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Elanso [62]
3 years ago
8

A proton is at the origin. One electron is at the point (2m, 4m)

Physics
1 answer:
zmey [24]3 years ago
7 0

Answer:

Correct answer:  F = 0 N

Explanation:

All three elementary charges are in the same straight line or direction.

Since the electron and proton are opposite charged, their force is attractive and directed toward the electron.

The forces are of the same intensity-magnitude and opposite directions and are canceled, so the resultant- net force by which the electrons act on the proton is zero.

God is with you!!!

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Mrac [35]

Hello there!

Physical changes do not change what a substance is, whereas a chemical change results in a new substance being formed, so they do change what a substance is.

I hope I could help you to solve this question. Have a great rest of your day! :)

5 0
4 years ago
What is the acceleration of a 0.30-kg volleyball when a player uses a force of 42 N to spike the ball?
quester [9]

Answer:

The acceleration will be 140 meter per second

Explanation:

Force F = mass m × acceleration a

If F = 42 N and m = 0.30 kg

Then  acceleration a = F/m

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Room temperature

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Read 2 more answers
A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th
photoshop1234 [79]

Answer:

The height is  h_c = 42.857

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

   From the question we are told that

           The height is h_s = 30 \ cm

            The angle of the slope is \theta = 15^o

According to the law of conservation of energy

     The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

                          mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2

Where I is the moment of inertia which is mathematically represented as this for  a sphere

                    I = \frac{2}{5} mr^2

  The angular velocity w is mathematically represented as

                         w = \frac{v}{r}

So the equation for conservation of energy becomes

               mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2

              mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]

             mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]

            gh_s =[\frac{7}{10} ] v^2

              v^2 = \frac{10gh_s}{7}

Considering a circular hoop

   The moment of inertial is different for circle and it is mathematically represented as

             I = mr^2

Substituting this into the conservation equation above

              mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2

Where h_c is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

                 mgh_c  = mv^2

                     gh_c = v^2

                     h_c = \frac{v^2}{g}

Recall that   v^2 = \frac{10gh_s}{7}

                    h_c= \frac{\frac{10gh_s}{7} }{g}

                      = \frac{10h_s}{7}

            Substituting values

                   h_c = \frac{10(30)}{7}

                       h_c = 42.86 \ cm    

       

     

                         

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BaLLatris [955]
4ms I'm just guessing by the way

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