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notsponge [240]
4 years ago
14

Salt water is denser than fresh water. a ship floats in both fresh water and salt water. compared to the fresh water, the volume

of water displaced in the salt water is
Physics
1 answer:
gavmur [86]4 years ago
7 0
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.  


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Maximum range of a projectile is 1.6 m. Then the velocity of projection will be..... (g=10m/s)
qaws [65]

Answer:

4 m/s

Explanation:

From the question given above, the following data were obtained:

Maximum range (Rₘₐₓ) = 1.6 m

Acceleration due to gravity (g) = 10 m/s²

Initial velocity (u) =?

The initial velocity of the projectile can be obtained as follow:

Rₘₐₓ = u² / g

1.6 = u² / 10

Cross multiply

u² = 1.6 × 10

u² = 16

Take the square root of both side

u = √16

u = 4 m/s

Therefore, the velocity of the projectile is 4 m/s

6 0
3 years ago
PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!
SVETLANKA909090 [29]

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 m/s^2

<u>Explanation:</u>

1)  Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2)  Average speed = Total distance covered / Time taken for that distance to cover.

    Total distance covered = 2+0.5+2.5 = 5 km

    Time taken = 3 hours

     Average speed = 5/3 = 1.66 km /hour

3)    Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

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   Time = 4 seconds

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A race car moves along a circular track at a speed of 0.512 m/s. If the car's centripetal acceleration is 15.4 m/s2, what is the
tiny-mole [99]

Answer:

The radius is  r = 0.0170 \ m  

Explanation:

From the question we are told that

      The speed at which the race car moves is v  =  0.512 \ m/s

       The centripetal acceleration is  a _r  =  15.4 \ m/s

Generally the centripetal acceleration is mathematically represented as

           a_r =  \frac{v^2 }{r}

=>        15.4  =  \frac{0.512^2 }{ r}

=>       r = 0.0170 \ m  

 

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