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3 years ago

Differentiate between atmospheric pressure and pressure.

Physics

2 answers:

Dovator [93]3 years ago

8 0

Answer: atmospheric is air by the earth and pressure is just someone or something doing it

Explanation:

RUDIKE [14]3 years ago

5 0

Answer:

Air pressure is the pressure exerted by the air around us while Atmospheric pressure is the pressure exerted by the atmosphere on the earth. Air pressure is measured by tore gauge while atmospheric pressure is measured using mercury barometer.

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What is your theory on frame of reference?

____ [38]

What is your theory on frame of reference?

Answer:

It is the idea of seeing something from a different perspective

Explanation:

Hope this helps!

8 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

What percentage of energy is available to organisms at each level of an energy pyramid?

Snowcat [4.5K]

The percentage of energy available to each organism is always 10 percent .

6 0

3 years ago

When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa

MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

I = Current = 0.33 A (series)

I = Current = 1.6 A (parallel)

In series

$\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2$

In parallel

$\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}$

$\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0$