Answer:
0.0268 m
Explanation:
Draw a free body diagram of the block. There are three forces: weight force mg pulling down, buoyancy of the oil B₁ pushing up, and buoyancy of the water B₂ pushing up.
Sum of forces in the y direction:
∑F = ma
B₁ + B₂ − mg = 0
ρ₁V₁g + ρ₂V₂g − mg = 0
ρ₁V₁ + ρ₂V₂ = m
ρ₁V₁ + ρ₂V₂ = ρV
ρ₁Ah₁ + ρ₂Ah₂ = ρAh
ρ₁h₁ + ρ₂h₂ = ρh
(930 kg/m³)h₁ + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)
Since the block is fully submerged, h₁ + h₂ = 4.93 cm.
(930 kg/m³) (4.93 cm − h₂) + (1000 kg/m³)h₂ = (968 kg/m³) (4.93 cm)
h₂ = 2.68 cm
h₂ = 0.0268 m
Answer:
Mass and Gravity
Explanation:
Objects with mass exert forces on each other via the force of gravity. This force is proportional to the mass of the two interacting objects and is inversely proportional to the square of the distance between them.
Answer:
the moment of inertia of the merry go round is 38.04 kg.m²
Explanation:
We are given;
Initial angular velocity; ω_1 = 37 rpm
Final angular velocity; ω_2 = 19 rpm
mass of child; m = 15.5 kg
distance from the centre; r = 1.55 m
Now, let the moment of inertia of the merry go round be I.
Using the principle of conservation of angular momentum, we have;
I_1 = I_2
Thus,
Iω_1 = I'ω_2
where I' is the moment of inertia of the merry go round and child which is given as I' = mr²
Thus,
I x 37 = ( I + mr²)19
37I = ( I + (15.5 x 1.55²))19
37I = 19I + 684.7125
37I - 19 I = 684.7125
18I = 684.7125
I = 684.7125/18
I = 38.04 kg.m²
Thus, the moment of inertia of the merry go round is 38.04 kg.m²
PERIOD AS WE SHOUUULLD WE GET TO MARRY WHOEVER WE WAAAANT WIN FOR US
Answer:
100m
Explanation:
100m
s=ut+1/2at^2
s= unknown, u=0, a=2, t=10
s=0*10+1/2(2)(10)^2
s=1/2(2)(100)
s=1(100)
displacement = 100 meters
