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3 years ago

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n the metric system the acceleration due to gravity near Earth's surface is about 9.81 meters per second per second. What is the

acceleration due to gravity as measured in the English system, that is, in feet per second per second? (Round your answer to one decimal place.)

1 answer:

Anna007 [38]3 years ago

3 0

Answer:

$32.2\frac{ft}{s^2}$

Explanation:

To do this conversion we have to know that:

$1ft=0.3048m$

(It is easy to find this value on physics textbooks or internet physics webpages)

So, we should convert the gravitational acceleration in the next way:

$9.81\frac{m}{s^{2}}=9.81\frac{m}{s^{2}}*\frac{1ft}{0.3048m}$

So, the meters cancel out:

$9.81\frac{m}{s^{2}}=9.81\frac{1}{s^{2}}*\frac{1ft}{0.3048}=32.18\frac{ft}{s^2}$

$9.81\frac{m}{s^{2}}=32.2\frac{ft}{s}$

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Answer:

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$a_{cB} = 6.075$ m/s²

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Explanation:

Normal or centripetal acceleration measures change in speed direction over time. Its expression is given by:

$a_{c} = \frac{v^{2} }{r}$ Formula 1

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$a_{c}$ : Is the normal or centripetal acceleration of the body ( m/s²)

v: It is the magnitude of the tangential velocity of the body at the given point

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r: It is the radius of curvature. (m)

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We replace data in formula (1) to calculate centripetal acceleration:

$a_{cA} = \frac{(27)^{2} }{120}$

$a_{cA} = 6.075$ m/s²

$a_{cB} = \frac{(27)^{2} }{120}$

$a_{cB} = 6.075$ m/s²

We replace data in formula (2) to calculate centripetal force Fc) :

$F_{cA} = m_{A} *a_{cA} = 1100kg*6.075\frac{m}{s^{2} }$

$F_{cA} = 6682.5 N$

$F_{cB} = m_{B} *a_{cB} = 1600kg*6.075\frac{m}{s^{2} }$

$F_{cB} = 9720 N$

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