The shadow of a pendulum cast on a flat board moves on a straight line. By placing the x-axis on the straight line with the origin at the middle of the total path, the x-coordinate of the shadow is given by the following function: x(t) = 57cos(πt), where t is in seconds and x is in centimeters.
How much distance, in centimeters, does the shadow travel in 20 sec?

22.8 m

Explanation:

The x-coordinate of the shadow as a function of time t is given by the expression

(1)

where

t is in seconds

x is in centimeters

Here we want to know the distance travelled by the shadown in 20 seconds.

First of all, we find the period of this oscillation. The periodic function can be rewritten as

(2)

where

A = 57 cm is the amplitude

is the angular frequency, with T being the period

By comparing (1) with (2), we notice that

So, the period is 2 seconds.

We also know that the amplitude of the motion is 57 cm: this means that in a quarter of period, the shadow covers a distance of 57 cm. So in one period, the distance covered by the shadow is 4 times the amplitude:

Here we want to find the distance covered by the shadow in 20 seconds, which corresponds to

So, 10 periods. Therefore, the distance covered in 10 periods is:

15.3 cos 65 degrees trig

energy stored in the capacitor= 9.72 x 10⁻⁴ j

explanation:

the energy stored in the capacitor is given by

e= 1/2 c v²

e= energy stored

c= capacitance= 1.5 µc= 1.5 x 10⁻⁶f

v= potential difference=36 v

so energy stored= e= 1/2 (1.5 x 10⁻⁶) (36)²

energy stored= 9.72 x 10⁻⁴j