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:
c)
Explanation:
As we know that resultant force is the net force that is acting on the system
As per Newton's II law we know that net force is product of mass and acceleration
so we will have
here we know
m = 80 kg
for circular motion acceleration is given as
now we have
Answer:
x = 1.6 + 1.7 t^2 omitting signs
a) at t = 0 x = 1.6 m
b) V = d x / d t = 3.4 t
at t = 0 V = 0
c) A = d^2 x / d t^2 = 3.4 (at t = 0 A = 3.4 m/s^2)
d) x = 1.6 + 1.7 * (4.4)^2 = 34.5 (position at 4.4 sec = 34.5 m)
Answer:
A:1.94
Explanation:
cause that the only one on there
Answer:
a.) 1567.2 m/s
b.) 149.4 m/s
Explanation:
Given that a 26 kg body is moving through space in the positive direction of an x axis with a speed of 350 m/s when, due to an internal explosion, it breaks into three parts. One part, with a mass of 7.8 kg, moves away from the point of explosion with a speed of 180 m/s in the positive y direction. A second part, with a mass of 8.8 kg, moves in the negative x direction with a speed of 640 m/s.
The x-component of the third part can be calculated by assuming that it moves in a positive x axis.
The third mass = 26 - ( 7.8 + 8.8)
The third mass = 26 - 16.6
The third mass = 9.4kg
since momentum is conserved, the momentum before explosion will be equal to sum of the momentum after explosion
26 x 350 = -8.8 x 640 + 9.4V
9100 = -5632 + 9.4V
9.4V = 9100 + 5632
9.4V = 14732
V = 14732/9.4
V = 1567.2 m/s
(b) y-component of the velocity of the third part will be
7.8 x 180 = 9.4 V
1404 = 9.4V
V = 1404/9.4
V = 149.4 m/s
