(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
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Answer:
Explanation:
Acceleration is given by
where
u is the initial velocity
v is the final velocity
t is the time interval
In this problem:
is the initial velocity
is the final velocity
t = 2 s is the time
Substituting, we find the acceleration:
G.P.E = mgh
Weight = mg = 200N
So G.P.E = 200 * 2 = 400 Joules
Is always virtual (meaning that the light rays do not actually come from the image), upright, and of the same shape and size as the object it is reflecting. A virtual image is a copy of an object formed at the location from which the light rays appear to come.
Answer: A
Explanation: Any short-duration exercise that is powered primarily by metabolic pathways that do not use oxygen. Examples
of anaerobic exercise include sprinting and weight lifting.
