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schepotkina [342]
3 years ago
8

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

y of the less massive object to it's final velocity
Physics
1 answer:
stepladder [879]3 years ago
7 0

Assume that the small-massed particle is m and the heavier mass particle is M.

Now, by momentum conservation and energy conservation:

   mv = mv_{m} + Mv_{M}

   mv^{2} = mv^{2}_{m} + Mv^{2}_{M}

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

   v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\

So now, we see that v_{m} < 0 and v_{M} > 0. So therefore, the smaller mass recoils out.

Hope this helps you!

Bye!

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Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
Debora [2.8K]

493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

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3 years ago
How does a toaster oven use micro waves
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Answer: I think Its the Height is 11.76 Meters (38.582677 Feet) between the bridge and the ground

Explanation: Supposing that where not counting air resistance in the equation, the equation D = ut+\frac{1}{2}  at^{2} states that 1/2 multiplied by earths gravitational acceleration multiplied by the amount of time to reach the bottom: 2.4 seconds equals 11.76 meters of height between the bridge and the ground.

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Answer:

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Explanation:

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