An airplane is flying in a horizontal circle of radius 1.5 km with a speed of 450 km/h. What is the magnitude of the centripetal acceleration of the plane?

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An airplane is flying in a horizontal circle of radius 1.5 km with a speed of 450 km/h. What is the magnitude of the centripetal acceleration of the plane?

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Answer 1 · 2018-04-03T17:57:26

$a_{c} = 10.4$ $a_c=10.4$ $m s^{- 2}$ $ms^-2$

Explanation:

v = 450 km/h= $450 \cdot \frac{5}{18} = 125$ $450*5/18=125$ m/s
r=1.5km = 1.5*1000 = 1500 m
centripetal acceleration $(a_{c}) = \frac{v^{2}}{r}$ $(a_c)=v^2/r$
$a_{c} = \frac{15625}{1500} = 10.4$ $a_c=15625/1500=10.4$ $m s^{- 2}$ $ms^-2$

Answer 2 · 2018-04-03T18:23:45

The magnitude of the centripetal acceleration is $10.4 \frac{m}{s^{2}}$ $10.4 m/s^2$ r.

Explanation:

The formula for centripetal acceleration, $a_{c}$ $a_c$ , is

$a_{c} = \frac{v^{2}}{r}$ $a_c = v^2/r$

Before using that, we should convert the units of our data to the standard m for the radius and m/s for the velocity.

$r = 1.5 cancel(km) * (1000 m)/(1 cancel(km)) = 1500 m$

$v = 450 cancel(km)/cancel(hr) * (1000 m)/(1 cancel(km)) * (1 cancel(hr))/(3600 s) = 125 m/s$

Going back to the formula for centripetal acceleration

$a_c = v^2/r = (125 m/s)^2/(1500 m) = (15625 m^2/s^2)/(1500 m) = 10.4 m/s^2$
(I tried to show cancelling of one of the meters in the numerator with the meter in the denominator. I could not make the formatter show that.)

I hope this helps,
Steve

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An airplane is flying in a horizontal circle of radius 1.5 km with a speed of 450 km/h. What is the magnitude of the centripetal acceleration of the plane?

2 Answers

Explanation:

Explanation:

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