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Ostrovityanka [42]
3 years ago
8

As shown above an adhesive has been applied to contacting faces of two blocks so that the blocks interact with an adhesive force

that has a magnitude, Fadhesion A tension, T, is exerted by a wire attached to the upper blocks causing the blocks to remain at rest The two blocks have weights, Wbottom and Wtop. Which of the following must be true? (1 point) ♡ O Wtop=T OT = Wtop - Fadhesion O Wbottom Fadhesion O T = Wbottom + Wtop + Fadhesion​
Physics
1 answer:
rusak2 [61]3 years ago
8 0

The magnitude of the adhesive force allows the top block to remain

attached to the bottom when the blocks and the wire are balanced.

The option that must be true is; \mathbf{W_{bottom} = F_{adhesion}}

Reason:

The tension exerted by the wire attached to the top block = T

Magnitude of the adhesive = F_{adhesion}

Weight of the top block = W_{top}

Weight of the bottom block = W_{bottom}

Given that with the exertion of the tension, the two blocks remain at rest, we have;

  • T =  W_{top} + W_{bottom}

The adhesive causes the bottom block to remain attached to the top block, we have;

Therefore, the magnitude of the adhesive force adds the bottom weight, to the top, weight, which gives;

The magnitude of the adhesive force = The weight of the bottom block

Therefore;

  • W_{bottom} = F_{adhesion}

Learn more here:

brainly.com/question/18907970

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F = 2.0*10^20 N,
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F=\dfrac{Gm_1m_2}{r^2}

With the given values of F,G,m_1,r, we have

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Try dealing with the powers of 10 first: On the right, we have

\dfrac{10^{-11}\times10^{24}}{(10^8)^2}=\dfrac{10^{24-11}}{10^{16}}=10^{-3}

Meanwhile, the other values on the right reduce to

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2.0\times10^{20}\,\mathrm N=\left(2.76\times10^{-3}\,\dfrac{\mathrm N}{\mathrm{kg}}\right)m_2

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