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Lyrx [107]
3 years ago
8

Star Twinkle but planet do not why​

Physics
1 answer:
V125BC [204]3 years ago
7 0

Answer:The total variation in the amount of light entering our eye is not dectiable therefore planets do not twinkle.

Explanation:Stars twinkle, while planets (usually) shine steadily. Why? Stars twinkle because … they're so far away from Earth that, even through large telescopes, they appear only as pinpoints. ... Planets shine more steadily because … they're closer to Earth and so appear not as pinpoints, but as tiny disks in our sky.As light from a star races through our atmosphere, it bounces and bumps through the different layers, bending the light before you see it. Since the hot and cold layers of air keep moving, the bending of the light changes too, which causes the star's appearance to wobble or twinkle.

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A ball at rest starts rolling down a hill with a constant acceleration of 3.2 meters/second. What is the final velocity of the b
a_sh-v [17]
19.2 meters/second^2 would be the correct answer.

Try multiplying 3.2 meters/second^2 by 6 and you will receive the answer provided above. If you have any further questions, let me know!
5 0
3 years ago
Do you think that nuclear fusion takes place in the atmospheres of stars? Why or why not?
vova2212 [387]

Answer:

Yes, why has been explained

Explanation:

Yes, Fusion takes place in the atmosphere of stars. Mostly stars are composed of hydrogen and helium, small quantity of other heavier elements.

The stars get their energy by the fusion reaction that continuously takes place inside them. The fusion  of two hydrogen molecules to form a helium molecule. This reaction generates enormous amount of heat and light, which why the stars are so shiny and hot.  

6 0
3 years ago
Identify the following as a suspension or a colloid. sand in water
sashaice [31]
I would say that sand in water is a suspension as you can see the particles with the naked eye or a microscope. Also, the solution will settle if left alone and the sand will sink to the bottom.
4 0
4 years ago
Read 2 more answers
starting from rest , a formula one car accelerates uniformly at 25m\s2 for 30secs. what distance does it cover in the last one s
Anestetic [448]

The distance covered in the last second of motion is 737.5 m

Explanation:

The motion of the car is a uniformly accelerated motion, so we can use the suvat equations.

First of all, we have to find the velocity of the car when the last second of motion starts, that is the velocity of the car after t = 29 s. We can use the equation:

v = u + at

where

u = 0 is the initial velocity

a=25 m/s^2 is the acceleration

Substituting t = 29 s,

v=0+(25)(29)=725 m/s

Now we can find the distance covered in the last second of motion by using

s=ut+\frac{1}{2}at^2

where

u = 725 m/s is the velocity at the beginning of the last second

t = 1 s is the time interval considered

a=25 m/s^2 is the acceleration

Substituting,

s=(725)(1)+\frac{1}{2}(25)(1)^2=737.5 m

Note that the acceleration of 25 m/s^2 is not realistic for a car, but I still have used the data of the problem.

Learn more about acceleration:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
3 years ago
The speed of a projectile when it reaches its maximum height is 0.58 times its speed when it is at half its maximum height. What
Ne4ueva [31]

Speed of the projectile at its maximum height is only along horizontal direction

so at highest point

v_1 = v_x

now when he is at half of the maximum height the speed will be in x and y direction both

v_2 = \sqrt{v_y^2 + v_x^2}

here it is given that

v_1 = 0.58 v_2

v_x = 0.58\sqrt{v_x^2 + v_y^2}

2.97 v_x^2 = v_x^2 + v_y^2

1.97 v_x^2 = v_y^2

also we know that

v_y^2 = v_{iy}^2 - 2 g \frac{H}{2}

here we know that maximum height is given as

H = \frac{v_{iy}^2}{2g}

v_y^2 = v_{iy}^2 - 2 g\frac{v_{iy}^2}{4g}

v_y^2 = \frac{v_{iy}^2}{2}

now from above

1.97 v_x^2 = \frac{v_{iy}^2}{2}

1.98 v_x = v_{iy}

also we know that angle of projection is

tan\theta = \frac{v_{iy}}{v_x}

tan\theta = \frac{1.98v_x}{v_x}

so angle is

\theta = tan^{-1} 1.98

\theta = 63.3 degree

6 0
3 years ago
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