Answer: (d)
Explanation:
Given
Mass of the first ram 
The velocity of this ram is 
Mass of the second ram 
The velocity of this ram 
They combined after the collision
Conserving the momentum
![\Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow 49\times (-7)+52\times (9)=(52+49)v\\\Rightarrow v=\dfrac{125}{101}\ m/s \quad[\text{east}]](https://tex.z-dn.net/?f=%5CRightarrow%20m_1v_1%2Bm_2v_2%3D%28m_1%2Bm_2%29v%5C%5C%5CRightarrow%2049%5Ctimes%20%28-7%29%2B52%5Ctimes%20%289%29%3D%2852%2B49%29v%5C%5C%5CRightarrow%20v%3D%5Cdfrac%7B125%7D%7B101%7D%5C%20m%2Fs%20%5Cquad%5B%5Ctext%7Beast%7D%5D)
Momentum after the collision will be
Therefore, option (d) is correct
Answer:
6400 m
Explanation:
You need to use the bulk modulus, K:
K = ρ dP/dρ
where ρ is density and P is pressure
Since ρ is changing by very little, we can say:
K ≈ ρ ΔP/Δρ
Therefore, solving for ΔP:
ΔP = K Δρ / ρ
We can calculate K from Young's modulus (E) and Poisson's ratio (ν):
K = E / (3 (1 - 2ν))
Substituting:
ΔP = E / (3 (1 - 2ν)) (Δρ / ρ)
Before compression:
ρ = m / V
After compression:
ρ+Δρ = m / (V - 0.001 V)
ρ+Δρ = m / (0.999 V)
ρ+Δρ = ρ / 0.999
1 + (Δρ/ρ) = 1 / 0.999
Δρ/ρ = (1 / 0.999) - 1
Δρ/ρ = 0.001 / 0.999
Given:
E = 69 GPa = 69×10⁹ Pa
ν = 0.32
ΔP = 69×10⁹ Pa / (3 (1 - 2×0.32)) (0.001/0.999)
ΔP = 64.0×10⁶ Pa
If we assume seawater density is constant at 1027 kg/m³, then:
ρgh = P
(1027 kg/m³) (9.81 m/s²) h = 64.0×10⁶ Pa
h = 6350 m
Rounded to two sig-figs, the ocean depth at which the sphere's volume is reduced by 0.10% is approximately 6400 m.
A front is a narrow region between two air masses of different densities.
