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Yuki888 [10]
3 years ago
10

The coefficient of kinetic friction between a suitcase and the floor is 2.76. if the suitcase has a mass of 71.5 kg, how far can

it be pushed across the level floor with 670 j of work?
Physics
1 answer:
mihalych1998 [28]3 years ago
8 0

suitcase can be pushed to a distance of 0.35 m

Explanation:

Work done= Ff d

w= 670 J

Ff= force of friction

d= distance

force of friction is given by Ff=μ mg

μ= coefficient of friction=2.76

m=mass=71.5 kg

so Ff= (2.76) (71.5) (9.8)

Ff=1934 N

so W= Ff d

670= 1934 d

d=0.35 m

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Answer:

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F(3) = sqrt{ F(3x)²+F(3y)²} =13.23 N

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3 years ago
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marysya [2.9K]

Answer:

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Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

E=1.36\times10^6\ \rm N/C

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\

Let consider a  sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let \phi be the flux of the Electric Field coming out\passing through it which is given  by

\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C

It can be observed that same amount of  flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

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