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Sergeeva-Olga [200]
2 years ago
6

A long, thin rod parallel to the y-axis is located at x = -1.0 cm and carries a uniform linear charge density of +1.0 nC/m. A se

cond long, thin rod parallel to the z-axis is located at x = +1.0 cm and carries a uniform linear charge density of -1.0 nC/m.

Physics
1 answer:
Bad White [126]2 years ago
8 0

Answer:

Hhere is the other part of the question ; What is the net electric field due to these rods at the origin? (epsilon not = 8.85*10^-12)

net electric field = 3600N/C

Explanation:

The detailed calculation is as shwon below

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If are spaced closely together on the map,there is a drastic temperature change over the distance
AlexFokin [52]

If <em>the isotherms</em> are spaced closely together over some portion of the map, there is a drastic temperature change over that portion.

6 0
3 years ago
Read 2 more answers
Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0
3 years ago
An aluminum wing on a passenger jet is 35 m long when its temperature is 17°C. At what temperature would the wing be 3 cm (0.03
Mnenie [13.5K]

Answer:

53.32°C

Explanation:

Length of the aluminium wing = 35 m

Change in length of aluminium wing = 0.03 m

The linear expansion coefficient of aluminium \alpha =23.6\times 10^{-6}/^{\circ}C

We know that change in length is given by \Delta L=L\alpha \Delta T

So 0.03=35\times 23.6\times 10^{-6}\Delta T

\Delta T=36.32^{\circ}C

So final temperature =T_I+\Delta T=17+36.32=53.3196^{\circ}C

5 0
3 years ago
Chicago Bears rookie running back Tarik Cohen stands 5 feet and 6 inches tall, which is 66 total inches. How many meters tall is
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1.6764 meters. You can also google inches to meters for your answer.

5 0
3 years ago
The charge density of a uniformly charged disk 0.420 m in diameter is 2.92 ✕ 10−2 C/m2. What is the magnitude of the electric fi
iragen [17]

Answer:

E = {(Charge Density/2e0)*(1 - [z/(sqrt(z^2 - R^2))]}

R is radius = Diameter/2 = 0.210m.

At z = 0.2m,

Put z = 0.2m, and charge density = 2.92 x 10^-2C/m2, and constant value e0 in the equation,

E can be calculated at distance 0.2m away from the centre of the disk.

Put z = 0.3m and all other values in the equation,

E can be calculated at distance 0.3m away from the centre of the disk

3 0
2 years ago
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