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Law Incorporation [45]
2 years ago
9

Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85

.0-kg grouper exert to stay submerged in salt water if its body density is
Physics
1 answer:
vichka [17]2 years ago
7 0

The force require to keep grouper submerged is 8.207N.

According to Archimedes principle  buoyant force of any object must equal to weight of fluid it displaced.

The expression for the force exerted to stay submerged in salt water is

           F = F(b) - w(fish)

 where F(b) = buoyant force

              w(fish) = weight

      now substitute w(b) for F(b)

   →  F = Vρg - w(fish)

where  V = volume of sea water

             ρ = density of sea water

Now by Archimedes principle   V = m(fish)  / ρ(fish)

    →    F = (m(fish) / ρ (fish) ) ρg - m(fish)g

           F =   (85 kg/1015 kg-m^-3) (1.025× 10³ kg-m^-3) (9.8 m/s^2)

                                  -   (85kg)  × 9.8 m/s^2

           F = 841.207N - 833N

           F = 8.207 N

  Hence, the force require to keep grouper submerged is 8.207N.

    Learn more about Archimedes Principle here:

         brainly.com/question/15076878

              #SPJ4

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Which is greater, the force exerted by the Earth on the Sun, or the force exerted by the Sun on the Earth? Why?
LekaFEV [45]

Answer:

There is no great force, the force exerted by the Earth on the Sun, and the force exerted by the Sun on the Earth are equal

Explanation:

By definition...

3 0
2 years ago
An initially motionless test car is accelerated to 115 km/h in 8.58 s before striking a simulated deer. The car is in contact wi
hoa [83]

Answer:

a)       a = 3.72 m / s², b)    a = -18.75 m / s²

Explanation:

a) Let's use kinematics to find the acceleration before the collision

             v = v₀ + at

as part of rest the v₀ = 0

             a = v / t

Let's reduce the magnitudes to the SI system

              v = 115 km / h (1000 m / 1km) (1h / 3600s)

              v = 31.94 m / s

              v₂ = 60 km / h = 16.66 m / s

l

et's calculate

             a = 31.94 / 8.58

             a = 3.72 m / s²

b) For the operational average during the collision let's use the relationship between momentum and momentum

            I = Δp

            F Δt = m v_f - m v₀

            F = \frac{m ( v_f - v_o)}{t}

            F = m [16.66 - 31.94] / 0.815

            F = m (-18.75)

Having the force let's use Newton's second law

            F = m a

            -18.75 m = m a

             a = -18.75 m / s²

4 0
2 years ago
A ball rolls forward in the grass slowing down as it rolls?
olga nikolaevna [1]

Yes it does, uh huh.  It slows down as it rolls.  That's a fact.

In order for the ball to roll forward, it has to push grass out of the way.  That takes energy.  To bend each blade of grass out of its way, the ball has to use a tiny bit of the kinetic energy that it has, so it gradually runs out of kinetic energy.  When its kinetic energy is all gone, it stops moving.

3 0
3 years ago
A 4 kg toy car moves horizontally on a rough road with coefficient of kinetic friction 0.2. It accelerates from rest to 20 m/s i
attashe74 [19]

The total work done on the car is 784Joule.

<h3>What's the acceleration of the car?</h3>
  • As per Newton's equation of motion, V= U+at
  • U= initial velocity= 0 m/s

V= vinal velocity= 20m/s

t= time = 10s

a= acceleration

  • So, 20= 0+ 10a

=> a= 20/10= 2m/s²

<h3>What's the distance covered by the car in 10 seconds?</h3>
  • As per Newton's equation of motion,

V²-U² = 2aS

  • S= distance covered by the car
  • So, 20²-0=2×2×S=4S

=> 400= 4S

=> S= 400/4= 100m

<h3>What's the work done on the car due to frictional force?</h3>

Work done by frictional force= frictional force × distance

= (0.2×4×9.8)×100

= 784Joule

Thus, we can conclude that the work done on the car is 784Joule.

Learn more about the work done here:

brainly.com/question/25573309

#SPJ1

4 0
1 year ago
Pls help
belka [17]

Answer:

2156 J

Explanation:

From the question,

Work done = Combined mass of the bucket and water×height×gravity.

W = (M+m)hg............................. Equation 1

Where M = mass of water, m = mass of the bucket, h = height, g = acceleration due to gravity.

Given: M = 20 kg, m = 2 kg, h = 10 m

Constant: g = 9.8 m/s²

Substitute these  value into equation 1

W = (20+2)×10×9.8

W = 22×98

W = 2156 J

4 0
2 years ago
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