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notsponge [240]
3 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]3 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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Tems11 [23]

Complete Question

The  complete question is shown on the first  uploaded image

Answer:

The  value  is  J_n  =  -0.864 \  A/cm^2

Explanation:

Generally for an n-type semiconductor the current density is mathematically represented  as

         J_n  =  -q * d_p  *  \frac{d p_{n_o}}{d_w}

Here  p_{n_o} is mathematically represented as

          p_{n_o}  = \frac{n_i^2}{n_d}

=>      p_{n_o}  = \frac{(1.5*10^{10})^2}{10^{16}}

=>      p_{n_o}  = 2.25 *10^{4} \  cm^{-3}

So

    d p_{n_o} =  P_n0 - p_{n_o}

From the diagram  P_n0  =  10^8 * p_{n_o}

=>  P_n0  =  10^8 * (2.25 *10^{4} )

So

   d p_{n_o} =  10^8 * (2.25 *10^{4} ) - 2.25 *10^{4}

      d p_{n_o} =  2.25 *10^{12} cm^{-3}

So  from  J_n  =  -q * d_p  *  \frac{d p_{n_o}}{d_w}

substitute

    1.60 *10^{-19} \  C for  q  and  w_2 =50 nm =  50*10^{-9} m  =  5*10^{-6} cm

and from the diagram  w_1 =0 \ cm

So

    J_n  =  -1.60 *10^{-19} *12 *  \frac{2.25 *10^{12} }{ 5*10^{-6} - 0 }

    J_n  =  -0.864 \  A/cm^2

6 0
3 years ago
We know that an electric field contains energy. What causes the electric field itself?
Art [367]

Electrical charges on one or more particles within the field cause the electric field

Each point in space has an electric field associated with it when a charge of any kind is present. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field

A region of space surrounding an electrically charged particle or object known as an electric field is one in which an electric charge would experience force. A vector quantity called an electric field can be represented by arrows pointing in the direction of or away from charges. The force per unit charge exerted on a positive test charge that is at rest at a given position is the force per unit charge that is used to define the electric field analytically.

To learn more about electric field please visit - brainly.com/question/15800304
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5 0
1 year ago
If you were traveling away from Earth at speed 0.5c, would you notice a change in your heartbeat? Would your mass, height, or wa
FrozenT [24]

Answer:

If you were traveling away from earth at speed 0.5c, you wouldn't notice any change in your heartbeat, you won't notice your mass, height and waistline change. This is because you are on the same frame of reference as the ship in spacetime and any measurement done from the ship will give normal readings from an observer on the ship.

For an observer on earth, your heartbeat will be seen to slowdown (because your time on the ship will be perceived to slow down to an

observer on earth). Also, your mass will be seen to increase, you height will also be seen to increase, and your waistline will be seen to decrease when viewed from earth.

8 0
3 years ago
A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will be its final velocity in m/s,
balu736 [363]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + 13 x 2

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

4 0
3 years ago
The temperature at a point (x, y) on a flat metal plate is given by T(x, y) = 54/(7 + x2 + y2), where T is measured in °C and x,
ANTONII [103]

Answer:

dt/dx = -0.373702

dt/dy =  -1.121107

Explanation:

Given data

T(x, y) = 54/(7 + x² + y²)

to find out

rate of change of temperature with respect to distance

solution

we know function

T(x, y) = 54 /( 7 + x² + y²)

so derivative it x and y direction i.e

dt/dx = -54× 2x / (7 +x² + y²)²    .........................1

dt/dy = -54× 2y / (7 + x² + y²)²      .........................2

now put the value point (1,3) as x = 1 and y = 3 in equation 1 and 2

dt/dx = -54× 2(1) / (7 +(1)² + (3)²)²  

dt/dx = -0.373702

and

dt/dy =  -54× 2(3) / (7 + (1)² + (3)²)²

dt/dy =  -1.121107

7 0
2 years ago
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