A 0.050 kilogram bullet is fired from a 4.0 kilogram rifle which is initially at
1 answer:
The recoil momentum of the rifle is (3) 20 kg-m/s
Explanation:
We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the bullet-rifle system must be conserved before and after the shot, in absence of external forces.
Before the shot, the total momentum of the system is zero, since both the bullet and the rifle are at rest:
(1)
After the shot, the total momentum of the system is:
(2)
where :
is the momentum of the bullet after the shot
is the recoil momentum of the rifle
Since momentum is conserved, (1) = (2):
Therefore we can find :
where the negative sign tells that the direction is opposite to the momentum of the bullet.
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Given :
A box weighing 12,000 N is parked on a 36° slope.
To Find :
What will be the component of the weight parallel to the plane that balances friction.
Solution :
The component of that will be parallel to the plane to balance friction is :
Therefore, component of force to balance friction is F sin 36° .
Hence, this is the required solution.
Answer:
4.62 s
Explanation:
We are given that
Initial angular speed,
Substitute the values
Hence, the wheel takes 4.62 s to come to rest.
Answer:
The volume of water that was in the kettle is 1170
Explanation:
Given:
Power, P = 2.0 kW = 2000 W, Mass of stainless steel, = 710 g = 0.71 kg at temperature of
Part A:
If it takes time, t = 3.5 minutes to reach boiling point of water , then from conservation of energy,
Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from to
i.e. Pt = (100 - 20 ) +
(100 - 20 )
=
= 1.17 kg
where = 4200 J/Kgk (specific heat capacity of water),
= 450 J/Kgk (specific heat capacity of steel)
But volume of water in the the kettle, v =
∴ v = 1170