Answer:
If we use the equation for the transformation of velocities for moving frames:
v' = (v - u) / (1 - u * v / c^2) where we measure the speed of v' approaching from the left where v is in a frame moving at -u towards v'
v' = (.6 c - (-.6 c)) / (1 - (-.6 c) * .6 c / c^2) = 1.2 c / (1 + .6 * .6)
or v' = 1.2 c / (1 + .36) = .88 c
v is approaching from the left at .6 c in the reference frame and the other frame approaches from the right at -.6 c with speed u (-.6 c) and we measure the speed of v as seen in the frame moving to the left
The convection going on with the magma in the asthenosphere
Answer:
i 5.3 cm ii. 72 cm
Explanation:
i
We know upthrust on iron = weight of mercury displaced
To balance, the weight of iron = weight of mercury displaced . So
ρ₁V₁g = ρ₂V₂g
ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?
V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³
So, the height of iron above the mercury is h = V₂/area of base iron block
= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm
ρ₁V₁g = ρ₂V₂g
ii
ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?
V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³
So, the height of column of water is h = V₃/area of base iron block
= 7200 cm³/10² cm² = 72 cm
