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3 years ago

Air is warmer and less dense than surrounding air at the equator because the equator receives more?

Physics

2 answers:

Hatshy [7]3 years ago

4 0

The answer is C: Solar energy

gladu [14]3 years ago

3 0

Solar energy

Given the fact that near the equator you have the sun at roughly the same level all year, you deal with very little to no temperature change, whereas North or South of the equator experiences temperature change due to increasing or decreasing distance from the sun when the earth tilts on its axis.

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I’ll mark you as brinlist please help.

Ede4ka [16]

Answer:

245 divided by 5.14=47.6653696 or 47.66

Explanation:

8 0

3 years ago

Read 2 more answers

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

3 years ago

You are riding a bicycle up a gentle hill. It is fairly easy to increase your potential

Dahasolnce [82]

True or false: while riding a bicycle up a gentle hill, it fairly easy to increase your potential energy, but to increase your kinetic energy would ...

3 0

2 years ago

A reconnaissance plane flies 605 km away from

kolezko [41]

Answer:

$v_{avg} =$ 355 m/s

Explanation:

Distance = 605 km

Initial speed = $v_{i}$ = 284 m/s

Final velocity = $v_{f}$ = 426 m/s

Average speed = ?

There is two method two find average speed. In first method, using 3rd equation of motion, we find acceleration.

$2as = v_{f}^{2}+v_{i}^{2}$

Then using first equation of motion, we find time

$v_{f} = v_{i}+at$

Then using the formula of average velocity, we find average velocity

$v_{avg}=\frac{total-distance}{total-time}$

Second method is very simple

$v_{avg}=\frac{v_{f}+v_{i} }{2}$

$v_{avg}=\frac{426+284}{2}$

$v_{avg} =$ 355 m/s

8 0

3 years ago

An object travels with velocity v = 4.0 meters/second and it makes an angle of 60.0° with the positive direction of the y-axis.

Galina-37 [17]

Vx=cos60(4)
x-component of velocity
If you think about it, it makes a right triangle when you combine all the different types of forces together such as v, vx and vy. Then, you can use trigonometry and soh cah toa in order to figure out vx.

8 0

3 years ago