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IRISSAK [1]
3 years ago
12

Pluto was first observed in 1930, and its largest moon, Charon, was discovered in 1978. A few years after Charon’s discovery, as

tronomers were able to observe a series of eclipses as Pluto and Charon passed in front of one another. By studying how the brightness of Pluto and Charon changed as they eclipsed each other, astronomers were able to measure the masses and radii of both Pluto and its moon. What did these measurements imply about the average densities of Pluto and Charon?
Physics
1 answer:
mylen [45]3 years ago
8 0

Answer:

The average densities of both matches the expected density for objects made from water ice.

Explanation:

Charon's density is 1.2 to 1.3 g / cm3, while Pluto's density is 1.8 to 2.1 g / cm3. This was discovered in many researches and measurements of these two celestial bodies, with the objective of understanding them and promoting efficient scientific knowledge.

With the measurements of the average densities between pluto and Charon it was possible to conclude several statements about them. Firstly, it is possible to see that the two formed independently and at different times, in addition to indicating the existence of few rocks in charon, which is consistent with the average density of objects made mostly of water ice.

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A sailboat is traveling to the right when a gust of wind causes the boat to accelerate leftward at 2.5​​​​m​​/s2 for 4s. After t
Pavlova-9 [17]
Taking right movement to be positive means leftward movement is negative.
Hence we have a deceleration of
a = - 2.5 {ms}^{ - 2}

t = 4s
v = 3.0 {ms}^{ - 1}
Using this 'suvat' equation
v = u + at
we can determine the initial velocity

3.0= u + -2.5(4)

3.0+2.5(4)=u

13.0 = u

Hence the initial velocity is 13.0 meters per seconds
3 0
3 years ago
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The luminous star Alnilam in the Orion belt is 1,340 light-years away from Earth. Use the conversion factor 1 parsec = 3.262 lig
svlad2 [7]

The answer is 410.8 pc.

8 0
3 years ago
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An object ends up at a final position of x=-55.25 meters after a displacement of -189.34 meters after a displacement of -189.34
Travka [436]

The initial position of the object was found to be 134.09 m.

<u>Explanation:</u>

As displacement is the measure of difference between the final and initial points. In other words, we can say that displacement can be termed as the change in the position of the object irrespective of the path followed by the object to change the path. So

Displacement = Final position - Initial position.

As the final position is stated as -55.25 meters and the displacement is also stated as -189.34 meters. So the initial position will be

Initial position of the object = Final position-Displacement

Initial position = -55.25 m - (-189.34 m) = -55.25 m + 189.34 m = 134.09 m.

Thus, the initial position for the object having a displacement of -189.34 m is determined as 134.09 m.

4 0
2 years ago
Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
2 years ago
Which two forces operate over the longest distances?
Harman [31]

Answer:

D Electromagnetic and gravitational

5 0
3 years ago
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