Most of the visible light we see coming from the sun originates from the

Physics

1 answer:

Diano4ka-milaya [45]2 years ago

7 0

Answer:

The answer for this is photosphere.

Explanation:

Most of the visible light we see coming from the sun originates from photosphere.The Photosphere is 300km dense and the temperature at the bottom of the Photosphere is 6400K and the top of the Photosphere is 4600K respectively.

Following are the feature of photosphere that is given below.

Limb Darkening: The edges are darker than the centre part of the sun.
Sunspots: The size of the sunspots is similar to the size of the Earth.

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You have a ball with a mass 1.3 kg tied to a rope, and you spin it in a circle of radius 1.8m. If the ball is moving at a speed

BlackZzzverrR [31]

Answer:

$6.5\; {\rm N}$ .

Explanation:

When an object travel at a speed of $v$ in a circle of radius $r$ , the (centripetal) acceleration of that object would be $a = (v^{2} / r)$ .

In this question, the ball is travelling at $v = 3\; {\rm m\cdot s^{-1}}$ in a circle of radius $r = 1.8\; {\rm m}$ . The (centripetal) acceleration of this ball would be:

$\begin{aligned} a &= \frac{v^{2}}{r} \\ &= \frac{(3\; {\rm m\cdot s^{-1}})^{2}}{1.8\; {\rm m}} \\ &= 5\; {\rm m\cdot s^{-2}}\end{aligned}$ .

By Newton's Laws of Motion, for an object of mass $m$ , if the acceleration of that object is $a$ , the net force on that object would be $m\, a$ . Since the acceleration of this ball is $a = 5\; {\rm m\cdot s^{-2}}$ , the net force on this ball would be:

$\begin{aligned} F &= m\, a \\ &= 1.3\; {\rm kg} \times 5\; {\rm m\cdot s^{-2}} \\ &= 6.5\; {\rm N} \end{aligned}$ .

7 0

2 years ago

Is the gravitational force between the two objects attractive, repelling, or both? Explain how you know

puteri [66]

Answer: attractive

Explanation:

According to Newton's law of Gravitation, the gravitational force $F$ exerted between two bodies of masses $m1$ and $m2$ and separated by a distance $r$ is equal to the product of their masses and inversely proportional to the square of the distance: