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eimsori [14]
3 years ago
11

Which objects have the most similar eccentricities?

Physics
2 answers:
abruzzese [7]3 years ago
7 0

Answer:

Jupiter and Uranus on edgenunity

Explanation:

mafiozo [28]3 years ago
4 0
It depends on the shape of ellipse of that object, if their shapes are quite similar then there eccentricities would be similar.

Real life example. - <span> Neptune, </span>Venus<span>, and </span>Earth<span> are the planets in our </span>solar system<span> with the least eccentric orbits (They all have approx. orbits with same eccentricities)

Hope this helps!</span>
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6 0
2 years ago
a plane is flying due east in still air at 395 km/h. suddenly, the plane is hit by wind blowing at 55km/h toward the west. what
Sphinxa [80]
Let's be clear:  The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.

Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.

After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.

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3 years ago
A northfield raider defensive back runs 20 m in 2.5 s what’s the average speed
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3 years ago
A 25 kh inductor carries a current of 60 ma. how much energy is stored in this system
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8 0
3 years ago
After being struck by a bowling ball, a 1.3 kg bowling pin sliding to the right at 5.0 m/s collides head-on with another 1.3 kg
GuDViN [60]

Answer:

a) 4.2m/s

b) 5.0m/s

Explanation:

This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.

The problem is also an illustration of elastic collision where there is no loss in kinetic energy.

Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses m_1 and m_2 whose respective velocities before collision are u_1 and u_2;

m_1u_1+m_2u_2=m_1v_1+m_2v_2..............(1)

where v_1 and v_2 are their respective velocities after collision.

Given;

m_1=1.3kg\\u_1=5m/s\\m_2=1.3kg\\u_2=0m/s

Note that u_2=0 because the second mass m_2 was at rest before the collision.

Also, since the two masses are equal, we can say that m_1=m_2=m so that equation (1) is reduced as follows;

mu_1+mu_2=mv_1+mv_2\\\\m(u_1+u_2)=m(v_1+v_2)..............(2)

m cancels out of both sides of equation (2), and we obtain the following;

u_1+u_2=v_1+v_2.............(3)

a) When v_1=0.8m/s, we obtain the following by equation(3)

5+0=0.8+v_2\\hence\\v_2=5-0.8\\v_2=4.2m/s

b) As m_1 stops moving v_1=0, therefore,

5+0=0+v_2\\v_2=5m/s

5 0
3 years ago
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