Answer:
Explanation:
As we know that tank itself has mass given as 13.4 kg
so it is given as
also the tank contains air in it and the mass of air inside the tank is also given as 4.47 kg
so it is
so total mass of the tank is the mass of the empty tank and mass of air in the tank
Answer:
The distance the block will slide before it stops is 3.3343 m
Explanation:
Given;
mass of bullet, m₁ = 20-g = 0.02 kg
speed of the bullet, u₁ = 400 m/s
mass of block, m₂ = 2-kg
coefficient of kinetic friction, μk = 0.24
Step 1:
Determine the speed of the bullet-block system:
From the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
where;
v is the speed of the bullet-block system after collision
(0.02 x 400) + (2 x 0) = v (0.02 + 2)
8 = v (2.02)
v = 8/2.02
v = 3.9604 m/s
Step 2:
Determine the time required for the bullet-block system to stop
Apply the principle of conservation momentum of the system
when the system stops, vf = 0
3.9604 -2.352t = 0
2.352t = 3.9604
t = 3.9604/2.352
t = 1.684 s
Thus, time required for the system to stop is 1.684 s
Finally, determine the distance the block will slide before it stops
From kinematic, distance is the product of speed and time
Now, recall that t = 1.684 s
S = 3.9604(1.684) - 1.176(1.684)²
S = 6.6693 - 3.3350
S = 3.3343 m
Thus, the distance the block will slide before it stops is 3.3343 m