Elastic potential energy is given by formula
![U = \frac{1}{2} kx^2](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20kx%5E2)
here we know that
![k = 800 N/m](https://tex.z-dn.net/?f=k%20%3D%20800%20N%2Fm)
![x = 6 m](https://tex.z-dn.net/?f=x%20%3D%206%20m)
Now using above formula we have
![U = \frac{1}{2}(800)(6^2)](https://tex.z-dn.net/?f=U%20%3D%20%5Cfrac%7B1%7D%7B2%7D%28800%29%286%5E2%29)
![U = 14400 J](https://tex.z-dn.net/?f=U%20%3D%2014400%20J)
So elastic potential energy in the chord is 14400 J
Answer:
X(t) = 9.8 *t - 4.9 * t^2
Explanation:
We set a frame of reference with origin at the hand of the girl the moment she releases the ball. We assume her hand will be in the same position when she catches it again. The positive X axis point upwards.The ball will be subject to a constant gravitational acceleration of -9.81 m/s^2.
We use the equation for position under constant acceleration:
X(t) = X0 + V0 * t + 1/2 * a *t^2
X0 = 0 because it is at the origin of the coordinate system.
We know that at t = 2, the position will be zero.
X(2) = 0 = V0 * 2 + 1/2 * -9.81 * 2^2
0 = 2 * V0 - 4.9 * 4
2 * V0 = 19.6
V0 = 9.8 m/s
Then the position of the ball as a function of time is:
X(t) = 9.8 *t - 4.9 * t^2
<span>The velocity of light is by far the greatest in air. There is nothing in air to stop or impede the movement of light in air. When light travels in air it will not stop unless running into clouds, smoke, smoke, droplets, or impurities. Any other substance is harder for light to travel through.</span>
It's average speed during that 26 seconds was about 4.77 m/s. Without seeing the graph, we can't tell if it was going faster or slower at any particular time during that period. All we can tell is its average for the full interval.
The speed of bullet =
850 m/s
Distance given = 1 km = 1000m
S = D/t
t • S = D/t • t
St = D
St/S = D/S
t = D/S
t = 1000m/850m/s
t = 1.176 s
It will take the bullet 1.176 or about 1.18 seconds to go 1 km.