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Ad libitum [116K]
3 years ago
9

An airplane has a velocity of 907 km/hr westward relative to the air. If the air is moving eastward at 82 km/hr, what is the vel

ocity of the airplane relative to the ground?
A.
989 km/hr eastward
B.
825 km/hr eastward
C.
989 km/hr westward
D.
825 km/hr westward
Physics
1 answer:
Lisa [10]3 years ago
7 0

Answer:

D.  825 km/hr westward

Explanation:

In order to solve the problem, we should define a positive and a negative direction for the velocities.

Let's take positive (+) as westward and negative (-) as eastward.

1) Relative to the ground, the air is moving eastward at 82 km/h, so the air velocity is v'=-82 km/h

2) Relative to the air, the airplane is moving westward at 907 km/h, so its velocity is v=+907 km/h

In the reference system of the ground, the velocity of the airplane will be given by the velocity of the airplane in the reference system of the air (v) plus the velocity of the reference system of the air relative to the ground (v'):

V=v+v'

Therefore, substituting:

V=907 km/h+(-82 km/h)=+825 km/h

And the sign (+) means its direction is westward.

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Answer:

E_r(6)=4.35614\ MPa

Explanation:

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\sigma _0 = 3.1 MPa

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Stress relation with time

\sigma=\sigma _0exp\left(-\frac{t}{\tau}\right)

at t = 32 s

0.41=3.1exp\left(-\frac{32}{\tau}\right)\\\Rightarrow exp\left(-\frac{32}{\tau}\right)=\frac{0.41}{3}\\\Rightarrow -\frac{32}{\tau}=ln\frac{0.41}{3}\\\Rightarrow \tau=-\frac{32}{ln\frac{0.41}{3}}\\\Rightarrow \tau=16.0787\ s

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At t = 6

\\\Rightarrow E_{r}(6)=\frac{\sigma(6)}{\epsilon_0}

From the first equation

\sigma(t)=\sigma _0exp\left(-\frac{t}{\tau}\right)\\\Rightarrow \sigma(6)=3.1exp\left(-\frac{6}{16.0787}\right)\\\Rightarrow \sigma(6)=2.13451

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E_r(6)=4.35614\ MPa

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Answer:

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Explanation:

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Explanation:

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From the question,

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