A 2-kg ball rolls at a speed of 10 m/s on the ground. Is this an example of Kinetic Energy, Potential Energy, or both?

1 answer:

3 0

If a 2kg ball rolls at a speed of 10 m/s on the ground, this is an example of kinetic energy. When a mass of the body moves at a specific rate, there is movement. The equation for kinetic energy is 1/2mv^2. Therefore the kinetic energy of the ball is (1/2)(2kg)(10m/s)^2 is 100 Joules.

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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?

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

<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.