Ask question

Ask question

All categories

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?

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.

You might be interested in

Stars combine Hydrogen to make Helium during nuclear fusion. Living things are made of heavier elements like Carbon, Oxygen, Iro

alex41 [277]

They were formed in the nuclear<span> fusion reaction inside older </span><span>stars.

As a star burns, fusion reactions inside its core create heavier elements. Those materials are released when the star dies of old age in an explosion.</span>

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

B. more accurate system of measurement

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

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

6 0

3 years ago