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babymother [125]
3 years ago
12

Imagine a sunny day at the pool. The sun is out and you are thinking about how the light travels from the sun and then hits the

water. How does the light react once it hits the water? Draw or insert a model (that you created not Googled) of what you think happens to the light as it hits the water in the space below. Make sure to include an explanation too. Use what you learned to help you.
Physics
2 answers:
Dovator [93]3 years ago
5 0

Answer:

it evaporats

Explanation:

because the sun is so hot that the water will turn into gas hope i helped

sesenic [268]3 years ago
4 0

Answer:

im pretty sure is reflects the water to objects appear blurrier than on land

Explanation:

PLS CORRECT ME IF I AM WRONG

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What occurs when light bends while passing from one medium to another?
Levart [38]

Answer:So the answer is B

Explanation: When the light rays either bend or change their direction while passing from one medium to another it is called refraction of light. The refraction of light takes place when light travels from air into glass, from glass into air, from air into water or from water into air.

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What is one result of meiosis?
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Where is the sun in the life cycle? HELP ITS A FINAL EXAM
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8 0
3 years ago
Three charges, q1 = +2.06 x 10-9 C, q2 = -3.27 x 10-9 C, and q3 = +1.05 x 10-9 C, are located on the x-axis at x1 = 0, x2 = 10.0
lisov135 [29]

Answer:

The resultant force on charge 3 is Fr= -2,11665 * 10^(-7)

Explanation:

Step 1: First place the three charges along a horizontal axis. The first positive charge will be at point x=0, the second negative charge at point x=10 and the third positive charge at point x=20. Everything is indicated in the attached graph.

Step 2: I must calculate the magnitude of the forces acting on the third charge.

F13: Force exerted by charge 1 on charge 3.

F23: Force exerted by charge 2 on charge 3.

K: Constant of Coulomb's law.

d13: distance from charge 1 to charge 3.

d23: distance from charge 2 to charge 3

Fr: Resulting force.

q1=+2.06 x 10-9 C

q2= -3.27 x 10-9 C

q3= +1.05 x 10-9 C

K=9-10^9 N-m^2/C^2

d13= 0,20 m

d23= 0,10 m

F13= K * (q1 * q3)/(d13)^2

F13=9,7335*10^(-8) N

F23=K * (q2 * q3)/(d23)^2

F23= -3,09 * 10^(-7)

Step 3: We calculate the resultant force on charge 3.

Fr=F13+F23= -2,11665 * 10^(-7)

3 0
3 years ago
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