Example 1: An 850-kg car is accelerating at a rate of 2.4m/s2 to the right along a ... 2) A nonzero net force ΣF acting on mass M causes an acceleration a in it such that ΣF = Ma. The acceleration has the same direction as the applied net force. ... (b) Knowing that the crate is being pushed to the left by a 53-N force
W = ∫ (x from 0.1 to +oo) F dx
= ∫ (x from 0.1 to +oo) A e^(-kx) dx
= A/k x [ - e^(-kx) ](between 0.1 and +oo)
= A/k x [ 0 + e^(-k * 0.1) ]
<span>
= A/k x e^(-k/10) </span>
You knew that this question is ridiculously easy. So, just to
make it harder, you decided not to let us see the picture, so
that we could not "examine the circuit".
The description is talking about a parallel circuit. The other
kind is a series circuit, and that one has no forks in the road.
Answer:
1) 50 facing towards the right
2) 150 facing right
3) 200 facing right
4) 0- no direction
5) 50- facing left
6) 50 facing right
Explanation:
forces in opposite directions and equal magnitudes counteract each other. in number 2 they face the same direction so they would just be added. in number 4 they oppose each other so would be subtracted
Answer:
Given that
speed u=4*10^6 m/s
electric field E=4*10^3 N/c
distance b/w the plates d=2 cm
basing on the concept of the electrostatices
now we find the acceleration b/w the plates to find the horizontal distance traveled by the electron when it hits the plate.
acceleration a=qE/m=
=
m/s
now we find the horizontal distance traveled by electrons hit the plates
horizontal distance
![X=u[2y/a]^{1/2}](https://tex.z-dn.net/?f=X%3Du%5B2y%2Fa%5D%5E%7B1%2F2%7D)
=![4*10^6[2*2*10^{-2}/7*10^{14}]^{1/2}](https://tex.z-dn.net/?f=4%2A10%5E6%5B2%2A2%2A10%5E%7B-2%7D%2F7%2A10%5E%7B14%7D%5D%5E%7B1%2F2%7D)
=
= 3 cm
