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Feliz [49]
3 years ago
12

An airplane travels 80 m/s as it makes a horizontal circular turn which has a 0.80-km radius. What is the magnitude of the resul

tant force on the 75-kg pilot of this airplane ?
Physics
1 answer:
Nadya [2.5K]3 years ago
6 0

Answer: 600N

Explanation:

Centripetal force is the force that causes a body to move in a circular path.

Centripetal force = MV²/r

M = 75kg v = 80m/s r = 0.80km = 800m

Substituting the values given in the formula;

F = 75 × 80²/800

F = 600N

The magnitude of the resultant force on the 75kg pilot is 600N.

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a)Amplitude ,A = 2 mm

b)f=95.49 Hz

c)V=  30 m/s  ( + x direction )

d)  λ = 0.31 m

e)Umax= 1.2 m/s

Explanation:

Given that

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

As we know that standard form of wave equation given as

y=A sin(\phi -\omega t)

A= Amplitude

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t=Time

Φ = Phase difference

y=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

So from above equation we can say that

Amplitude ,A = 2 mm

Frequency ,ω= 600 rad/s                     (2πf=ω)

ω= 2πf

f= ω /2π

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K= 20 rad/m

So velocity,V

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V = f λ

30 = 95.49 x  λ

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We know that speed is the rate of displacement

U=\dfrac{dy}{dt}

U=2\ mm\ sin[(20m^{-1})x-(600s^{-1})t]

U=1200\ cos[(20m^{-1})x-(600s^{-1})t]\ mm/s

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Umax = 1200 mm/s

Umax= 1.2 m/s

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