(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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Answer:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :


Let E is the kinetic energy of the plane. It is given by :


E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Answer:
The ocean is 6485.6m deep when measured from the vessel
Explanation:
v=1474m/s
t=8.88s
let d represent distance from the vessel to the ocean bottom.
an echo travels a distance equivalent to 2d, that is to and fro after it reflects from the obstacle.


d= 6485.6m