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cestrela7 [59]
3 years ago
8

Block A, with mass mA, is initially at rest on a horizontal floor. Block B, with mass mB, is initially at rest on the horizontal

top surface of A. The coefficient of static friction between the two blocks is μs. Block A is pulled with a horizontal force. It begins to slide out from under B if the force is greater than:
(A) mA*g
(B) mB*g
(C) μs*mA*g
(D) μs*mB*g
(E) μs(mA +mB)g
Physics
1 answer:
nexus9112 [7]3 years ago
8 0

Answer:

(E) μs(mA +mB)g

Explanation:

We can apply for mB:

∑ Fx = mB*a   (→)

⇒  Ffriction = mB*a   ⇒  a = Ffriction / mB = μs*N / mB

⇒ a = μs*(mB*g) / mB  ⇒   a = μs*g    (acceleration of the system)

Now, for mA we have

∑ Fx = mA*a   (→)

F - Ffriction = mA*a      ⇒  F = mA*a + Ffriction  

⇒     F = mA*(μs*g) + μs*(mB*g)   ⇒   F = μs*g*(mA + mB)

We must know that the friction acts only between the two blocks

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Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

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using formula

v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}

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Remark

One thing is certain: the ball has a mass of 101 grams wherever it is in the universe. That is not true of the force. The force on the moon is a whole lot less than it is on earth, and maybe planet x as well.

Givens

m = 101 g

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t = 2.91 s

d = 16 m

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d = vi*t + 1/2*a * t^2

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16 = 0 + 1/2 a * 2.91^2

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