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svlad2 [7]
3 years ago
5

Why is sky blue and why moon appear in daytime​

Physics
2 answers:
Nookie1986 [14]3 years ago
6 0
Bc it’s beautiful Mother Nature
Elena L [17]3 years ago
4 0

Answer:

The sky is blue because of the tiny molecules in the air and we see the moon in the day because the light of the sun is reflecting off the moon

Explanation:

Blue light is scattered in all directions by the tiny molecules of air in Earth's atmosphere. Blue is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time. Closer to the horizon, the sky fades to a lighter blue or white.

We can see the moon during the day for the same reason we see the moon at night. The surface of the moon is reflecting the sun's light into our eyes. ... "When we see the moon during the day it's because the moon is in the right spot in the sky and it's reflecting enough light to be as bright, or brighter, than the sky."

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A rotating object has an angular acceleration of α = 0 rad/s2. Which one or more of the following three statements is consistent
Murrr4er [49]

Answer:

A,B and C

Explanation:

Statement A  

At all times, angular velocity is \omega = 0\,{\rm{rad/s}  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 0\,{\rm{rad/s}}

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero hence statement A is valid.  

Statement B  

Angular acceleration is the rate of change in angular velocity with respect to time.  

Angular velocity and angular acceleration are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

Which when re-arranged becomes  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

There’s no change in angular velocity anytime when the angular velocity is \omega = 10\,{\rm{rad/s}}.The final and initial velocities remain the same.  

The equation can be modified as follows:  

\begin{array}{c}\\\alpha = \frac{{10\,{\rm{rad/s}} - 10\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement B is valid  

Statement C  

Angular velocity is defined as the change in the angular position with respect to time.  

Angular velocity and angular displacement are related by  

\theta = \omega t

Which can also be modified as:  

{\theta _{\rm{f}}} - {\theta _{\rm{i}}}

Note that the final position is {\theta _{\rm{f}}}and initial position is {\theta _{\rm{i}}}

Modifying the equation to find the angular velocity we obtain  

\omega = \frac{{{\theta _{\rm{f}}} - {\theta _{\rm{i}}}}}{t}

When the angular displacement has the same value at all times, the equation becomes  

\begin{array}{c}\\\omega = \frac{{{\theta _{\rm{i}}} - {\theta _{\rm{i}}}}}{t}\\\\ = 0\\\end{array}

The angular velocity becomes zero.  

Angular acceleration and angular velocity are related by  

{\omega _{\rm{f}}} = {\omega _{\rm{i}}} + \alpha t

The expression above can be rearranged as follows:  

\alpha = \frac{{{\omega _{\rm{f}}} - {\omega _{\rm{i}}}}}{t}

At all times, the angular velocity is \omega = 0\,{\rm{rad/s}} hence initial and final velocities remain the same  

We obtain  

\begin{array}{c}\\\alpha = \frac{{0\,{\rm{rad/s}} - 0\,{\rm{rad/s}}}}{t}\\\\ = 0\\\end{array}

Therefore, the angular acceleration becomes zero and statement C is valid.  

Therefore, statements A,B and C are consistent .

4 0
4 years ago
5. YOU SHOULD USUALLY DRIVE SLOWER
saveliy_v [14]

Answer:

The answer is C.

Explanation:

If there are brake lights then you should drive slower.

3 0
3 years ago
Read 2 more answers
Which of the following represents a covalent compound?<br><br> KF<br> NaCl<br> SBr3<br> Mg2O
slava [35]

Answer:

SBr3 is a covalent compound

5 0
4 years ago
Which most likely indicate a chemical change has occurred ?
Allushta [10]
If a new substance suddenly appears that wasn't there originally,
then a chemical change has occurred.

Like for example, (this is the only example I can think of right now):

-- You leave your bicycle outside in the rain, and it gets wet, and
a few days later there's some rust on it.

-- You scrape off some of the rust, take it to school, give it to the
Chemistry teacher, and ask her to analyze it and tell you what it
is.  Later that day, she tells you it's a substance called "Iron oxide".

-- Where did that come from ?  There was no iron oxide there.
There was only iron in the bicycle, and air, and water.

-- The iron oxide formed from a chemical change when the iron
on the surface of the bike combined with some of the oxygen in
the air, and molecules of a new substance were created.  (For
some reason, the presence of water makes this chemical reaction
go faster.)    


5 0
3 years ago
How long will it take a 2.3"x10^3 kg truck to go from 22.2 m/s to a complete stop if acted on by a force of -1.26x10^4 N.What wo
Greeley [361]

The stopping distance is 45.0 m

Explanation:

First of all, we find the acceleration of the truck, by using Newton's second law:

F=ma

where

F=-1.26\cdot 10^4 N is the net force on the truck

m=2.3\cdot 10^3 kg is the mass of the truck

a is its acceleration

Solving for a,

a=\frac{F}{m}=\frac{-1.26\cdot 10^4}{2.3\cdot 10^3}=-5.48 m/s^2

where the negative sign means the acceleration is opposite to the direction of motion.

Now, since the motion of the truck is at constant acceleration, we can apply the following suvat equation:

v^2-u^2=2as

where

v = 0 is the final velocity of the truck

u = 22.2 m/s is the initial velocity

a=-5.48 m/s^2 is the acceleration

s is the stopping distance

And solving for s,

s=\frac{v^2-u^2}{2a}=\frac{0-(22.2)^2}{2(-5.48)}=45.0 m

Learn more about accelerated motion:

brainly.com/question/9527152

brainly.com/question/11181826

brainly.com/question/2506873

brainly.com/question/2562700

#LearnwithBrainly

6 0
4 years ago
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