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kari74 [83]
2 years ago
7

Calculate The water pressure at the bottom of the Marianas Trench is approxi mately 1,100 kPa. With how much force would the wat

er pressure at the bottom of the Marianas Trench push on a fish with a surface area of 0.50 m^2?
Physics
1 answer:
miss Akunina [59]2 years ago
4 0

Answer:

Force, F = 550000 N

Explanation:

Given that,

Pressure at the bottom of the Marianas Trench, P=1100\ kPa=11\times 10^5\ Pa

Surface area, A=0.5\ m^2

We need to find the force with which the water pressure at the bottom of the Marianas Trench push on a fish. Mathematically, the pressure is given by :

P=\dfrac{F}{A}

F=P\times A

F=11\times 10^5\ Pa\times 0.5\ m^2

F = 550000 N

So, the force with which the water pressure at the bottom of the Marianas Trench push on a fish is 550000 N. Hence, this is the required solution.

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alex41 [277]
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8 0
3 years ago
A string is tied between two posts separated by 2.4 m. When the string is driven by an oscillator at frequency 567 Hz, 5 points
Alex787 [66]

Explanation:

The given data is as follows.

       Length (l) = 2.4 m

       Frequency (f) = 567 Hz

Formula to calculate the speed of a transverse wave is as follows.

                  f = \frac{5}{2l} \times v

Putting the gicven values into the above formula as follows.

                  f = \frac{5}{2l} \times v

                 567 Hz = \frac{5}{2 \times 2.4 m} \times v

                      v = 544.32 m/s

Thus, we can conclude that the speed (in m/s) of a transverse wave on this string is 544.32 m/s.

5 0
3 years ago
What are the advantages of using the metric system? SELECT ALL THAT APPLY
alexandr402 [8]

D. used by the entire scientific community

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3 0
2 years ago
Determine the minimum work per unit of heat transfer from the source reservoir that is required to drive a heat pump with therma
Sloan [31]

Answer:

The minimum work per unit heat transfer will be 0.15.

Explanation:

We know the for a heat pump the coefficient of performance (C_{HP}) is given by

C_{HP} = \dfrac{Q_{H}}{W_{in}}

where, Q_{H} is the magnitude of heat transfer between cyclic device and    high-temperature medium at temperature T_{H} and W_{in} is the required input and is given by W_{in} = Q_{H} - Q_{L}, Q_{L} being magnitude of heat transfer between cyclic device and low-temperature T_{L}. Therefore, from above equation we can write,

&& \dfrac{Q_{H}}{W_{in}} = \dfrac{Q_{H}}{Q_{H} - Q_{L}} = \dfrac{1}{1 - \dfrac{Q_{L}}{Q_{H}}} = \dfrac{1}{1 - \dfrac{T_{L}}{T_{H}}}

Given, T_{L} = 460 K and T_{H} = 540 K. So,  the minimum work per unit heat transfer is given by

\dfrac{W_{in}}{Q_{H}} = \dfrac{T_{H} - T_{L}}{T_{H}} = \dfrac{540 - 460}{540} = 0.15

8 0
3 years ago
A 5.6 cm diameter parallel-plate capacitor has a 0.58 mm gap. What is the displacement current in the capacitor if the potential
BARSIC [14]

Answer:

1.88\cdot 10^{-5} A

Explanation:

The capacitance of a parallel plate capacitor is given by:

C=\frac{\epsilon_0 A}{d} (1)

where

\epsilon_0 is the vacuum permittivity

A is the area of the plates

d is the separation between the plates

The charge stored on the capacitor is given by

Q=CV (2)

where C is the capacitance and V is the voltage across the capacitor.

The displacement current in the capacitor is given by

J=\frac{Q}{t} (3)

where t is the time elapsed

Substituting (1) and (2) into (3), we find an expression for the displacement current:

J=\frac{CV}{t}=\frac{\epsilon_0 A}{d} \frac{V}{t}

where we have

A=\pi (\frac{d}{2})^2=\pi (\frac{0.056 m}{2})^2=2.46\cdot 10^{-3} m^2

d = 0.58 mm = 5.8\cdot 10^{-4} m

\frac{V}{t}=500,000 V/s

Substituting into the equation, we find

J=\frac{(8.85\cdot 10^{-12} F/m)(2.46\cdot 10^{-3} m^2)}{5.8\cdot 10^{-4}m}(500,000 V/s)=1.88\cdot 10^{-5} A

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3 years ago
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