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Scilla [17]
3 years ago
13

A bowling ball has a mass of 5 kg what happens to its momentum when its speed increases from 1m/s to 2m/s?

Physics
1 answer:
Mama L [17]3 years ago
8 0
Here, Initial momentum = mu = 5*1 = 5 Kg m/s
Final momentum = mv = 5*2 = 10 Kg m/s

So, Momentum has been increased from 5 Kg m/s to 10 Kg m/s. Hence, Your Final answer is option B

Hope this helps!
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A pulley with a mechanical advantage of 5 will require you to pull _____ times the amount of rope.
Marat540 [252]

Answer:

It will require a force of 1/5, answer A.

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If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.

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6 0
3 years ago
A person's center of mass is very near the hips, at the top of the legs. Model a person as a particle of mass mm at the top of a
muminat

Answer:

\mathbf{v_{max} = \sqrt{gL}}

Explanation:

Considering an object that moving about in a circular path,  the equation for such centripetal force can be computed as:

\mathbf {F = \dfrac{mv^2}{2}}

The model for the person can be seen in the diagram attached below.

So, along the horizontal axis, the net force that is exerted on the person is:

mg cos \theta = \dfrac{mv^2}{L}

Dividing both sides by "m"; we have :

g cos \theta = \dfrac{v^2}{L}

Making "v" the subject of the formula: we have:

v^2 = g Lcos \theta

v=\sqrt{ gL cos \theta

So, when \theta = 0; the velocity is maximum

∴

v_{max} = \sqrt{gL \ cos \theta}

v_{max} = \sqrt{gL \ cos (0)}

v_{max} = \sqrt{gL \times 1}

\mathbf{v_{max} = \sqrt{gL}}

Hence; the maximum walking speed for the person  is \mathbf{v_{max} = \sqrt{gL}}

3 0
3 years ago
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