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Crazy boy [7]
3 years ago
13

NEED THE ANSWER ASAP When an object is not in motion, it can still have a form of energy. What form of energy does an object hav

e due to its
distance from Earth?
kinetic energy
thermal energy
gravitational potential energy
electrical potential energy
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Save and Exit
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Subm
Physics
1 answer:
mote1985 [20]3 years ago
7 0

Answer:

gravitational potential energy

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Jupiter is denser than water, yet composed for the most part of two light gases, hydrogen and helium. What makes Jupiter as dens
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it has a rocky core so the gravity from that compacts the gases extremly tight

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From the average distance of one of jupiter's moons to jupiter and its orbital period around jupiter, we can determine:
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Answer: Jupiter's mass

Explanation:

From Kepler's third law:

T^2=\frac{4\pi^2}{GM}a^3

where T is the orbital period of a satellite, a is the average distance of the satellite from the Planet, M is the mass of the planet, G is the gravitational constant.

If the average distance of one of Jupiter's moons to Jupiter and its orbital period around Jupiter is given then mass of the Jupiter can be found:

\Rightarrow M_J=\frac{4\pi^2}{GT_m^2}a_m^3

4 0
3 years ago
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galina1969 [7]

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3 0
2 years ago
You and a friend each hold a lump of wet clay. Each lump has a mass of 30 grams. You each toss your lump of clay into the air, w
Vesna [10]

Answer:

\ \text{m/s}

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u_1 = Velocity of one lump = 3x+3y-3z

u_2 = Velocity of the other lump = -4x+0y-4z

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The collision is perfectly inelastic as the lumps stick to each other so we have the relation

mu_1+mu_2=(m+m)v\\\Rightarrow m(u_1+u_2)=2mv\\\Rightarrow v=\dfrac{u_1+u_2}{2}\\\Rightarrow v=\dfrac{3x+3y-3z-4x+0y-4z}{2}\\\Rightarrow v=-0.5x+1.5y-3.5z=\ \text{m/s}

The velocity of the stuck-together lump just after the collision is \ \text{m/s}.

4 0
2 years ago
Please help Need good grade
ivann1987 [24]

Answer:

The other angle is 120°.

Explanation:

Given that,

Angle = 60

Speed = 5.0

We need to calculate the  range

Using formula of range

R=\dfrac{v^2\sin(2\theta)}{g}...(I)

The range for the other angle is

R=\dfrac{v^2\sin(2(\alpha-\theta))}{g}....(II)

Here, distance and speed are same

On comparing both range

\dfrac{v^2\sin(2\theta)}{g}=\dfrac{v^2\sin(2(\alpha-\theta))}{g}

\sin(2\theta)=\sin(2\times(\alpha-\theta))

\sin120=\sin2(\alpha-60)

120=2\alpha-120

\alpha=\dfrac{120+120}{2}

\alpha=120^{\circ}

Hence, The other angle is 120°

6 0
3 years ago
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