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3 years ago

11

Consider this situation: A baseball player dives head-first

1 answer:

siniylev [52]3 years ago

7 0

Of the forces listed I think the force of him diving and sliding across the infield acted on the player.

I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.

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A liquid of density 830 kg/m3 flows through a horizontal pipe that has a cross-sectional area of 1.20 x 10-2 m2 in region A and

kicyunya [14]

Answer:

A) volume flow rate = 0.047 m3/s

B) mass flow rate = 39.01 kg/s

Explanation:

Detailed explanation and calculation is shown in the image below

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3 years ago

Which of these will always produce a magnetic field?

vovikov84 [41]

Answer:

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3 years ago

Which statement is true about the Big Bang Theory?

zmey [24]

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3 years ago

The table shows data for the planet Uranus. A 2 column table with 4 rows. The first column is labeled Quantity with entries, Esc

prohojiy [21]

Answer:

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4 0

3 years ago

Consider a block on frictionless ice. Starting from rest, the block travels a distance din

sweet [91]

Answer:

<em>The distance is now 4d</em>

Explanation:

<u>Mechanical Force</u>

According to the second Newton's law, the net force exerted by an external agent on an object of mass m is:

F = m.a

Where a is the acceleration of the object.

The acceleration can be calculated by solving for a:

$\displaystyle a=\frac{F}{m}$

Once we know the acceleration, we can calculate the distance traveled by the block as follows:

$\displaystyle d = vo.t+\frac{at^2}{2}$

If the block starts from rest, vo=0:

$\displaystyle d = \frac{at^2}{2}$

Substituting the value of the acceleration:

$\displaystyle d = \frac{\frac{F}{m}t^2}{2}$

Simplifying:

$\displaystyle d = \frac{Ft^2}{2m}$

When a force F'=4F is applied and assuming the mass is the same, the new acceleration is:

$\displaystyle a'=\frac{4F}{m}$

And the distance is now:

$\displaystyle d' = \frac{4Ft^2}{2m}$

Dividing d'/d:

$\displaystyle \frac{d' }{d}=\frac{\frac{4Ft^2}{2m}}{\frac{Ft^2}{2m}}$

Simplifying:

$\displaystyle \frac{d' }{d}=4$

Thus:

d' = 4d

The distance is now 4d

3 0

2 years ago