As an object falls to the ground, its potential energy is being converted to kinetic energy.

Physics

1 answer:

alekssr [168]3 years ago

5 0

Answer:

<em>The statement is true</em>

Explanation:

<u>Energy Conversion </u>

When an object starts to fall in free air, it speeds up as it falls. The force of gravity acting on the object causes energy to be transferred from its gravitational potential energy to its kinetic energy. We can safely say the height converts to speed and vice-versa. If no external forces act on the system, we can easily calculate heights and speeds by knowing the total mechanical energy (gravitational potential plus kinetic) is conserved.

Answer:

$\boxed{\text{The statement is true}}$

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv

cestrela7 [59]

<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":

$P=\sigma A T^{4}$ (1)

Where:

$P=300J/min=5J/s=5W$ is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing $1W=\frac{1Joule}{second}=1\frac{J}{s}$

$\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}}$ is the Stefan-Boltzmann's constant.

$A=5cm^{2}=0.0005m^{2}$ is the Surface area of the body

$T=727\°C=1000.15K$ is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close. So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

$P=\sigma A \epsilon T^{4}$ (2)