Answer:
91.5 km
Explanation:
Hi!
If we are to ignore the variation in gravity we can use the formula for teh potential energy near the surface of a planet:
mgh
If the energy of the material ejected from the volcano on Io's surface is the same on earth's surface we have:
where subindexes Io and e means Jupiter's moon Io, and Earth, respectively
solving for h_e
The acceleration due to the gravity of a planet can be calculated as:
Where R and m are the radius and mass of the planet
Therefore:
m_e = 5,972 × 10^24 kg
R_e = 6 371 km
Replacing all given values:
h_e = 500 km *(0.183) = 91.515 990 km
Answer:
Static force of friction = 428.75 N
Kinetic Force of Friction = 281.75 N
Explanation:
The static force of friction is equal to Coefficient of static Friction * Mass * g
= 0.35 *125 * 9.8
= 428.75 N
The Kinetic force of friction is equal to Coefficient of kinetic Friction * Mass * g
= 0.23 *125 * 9.8
= 281.75 N
The change in kinetic energy of the car is equivalent to the change in its potential energy. Thus:
K.E = P.E
1/2 x mΔv² = mgΔh
h = (8.2² - 5²) / 2(9.81)
h = 2.15 meters
The ship floats in water due to the buoyancy Fb that is given by the equation:
Fb=ρgV, where ρ is the density of the liquid, g=9.81 m/s² is the acceleration of the force of gravity and V is volume of the displaced liquid.
The density of fresh water is ρ₁=1000 kg/m³.
The density of salt water is in average ρ₂=1025 kg/m³.
To compare the volumes of liquids that are displaced by the ship we can take the ratio of buoyancy of salt water Fb₂ and the buoyancy of fresh water Fb₁.
The gravity force of the ship Fg=mg, where m is the mass of the ship and g=9.81 m/s², is equal to the force of buoyancy Fb₁ and Fb₂ because the mass of the ship doesn't change:
Fg=Fb₁ and Fg=Fb₂. This means Fb₁=Fb₂.
Now we can write:
Fb₂/Fb₁=(ρ₂gV₂)/(ρ₁gV₁), since Fb₁=Fb₂, they cancel out:
1/1=1=(ρ₂gV₂)/(ρ₁gV₁), g also cancels out:
(ρ₂V₂)/(ρ₁V₁)=1, now we can input ρ₁=1000 kg/m³ and ρ₂=1025 kg/m³
(1025V₂)/(1000V₁)=1
1.025(V₂/V₁)=1
V₂/V₁=1/1.025=0.9756, we multiply by V₁
V₂=0.9756V₁
Volume of salt water V₂ displaced by the ship is smaller than the volume of sweet water V₁ because the force of buoyancy of salt water is greater than the force of fresh water because salt water is more dense than fresh water.
