(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.
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Good Day.
here's the first part but for the 2nd one all I know is that the word "compression" goes on the spirals that are closer together.
hope this helps!
First, we calculate the mass of Phosphorous present:
283.88 x 0.4364
= 123.88 amu
Atomic mass of P is 31 amu
moles of P = mass / Ar
= 123.88 / 31
= 4.0 moles
We know that one mole of substance has 6.02 x 10²³ particles
Atoms of P = 4 x 6.02 x 10²³
= 2.41 x 10²⁴ atoms