Answer: 3.2 * 10^-3 C
Explanation: In order to solve this problem we have to use the Coulomb force given by:
F=k*q^2/ d^2 where he consider the same charge for each point
so, we have
q^2= F*d^2/k= 1.137* (10 C*9*10m)^2/9*10^9 N*m^2/C^2)=3.2 * 10^-3 C.
Answer:
18.33 Ns
Explanation:
As the pitch back speed has the opposite direction as before, the change in velocity would be

So the change in momentum of the ball would be the product of its velocity change and its mass

This is equals to the impulse acted on the ball by the bat, which is 18.33 Ns
A manufacturer of printed circuit boards has a design
capacity of 1,000 boards per day. the effective capacity, however, is 700
boards per day. recently the production facility has been producing 950 boards
per day. The design capacity utilization is (950/100) *100 = 95 %
Wavelength, because of how it lengths out from the rest.
The frequency of the wave will not change. Since the change in medium doesn't affect the source of the waves, the frequency of those waves do not change.
Hope this helps! :)