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Ymorist [56]
4 years ago
5

He then takes a powerful magnet and positions it below the table and the keys. He finds that he is able to drag the keys along t

he table by moving the magnet. This demonstrates that A. gravitational forces are contact forces. B. gravity is proportional to mass. C. magnetic forces act at a distance. D. magnetic forces act on wooden objects. Reset Submit
Physics
2 answers:
irakobra [83]4 years ago
6 0
Answer is, C. Magnetic forces act at a distance, it is a non contact type of force, such that a body having similar or opposite charge with another body may experience this natural force with no physical contact whatsoever. The forces are categorized into two, force of attraction or repusion.
Murrr4er [49]4 years ago
3 0

answer is magnetic force act at a distance

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Answer:

<h2>42.67N</h2>

Explanation:

Step one:

<u>Given </u>

mass m= 0.32kg

intital velocity, u= 14m/s

final velocity v= 22m/s

time= 0.06s

Step two:

<u>Required</u>

Force F

the expression for the force is

F=mΔv/t

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F=(0.32*8)/0.06

F=2.56/0.06

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4 0
3 years ago
Calculate the mean free path of air molecules at a pressure of 7.00×10^−13 atm and a temperature of 303 K . (This pressure is re
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Answer:

82.8986 km

Explanation:

Given:

Pressure = 7.00×10⁻¹³ atm

Since , 1 atm = 101325 Pa

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Radius = 2.00×10⁻¹⁰ m

Diameter = 4.00×10⁻¹⁰ m (2× Radius)

Temperature = 303 K

The expression for mean free path is:

\lambda (Mean\ free\ path)=\frac {K (Boltzmann\ Constant)\times Temperature}{\sqrt {2}\times \pi\times (Diameter)^2\times Pressure}

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So,

\lambda (Mean\ free\ path)=\frac {1.38\times 10^{-23}\times 303}{\sqrt {2}\times \frac {22}{7}\times (4.00\times 10^{-10})^2\times 7.09275\times 10^{-8}}

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4 0
4 years ago
an object is moving forward with a constant velocity. which statement about this object must be true?
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Answer:

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from definition of acceleration:

{ \rm{acceleration =  \frac{change \: in \: velocity}{time} }} \\  \\ { \rm{acceleration =  \frac{ \delta  \: (velocity)}{time} }}

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{ \boxed{ \rm{ \delta  \: (velocity) = 0}}}

• therefore:

{ \rm{acceleration =  \frac{0}{time} }} \\  \\ { \boxed{ \tt{acceleration = 0 \: m {s}^{ - 2} }}}

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Closed is the correct answer :)
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