Answer:
The speed of the 8-ball is 2.125 m/s after the collision.
Explanation:
<u>Law Of Conservation Of Linear Momentum</u>
The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:
P=mv
If we have a system of masses, then the total momentum is the sum of all the individual momentums:

When a collision occurs, the velocities change to v' and the final momentum is:

In a system of two masses, the law of conservation of linear momentum is simplified to:

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.
After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.
The above equation is solved for v1':




The speed of the 8-ball is 2.125 m/s after the collision.
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Explanation:
for v vs t graph for t=9 to 11,v=15kmph
similarly v2=0,v3=60kmph&v4= -40kmph
The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge
C is placed which is the origin.
Let B be the point where the charge
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law,
,
= 



k is the Coulomb's law constant.
On substituting the values into the above equation, we get,

Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
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