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.
Answer:
P V = N R T ideal gas equation
P2 / P1 = T2 / T1 where the other variables are constant
P2 = (T2 / T1) * P1 = (313 / 293) * 40 psi = 42.7 psi
Answer: 1.77 s
Explanation: In order to solve this problem we have to use the kinematic equation for the position, so we have:
xf= xo+vo*t+(g*t^2)/2 we can consider the origin on the top so the xo=0 and xf=29 m; then
(g*t^2)/2+vo*t-xf=0 vo is the initail velocity, vo=7.65 m/s
then by solving the quadratric equation in t
t=1.77 s
Answer:
f = 485.62 N
Explanation:
Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:
Unbalanced Force = Horizontal Component Applied Force - Frictional Force
Unbalanced Force = Fx - f
But, from Newtons Second Law of Motion:
Unbalanced Force = ma
comparing the equations:
ma = Fx - f
f = F Cos θ - ma
where,
f = frictional force = ?
F = Applied force = 593 N
m = mass of person = 49 kg
a = acceleration = 0.57 m/s²
θ = Angle with horizontal = 30°
Therefore,
f = (593 N)(Cos 30°) - (49 kg)(0.57 m/s²)
f = 513.55 N - 27.93 N
<u>f = 485.62 N</u>
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