First we can say that since there is no external force on this system so momentum is always conserved.
now by the condition of elastic collision
![v_{2f} - v_{1f} = 0.8 - 0[\tex]now add two equations[tex]3*v_{2f} = 1.6](https://tex.z-dn.net/?f=v_%7B2f%7D%20-%20v_%7B1f%7D%20%3D%200.8%20-%200%5B%5Ctex%5D%3C%2Fp%3E%3Cp%3Enow%20add%20two%20equations%3C%2Fp%3E%3Cp%3E%5Btex%5D3%2Av_%7B2f%7D%20%3D%201.6)
also from above equation we have
So ball of mass 0.6 kg will rebound back with speed 0.267 m/s and ball of mass 1.2 kg will go forwards with speed 0.533 m/s.
Answer:
C) They walked in one direction, then they were still, then they walked in the opposite direction.
Explanation:
John and Caroline go out for a walk one day. This graph represents their distance from home.c According to the graph they walked with almost same in both the direction.
First they walked in one direction and travel 30 meter in same direction then they were still for some time. Then they walked in the opposite direction for 30 meter. They walked in both the direction with approximately equal speed.
Thus, Option C is correct.
Answer:
option c is correct
Explanation:
we know that
2as=vf^2-vi^2
vf=24 m/s
vi= 0 m/s
a=g= 9.8 m/s^2
s=vf^2-vi^2/2a
s=(24)²-(0)²/2*9.8
s=576/19.6
s=29.4 m
therefore option c is correct
Answer:
1.02*10^-5 C/m²
Explanation:
Given that
Radius of the smaller sphere, r = 0.05 m
Radius of the larger sphere, R = 0.12 m
Electric field, E = 2*10^5 V/m
Formula for the electric field is
E = Q/(4πεR²)
this then means that the surface charge density of the larger sphere is
Q/4πR² = Eε = 1.77*10^-6 C/m²
and
Q = 4πεER² = 3.203*10^-7 C
is the charge on the large sphere, which is the same as the charge on the small sphere since they are connected by the wire
so the surface charge density of the smaller sphere is
Q/4πr² = 4πεER²/4πr²
Q = εER²/r²
Q = 1.02*10^-5 C/m²
