(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
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The fiducial points of the Celsius<span> and the </span>Fahrenheit<span> temperature </span>scales<span> are the boiling and freezing </span>points<span> of pure water at 1 atm of pressure.
In short, Your Answer would be Option D
Hope this helps!</span>
Explanation:
see the above attachment to solve the question and get the answer.
hope this helps you.
Water and baking soda can be used, too.
Answer:
y(i) = h
v(y.i) = 0
Explanation:
See attachment for elaboration
