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vova2212 [387]
3 years ago
8

5. You are pulling a rubber sneaker along concrete

Physics
1 answer:
Ainat [17]3 years ago
3 0

Answer:

Fs = 17.66 N

Explanation:

The force of static friction is given by the following formula:

F_s = \mu R = \mu W\\

where,

Fs = The force of static friction = ?

μ = coefficient of static friction = 0.9

W = weight of sneakers = mg

m = mass of sneakers = 2 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

F_s = \mu mg\\F_s = (0.9)(2\ kg)(9.81\ m/s^2)\\

<u>Fs = 17.66 N</u>

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Peter, a 100 kg basketball player, lands on his feet after completing a slam dunk and then immediately jumps up again to celebra
Mars2501 [29]

Answer:

Explanation:

Impulse = change in momentum

Initial momentum = mass x initial velocity = 100 x 5 = 500 kg m/s

final momentum = mass x final velocity = 100 x - 4 = -400 ( - ve sign due to reversal of direction )

change in momentum = final momentum - initial momentum

= - 400 - 500 = - 900 kg m/s .

As it is - ve , it acts upwards .

So magnitude of  impulse on Perter = 900 kg m/s

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What are the4 spheres on earth
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northen henisphere,southern hemisphere, Eastern hemisphere, Western hemisphere.

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A uniform solid disk and a uniform hoop are placed side by side at the top of an incline of height h. If they are released from
svlad2 [7]

Answer:

) the uniform disk has a lower moment of inertia and arrives first.

Explanation:

(a) the uniform disk has a lower moment of inertia and arrives first.

(b) Let's say the disk has mass m and radius r, and

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disk: initial E = PE = mgh

I = ½mr², so KE = ½mv² + ½Iω² = ½mv² + ½(½mr²)(v/r)² = (3/4)mv² = mgh

m cancels, leaving v² = 4gh / 3

hoop: initial E = Mgh

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M cancels, leaving V² = gh

Vdisk = √(4gh/3) > Vhoop = √(gh)

3 0
4 years ago
What must be the distance between point charge q1 = 28.0 μC and point charge q2 = −57.0 μC for the electrostatic force between t
vlabodo [156]

Answer:

1.686 m

Explanation:

From coulomb's law,

F = kq1q2/r² ...................................... Equation 1

Where F = electrostatic force  between the two charges, q1 = first charge, q2 = second charge, r = distance between the charges.

making r the subject of the equation,

r = √(kq1q2/F).......................... Equation 2

Given: F = 5.05 N, q1 = 28.0 μC = 28×10⁻⁶ C, q2 = 57.0 μC = 57.0×10⁻⁶ C

Constant: k = 9.0×10⁹ Nm²/C².

Substituting into equation 2

r = √(9.0×10⁹×28×10⁻⁶×57.0×10⁻⁶/5.05)

r = √(14364×10⁻³/5.05)

r = √(14.364/5.05)

r = √2.844

r = 1.686 m

r = 1.686 m.

Thus the distance must be 1.686 m

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