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
Answer:
At point A, the cart has high potential energy. At point b, the cart is pulled down by gravity. At point c, the cart gains its highest kinetic energy. At point d, the cart returns back to the same state but with lower potential energy.
Answer:
a) 500
b)-500, north west
Explanation:
a) sum of F= F1+F2= 200+300= 500
b) sum of forces=0
so 200+300-500+0
Explanation:
Let us first calculate long does it take to go 12m at 30m/s( assumed speed)
12/30 = 0.4 seconds
horizontal distance the ball drop in that time
H= (0)(0.4)+1/2(-9.8)(0.4)2
H= -0.78m
negative sign shows that the height of the ball at the net from the top.
Height of the ball at the net and from the ground= H1-H=2.4-0.78=1.62m
As 1.62m>0.9m so the ball will clear the net.
H_1= V0y t’ + ½ g t’^2
-2.4= (0)t’ + ½ (-9.8) t’^2
t’= 0.69s
X’=V0x t’
X’=(30)(0.96)
X’= 20.7m
