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vichka [17]
2 years ago
7

A playground merry-go-round has radius 2.40 m and moment of inertia 2100 kg⋅m2 about a vertical axle through its center, and it

turns with negligible friction.
(a) A child applies an 18.0 N force tangentially to the edge of the merry-go-round for 15.0 s. If the merry-go-round is initially at rest, what is its angular speed after this 15.0 s interval?
(b) How much work did the child do on the merry-go-round?
(c) What is the average power supplied by the child?
Physics
1 answer:
daser333 [38]2 years ago
7 0

Answer:

a) 0.31 rad/s

b) 100 J

c) 6.67 W

Explanation:

(a) the force would generate a torque of:

T = FR = 18 * 2.4 = 43.2 Nm

According to Newton 2nd law, the angular acceleration would be

\alpha = \frac{T}{I} = \frac{43.2}{2100} = 0.021 rad/s^2

It starts from rest, then after 15s it would achieve a speed of

\omega = \alpha t = 0.021 * 15 = 0.31 rad/s

(b) The distance angle swept by it is:

\theta = \frac{\alpha t^2}{2} = \frac{0.021 * 15^2}{2} = 2.314 rad

Hence the work by the child

W = T\theta = 43.2 *2.314  \approx 100 J

c) Average power to work per time unit

P = \frac{W}{t} = \frac{100}{15} = 6.67 W

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A 500-N weight sits on the small piston of a hydraulic machine. The small piston has area 2.0 cm2. If the large piston has area
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Answer:

W₂= 10000 N

Explanation:

Pascal´s Principle can be applied in the hydraulic press:

If we apply a small force (F1) on a small area piston A1, then, a pressure (P) is generated that is transmitted equally to all the particles of the liquid until it reaches a larger area piston and therefore a force (F2) can be exerted that is proportional to the area (A2) of the piston:

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Data

W₁ = weight sits on the small piston

F₁ = W₁= 500 N

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A₂ = 40 cm²

Calculation of the weight  (W₂) can the large piston support

We replace data in the equation (1)

F_{2} = \frac{(500)*(40) }{2}

F₂ = 10000 N

W₂= F₂= 10000 N

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