This diagram shows two different forces acting on a skateboarder. The combined mass of the skateboard and the person is 81.5 kg. Based on this
information, what is the acceleration of the skateboarder?
Air resistance = 11.40 N
mi
Applied force = 52.80
A. 0.51 m/s2 to the left
O B. 1.94 m/s2 to the right
O C. 1.94 m/s2 to the left
O D. 0.51 m/s2 to the right

Option D. 0.51 m/s² to the right.

Explanation:

The following data were obtained from the question:

Force applied (Fₐ) = 52.8 N

Force resistance (Fᵣ) = 11.4 N

Mass = 81.5 kg

Acceleration (a) =.?

Next, we shall determine the net force (Fₙ). This can be obtained as follow:

Force applied (Fₐ) = 52.8 N

Force resistance (Fᵣ) = 11.4 N

Net force (Fₙ) =?

Fₙ = Fₐ – Fᵣ

Fₙ = 52.8 – 11.4

Fₙ = 41.4 N to the right

Finally, we shall determine the acceleration of the skateboarder as show below:

Net force (Fₙ) = 41.4 N to the right

Mass = 81.5 kg

Acceleration (a) =..?

F = ma

41.4 = 81.5 × a

Divide both side by 81.5

a = 41.4 / 81.5

a = 0.51 m/s² to the right.

Thus, the acceleration of the skateboarder is 0.51 m/s² to the right.

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j