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Yuki888 [10]
3 years ago
7

Is this right or they wrong definitions which ones are the right ones someone !!!!!

Physics
2 answers:
kap26 [50]3 years ago
5 0

Answer:

<em><u>Mechanical Energy : KE + PE</u></em>

<em><u></u></em>

<em><u>Conversion : "When energy transfers from one form to another"</u></em>

<em><u></u></em>

<em><u>Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'</u></em>

<em><u></u></em>

<em><u>Kinetic Energy: energy of an object in motion</u></em>

<em><u></u></em>

<em><u>Law of conservation of energy: KE+PE+friction=KE</u></em>

<em><u></u></em>

Explanation:

First of all mechanical energy is kinetic energy plus potential energy (it is the energy of movement) So:

Mechanical Energy : KE + PE

Conversion is when energy converts or becomes a different form. So:

Conversion : "When energy transfers from one form to another"

Potential energy is stored energy, in Physics I or AP Physics I, it is often due to it being at a height, but batteries, foods, etc. are also example of it, so:

Potential Energy: the energy possessed by a body by virtue of its position relative to others , stresses within itself, electric charge , and other factors .'

Kinetic energy is for objects in motion so you got it right!

Kinetic Energy: energy of an object in motion

The law of conservation of energy means there is the same amount of energy before, as there is after, so when you see an equation with energy on both sides, it is usually this. Also, this is the last question left, so this has to be the answer.

Law of conservation of energy: KE+PE+friction=KE

Alex73 [517]3 years ago
4 0

Answer:

They are right.

Explanation:

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A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g
7nadin3 [17]

Answer:

(a) r = 1.062·R_E = \frac{531}{500} R_E

(b) r = \frac{33}{25} R_E

(c) Zero

Explanation:

Here we have escape velocity v_e given by

v_e =\sqrt{\frac{2GM}{R_E} } and the maximum height given by

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}

Therefore, when the initial speed is 0.241v_e we have

v = 0.241\times \sqrt{\frac{2GM}{R_E} } so that;

v² = 0.058081\times {\frac{2GM}{R_E} }

v² = {\frac{0.116162\times GM}{R_E} }

\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r} is then

\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}

Which gives

-\frac{0.941919}{R_E} = -\frac{1}{r} or

r = 1.062·R_E

(b) Here we have

K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m  \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}

Therefore we put  \frac{0.241GM}{R_E} in the maximum height equation to get

\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}

From which we get

r = 1.32·R_E

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

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