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Inessa [10]
3 years ago
5

The maximum pressure most organisms can survive is about 1000 times the atmospheric pressure. Only small, simple organisms such

as bacteria can survive such high pressures. What then is the maximum depth at which these organisms can live under the sea (assuming that the density of seawater is 1025 kg/m3)
Physics
1 answer:
Schach [20]3 years ago
6 0

Answer:

h = 10000 m

Explanation:

The pressure applied at a depth of the liquid is given by:

P =ρgh

where,

P = Maximum Pressure to Survive = (1000)(Atmospheric Pressure)

P = (1000)(101325 Pa) = 1.01 x 10⁸ Pa

ρ = Density of sea water = 1025 kg/m³

g = 9.8 m/s²

h = maximum depth to survive = ?

Therefore,

1.01 x 10⁸ Pa = (1025 kg/m³)(9.8 m/s²)h

h = (1.01 x 10⁸ Pa)/(1025 kg/m³)(9.8 m/s²)

<u>h = 10000 m</u>

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A parallel-plate capacitor is charged by connecting it to a battery. If the battery is disconnected and then the separation betw
TEA [102]

Answer:

The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)

Explanation:

The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. Q, the amount of charge stored in this capacitor, will stay the same.

The formula \displaystyle Q = C\, V relates the electric potential across a capacitor to:

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While Q stays the same, moving the two plates apart could affect the potential V by changing the capacitance C of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:

\displaystyle C = \frac{\epsilon\, A}{d},

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  • d is the distance between the two plates.

Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of \epsilon. Neither will that change the area of the two plates.

However, as d (the distance between the two plates) increases, the value of \displaystyle C = \frac{\epsilon\, A}{d} will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.

On the other hand, the formula \displaystyle Q = C\, V can be rewritten as:

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The value of Q (charge stored in this capacitor) stays the same. As the value of C becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.  

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The total resistance of those two devices in parallel is 6 and 2/3 ohms. That doesn't depend on what they're plugged into. It would still be 6 and 2/3 ohms if they were connected in parallel, wrapped in tissue, sealed in a jar of chicken soup and stored in a box on a high shelf.

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