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>
Answer:
<h2>18150 J</h2>
Explanation:
The kinetic energy of the car can be found by using the formula

m is the Mass
v is the velocity
From the question we have

We have the final answer as
<h3>18150 J</h3>
Hope this helps you
A) Agreed.
<span>b) Value agreed but units should be W (watts). </span>
<span>c) Here's one method... </span>
<span>15 miles = 24140 m </span>
<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>
<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>
<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>
<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>
<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>
<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
The period of the pendulum is the reciprocal of the frequency:

The period of the pendulum is given by

where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can calculate the length of the pendulum L:
Answer:
frequency is .167/s for this wave