.9)
W = 55,977.93 J
<h3>Net work done by all the forces on the tractor</h3>
W(net) = work done by applied force - work done by friction force
W(net) = 79,968.47 J - 55,977.93 J
W(net) = 23,990.54 J
Learn more about work done here: brainly.com/question/25573309
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Given that:
Energy of bulb (Work ) = 30 J,
Time (t) = 3 sec
The power consumption = ?
We know that, Power can be defined as rate of doing work
Power (P) = Work(Energy supplied) ÷ time
= 30 ÷ 3
= 10 Watts
<em> The power consumption is 10 W.</em>
The answer is A. Or the first option. Pressure is changed by lowering the pressure, not reducing the volume. You would assume its C but its A.
D
Because the rest of the answers are illogical
