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fiasKO [112]
2 years ago
15

Assume that the cart is free to roll without friction and that the coefficient of static friction between the block and the cart

is μs. Derive an expression for the minimum horizontal force that must be applied to the block in order to keep it from falling to the ground. Express your answer in terms of acceleration of gravity g and some or all of the variables m, M, and μs.

Physics
2 answers:
Dahasolnce [82]2 years ago
6 0

Answer:F=\frac{(M+m)g}{\mu _s}

Explanation:

Given

Trolley of mass M is free to roll without friction

coefficient of friction between trolley and mass m is \mu _s

Force F is applied on mass m

Acceleration of the system is

a=\frac{F}{M+m}

friction Force will balance weight of block

friction force=\mu _sN

N=ma

\mu _sN=mg

\mu _sma=mg

\mu _s=\frac{g}{a}

F=\frac{(M+m)g}{\mu _s}

Ainat [17]2 years ago
3 0

Answer:<em> </em>N = <em>fa/g</em>

Explanation:

if the  block is not to fall,the friction force, <em>f , </em>must balance the block's weight :

<em>f = mg</em><em>. </em>But the horizontal motion of the block is given by <em>N = ma</em>

Therefore,

<em>f / N  = g/ a </em><em> </em>

<em>or </em><em>a = g/f / N</em>

Since the maximum value of <em>f / N  is μs</em><em>, </em>we must have <em>a</em> ≥ <em>g/μs,</em><em> </em>if<em> </em>the block is not to fall. Q.E.D

N = <em>fa/g</em>

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Alex_Xolod [135]

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2 years ago
A gas has an initial volume of 24.6 L at a pressure of 1.90 atm and a temperature of 335 K. The pressure of the gas increases to
Tatiana [17]

Answer:

the final temperature of the gas is 785.18 K

Explanation:

The computation of the final temperature of the gas is shown below:

Here we apply the gas law

= PV ÷ T

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T2 = ?

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(1.9 × 24.6) ÷ 335 = (3.5 × 31.3)/T2

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8 0
1 year ago
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2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
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Answer:

Explanation:

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Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

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the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

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