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Zinaida [17]
3 years ago
5

What are some examples of gravitational forces that keep an object in orbit? Pls give at least 2-3 example if you can.​

Physics
1 answer:
Lelu [443]3 years ago
3 0

Answer:

Without any external forces a moving object will continue to move in a straight line. The gravitational force between the two objects will provide the centripetal force to keep the objects moving around one another.

1. satellite in orbit around the earth             (motion of earth is negligible)

2. moon in orbit around the earth     (center of motion several thousand miles        

                                                             from center of earth)

3. earth in orbit around sun    (center of rotation close to center of sun)

4. binary stars         (if masses of stars are equal center of rotation is in middle)

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The magnetic field 0.02 m from a wire is 0.1 t. what is the magnitude of the magnetic field 0.01 m from the same wire?
juin [17]

Answer:

0.2 T

Explanation:

3 0
3 years ago
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    

       

     

                         

8 0
3 years ago
Which of the following statements is correct?
Mumz [18]

Answer: c. Generally, metals are ductile.

Explanation:

From the options given in the question, the correct statement is that"Generally, metals are ductile.

Ductility of a metal simply means that a metal can be plastically deform before it is then fractured. It implies that metals can be drawn to thin wires. The only exception we have in this case is mercury.

7 0
3 years ago
A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ
vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon M_{0K} = 494 MeV/c²

Rest mass of proton M_{0P}  = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy E_{0K} = 494c² MeV

for the proton, rest energy E_{0P} = 938c² MeV

Recall that the rest energy, and the total energy are related by..

E = γE_{0}

which can be written in this case as

E_{K} = γE_{0K} ...... equ 1

where E = total energy of the kaon, and

E_{0} = rest energy of the kaon

γ = relativistic factor = \frac{1}{\sqrt{1 - \beta ^{2} } }

where \beta = \frac{v}{c}

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

E_{K} = E_{0P} ......equ 2

where E_{K} is the total energy of the kaon, and

E_{0P} is the rest energy of the proton.

From E_{K} = E_{0P} = 938c²    

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = \frac{1}{\sqrt{1 - \beta ^{2} } } = 1.89

1.89\sqrt{1 - \beta ^{2} } = 1

squaring both sides, we get

3.57( 1 - \beta^{2}) = 1

3.57 - 3.57\beta^{2} = 1

2.57 = 3.57\beta^{2}

\beta^{2} = 2.57/3.57 = 0.72

\beta = \sqrt{0.72} = 0.85

but, \beta = \frac{v}{c}

v/c = 0.85

v = <em>0.85c </em>

7 0
3 years ago
A ball rolls down the hill which has a vertical height of 15 m. Ignoring friction what would be the gravitational potential ener
trasher [3.6K]

a) Potential energy: 147 m [J]

The gravitational potential energy of an object is given by

U=mgh

where

m is its mass

g=9.8 m/s^2 is the acceleration of gravity

h is the height of the object above the ground

In this problem,

h = 15 m

We call 'm' the mass of the ball, since we don't know it

So, the potential energy of the ball at the top of the hill is

U=(m)(9.8)(15)=147 m (J)

b) Velocity of the ball at the bottom of the hill: 17.1 m/s

According to the law of conservation of energy, in absence of friction all the potential energy of the ball is converted into kinetic energy as the ball reaches the bottom of the hill. Therefore we can write:

U=K=\frac{1}{2}mv^2

where

v is the final velocity of the ball

We know from part a) that

U = 147 m

Substituting into the equation above,

147 m = \frac{1}{2}mv^2

And re-arranging for v, we find the velocity:

v=\sqrt{2\cdot 147}=17.1 m/s

5 0
3 years ago
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