Mass of Jupiter=1.9×10
27
㎏=M
1
Mass of Sun=1.99×10
30
㎏=M
2
Mean distance of Jupiter from Sun=7.8×10
11
m=r
G=6.67×10
−11
N㎡㎏
−2
Gravitational Force, F=
r
2
GM
1
M
2
F=
(7.8×10
11
)
2
6.67×10
−11
×1.9×10
27
×1.99×10
30
F=4.16×10
23
N
Answer:
the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
Explanation:
This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law
B- W = 0
B = W
the thrust force is equal to the weight of the liquid that is dislodged
B = ρ g V
we substitute
ρ g V = m g
V = m /ρ_fluid 1
we can write the mass of the pot as a function of its density
ρ_body = m / V_body
m = ρ_body V_body
V_fluid / V_body = ρ_body / ρ _fluid 2
Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.
The effect appears the pot as if it had a lower apparent weight
Yes, the volume of the cylinder will remain constant. As the radius decreases, the height will increase to make sure that the volume is kept the same.
We have been given a value of dr/dt and are required to find dh/dt
Because the volume is constant, we can plug it into the formula for the volume of the cylinder and rearrange it to make h the subject:
128 = πr²h
h = 128/πr²
Now we differentiate both sides:
dh/dr = -256/πr³
Applying the chain rule:
dh/dt = dh/dr x dr/dt
dh/dt = (-256/πr³) x -0.05
dh/dt = 64/5πr³; substituting the value of r
dh/dt = 64/5π(1.5)³
dh/dt = 1.21 in/sec
