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2 years ago

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1 answer:

Montano1993 [528]2 years ago

5 0

Commercial airliners encounter very little turbulence above 25,000 feet because the air is so thin at altitudes above 25,000 feet that turbulence has much less effect. That is option A.

<h3>What is turbulence?</h3>

A turbulence is a sudden disturbance that an airline encounters when it hits a strong wind current that can push or pull the plane.

The depth at which the airline flies determines whether the turbulence will have little or more effects on it.

At altitudes above 25,000 feet the air is so thin and light than below 25,000 feet therefore, the airline will encounter little turbulence effect.

Learn more about turbulence here:

brainly.com/question/20389242

#SPJ1

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Please help on questions D AND E and please show working out as well so I can understand better thank you only QUESTIONS D AND E

tester [92]

Answer:

d. 87,500 J

e. 49,600 J

Explanation:

The total energy is the heat absorbed by the copper plus the heat absorbed by the water.

E = m₁C₁ΔT + m₂C₂ΔT

E = (1 kg) (390 J/kg/°C) (10 °C) + (2 kg) (4180 J/kg/°C) (10 °C)

E = 87,500 J

E = m₁C₁ΔT + m₂C₂ΔT

E = (2 kg) (390 J/kg/°C) (10 °C) + (1 kg) (4180 J/kg/°C) (10 °C)

E = 49,600 J

5 0

3 years ago

: 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:

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

$F_{No}= 7771.125 \ N$

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

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

$P_d__{D}} = 1828.2 * 79000$

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

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

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

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

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

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

6 0

4 years ago

An object traveling in a circular path is accelerating because its

Anton [14]

It is accelerating because the direction of the velocity vector is changing

8 0

4 years ago

What would be the new volume if the pressure of 600mL is increased from 100kPa to 150.56 kPa

Gelneren [198K]

ASSUMING that the substance is a gas . . .

It all depends on HOW the pressure was increased.

-- IF the gas was left in the same container, and the pressure was increased by heating the container, then the volume hasn't changed. It's still the inside volume of the container.

-- IF the pressure was increased by stuffing all of the gas into a smaller container, then the new volume is the volume of the smaller container.

No matter what the mass or temperature of the gas is, it always expands to fill whatever container you're keeping it in.

4 0

3 years ago

An example of constant velocity

pashok25 [27]

Some examples of constant velocity (or at least almost- constant velocity) motion include (among many others): • A car traveling at constant speed without changing direction. A hockey puck sliding across ice. A space probe that is drifting through interstellar space.

4 0

3 years ago