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Tasya [4]
3 years ago
13

A 0.203-kg plastic ball moves with a velocity of 0.30 m/s. it collides with a second plastic ball of mass 0.092 kg, which is mov

ing along the same line at a speed of 0.10 m/s. after the collision, both balls continue moving in the same, original direction. the speed of the 0.092-kg ball is 0.26 m/s. what is the new velocity of the 0.203-kg ball?
Physics
1 answer:
dolphi86 [110]3 years ago
7 0
<span><span>A 0.200 kg plastic ball moves with a velocity of0.30 m</span>s<span>A 0.205-kg plastic ball moves with a velocity of0.30 m</span><span>A 0.199 kg plastic ball moves with a velocity of0.30 m</span><span>A 0.204-kg plastic ball moves with a velocity of0.30 m</span><span>A 100 g ball moving to the right at 4.0 m</span>s collides<span>have less momentum if the velocities</span><span>the same</span><span>A ball with a momentum of 4.0 kg•<span>m</span></span></span>
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The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of
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Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Explanation: To find the answer, we need to know about the different equations of planetary motion.

<h3>How to find the initial speed of the rock as it left the astronaut's hand?</h3>
  • We have the expression for the initial velocity as,

                           v=\sqrt{2gh}

  • Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,

                       g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132

  • Now, the velocity will become,

                        v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s

<h3>How to find the speed of the satellite?</h3>
  • As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,

                       v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s

Thus, we can conclude that,

1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.

2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.

Learn more about the equations of planetary motion here:

brainly.com/question/28108487

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At a constant temperature, the volume of a gas doubles when the pressure is reduced to half of its original value. This is a sta
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<span>This is a statement of Boyle's Law.</span>
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In the experiment, a meter is hooked up to a speaker to monitor the amplitude of the received sound. Suppose the background sign
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The answer is "15.38\%"

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Yes

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