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icang [17]
3 years ago
12

A roundabout in a fairground requires an input power of 2.5 kW when operating at a constant angular velocity of 0.47 rad s–1 . (

a) Show that the frictional torque in the system is about 5 kN m. (3) 1. (b) When the power is switched off, the roundabout decelerates uniformly because the
Physics
1 answer:
natita [175]3 years ago
5 0

Answer:

Explanation:

Since the roundabout is rotating with uniform velocity ,

input power = frictional power

frictional power = 2.5 kW

frictional torque x angular velocity = 2.5 kW

frictional torque x .47 = 2.5 kW

frictional torque = 2.5 / .47 kN .m

= 5.32 kN . m

= 5 kN.m

b )

When power is switched off , it will decelerate because of frictional torque .

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Answer:

10

Explanation:

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The distance traveled by the object is the magnitude of vector r.

The magnitude of vector r is given by

r = \sqrt{(10 Cos 6t)^{2}+(10 Sin 6t)^{2}}

r =10 \sqrt{(Cos^{2} 6t)+(Sin^{2} 6t)

r = 10        

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The power in a circuit is 2 W and the voltage is 20 VDC. What's the circuit current?
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Answer:

Power in the circuit is 0.1 amp

Explanation:

The power in the circuit is given by the formula

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Where P is Power, V is voltage supplied and I is current in circuit.

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<u>https://brainly.in/question/1063947</u>

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Answer:

Explanation:

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P = h d g

Where, h is the depth, d be the density and g be the acceleration due to gravity.

If we talk about teh atmospheric pressure, the density of air goes on decreasing as we go up and up. o we cannot say that it is directly depends only on the depth of air, it also depends on the changing density of air.

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A ball filled with an unknown material starts from rest at the top of a 2 m high incline that makes a 28o with respect to the ho
Lady_Fox [76]

Answer:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>  

The ball rotates 6.78 revolutions.

     

Explanation:

<u>Searching in google I found the total mass and the radius of the ball (m = 1.5 kg and r = 10 cm) which are needed to solve the problem!</u>        

At the bottom the ball has the following angular speed:

\omega_{f} = \frac{v_{f}}{r} = \frac{4.9 m/s}{0.10 m} = 49 rad/s

Now, we need to find the distance traveled by the ball (L) by using θ=28° and h(height) = 2 m:

sin(\theta) = \frac{h}{L} \rightarrow L = \frac{h}{sin(\theta)} = \frac{2 m}{sin(28)} = 4.26 m

To find the revolutions we need the time, which can be found using the following equation:                

v_{f} = v_{0} + at  

t = \frac{v_{f} - v_{0}}{a} (1)

So first, we need to find the acceleration:

v_{f}^{2} = v_{0}^{2} + 2aL \rightarrow a = \frac{v_{f}^{2} - v_{0}^{2}}{2L}    (2)  

By entering equation (2) into (1) we have:

t = \frac{v_{f} - v_{0}}{\frac{v_{f}^{2} - v_{0}^{2}}{2L}}

Since it starts from rest (v₀ = 0):  

t = \frac{2L}{v_{f}} = \frac{2*4.26 m}{4.9 m/s} = 1.74 s

Finally, we can find the revolutions:  

\theta_{f} = \frac{1}{2} \omega_{f}*t = \frac{1}{2}*49 rad/s*1.74 s = 42.63 rad*\frac{1 rev}{2\pi rad} = 6.78 rev

Therefore, the ball rotates 6.78 revolutions.

I hope it helps you!                                                                                                                                                                                          

3 0
3 years ago
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