1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
skelet666 [1.2K]
3 years ago
12

PLEASE HELP! Explain how energy travels outward from the core and is emitted from the Sun. Include what happens in each layer of

the Sun.
Physics
2 answers:
xxMikexx [17]3 years ago
8 0
Fusion occurs in the Sun's core, releasing energy that is transferred outward. Once in the radiative zone, gamma rays are transferred by radiation. They are converted to other types of photons, which move into the convective zone, where they are transferred by convection. Finally, energy is emitted from the photosphere.
cricket20 [7]3 years ago
5 0

Answer:

fusion occurs in the core. it releases energy and its transferred outward. Gamma rays are transferred by radiation when in the radiative zone. then they are converted to other photons that move into the convective zone. there they are transferred by convection. lastly energy is emitted from the photosphere.

You might be interested in
PLEASE ANSWER THIS IF YOU HAVE AN OCULUS!!
yan [13]

Answer:

Ah good question um no u dont have to be near it its a vr thing ur gonna have it on u the whole time. You can only be a few feet away from ur tv if u chose to do it on there or u can do it on ur phone but its also the same distance.

Explanation:

8 0
3 years ago
When you go out into the sun, the UV light can give you a tan. UV light has a frequency of 7.89 x 1014 Hz. What is the wavelengt
Olin [163]

Answer:

Explanation:

Formula and givens

  • λ = c / f
  • λ is the wavelength
  • c = the speed of light
  • f = the frequency
  • c = 3*10^8
  • f = 7.89 * 10^14

λ = ?

Solution

λ = 3*10^8 / 7.89*10^14

λ = 3*10^8/7.89*10^14

λ = 2.36 * 10^7

λ = 236 nanometers. What you use as your solution depends on what what you have been taught.

6 0
3 years ago
A 62 kg skier is moving at 6.5 m/s on frictionless horizontal snow-covered plateau when she encounters a rough patch 3.50 m long
Degger [83]

Answer:

(A). The work done by friction in crossing the patch is -637.98 J.

(B). The speed of skier is 10.57 m/s.

Explanation:

Given that,

Mass of skier = 62 kg

Speed = 6.5 m/s

Length = 3.50 m

Coefficient kinetic friction = 0.30

Height = 2.5 m

(A) we need to calculate the work done by friction in crossing the patch

Using formula of work done

W=-\mu mg\times l

Put the value into the formula

W=-0.30\times62\times9.8\times3.50

W=-637.98\ J

The work done by friction in crossing the patch is -637.98 J.

(B) we need to calculate the speed of skier

Using conservation of energy

K.E_{i}+U_{i}-W_{friction}=K.E_{f}+U_{f}

\dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2+U_{f}

Final potential energy is zero

So, \dfrac{1}{2}mv_{1}^2+mgh-\mu mgl=\dfrac{1}{2}mv_{2}^2

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}v_{1}^2+gh-\mu gl

Put the value into the formula

\dfrac{1}{2}v_{2}^2=\dfrac{1}{2}\times6.5^2+9.8\times2.5+0.30\times9.8\times3.50

v_{2}=\sqrt{2\times55.915}

v_{2}=10.57\ m/s

The speed of skier is 10.57 m/s.

Hence,  (A).The work done by friction in crossing the patch is -637.98 J.

(B).The speed of skier is 10.57 m/s.

6 0
3 years ago
If the mass of an object is 10 kg and the
ASHA 777 [7]

Answer:

B. -40 kgm/s is the answer

6 0
3 years ago
You are standing on a sidewalk. There is a car in the distance. The horn on the car sounds. You hear it at a pitch that correspo
AnnyKZ [126]

Answer: The answer: The car is moving away from you.

Both A and C are true as Car can be moving in line away from you or has component of velocity in opposite direction.

Explanation:The decrease in the frequency of the sound is the result of Doppler's effect. A/c to Doppler's effect the frequency of received sound of source is changed if it is moving relative to the receiver, i.e. the distance between them is changing due to motion.

The general formula of Doppler's Effect is attached as the picture.

In this formula v_D is the velocity of Detector i.e the receiver relative to wind. While v_s is the velocity of source relative to wind and v is the velocity of sound.

The Doppler's effect is not effected by the velocity of wind as the wind itself could not change the distance between the two objects i.e. you and the car. Wind velocity can change the speed of sound and its wavelength but the change does not effect the frequency.

Hence if we assume the car to be moving with velocity v_c and you are stationary

f'=f_s*\frac{v}{v-v_c}

hence the frequency is reduced.

4 0
4 years ago
Other questions:
  • How does fertilizer run off pollution?
    8·1 answer
  • Qroups of specialized cells that work together are called____
    11·1 answer
  • 12. What does positive and negative acceleration indicate?
    6·1 answer
  • What are the benefits and drawbacks of using renewable energy sources to generate electricity?
    11·1 answer
  • What is stellar parallax?a.It describes the fact that stars are actually moving relative to one another, even though to our eyes
    13·1 answer
  • The smallest unit of an element that maintains the properties of that element
    6·1 answer
  • Which is not a characteristic of an ideal fluid?
    9·1 answer
  • What does area under a velocity time graph represent
    11·1 answer
  • A truck initially traveling at a speed of 22 m/s increases at a constant rate of 2.4 m/s^2 for 3.2s. What is the total distance
    12·1 answer
  • . If block A has a velocity of 0.6 m/s to the right, determine the velocity of cylinder​
    14·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!