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3 years ago

12

Help me?!?!?!?!?!?!?!?!?!?!?!?!?!

2 answers:

fredd [130]3 years ago

8 0

Before you can pick the correct one, you have to review in your mind what
you know about the electric field. (Or what you SHOULD know about it.)

-- The field lines show where a teeny tiny POSITIVE charge
wants to go when you put it down anywhere.

-- A teeny tiny POSITIVE charge always wants to get AWAY from other
positive charges, and it wants to go TOWARD negative charges.

That's all you need. Now let's look at the pictures.
I want to look at them from the bottom up:

(D). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
the negative (blue) charge and TOWARD the positive (red) charge.
That's wrong.

(C). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
any negative charges. That's wrong.

(B). ... No.
The arrows show that the tiny positive test charge wants to go AWAY from
BOTH the negative charge AND the positive charge. That's wrong.

(A). ... YES !
The arrows show that the tiny test charge wants to go away from the positive
charge, and toward the negative charge. That's exactly right.

(That's why I wanted to do them backwards ... because
the first picture on the page is the correct one.)

marishachu [46]3 years ago

3 0

The correct field line would be A.

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Vsevolod [243]

Answer:

a The diagram of the situation is shown on the first uploaded image

b the angle of incidence beam striking the water is $\theta = 49.63^o$

c the angle of refraction beam striking the water is $r = 59.7^o$

d The angle the refracted beam make with respect to the horizontal is $= 30.3^o$

e The height of the target above sea level is

$h= 125.05m$

Explanation:

From the diagram we see that the angle of the beam striking the water is

$tan \theta = \frac{100}{85}$

$\theta = tan^{-1}(\frac{100}{85} )$

$= 49.63^o$

According to Snell's law

$\mu_{water} *sin(i) = \mu_{air} *sin(r)$

Where $\mu_{water }$ is the refractive index of water = 1.333

$i$ is the angle of incidence

$\mu_{air}$ is the refractive index of air = 1

r is the angle of refraction

Substituting values accordingly

$1.33 * sin (40.37) = 1 * sin(r)$

Making r the subject of the formula

$r = sin^{-1}(\frac{1.333 *sin(40.37)}{1})$

$= 59.7^o$

looking at the diagram we can see that to obtain the angle the refraction beam makes with the horizontal by subtracting the angle refraction from 90°

i.e $90 -59.7 = 30.3$ °

From the diagram we see that the height target above sea level can be obtained by this relation

$tan \theta = \frac{h}{214}\\$

Where h is the is the height

$tan(30.3) = \frac{h}{214}$

$h = 214 * tan (30.3)$

$=125.05m$

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