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PSYCHO15rus [73]
1 year ago
15

A 600g toy train completes 10 laps of its circular track in 1 min 20s. If the radius of the track is 1.2 m, Find the centripetal

acceleration of the train in m/s2
Physics
1 answer:
Lynna [10]1 year ago
4 0

Wow !  This will take more than one step, and we'll need to be careful
not to trip over our shoe laces while we're stepping through the problem.

The centripetal acceleration of any object moving in a circle is

                          (speed-squared)  /  (radius of the circle)  .

Notice that we won't need to use the mass of the train.

We know the radius of the track.  We don't know the trains speed yet,
but we do have enough information to figure it out.  That's what we
need to do first.

Speed  =  (distance traveled) / (time to travel the distance).

Distance = 10 laps of the track.   Well how far is that ? ? ?

1 lap = circumference of the track = (2π) x (radius) =  2.4π  meters

10 laps =  24π  meters.

Time = 1 minute 20 seconds  =  80 seconds

The trains speed is  (distance) / (time)

                               =  (24π meters) / (80 seconds)

                               =        0.3 π  meters/second .

NOW ... finally, we're ready to find the centripetal acceleration.

                                 <span> (speed)²  /  (radius)

                           =    (0.3π m/s)²  /  (1.2 meters)

                           =    (0.09π m²/s²)  /  (1.2 meters)

                           =    (0.09π  /  1.2)   m/s²

                           =          0.236  m/s²  .        (rounded)

If there's another part of the problem that wants you to find
the centripetal FORCE ...

Well,       Force = (mass) · (acceleration) .

We know the mass, and we ( I ) just figured out the acceleration,
so you'll have no trouble calculating the centripetal force.       </span>
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Answer:

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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
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Answer:

Explanation:

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Explanation:

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Height = 4.0 m

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The angle at which the forester aims his gun are,

\tan\theta=\dfrac{4}{15}

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Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

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Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

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y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2

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Hence, The distance between knothole and the paint ball is 0.483 m.

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