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2 years ago

6

conductors of resistance 2 ohm 3 ohm and 6 ohm are connected (1) in series and (2) in parallel. calculate the equivalent resista

nce of each combination

1 answer:

Georgia [21]2 years ago

4 0

Answer
For conductors in series 1/2+1/3+1/6=1
And in parallel 2+3+6= 11
For the combination of series and parallel you get 2+11=13
Explanation:

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True or false in order for light to be emitted electrons must move from high energy state to a lower energy state?

lara31 [8.8K]

Answer:

B

I think it is false.

It does not move to lower energy state.

5 0

3 years ago

Read 2 more answers

A cube of wood 15.0 cm on each side is tied to the bottom of a tank filled with water to a depth of 50 cm. The tension in the st

Elina [12.6K]

Answer:

(a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

Explanation:

Given that,

Side of cube = 15.0 cm

Depth = 50 cm

Tension = 6.615 N

We need to calculate the volume of the wood

Using formula of volume

$V = a^3$

$V=(15.0\times10^{-2})^3$

$V=0.003375\ m^3$

We need to calculate the density of the wood

Using buoyant force

$\rho_{w}gh=mg+T$

$\rho_{w}gh=\rho_{c}Vg+T$

Put the value into the formula

$\rho_{c}=\dfrac{\rho_{w}gh-T}{Vg}$

Put the value into the formula

$\rho_{c}=\dfrac{1000\times9.8\times50\times10^{-2}-6.615}{0.003375\times9.8}$

$\rho_{c}=1479.48\times10^{2}\ Kg/m^3$

(b). We need to calculate the absolute pressure at the bottom of the tank

Using formula of absolute pressure

$P=P_{atm}+\rho gh$

Put the value into the formula

$P=1.013\times10^{5}+1000\times9.8\times0.5$

$P=1.06200\times10^{5}\ Pa$

Hence, (a). The density of the wood is $1479.48\times10^{2}\ Kg/m^3$

(b). The absolute pressure at the bottom of the tank is $1.06200\times10^{5}\ Pa$ .

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3 years ago

The formation of volcanos and mountain ranges can be explained by the theroy of _____

Alenkasestr [34]

It can be explained by the theory of Plate Tectonics.

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3 years ago

A ray diagram for a refracted light ray is shown.

Rina8888 [55]

The dashed line represents the<u> normal </u>to the boundary

Explanation:

Refraction is a phenomenon that occurs when a ray of light crosses the interface between two mediums.

When this occurs, the ray of light changes direction and speed.

The new direction of the ray of light is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where

$n_1$ is the index of refraction of the 1st medium

$n_2$ is the index of refraction of thre 2nd medium

$\theta_1$ is the angle of incidence, which is the angle between the direction of the incident ray and the normal to the boundary

$\theta_2$ is the angle of refraction, which is the angle between the direction of the refracted ray and the normal to the boundary

The index of refraction of a medium is the ratio between the speed of light in a vacuum (c) and the speed of light in that medium (v):

$n=\frac{c}{v}$

The figure in the question is missing, however you can find it in attachment.

We see that the dashed line is perpendicualr to the boundary between the two mediums, so it represents the normal.

So, the correct answer is

Normal

We note that in this case the ray of light bends towards the normal, this means that $\theta_2$ , therefore the second medium has a larger index of refraction ( $n_2>n_1$ ).

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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3 years ago

The distance between the first and fifth minima of a single-slit diffraction pattern is 0.500 mm with the screen 37.0 cm away fr

WARRIOR [948]

In the single-slit experiment, the displacement of the minima of the diffraction pattern on the screen is given by
$y_n= \frac{n \lambda D}{a}$ (1)
where
n is the order of the minimum
y is the displacement of the nth-minimum from the center of the diffraction pattern
$\lambda$ is the light's wavelength
D is the distance of the screen from the slit
a is the width of the slit

In our problem,
$D=37.0 cm=0.37 m$
$\lambda=530 nm=5.3 \cdot 10^{-7} m$
while the distance between the first and the fifth minima is
$y_5-y_1 = 0.500 mm=0.5 \cdot 10^{-3} m$ (2)

If we use the formula to rewrite $y_5, y_1$ , eq.(2) becomes
$\frac{5 \lambda D}{a} - \frac{1 \lambda D}{a} =\frac{4 \lambda D}{a}= 0.5 \cdot 10^{-3} m$
Which we can solve to find a, the width of the slit:
$a= \frac{4 \lambda D}{0.5 \cdot 10^{-3} m}= \frac{4 (5.3 \cdot 10^{-7} m)(0.37 m)}{0.5 \cdot 10^{-3} m}= 1.57 \cdot 10^{-3} m=1.57 mm$

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3 years ago