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Mumz [18]
3 years ago
5

An astronaut in a space craft looks out the window and sees an asteroid move pas a backward direction at 68 mph relative to the

space craft. If the velocity of the space craft is 126 mph relative to the position of the sun, what is the velocity of the asteroid relative to the sun?
Physics
1 answer:
Annette [7]3 years ago
6 0

Answer: -194 mph

Explanation:

Taking into account the <u>Sun as the center </u>(origin, point zero) <u>of the reference system</u>, the velocity of the spacecraft relative to the Sun V_{R-S} is:

V_{R-S}=126mph  Note it is <u>positive</u> because the spacecraft is moving <u>away</u> from the Sun

Taking into account the <u>spacecraft as the center of another reference system</u>, the velocity of the asteroid relative to the spacecraft V_{A-R} is:

V_{A-R}=-68mph Note it is <u>negative</u> because the asteroid is moving<u> towards</u> the spacracft.

Now, the velocity of the asteroid relative to the Sun V_{A-S} is:

V_{A-S}=V_{A-R}-V_{R-S}

V_{A-S}=-68mph-126mph

Finally:

V_{A-S}=-194mph This is the velocity of the asteroid relative to the Sun and its negative sign indicates it is <u>moving towards the Sun</u>.

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Answer:

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8 0
3 years ago
A coil of wire containing N turns is in an external magnetic field that is perpendicular to the plane of the coil and it steadil
krok68 [10]

Answer:

The Resultant Induced Emf in coil is 4∈.

Explanation:

Given that,

A coil of wire containing having N turns in an External magnetic Field that is perpendicular to the plane of the coil which is steadily changing. An Emf (∈) is induced in the coil.

To find :-

find the induced Emf if rate of change of the magnetic field and the number of turns in the coil are Doubled (but nothing else changes).

So,

   Emf induced in the coil represented by formula

                          ∈  =   -N\frac{d\phi}{dt}                                  ...................(1)

                                          Where:

                                                    .   \phi = BAcos\theta     { B is magnetic field }

                                                                                 {A is cross-sectional area}

                                                    .  N = No. of turns in coil.

                                                    .  \frac{d\phi}{dt} = Rate change of induced Emf.

Here,

Considering the case :-

                                    N1 = 2N  &      \frac{d\phi1}{dt} = 2\frac{d\phi}{dt}

Putting these value in the equation (1) and finding the  new emf induced (∈1)

                           

                                      ∈1 =-N1\times\frac{d\phi1}{dt}

                                      ∈1 =-2N\times2\frac{d\phi}{dt}

                                       ∈1 =4 [-N\times\frac{d\phi}{dt}]

                                        ∈1 = 4∈             ...............{from Equation (1)}      

Hence,

The Resultant Induced Emf in coil is 4∈.        

                           

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