Answer:
=24.25 ^−1
Explanation:
Let and be initial and final velocity of the body respectively,
be acceleration due to gravity ( 9.8^−2 ), ℎ be the height of the body.
=0 ^ −1
ℎ=30
we know that, ^2−^ 2=2ℎ
^2=2∗9.8∗30
^2=588
=24.25 ^−1
Answer:
a) 3.43 m/s
Explanation:
Due to the law of conservation of momentum, the total momentum of the bullet - rifle system must be conserved.
The total momentum before the bullet is shot is zero, because they are both at rest, so:
Instead the total momentum of the system after the shot is:
where:
m = 0.006 kg is the mass of the bullet
M = 1.4 kg is the mass of the rifle
v = 800 m/s is the velocity of the bullet
V is the recoil velocity of the rifle
The total momentum is conserved, therefore we can write:
Which means:
Solving for V, we can find the recoil velocity of the rifle:
where the negative sign indicates that the velocity is opposite to direction of the bullet: so the recoil speed is
a) 3.43 m/s
Answer:
- This means that the integral of the square modulus over the space is dimensionless.
Explanation:
We know that the square modulus of the wavefunction integrated over a volume gives us the probability of finding the particle in that volume. So the result of the integral
must be dimensionless, as represents a probability.
As the differentials has units of length
for the integral to be dimensionless, the units of the square modulus of the wavefunction has to be:
taking the square root this gives us :
Answer:
60 kg m/s
Explanation:
Let be the acceleration of the object.
As the acceleration of the object is constant, so
Given that applied force, F=6.00 N,
From Newton's second law, we have
,
[from equation (i)]
[given that time, t=10 s and F=6 N]
Here mv is the final momentum of the object and mu is the initial momentum of the object.
So, the change in the momentum of the object is mv-mu.
Hence, the change in the momentum of the object is 60 kg m/s.