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Zinaida [17]
3 years ago
10

Aliens come blasting into our solar system and wipe out everything but the Sun, the Earth, and Jupiter. Discuss (conceptually) w

here the center of mass of the new, smaller solar system is now roughly located. Why
Physics
1 answer:
KATRIN_1 [288]3 years ago
4 0

Answer:

There is no short answer.

Explanation:

In theory, the gravitational pull between the planets depends on their total mass and the distance between them. The sun has a mass of 1.989x10^{30} kg, Jupiter has a mass of 1.898x10^{27} kg and Earth has a mass of 5.972x10^{24} kg.

Earth is closer to Jupiter than it is to the sun. The other planets that are behind jupiter are gone so that means the sun's pulling force's effect on jupiter is increased drastically since there is no force the balance that and that means that jupiter is going to get closer to earth.

The new center of mass of the new solar system is roughly located between earth and the sun, closer to earth.

I hope this answer helps.

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Momentum is a vector quantity because it includes velocity, which is also a
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8TH GRADE SCIENCE I NEED NOW DO NOT SKIP
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Explanation:

1. Force=mass*acceleration

acceleration=force/mass

=100/50

=2m/s^2

2. Gravitational force for downward acceleration= mg-ma=m(g-a) , since a is less than g,

So it will be= 50(9.8-2)

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The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Dete
jeka57 [31]

Answer:

The answer is below

Explanation:

a) The location ӯ of the center of mass G of the pendulum is given as:

y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m

b)  the mass moment of inertia about z axis passing the rotation center O is:

I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2

c) The mass moment of inertia about z axis passing the rotation center O is:

I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2

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3 years ago
How are animals of the coniferous forest well adapted to long, cold winters?
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