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3 years ago

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A push on a 1-kilogram brick accelerates the brick. Neglecting friction, to equally accelerate a 10-kilogram brick, one would ha

ve to push
A. with 100 times as much force.
B. with 10 times as much force.
C. with just as much force.
D. with the amount of force.
E. none of the above

1 answer:

olga55 [171]3 years ago

4 0

I believe the answer is C. With just as much force.

Hope this helps

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Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, while a

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a) $F_{g}=735 N$ and $n=732.47 N$ , hence $F_{g} > n$

b) $n_{poles}=735 N$ $n_{equator}=732.47 N$

Explanation:

a) At the equator, both the <u>centripetal force</u> $F_{c}$ and the <u>gravitational force</u> $F_{g}$ (also called true weight) are directed "downward", while the <u>normal force</u> $n_{equator}$ (also called apparent weight) is directed "upward". Therefore we have the following equation:

$n_{equator}-F_{g}=-F_{c}$ (1)

Where:

$F_{g}=m g$ being $m=75 kg$ the mass and $g=9.8 m/s^{2}$ the acceleration due gravity

$F_{c}=m a_{c}$ being $a_{c}=0.0337 m/s^{2}$ the centripetal acceleration at the equator

According to this (1) is rewritten as:

$n_{equator}-mg=-m a_{c}$ (2)

Isolating $n_{equator}$ :

$n_{equator}=-m a_{c} + mg$ (3)

$n_{equator}=m(-a_{c}+g)$ (4)

$n_{equator}=75 kg (-0.0337 m/s^{2}+9.8 m/s^{2})$ (5)

$n_{equator}=732.47 N$ (6) This is the apparent weight at the equator

The true weight is given by $F_{g}=m g=75 kg (9.8 m/s^{2})$

Hence: $F_{g}=735 N$ (7)

As we can see $F_{g} > n_{equator}$

b) Now we have to calculate the apparent weight at the poles $n_{poles}$ :

$n_{poles}-F_{g}=-F_{c-poles}$ (8)

Since $F_{c-poles}=0$ (8) is rewritten as:

$n_{poles}=F_{g}$ (9)

$n_{poles}=m g$ (10)

$n_{poles}=(75 kg)(9.8 m/s^{2})=735 N$ (11)

So, the apparent weight of the person at the poles is 735 N and at the equator is 732.47 N

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