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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.

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A stereo speaker produces a pure \"E\" tone, with a frequency of 329.6 Hz. What is the period of the sound wave produced by the
olya-2409 [2.1K]

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

f = frequency of the tone = 329.6 Hz

T = Time period of the sound wave

we know that, Time period and frequency are related as

T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s

v = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m

v = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m

8 0
3 years ago
HELP FAST!!!!!
Paul [167]
Mirror: Reflects off of
Glass of water: Goes through
Dark Fabric: Absorbs into
4 0
3 years ago
Read 2 more answers
A roller coaster cart of mass m = 223 kg starts stationary at point A, where h1 = 26.8 m and a while later is at B, were h2 = 14
Tresset [83]

Answer:

vB = 15.4 m/s

Explanation:

Principle of conservation of energy:

Because there is no friction the mechanical energy is conserve

ΔE = 0

ΔE : mechanical energy change (J)

K : Kinetic energy (J)

U: Potential energy (J)

K = (1/2)mv²

U = m*g*h

Where :

m: mass (kg)

v : speed (m/s)

h : hight (m)

Ef - Ei = 0

(K+U)final - (K+U)initial =0

(K+U)final = (K+U)initial

((1/2)mv²+m*g*h)final = ((1/2)mv²+m*g*h)initial , We divided by m both sides of the equation:

((1/2)vB² + g*hB = (1/2 )vA²+ g*hA

(1/2) (vB)² + (9.8)*(14.7) =  0 + (9.8)(26.8 )

(1/2) (vB)² = (9.8)(26.8 ) - (9.8)*(14.7)

(vB)² = (2)(9.8)(26.8 - 14.7)

(vB)² = 237.16

v_{B} = \sqrt{237.16}

vB = 15.4 m/s : speed of the cart at B

4 0
3 years ago
The information on a can of soda indicates that the can contains 355 mL. The mass of a full can of soda is 0.369 kg, while an em
Vilka [71]

Answer:

\rho=995.50\ kg.m^{-3}

\bar w=9765.887\ N.m^{-3}

s=0.9955

Explanation:

Given:

  • volume of liquid content in the can, v_l=0.355\ L=3.55\times 10^{-4}\ L
  • mass of filled can, m_f=0.369\ kg
  • weight of empty can, w_c=0.153\ N

<u>So, mass of the empty can:</u>

m_c=\frac{w_c}{g}

m_c=\frac{0.153}{9.81}

m_c=0.015596\ kg

<u>Hence the mass of liquid(soda):</u>

m_l=m_f-m_c

m_l=0.369-0.015596

m_l=0.3534\ kg

<u>Therefore the density of liquid soda:</u>

\rho=\frac{m_l}{v_l} (as density is given as mass per unit volume of the substance)

\rho=\frac{0.3534}{3.55\times 10^{-4}}

\rho=995.50\ kg.m^{-3}

<u>Specific weight of the liquid soda:</u>

\bar w=\frac{m_l.g}{v_l}=\rho.g

\bar w=995.5\times 9.81

\bar w=9765.887\ N.m^{-3}

Specific gravity is the density of the substance to the density of water:

s=\frac{\rho}{\rho_w}

where:

\rho_w= density of water

s=\frac{995.5}{1000}

s=0.9955

3 0
3 years ago
Read 2 more answers
A ball traveling at a speed ν0 rolls off a desk and lands at a horizontal distance x0 away from the desk, as shown in the figure
klasskru [66]

Answer:

3x_0

Explanation:

The horizontal distance covered by the ball in the falling is only determined by its horizontal motion - in fact, it is given by

d=v_x t

where

v_x is the horizontal velocity

t is the time of flight

The time of flight, instead, is only determined by the vertical motion of the ball: however, in this problem the vertical velocity is not changed (it is zero in both cases), so the time of flight remains the same.

In the first situation, the horizontal distance covered is

d=v_0 t = x_0

in the second case, the horizontal velocity is increased to

v_x' = 3v_0

And so the new distance travelled will be

d' = v_x' t = 3 v_0 t = 3 x_0

So, the distance increases linearly with the horizontal velocity.

5 0
3 years ago
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