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3 years ago

9

We have a 10kg ball and a 15kg ball. We drop them from the top of the science building while objects are at rest and there is no

wind resistance. Which one hits the ground first?

1 answer:

Mashcka [7]3 years ago

3 0

They will hit the ground at the same time, as mass is negligible when calculating the acceleration of gravity when there is no air resistance

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An alien spaceship traveling at 0.600 c toward the Earth launches a landing craft. The landing craft travels in the same directi

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The kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

To find the answer, we have to know about the Lorentz transformation.

<h3>What is its kinetic energy as measured in the Earth reference frame?</h3>

It is given that, an alien spaceship traveling at 0.600 c toward the Earth, in the same direction the landing craft travels with a speed of 0.800 c relative to the mother ship. We have to find the kinetic energy as measured in the Earth reference frame, if the landing craft has a mass of 4.00 × 10⁵ kg.

$V_x'=0.8c\\V=0.6c\\m=4*10^5kg$

Let us consider the earth as S frame and space craft as S' frame, then the expression for KE will be,

$KE=m_0c^2=\frac{mc^2}{1-(\frac{v_x^2}{c^2} )}$

So, to $V_x=(0.8+0.6)c-[\frac{0.6c*(0.8c)^2}{c^2}]=1.016$ find the KE, we have to find the value of speed of the approaching landing craft with respect to the earth frame.
We have an expression from Lorents transformation for relativistic law of addition of velocities as,

$V_x'=\frac{V_x-V}{1-\frac{VV_x}{c^2} } \\thus,\\V_x=V_x'(1-\frac{VV_x}{c^2} )+V$

Substituting values, we get,

$V_x=0.8c(1-\frac{0.8c*0.6c}{c^2} )+0.6c=(0.8c*0.52)+0.6c=1.016c$

Thus, the KE will be,

$KE=\frac{4*10^5*(3*10^8)^2}{\sqrt{1-\frac{(1.016c)^2}{c^2} } } =\frac{1.2*10^{22}}{0.179}=6.704*10^{22}J$

Thus, we can conclude that, the kinetic energy as measured in the Earth reference frame is 6.704*10^22 Joules.

Learn more about frame of reference here:

brainly.com/question/20897534

SPJ4

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2 years ago