What happens to the orbit time as the distance increases.

Physics

1 answer:

Maslowich2 years ago

6 0

The orbital period increases if the orbital distance is increased.

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A coin is dropped in a 15.0 m deep well.

labwork [276]

Answer:

t = 1.75

t = 0.04

Explanation:

For part 1 we want to use a kenamatic equation with constant acceleration:

X = 1/2*a*t^2

isolate time

t = sqrt(2X / a)

Plugin known variables. Acceleration is the force of gravity which is 9.8 m/s^2

t = sqrt(2*15m / 9.8m/s^2)

t = 1.75 s

The speed of sound travels at a constant speed therefore we don't need acceleration and can use the equation:

v = d / t

isolate time

t = d / v

plug in known variables

t = 15m / 340m/s

t = 0.04 s

7 0

3 years ago

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)