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Charra [1.4K]
3 years ago
5

1. Suppose a 10 N force is applied to the side of a 4.0 kg block that is sitting on a table what would be the minimum value of t

he coefficient of static friction in order for the block to remain motionless?
Physics
1 answer:
vlada-n [284]3 years ago
3 0

Answer:

\mu= 0.25

Explanation:

Given data

Force= 10N

mass= 4kg

r= 4*9.81

r= 39.24N

The expression for the force acting is expressed as

F=\mu*r

substitute

\mu=F/r

\mu= 10/39.24

\mu= 0.25

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An initially uncharged 3.47-μF capacitor and a 6.43-kΩ resistor are connected in series to a 1.50-V battery that has negligible
harkovskaia [24]

Answer: a) io=233.28 A ( initial current); b) τ=R*C= 22.31 ms; c) 81.7 ms

Explanation:  In order to explain this problem we have to use, the formule for the variation of the current in a RC circuit:

I(t)=io*Exp(-t/τ)

and also we consider that io=V/R=(1.5/6.43*10^3)

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then the time constant for the RC circuit is τ=R*C=6.43*10^3*3.47*10^-6

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Finally the time to reduce the current to 2.57% of its initial value is obtained from:

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t=-ln(0.0257)*τ=81.68 ms

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3 years ago
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Answer:6.2

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I am boring and you can talk to me?
sladkih [1.3K]
SURE HI



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