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3 years ago

10

What is the power of a parallel circuit with a resistance of 1,000 and a current of 0.03 A?

1 answer:

Kobotan [32]3 years ago

7 0

i squared r = 0.03x0.03x1000=3x0.03x10=.9W

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What do nuclear fission and nuclear fusion have in common?

adoni [48]

Answer:

Both fission and fusion are nuclear reactions that produce energy, but the applications are not the same. Fission is the splitting of a heavy, unstable nucleus into two lighter nuclei, and fusion is the process where two light nuclei combine together releasing vast amounts of energy

Explanation:

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3 years ago

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The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he

schepotkina [342]

Answer:

The change in volume is $6.885\times 10^{- 5}\$

Solution:

As per the question:

Coefficient of linear expansion of Copper, $\alpha = 17\times 10^{- 6}\ K^{- 1}$

Initial Temperature, T = $0^{\circ}$ = 273 K

Final Temperature, T' = $100^{\circ}$ = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = $30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}$

$V' = V(1 + \gamma \Delta T)$

$\gamma = 3\alpha$

$V' = V(1 + 3\alpha \Delta T)$

where

V' = Final volume

$V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))$

$\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\$

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3 years ago

Which of the following is an example of an electromagnetic wave ?

Alex

Answer:

Its a wave

Explanation:

THis is one

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3 years ago

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light travels approximately 982,080,000 ft/s, and one year has approximately 32,000,000 seconds. A light year is the distance li

lapo4ka [179]

Answer:

The distance traveled in 1 year is: $3.143*10^{16}ft$

Explanation:

Given

$s = 982,080,000 ft/s$ --- speed

$t = 32,000,000 s$ --- time

Required

The distance traveled

This is calculated as:

$Speed = \frac{Distance}{Time}$

So, we have:

$Distance = Speed * Time$

This gives:

$Distance = 982,080,000 ft/s * 32,000,000 s$

$Distance = 982,080,000 * 32,000,000ft$

$Distance = 3.143*10^{16}ft$ -- approximated

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2 years ago

The refractive index n of transparent acrylic plastic (full name Poly(methyl methacrylate)) depends on the color (wavelength) of

Novosadov [1.4K]

Answer:

The angle between the blue beam and the red beam in the acrylic block is

$\theta _d =0.19 ^o$

Explanation:

From the question we are told that

The refractive index of the transparent acrylic plastic for blue light is $n_F = 1.497$

The wavelength of the blue light is $F = 486.1 nm = 486.1 *10^{-9} \ m$

The refractive index of the transparent acrylic plastic for red light is $n_C = 1.488$

The wavelength of the red light is $C = 656.3 nm = 656.3 *10^{-9} \ m$

The incidence angle is $i = 45^o$

Generally from Snell's law the angle of refraction of the blue light in the acrylic block is mathematically represented as

$r_F = sin ^{-1}[\frac{sin(i) * n_a }{n_F} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_F = sin ^{-1}[\frac{sin(45) * 1 }{ 1.497} ]$

$r_F = 28.18^o$

Generally from Snell's law the angle of refraction of the red light in the acrylic block is mathematically represented as

$r_C = sin ^{-1}[\frac{sin(i) * n_a }{n_C} ]$

Where $n_a$ is the refractive index of air which have a value of $n_a = 1$

So

$r_C = sin ^{-1}[\frac{sin(45) * 1 }{ 1.488} ]$

$r_F = 28.37^o$

The angle between the blue beam and the red beam in the acrylic block

$\theta _d = r_C - r_F$

substituting values

$\theta _d = 28.37 - 28.18$

$\theta _d =0.19 ^o$

4 0

3 years ago