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brilliants [131]
3 years ago
13

A very large magnet applies a repulsive force to a smaller (0.05 Kg) magnet. If the smaller magnet accelerates across a friction

less table at 2m/s2, then what is the magnitude of the repulsive force?
Physics
1 answer:
makkiz [27]3 years ago
5 0

Answer:

0.1 N

Explanation:

using F = ma yields 0.1 N.

net

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A banked circular highway curve is designed for traffic moving at 58 km/h. The radius of the curve is 201 m. Traffic is moving a
skelet666 [1.2K]

Answer:0.077

Explanation:

Given

banked designed for traffic moving at 58 km/h\approx 16.11 m/s

Radius of the curve 201 m

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For banking of road

tan\theta =\frac{v^2}{rg}

tan\theta =\frac{16.11^2}{201\times 9.81}

\theta =7.49^{\circ}

Centripetal acceleration is given by

a=\frac{v^2}{r}

Taking component of centripetal acceleration

along and perpendicular to surface

a_{parallel}=\frac{v^2cos\theta }{r}

a_{perpendicular}=\frac{v^2sin\theta }{r}

From FBD

mgsin\theta -f_s=ma_{parallel}

f_s=mgsin\theta -ma_{parallel}----1

where f_s is frictional force

N-mgcos\theta =ma_{perpedicular}

N=mgcos\theta +ma_{perpedicular}----2

and we know coefficient of friction is given by

\mu =\frac{f_s}{N}

\mu =\frac{mgsin\theta -ma_{parallel}}{mgcos\theta +ma_{perpedicular}}

\mu =\frac{gsin\theta -\frac{v^2cos\theta }{r}}{gcos\theta +\frac{v^2sin\theta }{r}}

\mu =\frac{1.2804-0.5202}{9.726+0.068}

\mu =0.077

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Which statement best describes the law of conservation of energy?
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