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3 years ago

A beam of red and blue light in air is incident at 30º on a glass surface. Which is true?

Physics

1 answer:

Alex17521 [72]3 years ago

4 0

Answer:

a. The blue light is bent toward the normal more than the red light.

Explanation:

When light travels from a less dense medium, in this case air into a denser medium (glass) it is bent towards the normal because of the reduction in speed due to the dense medium.In this scenario we have the same incident angle of 30º. therefore, we consider the wavelength of the light because at shorter wavelengths we have more refraction occurring as compared to longer wavelengths. We know that Blue light has a shorter wavelength as compared to red light, which means it will bend toward the normal more than the red light.

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A magnetic field is uniform over a flat, horizontal circular region with a radius of 1.50 mm, and the field varies with time. In

Daniel [21]

Answer: $81.57\mu V$

Explanation:

Given

radius of circular region r=1.50 mm

$A=\pi r^2=7.069\times 10^{-6}\ m^2$

Magnetic Field $B=1.50\ T$

time t=130 ms

Flux is given by

$\phi =B\cdot A$

change in Flux $d\phi =(B_f-B_i)A$

Emf induced is $e=\frac{\mathrm{d} \phi}{\mathrm{d} t}$

$e=\frac{(1.5)\cdot 7.069\times 10^{-6}}{130\times 10^{-3}}$

$e=81.57 \mu V$

3 0

3 years ago

Two charged bees land simultaneously on flowers that are separated by a finite distance. For a few moments, the charged bees res

damaskus [11]

Answer:

Same, the electric fields point in opposite directions and therefore cancel at some midpoint.

Explanation:

The Electric field net at some point between them is zero, only if they point in opposite direction (they cancel to the each other). In order the electric fields have opposite direction, at some point between the bees , the bees must have the same sign of electric charge

3 0

3 years ago

Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion

stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas = 240 K

= initial volume of gas =

= final volume of gas =

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0

2 years ago

Consider three planets. All have the same mass as Earth, but with different radii (from largest to smallest: Planet 1, 2, 3). Fo

LuckyWell [14K]

Answer:

option C

Explanation:

given,

mass of the three planet is same

radius of the planets are

R₁ > R₂ > R₃

expression of escape velocity

$v = \sqrt{\dfrac{2GM}{R}}$

G is the gravitational constant

M is the mass of the planet

R is the radius of the planet

from the above expression we can clearly conclude that the escape velocity is inversely proportional to the radius of the Planet.

radius of planet increases escape velocity decreases.

Hence planet 3 has the smallest radius so the escape velocity of the third planet will be maximum.

The correct answer is option C

3 0

3 years ago

A 0.89 m aqueous solution of an ionic compound with the formula mx has a freezing point of -3.0 ∘c . van't hoff factor?

BigorU [14]

Answer is: V<span>an't Hoff factor (i) for this solution is 1,81.
Change in freezing point from pure solvent to solution: ΔT =i · Kf · b.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
b - molality, moles of solute per kilogram of solvent.
</span><span>b = 0,89 m.
ΔT = 3°C = 3 K.
i = </span>3°C ÷ (1,86 °C/m · 0,89 m).

i = 1,81.

5 0

3 years ago