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3 years ago

Which physical phenomena is responsible for the earth’s sky appearing blue? scattering reflection dispersion refraction

Physics

2 answers:

Bogdan [553]3 years ago

8 0

Answer:

scattering

Explanation:

oddessy ware

klio [65]3 years ago

7 0

Answer: Scattering reflection

Sunlight reaches earth's atmosphere and is scattered in all directions by all the gasses and particles in the air. Blue light is seen more than others because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

Explanation:

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Two scientists interpret the same set of data differently. How would the scientific community deal with this problem?

alexdok [17]

The correct answer would be D. A new experiment would be needed to be done in order to test the conclusions. In science there is no authority, data is the only thing that matters. So if we have two different conclusions from the same date the only solution is to perform more tests and more experiments to see what is correct.

7 0

3 years ago

The force exerted by the wind on the sails of a sailboat is Fsail = 330 N north. The water exerts a force of Fkeel = 210 N east.

Elena L [17]

Answer:

The magnitude of the acceleration is $a_r = 1.50 \ m/s^2$

The direction is $\theta = 32.5 6^o$ north of east

Explanation:

From the question we are told that

The force exerted by the wind is $F_{sail} = (330 ) \ N \ north$

The force exerted by water is $F_{keel} = (210 ) \ N \ east$

The mass of the boat(+ crew) is $m_b = 260 \ kg$

Now Force is mathematically represented as

$F = ma$

Now the acceleration towards the north is mathematically represented as

$a_n = \frac{F_{sail}}{m_b}$

substituting values

$a_n = \frac{330 }{260}$

$a_n = 1.269 \ m/s^2$

Now the acceleration towards the east is mathematically represented as

$a_e = \frac{F_{keel}}{m_b }$

substituting values

$a_e = \frac{210}{260}$

$a_e =0.808 \ m/s^2$

The resultant acceleration is

$a_r = \sqrt{a_e^2 + a_n^2}$

substituting values

$a_r = \sqrt{(0.808)^2 + (1.269)^2}$

$a_r = 1.50 \ m/s^2$

The direction with reference from the north is evaluated as

Apply SOHCAHTOA

$tan \theta = \frac{a_e}{a_n}$

$\theta = tan ^{-1} [\frac{a_e}{a_n } ]$

substituting values

$\theta = tan ^{-1} [\frac{0.808}{1.269 } ]$

$\theta = tan ^{-1} [0.636 ]$

$\theta = 32.5 6^o$

5 0

3 years ago

Can someone explain which of Newton’s Law is demonstrated in part 1 and which is demonstrated in part 2? (Picture)

rewona [7]

Answer:

Every action has an equal and opposite reaction. If the student doesn't push, nothing moves, is one student pushes, both move which is an example of newtons third law.

Explanation:

3 0

3 years ago

When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.a)

RideAnS [48]

Answer:

a) True.

Explanation:

If you turn the wheel in the direction of the turn before beginning the turning maneuver then it's possible that there might be not enough space available for turning and also if you are waiting for the traffic to get clear with rear ended then it will get pushed forward onto the coming traffic.

7 0

3 years ago

A particle R moves with a velocity of 6 m/s due East and another particle S moves at 8 m/s due South. What is the magnitude of t

cluponka [151]

Answer:

Explanation:

We can look at this problem as a triangle, r is the hypotenuse so if we take the square root of 6^2+8^2 we get 10

6 0

2 years ago