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NeTakaya
3 years ago
15

An airplane is flying at a constant speed in a positive direction. It slows down when it approaches the airport where it's going

to land. Which term describes the slowing of the plane?
A) stationary
B) positive velocity
C) positive acceleration
D) negative acceleration
E) constant velocity
Physics
1 answer:
guapka [62]3 years ago
6 0
Negative acceleration d. I think
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Explanation:

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An overhead door is guided by wheels at a and b that roll in horizontal and vertical tracks. when θ = 40°, the velocity of wheel
Karo-lina-s [1.5K]

I think the situation is modeled by the scenario in the attached image. Some specific values seem to be missing (like the height of door d)...

The door forms a right triangles that satisfies

\tan\theta=\dfrac ab\implies\sec^2\theta\dfrac{\mathrm d\theta}{\mathrm dt}=\dfrac{b\frac{\mathrm da}{\mathrm dt}-a\frac{\mathrm db}{\mathrm dt}}{b^2}

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\tan\theta=\dfrac ab\implies\cos\theta=\dfrac bd

so if you happen to know the height of the door, you can solve for b and a.

d is fixed, so

a^2+b^2=d^2\implies2a\dfrac{\mathrm da}{\mathrm dt}+2b\dfrac{\mathrm db}{\mathrm dt}=0\implies\dfrac{\mathrm da}{\mathrm dt}=-\dfrac ba\dfrac{\mathrm db}{\mathrm dt}

We can solve for the angular velocity \dfrac{\mathrm d\theta}{\mathrm dt}:

\dfrac{\mathrm d\theta}{\mathrm dt}=\cos^2\theta\dfrac{b\left(-\frac ba\frac{\mathrm db}{\mathrm dt}\right)-a\frac{\mathrm db}{\mathrm dt}}{b^2}=-\dfrac1a\dfrac{\mathrm db}{\mathrm dt}

At the point when \theta=40^\circ and \dfrac{\mathrm db}{\mathrm dt}=1.8 ft/s, we get

\dfrac{\mathrm d\theta}{\mathrm dt}=-\dfrac{1.8}a\dfrac{\rm deg}{\rm s}=-\dfrac{1.8}{d\sin40^\circ}\dfrac{\rm deg}{\rm s}

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3 years ago
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Explanation:

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3 years ago
The motion of a free falling body is an example of __________ motion​
swat32

Answer:

accelerated

Explanation:

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3 years ago
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