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:
80km/hr
Explanation:
Total distance: 400km
Total time 5 hours
Average speed= distance/time
400/5= 80km/hr
The answer is B-plus.
The object is accelerating to the right, and up.
5.7 km/h north and 5.8 km/h west are instantaneous velocities, while 8.1 km/h northwest is the average velocity.<span>
The answer choice above is correct.
The instantaneous velocities are the actual </span>velocities while traveling ( the velocity during that instant ). The average velocity is the average of the instantaneous velocities ( the speed in one direction equivalent to the two speeds <span>in different directions ).
I used speed in the explanation because velocity is speed with direction.</span>
The velocity equation is 
Known facts:
- t = 3.83s
- a= -3.04
- intial velocity = 0
Plug into equation known quantities:
Thus the final velocity is -11.6432m/s
Hope that helps!
