(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:
Ae/A* = 1.115
Explanation:
Let the reservoir pressure be 
Let the exit pressure be 
Ratio of reservoir pressure and exit pressure

= 3.182
For the above value of pressure ratio
Obtain the area ratio from the isentropic flow table
Ae/A* = 1.115
The value of pressure ratio is Ae/A* = 1.115
Answer:
The force will have to increase
Explanation:
Since Juan has upgraded from a sports car to a large truck, based on Newton's second law of motion, the force needed to keep the truck going at the same speed will have to increase.
According to Newton's second law "the force on an object is equal to the product of its mass and acceleration".
Force = mass x acceleration
A truck has a larger mass compared to a sports car.
By virtue of this, to make sure both automobiles attain the same speed, the force powering them to accelerate must be the same.
Therefore, the force from the engine must increase.
Answer:
lipids are insoluble in water which is why lipids are often found in biological membranes and other waterproof coverings.