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olganol [36]
3 years ago
9

A distant large asteroid is detected that might pose a threat to Earth. If it were to continue moving in a straight line at cons

tant speed, it would pass 24000 km from the center of Earth. However, it will be attracted to Earth and might hit our planet. What is the minimum speed the asteroid should have so it will just graze the surface of the Earth?
Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

The minimum speed required is 5.7395km/s.

Explanation:

To escape earth, the kinetic energy of the asteroid must be greater or equal to its gravitational potential energy:

K.E\geq P.E

or

\dfrac{1}{2}mv^2 \geq  G\dfrac{Mm}{R}

where m is the mass of the asteroid, R= 24,000,000\:m is its distance form earth's center, M = 5.9*10^{24}kg is the mass of the earth, and G = 6.7*10^{-11}m^3/kg\: s^2 is the gravitational constant.

Solving for v we get:

v \geq \sqrt{\dfrac{2GM}{R} }

putting in numerical values gives

v \geq \sqrt{\dfrac{2(6.7*10^{-11})(5.9*10^{24})}{(24,000,000)} }

\boxed{v\geq 5739.5m/s}

in kilometers this is

v\geq5.7395m/s.

Hence, the minimum speed required is 5.7395km/s.

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Answer:

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Explanation:

The thermal efficiency of an engine is given by:

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Here we have:

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A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
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Explanation:

Given that,

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After the collision,

Final speed of first ball, v = 50 cm/s

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Let us consider that both balls have same mass i.e. m

Initial kinetic energy of the system is :

K_i=\dfrac{1}{2}mu^2+\dfrac{1}{2}mu'^2\\\\K_i=\dfrac{1}{2}m(u^2+u'^2)\\\\K_i=\dfrac{1}{2}m((30)^2+(40)^2)\\\\K_i=1250m\ J

Final kinetic energy of the system is :

K_f=\dfrac{1}{2}mv^2+\dfrac{1}{2}mv'^2\\\\K_f=\dfrac{1}{2}m(v^2+v'^2)\\\\K_f=\dfrac{1}{2}m((50)^2+(0)^2)\\\\K_f=1250m\ J

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\Delta K=K_f-K_i\\\\\Delta K=1250m-1250m\\\\\Delta K=0

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