Answer:
<em>Correct choice: b 4H</em>
Explanation:
<u>Conservation of the mechanical energy</u>
The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):
E = U + K
The GPE is calculated as:
U = mgh
And the kinetic energy is:
Where:
m = mass of the object
g = gravitational acceleration
h = height of the object
v = speed at which the object moves
When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:
When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:
Since the energy is conserved, U1=U2
![\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%3D%5Cfrac%7B1%7D%7B2%7Dmv%5E2%20%20%20%20%5Cqquad%5Cqquad%20%5B1%5D)
For the speed to be double, we need to drop the snowball from a height H', and:
Operating:
![\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20mgH%27%3D4%5Cfrac%7B1%7D%7B2%7Dm%28v%29%5E2%20%5Cqquad%5Cqquad%20%5B2%5D)
Dividing [2] by [1]
Simplifying:
Thus:
H' = 4H
Correct choice: b 4H
The rate of change of velocity per unit time is called acceleration.
Its SI unit is m/s².
Answer:
B. He should change the lengths of the vectors that point tangent to the circle so that each is the same length.
Explanation:
A uniform circular motion is a motion in a circle where the tangential speed of the object is constant.
In the motion map:
- The arrows pointing towards the centre of the circle represent the centripetal acceleration, and their length represent the magnitude of the acceleration
- The arrows pointing tangential to the circle represent the tangential speed, and their length represent the magnitude of the speed
In this motion map, we see that the length of the vectors pointing tangent to the circle is not constant: this means that the speed is not constant. In order to have a uniform circular motion, the speed must be constant, therefore the lengths of the vectors that point tangent to the circle must be the same.
Answer:
The drill's angular displacement during that time interval is 24.17 rad.
Explanation:
Given;
initial angular velocity of the electric drill, 
= 5.21 rad/s
angular acceleration of the electric drill, α = 0.311 rad/s²
time of motion of the electric drill, t = 4.13 s
The angular displacement of the electric drill at the given time interval is calculated as;
Therefore, the drill's angular displacement during that time interval is 24.17 rad.
