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Alenkasestr [34]
3 years ago
8

Ellen needs to move a heavy box across the floor and then place it on a shelf that is four feet above the floor. Which use of ma

chines would most likely require the least amount of force in order to get the box on the shelf?
sliding the box up an inclined plane to the shelf
rolling the box up an inclined plane on a cart
sliding the box across the floor and then lifting it using a pulley
rolling the box across the floor in a cart and then using a pulley
Physics
2 answers:
Naily [24]3 years ago
5 0

Option B is correct: Rolling the box up an inclined plane on a cart.

When using the inclined, Ellen needs to overcome frictional force and the weight component of the box along the ramp which will be less than the total weight of the heavy box. Now, in sliding and rolling the weight of the heavy box along the ramp will be same but the difference will come in the type of the frictional force. Since, the rolling friction is less than the sliding friction. Hence, less force will be required when the box is rolled up an inclined plane on a cart.

OleMash [197]3 years ago
3 0

The correct answer is:

B. Rolling the box up an inclined plane on a cart

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6 0
3 years ago
To have the highest magnification in a telescope, the focal length of the objective lens should be _________ and the focal lengt
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Large; small.

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Two asteroids identical to those above collide at right angles and stick together; i.e, their initial velocities were perpendicu
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Answer:

velocity = 62.89 m/s  in 58 degree measured from the x-axis

Explanation:

Relevant information:

Before the collision, asteroid A of mass 1,000 kg moved at 100 m/s, and asteroid B of mass 2,000 kg moved at 80 m/s.

Two asteroids moving with velocities collide at right angles and stick together. Asteroid A initially moving to right direction and asteroid B initially move in the upward direction.

Before collision Momentum of A = 1000 x 100 = $ 10^5$ kg - m/s in the right direction.

Before collision Momentum of B = 2000 x 80 = 1.6 x $ 10^5$  kg - m/s in upward direction.

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Now applying the Momentum Conservation, we get

Initial momentum in right direction = final momentum in right direction = $ 10^5$

And, Initial momentum in upward direction = Final momentum in upward direction = 1.6 x $ 10^5$

So, $ V_x = \frac{10^5}{3000} $  = $ \frac{100}{3} $  m/s

and $ V_y=\frac{160}{3}$  m/s

Therefore, velocity is = $ \sqrt{V_x^2 + V_y^2} $

                                   = $ \sqrt{(\frac{100}{3})^2 + (\frac{160}{3})^2} $

                                   = 62.89 m/s

And direction is

tan θ = $ \frac{V_y}{V_x}$     = 1.6

therefore, $ \theta = \tan^{-1}1.6 $

                   = $ 58 ^{\circ}$  from x-axis

4 0
3 years ago
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