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lina2011 [118]
3 years ago
8

The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth?AU

Physics
1 answer:
Ipatiy [6.2K]3 years ago
8 0

Answer:

Depending on the relative position of the Earth the Sun and Neptune in the Earths orbit the distances are;

The closest (minimum) distance of Neptune from the Earth is 29 AU

The farthest (maximum) distance of Neptune fro the Earth is 31 AU

Explanation:

The following parameters are given;

The distance from the Earth to the Sun = 1 AU

The distance of Neptune from the Earth = 30 AU

We have;

When the Sun is between the Earth and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 + 1 = 31 AU

When the Earth is between the Sun  and Neptune, the distance is found by the relation;

Distance from the Earth to Neptune = 30 - 1 = 29 AU

Therefore, the closest distance from Neptune to the Earth in the Earth's Orbit is 29 AU

The farthest distance from Neptune to the Earth in the Earth's orbit is 31 AU.

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3 years ago
A satellite orbits earth with a mean altitude of 361 km. If the orbit is circular, what are the satellite's time period and spee
Advocard [28]

Answer:

v = 7.69 x 10³ m/s = 7690 m/s

T = 5500 s = 91.67 min = 1.53 h

Explanation:

In order for the satellite to orbit the earth, the force of gravitation on satellite must be equal to the centripetal force acting on it:

F_{gravitation}= F_{centripetal}\\\\\frac{GM_{s} M_{E}}{r^2}  = \frac{M_{s} v^2}{r}\\\\\frac{GM_{E}}{r} = v^2\\\\v = \sqrt{\frac{GM_{E}}{r} } \\\\

where,

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Me = Mass of Earth = 5.97 x 10²⁴ kg

r = distance between the center of Earth and Satellite = Radius of Earth + Altitude = 6.371 x 10⁶ m + 0.361 x 10⁶ m = 6.732 x 10⁶ m

v = orbital speed = ?

Therefore,

v = \sqrt{\frac{(6.67 x 10^{-11}N.m^2/kg^2)(5.97 x 10^{24} kg)}{6.732 x 10^6 m} }\\\\

<u>v = 7.69 x 10³ m/s</u>

For time period satellite completes one revolution around the earth. It means that the distance covered by satellite is equal to circumference of circle at the given altitude.

So, its orbital speed can be given as:

v = \frac{Circumference of Circle at Given Altitude}{T}\\\\v =  \frac{2\pi r}{T}\\\\

where,

T = Time Period of Satellite = ?

Therefore,

T = \frac{2\pi r}{v}\\\\T = \frac{(2\pi )(6.732 x 10^6 m}{7.69 x 10^3 m/s}\\\\

<u>T = 5500 s = 91.67 min = 1.53 h</u>

7 0
3 years ago
Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is th
katrin [286]

Complete Question

Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at 15.0 m/s and brought to rest in 2.80 mm. How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long.What Average force is excreted on the Nail

Answer:

F=2*10^{4}N

Explanation:

From the question we are told that:

Mass m=0.500kg

Initial Velocity V=15.0m/s

Distance x=2.80mm=>0.00280m

Diameter d=2.50mm=>0.00250m

Length l=6.00cm=>0.6m

Generally the equation for Force is mathematically given by

 F=\frac{mv^2}{2d}

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6 0
2 years ago
2. A car traveled 240 km in 3 hours. What was its average speed in km per hour?​
vekshin1

Answer:

80 km

Explanation:

Since 240 divided by three is 80, the velocity the car is traveling at is 80 kilometers per hour.

8 0
2 years ago
Read 2 more answers
Can a body have increasing velocity with decreasing acceleration?
Sergio039 [100]
Sure.  The acceleration may be decreasing, but as long as it stays
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I think you meant to ask whether the body can have increasing velocity
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BUT ... If the body's velocity is in the negative direction, then positive
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speeding up.

I know that's confusing. 

-- Take a piece of scratch paper, write a 'plus' sign at one edge and
a 'minus' sign at the other edge.  Those are the definitions of which
direction is positive and which direction is negative. 

-- Then sketch some cars ... one traveling in the positive direction, and
one driving in the negative direction.  Those are the directions of the
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-- Now, one car at a time:
. . . . . first push on the back of the car, in the direction it's moving;.
. . . . . then push on the front of the car, against its motion.
Each push causes the car to accelerate in the direction of the push.

When you see it on paper, all the positive and negative velocities
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3 0
3 years ago
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