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jeka94
2 years ago
7

Which change will cause the gravitational force between a baseball and a soccer ball to increase? Select three options. Moving t

he balls closer together moving the balls farther apart wrapping the soccer ball in layers of duct tape using a baseball with a thinner leather removing the outer layer of the baseball putting the baseball in a plastic case.
Physics
1 answer:
almond37 [142]2 years ago
3 0

The three options are 1,3 and 5 Moving the balls closer together. The distance decreases and forces increase.

<h3>What is the gravitational force?</h3>

The gravitational force ia s type of force has the ability attracts any two objects with mass. Gravitational force tries to pull two masses towards each other.

The value of the gravitational force is given by;

F_g=\frac{ Gm_ 1m_2}{r^2}

It is observed that the gravitational force is directly propotional to the product of mass and inversly propotional to the square of the distance between them.

So, the correct options are :

(1) Moving the balls closer together. The distance decreases and forces increase.

(3) Wrapping the soccer ball in layers of duct tape will also increase force because the mass will increase.

(5) Similarly, putting the baseball in a plastic case will increase mass, and hence force increases.

Hence the three options are 1,3 and 5.

To learn more about the gravitational force refer to the link;

brainly.com/question/24783651

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A tennis player hits a ball at an angle of 50 degrees above horizontal so that it has an acceleration of 12 m/s2. What is the ho
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4 years ago
Two objects, one 4 times as massive as the other, are approaching each other under their mutual gravitational
marusya05 [52]

To Find :

The acceleration of the heavier object.

Solution :

Force of gravitation on lighter object by heavier object is :

F = \dfrac{Gm(4m)}{100^2}

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a_s = \dfrac{4Gm}{100^2}\\\\m = \dfrac{50^2}{G}          .....1)

Now, acceleration of heavier object when separation between them is 25 km :

4ma_h = \dfrac{Gm(4m)}{25}\\\\a_h = \dfrac{Gm}{25}.....2)

Putting value of m in equation 2, we get :

a_h = \dfrac{G\times 50^2}{G \times 25}\\\\a_h = \dfrac{2500}{25}\\\\a_h = 100\ m/s^2

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6 0
3 years ago
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3 0
3 years ago
Expressions have been found for the vertical acceleration of the cylinder ay and the angular acceleration α of the cylinder in t
azamat

Complete Question

A cylinder with a moment of inertia I (about its axis of symmetry), mass m, and radius r has a mass-less string wrapped around it which is tied to the ceiling .

At time t=0 the cylinder is released from rest at height h above the ground. Use g for the magnitude of the acceleration of gravity. Assume that the string does not slip on the cylinder. Let v  represent the instantaneous velocity of the center of mass of the cylinder, and let \omega  represent the instantaneous angular velocity of the cylinder about its center of mass. Note that there are no horizontal forces present, so for this problem \= v =-  v\r j \ and \ \omega =-\omega \r k.

Expressions have been found for the vertical acceleration of the cylinder a_y and the angular acceleration α of the cylinder in the k^ direction; both expressions include an unknown variable, namely, the tension T in the vertical section of string. The string constrains the rotational and vertical motions, providing a third equation relating a_y and α. Solve these three equations to find the vertical acceleration, a_y, of the center of mass of the cylinder. Express a_y in terms of g, m, r, and I; a positive answer indicates upward acceleration.

Answer:

The vertical acceleration is  a_y = \frac{mg}{[m+ \frac{I}{r^2} ]}

Explanation:

The equation of motion is mathematically represented as

        ma_y = T -mg ---(1)

The relation between the tension and the moment of inertia is

        I \alpha  = -Tr

         T = \frac{-I\alpha }{r} ---(2)

Now angular acceleration can be mathematically represented as

                 \alpha  = \frac{a_y}{r}

Now substituting this into equation 2

              T = \frac{-Ia_y}{r^2} ---(3)

Now substituting these into equation 1

          ma_y = \frac{-Ia_y}{r^2} - mg

            a_y [m + \frac{I}{r^2} ] = -mg

Hence the vertical acceleration is evaluates as

            a_y = \frac{mg}{[m+ \frac{I}{r^2} ]}

 

   

           

7 0
3 years ago
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