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
I believe that the answer is C<span />
Answer:
40 N/m
Explanation:
The diagram attached is used to answer the question
We know from Hooke's law that extension is directly proportional to the applied force hence
F=kx where x is extension, F is applied force and k is the spring constant. Making k the subject of the formula then

From the attached diagram extension is given by subtracting unstretched spring from stretched spring hence extension, x=1-0.5=0.5m
Substituting 20 N for F and 0.5 m for x then

If ball remains in air for total time T = 0.85 s
this is also known as time of flight
In order to find the time of flight we can use kinematics

so for complete motion its displacement in y direction will be zero



now we know that net velocity of the ball is 8 m/s
while is y direction component we got is vy = 4.165 m/s
now by component method we can say




so it is projected at an angle of 31.4 degree above horizontal
because they are floating on top of the liquid mantle.