Answer:
5295.3 N
Explanation:
According to law of momentum conservation, the change in momentum of the ball shall be from the momentum generated by the batter force
mv + P = mV
P = mV - mv = m(V - v)
Since the velocity of the ball before and after is in opposite direction, one of them is negative
P = 0.14(44.8 - (-19.5)) = 9 kg m/s
Hence the force exerted to generate such momentum within 1.7ms (0.0017s) is
F = P/t = 9/0.0017 = 5295.3 N
Answer:
<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg
k = 0.4
F = 200 N
<u>To </u><u>find </u><u>-</u><u> </u> acceleration
<u>Solution </u><u>-</u><u> </u>
F= kMA
200 = 0.4 * 20 * acceleration
200 = 8 * a
a = 8/200
a = 0.04 m s²
<h3>a = 0.04 m s²</h3>
Answer:
1. The final velocity of the truck is 15 m/s
2. The distance travelled by the truck is 37.5 m
Explanation:
1. Determination of the final velocity
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Final velocity (v) =?
The final velocity of the truck can be obtained as follow:
v = u + at
v = 0 + (3 × 5)
v = 0 + 15
v = 15 m/s
Therefore, the final velocity of the truck is 15 m/s
2. Determination of the distance travelled
Initial velocity (u) = 0 m/s
Acceleration (a) = 3 m/s²
Time (t) = 5 s
Distance (s) =?
The distance travelled by the truck can be obtained as follow:
s = ut + ½at²
s = (0 × 5) + (½ × 3 × 5²)
s = 0 + (½ × 3 × 25)
s = 0 + 37.5
s = 37.5 m
Therefore, the distance travelled by the truck is 37.5 m
Answer:
Option a)
Explanation:
In the process of charging anything by the method of induction, a charged body is brought near to the body which is neutral or uncharged without any physical contact and the ground must be provided to the uncharged body.
The charge is induced and the nature of the induced charge is opposite to that of the charge present on the charged body.
So when a positively charged rod is used to charge an electroscope, the rod which is positive attracts the negative charge in the electroscope and the grounding of the electroscope ensures the removal of the positive charge and renders the electroscope negatively charged.
