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1 year ago

A light wave moves from diamond (n= 2.4) into water (n= 1.33) at an angleof 24°. what angle does it have in water?

Physics

1 answer:

worty [1.4K]1 year ago

3 0

Answer:

n1 sin θ1 = n2 sin θ2 Snell's Law (θ1 is the angle of incidence)

sin θ2 = n1 / n2 * sin θ1

sin θ2 = 2.4 / 1.33 * sin θ1

sin θ2 = 1.80 * .407 = .734

θ2 = 47.2 deg

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What are the density, specific gravity and mass of the air in a room whose dimensions are 4 m * 6 m * 8 m at 100 kPa and 25 C.

Volgvan

Answer:

Density = 1.1839 kg/m³

Mass = 227.3088 kg

Specific Gravity = 0.00118746 kg/m³

Explanation:

Room dimensions are 4 m, 6 m & 8 m. Thus, volume = 4 × 6 × 8 = 192 m³

Now, from tables, density of air at 25°C is 1.1839 kg/m³

Now formula for density is;

ρ = mass(m)/volume(v)

Plugging in the relevant values to give;

1.1839 = m/192

m = 227.3088 kg

Formula for specific gravity of air is;

S.G_air = density of air/density of water

From tables, density of water at 25°C is 997 kg/m³

S.G_air = 1.1839/997 = 0.00118746 kg/m³

4 0

2 years ago

The wavelength of red light is 7 x 10^-7 meter. Express this value in nanometers

3241004551 [841]

The wavelength of the red light in "nanometer" is 7× $10^{2}$

Wavelength is given as : 7× $10^{-7}$ meter

1 nanometer = ( $10^{-9}$ meter)

Let X= value of the wavelength in nanometer.

1 nanometer = $10^{-9}$ meter

X nanometer = 7× $10^{-7}$ meter

<em>If we Cross multiply</em>

X nanometer = ( $\frac{ 7* 10^{-7} }{ 10^{-9} }$ )

X= 7× $10^{2}$ nanometer

Therefore, the wavelength in "nanometer" is 7× $10^{2}$

Learn more at :brainly.com/question/12924624?referrer=searchResults

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2 years ago

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

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2 years ago

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djyliett [7]

heya mate

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VashaNatasha [74]

Answer:

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2 years ago