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Gnesinka [82]
3 years ago
13

You wish to move a heavy box on a rough floor. The coefficient of kinetic friction between the box and the floor is 0.9. What is

the best way to reduce the force of friction between the box and the floor while the box is moving? A. Push or pull the box with a force parallel to the floor. B. Pull the box with a force directed up at an angle of 30° from the horizontal. C. Push the box with a force directed down at an angle of 30° from the horizontal. D. The coefficient of friction does not change so the force of kinetic friction acting on the block must be constant E. Both B and C result in the same force of friction.
Physics
1 answer:
Maksim231197 [3]3 years ago
5 0

Answer:

8%

Explanation:

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What the question for this assessment
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A miniature quadcopter is located at x = -2.25 m and y, - 5.70 matt - 0 and moves with an average velocity having components Vv,
kupik [55]

Recall that average velocity is equal to change in position over a given time interval,

\vec v_{\rm ave} = \dfrac{\Delta \vec r}{\Delta t}

so that the <em>x</em>-component of \vec v_{\rm ave} is

\dfrac{x_2 - (-2.25\,\mathrm m)}{1.60\,\mathrm s} = 2.70\dfrac{\rm m}{\rm s}

and its <em>y</em>-component is

\dfrac{y_2 - 5.70\,\mathrm m}{1.60\,\mathrm s} = -2.50\dfrac{\rm m}{\rm s}

Solve for x_2 and y_2, which are the <em>x</em>- and <em>y</em>-components of the copter's position vector after <em>t</em> = 1.60 s.

x_2 = -2.25\,\mathrm m + \left(2.70\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{x_2 = 2.07\,\mathrm m}

y_2 = 5.70\,\mathrm m + \left(-2.50\dfrac{\rm m}{\rm s}\right)(1.60\,\mathrm s) \implies \boxed{y_2 = 1.70\,\mathrm m}

Note that I'm reading the given details as

x_1 = -2.25\,\mathrm m \\\\ y_1 = -5.70\,\mathrm m \\\\ v_x = 2.70\dfrac{\rm m}{\rm s}\\\\ v_y=-2.50\dfrac{\rm m}{\rm s}

so if any of these are incorrect, you should make the appropriate adjustments to the work above.

8 0
3 years ago
A 40kg dog is sitting on top of a hillside and has a potential energy of 1,568 J. What is the height of the hill side?
bagirrra123 [75]

Answer:

<h2>39.2 m</h2>

Explanation:

The height of the hill side can be found by using the formula

h =  \frac{p}{m}  \\

p is the potential energy

m is the mass

From the question we have

h =  \frac{1568}{40}  =  \frac{196}{5}  \\

We have the final answer as

<h3>39.2 m</h3>

Hope this helps you

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Work = Force x Distance

Assuming that this work is being done parallel to the displacement that is, but under that assumption:

W = (50)(10)

W = 500 J

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Answer:

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