Answer:
you could go 12 miles paying $7.80 and $1.75
So in total being $9.55
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:
That's almost the true
Explanation:
it does not happen all the time
Answer:
Therefore the correct statement is B.
Explanation:
In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.
The sum of all these waves results in the diffraction curve of the slit that has the shape
I = Io sin² θ /θ²
Where the angle is a function of the wavelength and the width of the slit.
From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)
Therefore the correct statement is B
