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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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7 0
4 years ago
The suspension system of a 1700 kg automobile "sags" 7.7 cm when the chassis is placed on it. Also, the oscillation amplitude de
spin [16.1K]

Answer:

the spring constant k = 5.409*10^4 \ N/m

the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

Explanation:

From Hooke's Law

F = kx\\\\k =\frac{F}{x}\\\\where \ F = mg\\\\k = \frac{mg}{x}\\\\given \ that:\\\\mass \ of \ each \ wheel = 425 \ kg\\\\x = 7.7cm = 0.077 m\\\\g = 9.8 \ m/s^2\\\\Then;\\\\k = \frac{425 \ kg * 9.8 \ m/s^2}{0.077 \ m}\\\\k = 5.409*10^4 \ N/m

Thus; the spring constant k = 5.409*10^4 \ N/m

The amplitude is decreasing 37% during one period of the motion

e^{\frac{-bT}{2m}}= \frac{37}{100}\\\\e^{\frac{-bT}{2m}}= 0.37\\\\\frac{-bT}{2m} = In(0.37)\\\\\frac{-bT}{2m} = -0.9943\\\\b = \frac{2m(0.9943)}{T}\\\\b = \frac{2m(0.9943)}{\frac{2 \pi}{\omega}}\\\\b = \frac{m(0.9943) \ ( \omega) )}{ \pi}

b = \frac{m(0.9943)(\sqrt{\frac{k}{m})}}{\pi}\\\\b = \frac{425*(0.9943)(\sqrt{\frac{5.409*10^4}{425}) }    }{3.14}\\\\b = 1518.24 \ kg/s\\\\b = 1.518 *10^3 \ kg/s

Therefore; the value for the damping constant \\ \\b = 1.518 *10^3 \ kg/s

5 0
3 years ago
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Answer:

Pneumatics use easily compressible gas like air or pure gas. Meanwhile, hydraulics utilize relatively-incompressible liquid media like mineral oil, ethylene glycol, water, synthetic types, or high-temperature fire-resistant fluids to make power transmission possible.

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hope this helps

6 0
3 years ago
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