To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzmann law which establishes that a black body emits thermal radiation with a total hemispheric emissive power (W / m²) proportional to the fourth power of its temperature.
Heat flow is obtained as follows:

Where,
F =View Factor
A = Cross sectional Area
Stefan-Boltzmann constant
T= Temperature
Our values are given as
D = 0.6m

The view factor between two coaxial parallel disks would be


Then the view factor between base to top surface of the cylinder becomes
. From the summation rule


Then the net rate of radiation heat transfer from the disks to the environment is calculated as





Therefore the rate heat radiation is 780.76W
The answer is D. <span> There would be a decrease in the population of marine organisms.
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Answer: you subtract the number of protons from the mass number, on the periodic table your atomic number is your protons and your atomic mass is the mass number
Explanation:
Let the key is free falling, therefore from equation of motion
.
Take initial velocity, u=0, so
.

As velocity moves with constant velocity of 3.5 m/s, therefore we can use formula

From above substituting t,
.
Now substituting all the given values and g = 9.8 m/s^2, we get
.
Thus, the distance the boat was from the point of impact when the key was released is 10.60 m.
Answer:
2,400kg * m/s
Explanation:
You are missing some information in the question but the rest could be found some where else.
The question gives the masses and starting velocity of each car.
Car 1: m = 600kg and sv = 4m/s
Car 2: m 400kg and sv = 0m/s
Find the momentum of both cars.
Car 1: 600 * 4 = 2400
Car 2: 400 * 0 = 0
Add both.
2400 + 0 = 2400
Best of Luck!