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: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

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Which of the following describes the least amount of work being done? A. wind turning a windmill B. a batter hitting a baseball

Mandarinka [93]

Answer:

a ball hanging from a tall pole

Explanation:

work is force times a distance in the direction of the force. The ball just hanging there has no motion, so is associated with no work. Work was probably done in the hanging process, but none after.

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How do we know the sun is supported by nuclear fusion?

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we know that the sun is supported by nuclear fusion because the sun is a main-sequence star, meaning it has to use nuclear fusion to keep itself going, no nuclear fusion, no sun.

i hope this helps!

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4 years ago

A retaining wall is 5m long (into the plane of the paper), and it is supported by an anchor rod. The soil it supports has a weig

vekshin1

Knowing that the greatest horizontal pressure is the pressure made by the material on the wall -considering for that the total height of the wall-, then P = 4.375 t/m

---------------------

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<u>Available data</u>:

The wall is 5m long into the plane ⇒ H

The soil density is 1.4 metric tonnes per cubic metre ⇒ γ

K = 0.25 ⇒ Coefficient of earth pressure. Depends on the angle.

From this information, we need to calculate the greatest horizontal pressure.

Horizontal pressure refers to the force made by the supported soil on the retaining wall that tends to deflect the wall outward.

We can calculate horizontal pressure at different heights from the top. And since we need to calculate the greatest horizontal pressure, we need to consider the total height that equals the wall height, H.

To do it, we will use the following formula, and replace the terms.

P = [ k γ H² ] / 2

P = [ 0.25 x 1.4 t/m³ x 5m² ] / 2

P = 8.75/2

P = 4.375 t/m

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3. If you were in a car collision, would you rather be hit by Vehicle 1 (2,027 kg) or Vehicle 2 (1,415 kg) from Activity 73? Exp

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Answer:

I say vehicle one, because both will apply the same amount of force but, because vehicle two has less mass, it will accelerate more rather than vehicle one which will accelerate less.

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