Answer:
the velocity of the bullet-wood system after the collision is 2.48 m/s
Explanation:
Given;
mass of the bullet, m₀ = 20 g = 0.02 kg
velocity of the bullet, v₀ = 250 m/s
mass of the wood, m₁ = 2 kg
velocity of the wood, v₁ = 0
Let the velocity of the bullet-wood system after collision = v
Apply the principle of conservation of linear momentum to calculate the final velocity of the system;
Initial momentum = final momentum
m₀v₀ + m₁v₁ = v(m₀ + m₁)
0.02 x 250 + 2 x 0 = v(2 + 0.02)
5 + 0 = v(2.02)
5 = 2.02v
v = 5/2.02
v = 2.48 m/s
Therefore, the velocity of the bullet-wood system after the collision is 2.48 m/s
Answer:

Explanation:
q = Charge in the potassium ion = 
e = Charge of electron = 
= Change in potential = 
Change in electric potential is given by

The energy is 
Answer:
0.5 kg
Explanation:
» <u>Concepts</u>
Newton's second law, the Law of Acceleration, states that F = ma, where F = Force in Newtons, m = mass in kg, and a = acceleration in m/s^2.
» <u>Application</u>
We are asked to find the mass of the ball using the equation F = ma. We're also given the force and acceleration, so the equation looks like 5 = 10(m).
» <u>Solution</u>
Step 1: Divide both sides by 10.
Thus, the mass of the ball is 0.5 kg.