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PSYCHO15rus [73]
3 years 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]3 years 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:

9 cm.

Explanation:

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The energy used for stretch the spring from 21 cm to 27 cm will be , E_{2}=45J

using the energy of spring formula ,we find that

27 = \frac{1}{2}K((21-L^{2})-(15-L^{2}))

45 = \frac{1}{2}K((27-L^{2})-(21-L^{2}))

Dividing both the equation will get,

\frac{3}{5}=\frac{(21-L)^{2}-(15-L)^{2}}{(27-L)^{2}-(21-L)^{2}}\\5((21-L)^{2}-(15-L)^{2})=3((27-L)^{2}-(21-L)^{2})\\3(729 - 54L + L^{2}- 441 + 42L - L^{2} ) = 5(441 - 42L + L^{2} - 225 + 30L - L^{2} )\\3(288 - 12L) = 5(216 - 12L)\\24L = 216\\L = 9 cm

Therefore, the natural length of the spring is, 9 cm.

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Answer:

20.96 m/s

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A roller coaster car moving at a velocity of 30 meters/second has a momentum of 2.5 × 104 kilogram meters/second. what is its ma
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The mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

<h3>How to calculate mass?</h3>

The mass of the roller coaster car can be calculated using the following formula:

P = m × v

Where;

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Therefore, the mass of a rollercoaster car moving at a velocity of 30 meters/second and has a momentum of 2.5 × 104 kilogram meters/second is 8.3 × 10²kg.

Learn more about mass at: brainly.com/question/19694949

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t=480.83/34
t=14.142 s
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