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Airida [17]
3 years ago
5

How does radiation differ from conduction?

Physics
1 answer:
Natali5045456 [20]3 years ago
3 0

Answer:

b

Explanation:

You might be interested in
Consider two insulating balls with evenly distributed equal and opposite charges on their surfaces, held with a certain distance
siniylev [52]

Answer:

interest point:

1) Point on the left side

2) Point within the radius r₁ of the first sphere

3) Point between the two spheres

4) point within the radius r₂ of the second sphere

5) Right side point

Explanation:

In this case, the total electric field is the vector sum of the electric fields of each sphere, to simplify the calculation on the line that joins the two spheres

       

We will call the sphere on the left 1 and it has a positive charge Q with radius r1, the sphere on the right is called 2 with charge -Q with radius r2. The total field is

          E_ {total} = E₁ + E₂

          E_{ total} = k \frac{Q}{x_1^2} + k  \frac{Q}{x_2^2}

the bold indicate vectors, where x₁ and x₂ are the distances from the center of each sphere. If the distance that separates the two spheres is d

          x₂ = x₁ -d

          E total = k  \frac{Q}{x_1^2} - k \frac{Q}{(x_1 - d)^2}

Let's analyze the field for various points of interest.

1) Point on the left side

in this case

            E_ {total} = k Q \ ( \frac{1}{x_1^2} - \frac{1}{(x_1 +d)2} )

            E_ {total} = k \frac{Q}{x_1^2}   ( 1 - \frac{1}{(1 + \frac{d}{x_1} )^2 } )

We have several interesting possibilities:

* We can see that as the point is further away the field is more similar to the field created by two point charges

* there is a point where the field is zero

            E_ {total} = 0

             x₁² =  (x₁ + d)²

           

2) Point within the radius r₁ of the first sphere.

In this case, according to Gauus' law, the charge is on the surface of the sphere at the point, there is no charge inside so this sphere has no electric field on its inner point

              E_ {total} = -k \frac{Q}{x_2^2} = -k \frac{Q}{((d-x_1)^2}

this expression holds for the points located at

                  -r₁ <x₁ <r₁

3) Point between the two spheres

                E_ {total} = k \frac{Q}{x_1^2} + k \frac{Q}{(d+x_1)^2}

This champ is always different from zero

4) point within the radius r₂ of the second sphere, as there is no charge inside, only the first sphere contributes

                  E_ {total} = + k \frac{Q}{(d-x_1)^2}+ k Q / (d-x1) 2

point range

                  -r₂ <x₂ <r₂

             

5) Right side point

            E_ {total} = k \frac{Q}{(x_2-d)^2} - k \frac{Q}{x_2^2}

             E_ {total} = - k \frac{Q}{x_2^2} ( 1- \frac{1}{(1- \frac{d}{x_2})^2 } )- k Q / x22 (1- 1 / (x1 + d) 2)

we have two possibilities

* as the distance increases the field looks more like the field created by two point charges

* there is a point where the field is zero

8 0
2 years ago
Which state(s) of matter have no definite shape or volume? liquids and solids liquids and gasses gasses solids
Alex777 [14]

Answer:

Liquid has a defined volume but undefined shape

Explanation:

Explanation:

Solid has specific shape and volume

Liquid has specific volume but no specific shape

Gases have no specific volume and no specific shape

So correct answer is liquid.

8 0
3 years ago
Read 2 more answers
Jane has a mass of 55 kg and his body covers 700 nails with a surface area of 1.00 mm,
Oksana_A [137]

We have,

  • Jane mass is 55 kg
  • His body covered with 700 nails all of them having a surface area of 1.00 mm² each = 700 × 1 = 700 mm² = 700/1000000 = 7/10000

We know that,

  • Pressure = Force/Area

Let's calculate force as we already have area;

  • F = ma
  • F = 55 × 9.8 { Acceleration due to gravity }
  • F = 539 N

Now, if should she would be on 700 nails then pressure will be;

  • P = F/A
  • P = 539/7 × 10000
  • P = 5390000/7
  • P = 770,000 Pascal

And if should would be on a 1 nail only,

  • P = F/A
  • P = 539/1 × 1000000
  • P = 539000000 Pascal

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6 0
2 years ago
Which of the following criteria can help Linda classify gases as greenhouse gases? gas molecules having three or more atoms gas
ratelena [41]
I think the correct answer would be the third  option. The criteria that could help Linda in classifying whether the gases are greenhouse gases would be gas molecules having at least one oxygen atom. Most of the greenhouse gases has an oxygen atom in their structures especially those that naturally occurs. These gases are CO2, H2O vapor and nitrous oxide. 
7 0
3 years ago
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What does Charles Darwin theory of natural selection state about traits
Maslowich
That only the best traits would go through. The weaker trait in the animals would eventually die off.
8 0
3 years ago
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