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2 years ago

Explain why the energy in the sun moves out toward the surface

Physics

1 answer:

IgorLugansk [536]2 years ago

5 0

energy moves through the sun in two main way, Radiation and Convection.

Explanation:

as the energy moves outward from the sun's core, 1st enters rhe radiation zone. the radiation zone is a region of highly compressed gas. here the energy is transferred by the absorption and irradiation of electromagnetic waves

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. A telescope is constructed with two lenses separated by a distance of 25 cm. The focal length of the objective is 20 cm. The f

beks73 [17]

Answer:

the magnitude of the angular magnification of the telescope. is 4

Explanation:

Calculate the magnitude of the angular magnification of the telescope.

Given that,

distance = 25cm

focal length from the objective f₀ = 20cm

focal length from eye piece f₁ = 5cm

The angular magnification of the telescope is

$M = \frac{f_0}{f_1}$

Magnification = 20 / 5

magnification = 4

Hence, the magnitude of the angular magnification of the telescope. is 4

5 0

2 years ago

Use the work—energy theorem to solve each of these problems. You can use Newton's laws to check your answers. Neglect air resist

andreyandreev [35.5K]

Answer:

a) It is moving at $43.15\frac{m}{s^{2}}$ when reaches the ground.

b) It is moving at $101.44\frac{m}{s^{2}}$ when reaches the ground.

Explanation:

Work energy theorem states that the total work on a body is equal its change in kinetic energy, this is:

$W=K_f-K_i$ (1)

with W the total work, Ki the initial kinetic energy and Kf the final kinetic energy. Kinetic energy is defined as:

$K=\frac{mv^2}{2}$ (2)

with m the mass and v the velocity.

Using (2) on (1):

$W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (3)

In both cases the total work while the objects are in the air is the work gravity field does on them. Work is force times the displacement, so in our case is weight (w=mg) of the object times displacement (d):

$W=Fd=wd=mgd$ (4)

Using (4) on (3):

$mgd=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}$ (5)

That's the equation we're going to use on a) and b).

a) Because the branch started form rest initial velocity (vi) is equal zero, using this and solving (5) for final velocity:

$v_f=\sqrt{\frac{2mgd}{m}}=\sqrt{2gd}=\sqrt{2*9.8*95}$

$v_f=43.15\frac{m}{s^{2}}$

b) In this case the final velocity of the boulder is instantly zero when it reaches its maximum height, another important thing to note is that in this case work is negative because weight is opposing boulder movement, so we should use -mgd:

$-mgd=-\frac{mv_i^2}{2}$