Answer: I think the answer is D.
Positively charged particles.
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
Explanation:
1. Mass of the proton, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :



V = 45.83 volts
2. Mass of the electron, 
Wavelength, 
We need to find the potential difference. The relationship between potential difference and wavelength is given by :




V = 84109.27 volt
Hence, this is the required solution.
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From 1st EQ. of uniform motion
1)20.9-3.44/6=t=2.91 sec
2)28= 0+4.22a
a=6.66m/sec^2
3) Instantaneous acceleration