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jeka94
3 years ago
9

What factors affect the amount of solar energy that reaches earth's surface

Physics
2 answers:
zalisa [80]3 years ago
8 0
Weather
Particles in air such as smoke smog
bazaltina [42]3 years ago
7 0
Latitude~ due to earth spherical shape the solar rays have more intensity around the equatorial regions

 cloud cover: clouds have a big impact on the amount of solar radition reaching earth surface
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14. What is the standard deviation for the numbers 75,83,96,100,121 and 125?
seraphim [82]

Answer:

2√109 about 20.88

Explanation:

The mean of this set of numbers is (75+83+96+100+121+125)/6 is 100. Next step is ((75-100)^2+(83-100)^2+(96-100)^2+(100-100)^2+(121-100)^2+(125-100)^2)/6 this result is called deviation. Square root the deviation(436) then you set the standard deviation.

3 0
3 years ago
Give three practical uses of electromagnets
PtichkaEL [24]
Three practical uses of electromagnets would be
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7 0
3 years ago
explain why conductors and insulators are both required to construct the electrical wiring in your home
olchik [2.2K]
Conductors (something that allows electricity to flow easily) allow for electricity to flow easily. This would be the wires. If you don't have conductors, then you cannot have electricity flow.

Insulators (something that doesn't allow electricity to flow through it) is important because it allows us to be able to touch the cables or place them next to one another and not shock ourselves

Hope this helps
6 0
3 years ago
Read 2 more answers
John decided to cycle to his friend's house at a speed of 5km/h and the journey took 2
Marianna [84]

Answer:100 miles

Explanation:

5 0
3 years ago
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A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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