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2 years ago

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A 20 kg box is being pulled across a floor by a horizontal rope. The tension in the rope is 99 Newtons. The coefficient of frict

ion is 0.25. What is the force of friction on the box? What is the acceleration of the box?

2 answers:

fgiga [73]2 years ago

5 0

Answer:

friction = 49 N, a = 2.5 m/s/s

Explanation:

big brain

Burka [1]2 years ago

4 0

Yeah what that guy said lol

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A wheel rotates about a fixed axis with a constant angular acceleration of 3.3 rad/s2. The diameter of the wheel is 21 cm. What

AleksandrR [38]

Answer:

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Explanation:

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We have to find the linear speed of a at an instant when that point has a total linear acceleration with a magnitude of 1.7 m/s2.

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$a_t=3.3\times \frac{21\times 10^{-2}}{2}$

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We know that

$a=\sqrt{a^2_t+a^2_r}$

Using the formula

$1.7=\sqrt{(34.65\times 10^{-2})^2+(\frac{v^2}{r})^2}$

Squaring on both sides

we get

$2.89=1200.6225\times 10^{-4}+\frac{v^4}{r^2}$

$\frac{v^4}{r^2}=2.89-1200.6225\times 10^{-4}$

$v^4=r^2\times 2.7699$

$v^4=(10.5\times 10^{-2})^2\times 2.7699$

$v=((10.5\times 10^{-2})^2\times 2.7699)^{\frac{1}{4}}$

$v=0.418 m/s$

Hence, the the linear speed (in m/s) of a point on the rim of this wheel at an instant=0.418 m/s

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