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Roman55 [17]
2 years ago
5

The distance from the dwarf planet Pluto to the sun is 4.43 × 109 km at perihelion and 7.37 × 109 km at aphelion. Find the eccen

tricity e of Pluto's orbit. (Round your answer to three decimal places.) e =
Physics
1 answer:
kaheart [24]2 years ago
3 0

Answer:

0.249

Explanation:

Perihelion = 4.43 x 10^9 km

Aphelion = 7.37 x 10^9 km

Let e be the eccentricity and a be the length of semi major axis.

The relation between the semi major axis, perihelion and aphelion s given by

Semi major axis = half of sum of perihelion and aphelion

a = \frac{4.43+7.37}{2}\times10^{9}

a = 5.9 x 10^9 km

The relation between the perihelion, semi major axis and the eccentricity is given by

Perihelion = a (1 - e)

4.43 x 10^9 = 5.9 x 10^9 (1 - e)

0.751 = 1 - e

e = 0.249

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Part A.)Six boxes held at rest against identical walls.
krok68 [10]

Answer:

Explanation:

When a body is held against a vertical wall , to keep them in balanced position , normal force is applied on their surface . this force creates normal reaction which acts against the normal force and it is equal to the normal force as per newton's third law . Ultimately friction force is created which is proportional to normal force and it acts in vertically upward direction . It prevents the body from falling down .

Hence normal force = reaction force .

From second law also net force is zero , so if  normal force is N and reaction force is R

R - N = mass x acceleration = mass x 0 = 0

R = N .

Ranking normal force from  highest  to smallest

150 N , 130 N , 120 N

B )

Frictional force is equal to the weight of the body because the body is held at rest .

Ranking of frictional force form largest to smallest

7 kg , 5 kg , 3 kg , 1 kg .

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6 0
3 years ago
Calculate the orbital period for Jupiter's moon Io, which orbits 4.22×10^5km from the planet's center (M=1.9×10^27kg) .
Verdich [7]

According to the <u>Third Kepler’s Law of Planetary motion</u> “<em>The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”.</em>



In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.



This Law is originally expressed as follows:



<h2>T^{2} =\frac{4\pi^{2}}{GM}a^{3}    (1) </h2>

Where;


G is the Gravitational Constant and its value is 6.674(10^{-11})\frac{m^{3}}{kgs^{2}}



M=1.9(10^{27})kg is the mass of Jupiter


a=4.22(10^{5})km=4.22(10^{8})m  is the semimajor axis of the orbit Io describes around Jupiter (assuming it is a circular orbit, the semimajor axis is equal to the radius of the orbit)



If we want to find the period, we have to express equation (1) as written below and substitute all the values:



<h2>T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2) </h2>

T=\sqrt{\frac{4\pi^{2}}{6.674(10^{-11})\frac{m^{3}}{kgs^{2}}1.9(10^{27})kg}(4.22(10^{8})m)^{3}}    



T=\sqrt{\frac{2.966(10^{27})m^{3}}{1.268(10^{17})m^{3}/s^{2}}}    



T=\sqrt{2.339(10^{10})s^{2}}    



Then:


<h2>T=152938.0934s    (3) </h2>

Which is the same as:



<h2>T=42.482h     </h2>

Therefore, the answer is:



The orbital period of Io is 42.482 h



7 0
3 years ago
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liberstina [14]
If F =m*a
and the question says how much force the s needed to accelerate a 68kg skier to a rate of 1.2ms^-2
Then F = 68*1.2
7 0
3 years ago
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