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kodGreya [7K]
2 years ago
14

3. The space shuttle aims for an orbit about 250 km above the surface of the earth. In orbit, the mass

Physics
1 answer:
Contact [7]2 years ago
7 0

The orbital speed of the space shuttle is 2.12 x 10⁴ m/s.

<h3>Orbital speed of the space shuttle</h3>

The orbital speed of the space shuttle is calculated as follows;

V = \sqrt{\frac{GM}{r} }

where;

  • G is gravitational constant
  • M is mass of the Earth
  • r is the distance from the center point of the planet

V = \sqrt{\frac{6.67 \times 10^{-11} 5.98 \times 10^{24}}{(250 + 638) \times 10^3} }\\\\V = 2.12 \times 10^4 \ m/s

Thus, the orbital speed of the space shuttle is 2.12 x 10⁴ m/s.

Learn more about orbital speed here: brainly.com/question/22247460

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A piece of toast weighing 8 grams is flying through the air at 15
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The kinetic energy of toast is 0.06 J.

<u>Explanation:</u>

Kinetic energy is the way to determine the energy released when an object is in motion. In other times, it can be the energy required to move any object and to make it in motion.

As the mass of the toast is given as 8 g and speed is given as 15 m/s, if we ignore the friction caused by air molecules. Then the kinetic energy is the product of mass and square of velocity.

K.E. = \frac{1}{2} × mass × v²

Kinetic energy =\frac{1}{2} \times \frac{8}{1000} \times 15

Since, the weight is given in grams , it needed to be converted into kg.

Kinetic energy = 0.06 J

Thus, the kinetic energy of toast is 0.06 J.

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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of
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\underline {\huge \boxed{ \sf \color{skyblue}Answer :  }}

<u>Given :</u>

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{D} = 5mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  I_{DSS} = 10mA

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS(off)} = -6V

\:  \:

\tt \large {\color{purple}     ↬ }  \:  \:  \:  \:  \:  V_{GS} =   {?}

\:  \:  \:

<u>Let's Slove :</u><u> </u>

  • \tt \large  I_{D} = I_{(DSS)}  (1 -   \frac {V_{GS}}{V_{GS(off)}} )^{2}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times  V_{GS(off)}

\:  \:  \:

  • \tt \large \: V_{GS} = (1 -  \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times  { - 6}

\:  \:

  • \underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}
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Why does ice melt faster in water than in oil when both liquids are at the same temperature​
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