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3 years ago

Select the correct answer.

Physics

1 answer:

satela [25.4K]3 years ago

7 0

Answer:

all of the above

Explanation:

muscular endurance is the ability to be able to do muscular activities for a long period of time. The longer you do it, the better you can handle it and for longer.

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2 years ago

Three blocks are placed in contact on a horizontal frictionless surface. A constant force of magnitude F is applied to the box o

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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:

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

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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