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2 years ago

Can someone help me? Idk why I can't understand this

Physics

1 answer:

olya-2409 [2.1K]2 years ago

5 0

It's been a while since I've studied this, but my answers would be:

13. 5730 years. The half-life of a substance is the amount of time it takes for half of it to decay, and, according to the graph, half of the substance remained at 5730 years.

14. 10740 years. According to the graph, only 25% of the carbon remained after 10740 years.

15. 15 atoms. According to the graph, only 12.5% of the carbon remained after 16110 years. 12.5% of 120 atoms is 15 atoms.

16. 1600 atoms. According to the graph, if a sample of carbon is 10740 years old, only 25% of it remains. To find the original amount, multiply the current amount by (100% / 25%), which equals 4. So, 4. 400 atoms * 4 = 1600 atoms is the original amount.

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Two parallel disks of diameter D 5 0.6 m separated by L 5 0.4 m are located directly on top of each other. Both disks are black

oksian1 [2.3K]

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:

$Q = FA\sigma\Delta T^4$

Where,

F =View Factor

A = Cross sectional Area

$\sigma =$ Stefan-Boltzmann constant

T= Temperature

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

5 0

2 years ago

1. What are three questions you should ask before joining a club?

lara [203]

Answer:

what time does it start.

what do I need to join.

what are your expectations.

7 0

2 years ago

a rock is dropped from a height of 80 m and is in free fall what is the velocity of as it reaches the ground 4.0 seconds later

Marta_Voda [28]

Velocity = displacement (distance)/time

v=80m/4s

v=20m/s

velocity = 20 meters per second

8 0

2 years ago

There is a 90 kg woman attached at the end of a bungee cord (k = 35 N/m) that is experiencing simple harmonic motion. How long d

Cloud [144]

Answer:

The value is $T =10.1 \ s$

Explanation:

From the question we are told that

The mass of the woman is $m = 90 \ kg$

The spring constant of the bungee cord is $k = 35 \ N/ m$

Generally the period of the oscillation (i,e time taken to complete on cycle ) is mathematically represented as

$T = 2 \pi * \sqrt{ \frac{m}{k} }$

=> $T = 2 * 3.142 * \sqrt{ \frac{90 }{ 35} }$

=> $T =10.1 \ s$

3 0

2 years ago

a whistle you use to call your hunting dog has a frequency of 21 khz, but your dog is ignoring it. you suspect the whistle may n

laila [671]

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$f = f_0 (\frac{v-v_0}{v})$