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2 years ago

PLEASE HELP WILL GET BRAINIEST, ONLY 2 QUESTIONS. WILL GET AN EXTRA 40 POINTS

1 answer:

8 0

1. They have more mass, thus more gravity.
1a.) <span>They are more distant from the Sun, so the solar winds that have stripped the atmospheres of Mercury and Mars, for example, down to practically nothing, are not as strong.

2.) Inner planets have hard, rocky surfaces and no rings. Outer planets have rings and are gassy!</span>

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A 1 036-kg satellite orbits the Earth at a constant altitude of 98-km. (a) How much energy must be added to the system to move t

Veronika [31]

Answer:

a) The Energy added should be 484.438 MJ

b) The Kinetic Energy change is -484.438 MJ

c) The Potential Energy change is 968.907 MJ

Explanation:

Let 'm' be the mass of the satellite , 'M'(6× $10^{24}$ be the mass of earth , 'R'(6400 Km) be the radius of the earth , 'h' be the altitude of the satellite and 'G' (6.67× $10^{-11}$ N/m) be the universal constant of gravitation.

We know that the orbital velocity(v) for a satellite -

v= $\sqrt{\frac{Gmm}{R+h} }$ [(R+h) is the distance of the satellite from the center of the earth ]

Total Energy(E) = Kinetic Energy(KE) + Potential Energy(PE)

For initial conditions ,

h = $h_{i}$ = 98 km = 98000 m

∴Initial Energy ( $E_{i}$ ) = $\frac{1}{2}$ m $v^{2}$ + $\frac{-GMm}{(R+h_{i} )}$

Substituting v= $\sqrt{\frac{GMm}{R+h_{i} } }$ in the above equation and simplifying we get,

$E_{i}$ = $\frac{-GMm}{2(R+h_{i}) }$

Similarly for final condition,

h= $h_{f}$ = 198km = 198000 m

∴Final Energy( $E_{f}$ ) = $\frac{-GMm}{2(R+h_{f}) }$

a) The energy that should be added should be the difference in the energy of initial and final states -

∴ ΔE = $E_{f}$ - $E_{i}$

= $\frac{GMm}{2}$ ( $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ )

Substituting ,

M = 6 × $10^{24}$ kg

m = 1036 kg

G = 6.67 × $10^{-11}$

R = 6400000 m

$h_{i}$ = 98000 m

$h_{f}$ = 198000 m

We get ,

ΔE = 484.438 MJ

b) Change in Kinetic Energy (ΔKE) = $\frac{1}{2}$ m[ $v_{f} ^{2}$ - $v_{i} ^{2}$ ]

= $\frac{GMm}{2}$ [ $\frac{1} {R+h_{f} }$ - $\frac{1} {R+h_{i} }$ ]

= -ΔE

= - 484.438 MJ

c) Change in Potential Energy (ΔPE) = GMm[ $\frac{1}{R+h_{i} }$ - $\frac{1}{R+h_{f} }$ ]

= 2ΔE

= 968.907 MJ

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3 years ago

The increase of kinetic energy at the square of the speed of your vehicle has a major influence on all motor vehicles in three p

Viktor [21]

This implies that stopping distance and impact force grow as a function of speed. The best ways to improve manoeuvrability and lessen crash severity are to drive at an appropriate pace and to slow down as soon as you spot dangers in front of you.

Keep in mind that stopping distance increases with speed; at 50 mph, it is four times longer than at 25 mph, and at 75 mph, the force of impact is nine times greater.

<h3>What is the impact of speed on kinetic energy ?</h3>

When your car expends or absorbs energy to speed up or slow down, you may feel a pull or a jolt, called impulse. Impulse increases as the energy or force increases, and increases as the duration of the force decreases. You'll feel a harder jolt if you speed up or slow down suddenly.

Consider: coming to a stop from 60 mph in ten seconds doesn't hurt you or your vehicle because the force of this event is spread out over a long time. But if you hit a wall and come to a stop in just half a second, you'll feel twenty times the impulse, causing severe damage.

Learn more about Kinetic energy here:

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5 0

2 years ago