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satela [25.4K]
3 years ago
10

Is the expression "The bigger they are, the harder they fall" a generally true statement since, in the absence of air resistance

, all objects fall with the same acceleration due to gravity? Pick the most correct answer.
Yes, because as a heavier man falls, he comes closer to the center of the earth and he falls faster.
No, since the acceleration is the same, each man will be affected the same by it as he falls.
Yes, since F = ma, therefore "a" will be the same in each case, but the mass of each will be different.
Yes, since F = ma, the heavier an object is, the more it will accelerate and therefore fall with more force.
Physics
2 answers:
lions [1.4K]3 years ago
6 0
No, because in oxygen depraved rooms, if you drop a feather and a bowling ball at the same height and time, they will fall at the same speed and have the same amount of impact.
vaieri [72.5K]3 years ago
4 0

Answer:

No, since the acceleration is the same, each man will be affected the same by it as he falls.

Explanation:

In the problem, we are told that air resistance is gone. The reason a heavier man would fall harder is because the heavier a man, the less the air resistance would affect him. The air resistance is arguably the only factor that applies in what makes heavier men fall harder. So, if the aire resistance is gone and both mean are falling at exactly the same speed, then they will hit the ground equally as hard, making the correct answer B.

Hope this helps and makes sense.

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Heat flow is obtained as follows:

Q = FA\sigma\Delta T^4

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A = Cross sectional Area

\sigma = Stefan-Boltzmann constant

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Our values are given as

D = 0.6m

L = 0.4m\\T_1 = 450K\\T_2 = 450K\\T_3 = 300K

The view factor between two coaxial parallel disks would be

\frac{L}{r_1} = \frac{0.4}{0.3}= 1.33

\frac{r_2}{L} = \frac{0.3}{0.4} = 0.75

Then the view factor between base to top surface of the cylinder becomes F_{12} = 0.26. From the summation rule

F_{13} = 1-0.26

F_{13} = 0.74

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

\dot{Q_3} = \dot{Q_{13}}+\dot{Q_{23}}

\dot{Q_3} = 2\dot{Q_{13}}

\dot{Q_3} = 2F_{13}A_1 \sigma (T_1^4-T_3^4)

\dot{Q_3} = 2(0.74)(\pi*0.3^2)(5.67*10^{-8})(450^4-300^4)

\dot{Q_3} = 780.76W

Therefore the rate heat radiation is 780.76W

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