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ziro4ka [17]
1 year ago
8

Sherry and Ellen are talking about the factors that affect light scattering in the atmosphere. Sherry says one factor is the len

gth of the path of sunlight. Ellen says one factor is altitude. Who is correct
Physics
1 answer:
Delvig [45]1 year ago
3 0

Sherry who says one factor is the length of the path of sunlight is correct.

<h3>Factors affecting light scattering</h3>

There are two main factors which affects light scattering, and they include the following;

  1. the size of the particles
  2. wavelength of the light

length of the path of sunlight is equivalent to wavelength of the light.

Thus, we can conclude that Sherry who says one factor is the length of the path of sunlight is correct.

Learn more about light scattering here: brainly.com/question/1381101

#SPJ1

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AleksandrR [38]
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Since specific heat is part of the equation. A smaller specific heat will create a smaller heat gain or loss. </span>
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6 0
3 years ago
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Polar molecules have _____. A) an uneven distribution of charges around the molecule B) atoms that are the same element C) an ev
Keith_Richards [23]

Answer: ionic bonds

Explanation:

7 0
3 years ago
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A cadillac escalade has a mass of 2 569.6 kg, if it accelerate at 4.65m/s2 what is the net force of the car
anygoal [31]
Force is equal to mass multiplied by acceleration, therefore 
F=ma
m=2569.6 kg
a=4.65m/s^2
therefore F=2569.6*4.65=11948.6 (correct to 1 d.p.)
3 0
3 years ago
Which car is harder to stop and why?
kondor19780726 [428]

Answer:

The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.

Explanation:

Momentum (P) =mass(m) × velocity (v)

For car A,

                    P = m × v = 1000 × 40 = 40000 Ns

For car B,

                    P = m × v = 4000 × 10 = 40000 Ns

Force (F) = Momentum change(ΔΡ)/ time taken(t)

            F = ΔΡ/t

When stopping the car the momentum changes from 40000 Ns to 0

So momentum change in both cars is the same. So to stop the two cars in a  given time (t) you need the same force, which means you will feel same difficulty.

             

8 0
3 years ago
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A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

4 0
3 years ago
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