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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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Alinara [238K]

Answer:

(a) 5.88\times 10^{12}mi

Explanation:

We have given speed of light c=3\times 10^8m/sec

Given time = 1 year =365 days

We know that in 1 minute = 60 sec

1 hour = 60×60 = 3600 sec

In one day = 24 hour = 24×60×60=86400 sec

So in 365 days = 365×86400=3.1536\times 10^7sec

We know that distance = speed ×time=3.1536\times 10^7\times 3\times 10^8=9.46\times 10^{15}m

We have given that 1 mi = 1609 m

So 9.46\times 10^{15}m=\frac{9.46\times 10^{15}}{1609}=5.88\times 10^{12}miSo option (a) is correct

8 0
3 years ago
A jet engine gets its thrust by taking in air, heating and compressing it, and
nadya68 [22]

Answer:

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

Explanation:

Thrust is known as a reaction force which appears when a system expels or accelerates mass in one specific direction. If we know the acceleration and the mass of the air expelled by the jet engine, we can compute the thrust .

The acceleration is calculated by using the dynamics formula

\displaystyle a=\frac{v_f-v_o}{t}

The values are  

v_f=500\ m/s,\ v_o=100\ m/s,\ t=1\ sec

\displaystyle a=\frac{500-100}{1}=400\ m/s^2

The thrust is

F=ma=20\ Kg\ 400\ m/s^2=8,000\ Nw

4 0
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A student is throwing a ball. She throws the ball up with her left hand, it passes over her head, and she catches it with her ri
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Answer:

(c) at point 2, the ball is at its highest height do its PE is max. Also at ms height, velocity is zero therefore KE is zero.

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A divot is created _____.
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Examine the roller coaster track above. Assume there is negligible friction as the roller coaster moves from position A to posit
anyanavicka [17]

At point E

  • the kinetic energy of the rollercoaster is small compared to the potential energy
  • the potential energy is greater than the kinetic energy
  • the total energy is a mixture of potential and kinetic energy

<h3>What is the energy of the roller coaster at point E?</h3>

The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.

Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.

At point E

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  • the total energy is a mixture of potential and kinetic energy

In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,

Learn more about potential and kinetic energy at: brainly.com/question/18963960

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