(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
Learn more about work done here: brainly.com/question/8119756
#SPJ1
Answer: an ectomorph is a body type that struggles to gain weight and muscle
Answer:
The answer is below
Explanation:
Newton's law of gravity states that the force between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The law is expressed by the formula:

The masses and distances for this question is in common units, Therefore the result would be in ratios
a) 4 MEarth / 2 MSolar / 3 AU
The force (F) = (4 * 3) / 3² = 4/3
b) 1 MEarth / 1 MSolar / 1 AU
The force (F) = (1 * 1) / 1² = 1
c) 1 MEarth / 2 MSolar / 2 AU
The force (F) = (1 * 2) / 2² = 1/2
Divide 24 by 12.
24/t = 12
24/12 = t
24/12 = 2
t = 2