Calculate the period (T) of uniform circular motion if the velocity is 40.0 m/s and centripetal acceleration is 20.0 m/s2.

1 answer:

3 0

T is the time for a whole round.

centripetal acceleration = V^2/R,

20 = 40^2 / R, find R = 40^2/20 = 40*40/20 = 80 m, right?

Now, one round is L = 2*pi*R = 2*pi*80 = 160*pi

And T = L/v (distance/speed) = 160*pi/40 = 4*pi seconds, or ~ 12.57 s

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A loop of wire is carrying current of 2 A . The radius of the loop is 0.4 m. What is the magnetic field at a distance 0.09 m alo

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Answer:

$B=2.91\ \mu T$

Explanation:

Given that,

The current in the loop, I = 2 A

The radius of the loop, r = 0.4 m

We need to find the magnetic field at a distance 0.09 m along the axis and above the center of the loop. The formula for the magnetic field at some distance is given as follows :

$B=\dfrac{\mu_o}{4\pi }\dfrac{2\pi r^2 I}{(r^2+d^2)^{3/2}}$