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nlexa [21]
1 year ago
5

if the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspe

ed will
Physics
1 answer:
Mazyrski [523]1 year ago
5 0

If the outside air temperature increases during a flight at constant power and at a constant indicated altitude, the true airspeed will Increase and true altitude will increase.

<h3>What is Altitude?</h3>

This refers to the vertical elevation of  a body and is the measurement of the height above the sea or ground level. An airplane's altitude is directly proportional to the airspeed as long as the power is constant.

This means that increase in airspeed will lead to an increase in altitude and vie versa. This is the reason why as it runs fast, it can be observed that it begins to elevate.

Read more about Altitude here brainly.com/question/1159693

#SPJ1

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Answer:

2.73414 seconds

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Explanation:

t = Time taken

u = Initial velocity

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v=60\times \dfrac{1609.34}{3600}=26.822\ m/s

mgh=\dfrac{1}{2}mv^2\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{26.822^2}{2\times 9.81}\\\Rightarrow h=36.66766\ m

s=ut+\frac{1}{2}at^2\\\Rightarrow 36.66766=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{36.66766\times 2}{9.81}}\\\Rightarrow t=2.73414\ s

or

v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{26.822-0}{9.81}\\\Rightarrow t=2.73414\ s

The time taken is 2.73414 seconds

The potential energy is given by

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The change in potential energy is 467622.66798 J

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3 years ago
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A piece of aluminum has a volume of 1.50 10-3 m3. the coefficient of volume expansion for aluminum is β = 69 ✕ 10-6 (°c)-1. the
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Answer:

W = 3.12 J

Explanation:

Given the volume is 1.50*10^-3  m^3 and the coefficient of volume for aluminum is β = 69*10^-6 (°C)^-1. The temperature rises from 22°C to 320°C. The difference in temperature is 320 - 22 = 298°C, so ΔT = 298°C. To reiterate our known values we have:

β = 69*10^-6 (°C)^-1       V = 1.50*10^-3  m^3       ΔT = 298°C

So we can plug into the thermal expansion equation to find ΔV which is how much the volume expanded (I'll use d instead of Δ because of format):

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So ΔV = 3.0843*10^-5 m^3

Now we have ΔV, next we have to solve for the work done by thermal expansion. The air pressure is 1.01 * 10^5 Pa

To get work, multiply the air pressure and the volume change.

W = P * dV = (1.01 * 10^5)Pa * (3.0843*10^{-5})m^3\\W = 3.115143J

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Hope this helps!

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