H<span>igh voltage demonstrations!</span>
Here We can use principle of angular momentum conservation
Here as we know boy + projected mass system has no external torque
Since there is no torque so we can say the angular momentum is conserved

now we know that
m = 2 kg
v = 2.5 m/s
L = 0.35 m
I = 4.5 kg-m^2
now plug in all values in above equation

![1.75 = [4.5 + 0.245]\omega](https://tex.z-dn.net/?f=1.75%20%3D%20%5B4.5%20%2B%200.245%5D%5Comega)


so the final angular speed will be 0.37 rad/s
Answer:
Consider the diagram. We are effectively being asked to prove that $\alpha=i_1$, for any value of $i_1$. Now, from trigonometry,
Explanation:
Answer:
medium
Explanation:
<em>A sound </em><em>medium</em><em> is defined as channel through which sound can travel or be transmitted. </em>
Sound medium could be in the form gases, liquids, solids or plasmas. Space is made up of vacuum and therefore, has no medium within it. Hence, space cannot transmit sound in any form or allows sound to travel through it.
Answer:
<u><em>The truck was moving 16.5 m/s during the time it took to stop, which was 3 seconds. </em></u>
- <u><em>Initial velocity = 33 m/s</em></u>
- <u><em>Final velocity = 0 m/s</em></u>
- <u><em>Average velocity = (33 + 0) / 2 m/s = 16.5 m/s</em></u>
Explanation:
- <u><em>First, how long does it take the truck to come to a complete stop?</em></u>
- <u><em>( 33 m/s ) / ( 11 m / s^2 ) = 3 seconds</em></u>
- <u><em>Then we can look at the average velocity between when the truck started decelerating and when it came to a complete stop. Because the deceleration is constant (always 11m/s^2) we can use this trick.</em></u>