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3 years ago

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Tom has a mass of 67.1 kg and Sally has a mass of 58.6 kg. Tom and Sally are standing 32.3 m apart on a massless dance floor. Sa

lly looks up and she sees Tom. She feels an attraction. If the attraction is gravitation, find its magnitude. Assume both can be replaced by point masses and that the gravitational constant is 6.67259 × 10−11 N · m2 /kg2 . Answer in units of N.

1 answer:

Gnom [1K]3 years ago

3 0

Answer: $F=25.14\times 10^{-11} N$

Explanation:

Given

mass of Tom $(m_t)=67.1 kg$

mass of sally $(m_s)=58.6 kg$

Distance between them(d)=32.3 m

Gravitational constant(G) $=6.67\times 10^{-11} N.m^2/kg^2$

Gravitational attraction is given by

$F=\frac{Gm_1m_2}{d^2}$

$F=\frac{6.67\times 10^{-11}\times 67.1\times 58.6}{32.3^2}$

$F=25.14\times 10^{-11} N$

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Answer:

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3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

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Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

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The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

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$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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2 years ago

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Explanation:

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We have

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