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Ahat [919]
4 years ago
14

A large box of mass M is pulled across a horizontal, frictionless surface by a horizontal rope with tension T. A small box of ma

ss m sits on top of the large box. The coefficients of static and kinetic friction between the two boxes are μs and μk, respectively.Find an expression for the maximum tension T max for which the small box rides on top of the large box without slipping.
Physics
1 answer:
Alexxx [7]4 years ago
6 0

Answer:

Explanation:

Given

Mass of big box is M and small box is m

Tension T will cause the boxes to accelerate

T=(M+m)a

where a=acceleration of the boxes

Now smaller box will slip over large box if the acceleration force will exceed the static friction

i.e. for limiting value

\mu _smg=ma

a=\mu _s\cdot g

thus maximum tension

T=\mu _s(M+m)g

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A student of weight 652 N rides a steadily rotating Ferris wheel (the student sits upright). At the highest point, the magnitude
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Explanation:

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The drift velocity should be in the same direction as the applied force. <br> a. true <br> b. false
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2 years ago
A nonconducting rod of length L = 8.15 cm has charge –q = -4.23 fC uniformly distributed along its length.(a) What is the linear
vichka [17]

Answer:

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Explanation:

a) the linear density defined as the ratio between the charge per unit length

       λ = q / l

Let's start by reducing the units to the SI system

     L = 8.15 cm (1m / 100cm) = 8.15 10⁻² m

     a = 12 cm (1m / 100cm) = 12 10⁻² m

    q = -4.23 fC (1 C / 10¹⁵ ft) = -4.23 10⁻¹⁵ C

    λ = -4.23 10⁻¹⁵ C / 8.15 10⁻²

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b) Let's look for the electric field for a point at a distance a from the end of the bar

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To simplify the notation, suppose the bar is the x axis. Since the density is constant, we can write it differentially

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Let's change the density for its value

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     E = 8.99 10⁹ 4.23 10⁻¹⁵ / 0.5²

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