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3 years ago

Which processes transfer energy from the core to the photosphere?

Physics

2 answers:

nikklg [1K]3 years ago

7 0

<span>-radiation and convection

Hope this helps :)
</span>

Elodia [21]3 years ago

4 0

Answer:

conduction and convection

Explanation:

Sun is a huge ball of gases- Hydrogen and Helium. Nuclear fusion reaction in the core powers the sun. The heat from the core is transferred to photo-sphere via two modes- convection and conduction.

Conduction is the mode of heat transfer when the particles vibrate at a their fixed position. The particles collide with the neighboring particles and transfer heat. More the temperature, with particles vibrate with greater energy. There is direct contact between particles. conduction happens in solids, liquids and gases.

Convection is the mode of heat transfer in fluids. The fluid heats up near the source and expands. Due to its less density, it rises up and the cool fluid from top rushes to take its place. This process continues.

Convection currents can be observed on the surface of the sun. Due to convection currents, sun has granular texture.

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Si se aplic una fuerza de 150N en un área de de 0.4m2¿cual será la preciosa ejercida?

natali 33 [55]

Answer:

P = 375 Pa

Explanation:

The question says that,"If a force of 150N was applied in an area of 0.4m², what will be the precious exerted?"

We have,

Force, F = 150 N

Area, A = 0.4m²

We need to find the pressure exerted. We know that,

Pressure = forece/area

So,

$P=\dfrac{150\ N}{0.4\ m^2}\\\\P=375\ Pa$

So, the required pressure is equal to 375 pa.

7 0

3 years ago

A rocket is fired vertically upwards with initial velocity 92 m/s at the ground level. Its engines then fire and it is accelerat

Readme [11.4K]

Answer:

The rocket above the ground is in 44 sec.

Explanation:

Given that,

Initial velocity = 92 m/s

Acceleration = 4 m/s²

Altitude = 1200 m

Suppose, How long was the rocket above the ground?

We need to calculate the time

Using equation of motion

$s=ut-\dfrac{1}{2}at^2$

Put the value into the formula

$1200=92t+\dfrac{1}{2}\times4t^2$

$2t^2+92-1200=0$

$t=10\ sec$

We need to calculate the velocity

Using equation of motion

$v=u+at$

Put the value into the formula

$v=92+4\times10$

$v=132\ m/s$

When the rocket hits the ground,

Then, h'=0

We need to calculate the time

Using equation of motion

$h'=h+ut-\dfrac{1}{2}at^2$

Put the value into the formula

$0=1200+132t-\dfrac{1}{2}\times9.8t^2$

$4.9t^2-132t-1200=0$

$t=34\ sec$

When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground

So, the total time will be

$t'=34+10$

$t'=44\ sec$

Hence, The rocket above the ground is in 44 sec.

7 0

4 years ago

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0

3 years ago

A snowboarder is sliding down a hill to the right. What are the 3 forces being used?

Softa [21]

Answer:

ANswers

Explanation:

Friction, knetic, and physics

8 0

3 years ago

A brass statue with a mass of 0.40 kg and a density of 8.00×103kg/m3 is suspended from a string. When the statue is completely s

Mars2501 [29]

Answer:

ρ = 1469 kg/m³

Explanation:

given,

mass of statue = 0.4 Kg

density of statue = 8 x 10³ kg/m³

tension in the string = 3.2 N

density of the fluid = ?

Volume of the statue

$V = \dfrac{0.4}{8\times 10^3}$

V = 5 x 10⁻⁵ m³

W = ρ g V

W = ρ x 9.8 x 5 x 10⁻⁵

now, tension on the string will be equal to

T = mg - W

3.2 = 0.4 x 9.8 - ρ x 9.8 x 5 x 10⁻⁵

ρ x 9.8 x 5 x 10⁻⁵ = 0.72

ρ = 1469 kg/m³

8 0

3 years ago