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amid [387]
3 years ago
13

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.
Physics
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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600 N upward

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What are two ways you can change atomic mass
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8.
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3 years ago
Motion equation of object is
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5 0
3 years ago
An airplane wing is designed so that the speed of the air across the top of the wing is 255 m/s when the speed of the air below
grin007 [14]
<h2>Answer:442758.96N</h2>

Explanation:

This problem is solved using Bernoulli's equation.

Let P be the pressure at a point.

Let p be the density fluid at a point.

Let v be the velocity of fluid at a point.

Bernoulli's equation states that P+\frac{1}{2}pv^{2}+pgh=constant for all points.

Lets apply the equation of a point just above the wing and to point just below the wing.

Let p_{up} be the pressure of a point just above the wing.

Let p_{do} be the pressure of a point just below the wing.

Since the aeroplane wing is flat,the heights of both the points are same.

\frac{1}{2}(1.29)(255)^{2}+p_{up}= \frac{1}{2}(1.29)(199)^{2}+p_{do}

So,p_{up}-p_{do}=\frac{1}{2}\times 1.29\times (25424)=16398.48Pa

Force is given by the product of pressure difference and area.

Given that area is 27ms^{2}.

So,lifting force is 16398.48\times 27=442758.96N

6 0
3 years ago
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