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GREYUIT [131]
3 years ago
9

An electric model train is accelerated at a rate of 8 m/s^2 by a 12 N force? What is the

Physics
1 answer:
jok3333 [9.3K]3 years ago
4 0

Answer:

m = 1.5 kg

Explanation:

Data:

  • Aceleration (a) = 8 m/s²
  • Force (F) = 12 N
  • Mass (m) = ?

Use formula:

  • \boxed{m = \frac{F}{a}}

Replace in the formula:

  • \boxed{m = \frac{12N}{8\frac{m}{s^{2}}}}

Equate the newtons:

  • \boxed{m = \frac{12kg*\frac{m}{s^{2}}}{8\frac{m}{s^{2}}}}

Simplify m/s²:

  • \boxed{m = \frac{12kg}{8}}

It divides:

  • \boxed{m =1.5kg}

What is the mass of the train?

The mass of the train is <u>1.5 kilograms.</u>

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The form of energy that can move from place to place across the universe is light energy. On earth, the main source of this energy is Sun. Most of the light energy comes from the sun because it is the primary source of all the energies. The food, fossil fuels, movement of winds, etc all exists due to Sun. Without sun, there won't be any light energy on the earth. In all the processes which occur on earth has a direct or indirect involvement of light energy which comes from sun.

5 0
3 years ago
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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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yawa3891 [41]
1. circle graph
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AlladinOne [14]

Answer:

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Explanation:

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mihalych1998 [28]
I believe this is known as wave period.
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