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3 years ago

7

The circuit shown is used to control the brightness of two identical lamps. The variable resistor is adjusted until the lamps op

erate at their correct voltage of 3.0V
1) When the lamps operate at the correct voltage, the reading on the ammeter is 1.2A. Calculate the current in one lamp.

2 answers:

tamaranim1 [39]3 years ago

6 0

Since the lamps are identical, the current divides equally through them... 0.6A through each lamp.

kati45 [8]3 years ago

3 0

You don't have to worry about voltage drop or the value of the resistor to make that 9 volt drop to happen. The current is divided evenly by the two lamps. If the ammeter reads 1.2 amps before the current goes into the lamps then since this is a parallel circuit, the lamps will divide up the current evenly between them.

That's another way of saying each one will get 1.2 amps /2 = 0.6 amps.

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4 years ago

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Consider a particle with initial velocity v⃗ that has magnitude 12.0 m/s and is directed 60.0 degrees above the negative x axis.

lina2011 [118]

Answer:

-6.0 m/s, 10.4 m/s

Explanation:

To find the x- and y- components, we have to apply the formulas:

$v_x = v cos \theta$

$v_y = v sin \theta$

where

v = 12.0 m/s is the magnitude of the vector

$\theta$ is the angle between the direction of the vector and the positive x-axis

Here, the angle given is the angle above the negative x-axis; this means that the angle with respect to the positive x-axis is

$\theta=180^{\circ} - 60^{\circ} = 120^{\circ}$

So, the two components are:

$v_x = (12.0 m/s) cos 120^{\circ}=-6.0 m/s$

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3 years ago

Refraction occurs through a glass slab why

masha68 [24]

Answer:

One when it enters the glass slab from air and second time when it enters the air through glass slab. When light rays travelling through air enters glass slab, they get refracted and bend towards the normal. Now the direction of refracted ray changes again when it comes out of the glass slab into air.

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3 years ago

When water (H₂O) freezes into ice, some of the properties have changed. What stays the same?

erastova [34]

Answer:

1st one

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Hope This Helps

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2 years ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

<u>Distance of the center from each corners</u> $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

<u>For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:</u>

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

<u>Force by each of the charges at the remaining corners:</u>

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

<u> Now, net electric field in the vertical direction:</u>

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

<u>Now, net electric field in the horizontal direction:</u>

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

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3 years ago