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Katen [24]
3 years ago
12

Which of the following is an example of conduction?

Physics
1 answer:
Jet001 [13]3 years ago
7 0

the answer was C because Heat is transferred by conduction when a hot object is in direct contact with a cold object. So, a metal spoon becoming warm when placed in a cup of hot tea is an example of conduction because the spoon gains heat by being in direct contact with the hot tea.

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I think it’s B because we aren’t looking for Vf
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Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o
Lina20 [59]

Answer:

A) M

Explanation:

The three blocks are set in series on a horizontal frictionless surface, whose mutual contact accelerates all system to the same value due to internal forces as response to external force exerted on the box of mass M (Newton's Third Law). Let be F the external force, and F' and F'' the internal forces between boxes of masses M and 2M, as well as between boxes of masses 2M and 3M. The equations of equilibrium of each box are described below:

Box with mass M

\Sigma F = F - F' = M\cdot a

Box with mass 2M

\Sigma F = F' - F'' = 2\cdot M \cdot a

Box with mass 3M

\Sigma F = F'' = 3\cdot M \cdot a

On the third equation, acceleration can be modelled in terms of F'':

a = \frac{F''}{3\cdot M}

An expression for F' can be deducted from the second equation by replacing F'' and clearing the respective variable.

F' = 2\cdot M \cdot a + F''

F' = 2\cdot M \cdot \left(\frac{F''}{3\cdot M} \right) + F''

F' = \frac{5}{3}\cdot F''

Finally, F'' can be calculated in terms of the external force by replacing F' on the first equation:

F - \frac{5}{3}\cdot F'' = M \cdot \left(\frac{F''}{3\cdot M} \right)

F = \frac{5}{3} \cdot F'' + \frac{1}{3}\cdot F''

F = 2\cdot F''

F'' = \frac{1}{2}\cdot F

Afterwards, F' as function of the external force can be obtained by direct substitution:

F' = \frac{5}{6}\cdot F

The net forces of each block are now calculated:

Box with mass M

M\cdot a = F - \frac{5}{6}\cdot F

M\cdot a = \frac{1}{6}\cdot F

Box with mass 2M

2\cdot M\cdot a = \frac{5}{6}\cdot F - \frac{1}{2}\cdot F

2\cdot M \cdot a = \frac{1}{3}\cdot F

Box with mass 3M

3\cdot M \cdot a = \frac{1}{2}\cdot F

As a conclusion, the box with mass M experiments the smallest net force acting on it, which corresponds with answer A.

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Answer:

Image result for Why does the Moon appear red during a lunar eclipse?

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