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Dmitry [639]
3 years ago
9

PREDICTION: What do you predict will happen to the distance that the box moves if the height of the pendulum increases to 20cm?

Explain.
Physics
1 answer:
castortr0y [4]3 years ago
6 0

Answer:

sorry sir this makes no sense.. at all

but ig you could put:

I predict that the distance that the box will move if the height of the pendulum increases to 20cm is going to close  

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A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I_1 is the m
Paraphin [41]

Answer:

I2>I1

Explanation:

This problem can be solved by using the parallel axis theorem. If the axis of rotation of a rigid body (with moment of inertia I1 at its center of mass) is changed, then, the new moment of inertia is gven by:

I_2=I_{1}+Md^2

where M is the mass of the object and d is the distance of the new axis to the axis of the center of mass.

It is clear that I2 is greater than I1 by the contribution of the term Md^2.

I2>I1

hope this helps!!

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3 years ago
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Schach [20]
To determine the mass plated, we use Faraday's Law of Electrolysis. We calculate as follows:

q = It
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The drawing shows a tire of radius R on a moving car
masha68 [24]

Answer:

 H / R = 2/3

Explanation:

Let's work this problem with the concepts of energy conservation. Let's start with point P, which we work as a particle.

Initial. Lowest point

          Em₀ = K = 1/2 m v²

Final. In the sought height

         Em_{f} = U = mg h

Energy is conserved

        Em₀ =  Em_{f}

        ½ m v² = m g h

        v² = 2 gh

Now let's work with the tire that is a cylinder with the axis of rotation in its center of mass

Initial. Lower

        Em₀ = K = ½ I w²

Final. Heights sought

        Emf = U = m g R

        Em₀ =  Em_{f}

       ½ I w² = m g R

The moment of inertial of a cylinder is

       I = I_{cm} + ½ m R²

      I= ½ I_{cm}  + ½ m R²

Linear and rotational speed are related

       v = w / R

       w = v / R

We replace

      ½ I_{cm}  w² + ½ m R² w²  = m g R

moment of inertia of the center of mass      

      I_{cm}  = ½ m R²

     ½ ½ m R² (v²/R²) + ½ m v² = m gR

    m v² ( ¼ + ½ ) = m g R

   

       v² = 4/3 g R

As they indicate that the linear velocity of the two points is equal, we equate the two equations

        2 g H = 4/3 g R

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The force applied to a moving object was 100 N. The object moved 5 m. How much work was done on the object
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Answer:

500J

................

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