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strojnjashka [21]
3 years ago
5

There is less difference between the speed of light in glass and the speed of light in water than there is between the speed of

light in glass and the speed of light in air. Does this mean that a magnifying glass will magnify more or magnify less when it is used under water rather than in air?
Physics
1 answer:
CaHeK987 [17]3 years ago
8 0

Answer: no, the magnifying glass has nothing to do with the speed of light.

Explanation: When light moves between 2 media (refraction), part of it properties that changes during this process is it speed.

The change in speed is dependent on the refractive index of the 2 media and as given by snell's law, the refractive index is inversely proportional to wave speed.

This implies that moving from dense to a more dense medium reduces wave speed and moving from dense to less dense medium increases wave speed.

For the first statement, light moved from glass to water, it implies that it moves from a dense to a less dense medium, it wave speed increases in water.

For the second statement, light moved from glass to air, it implies that it also moves from a dense to a less dense medium and it wave speed will also increase in air.

Looking at both second medium, for the first statement, the second medium is water, and for the second statement, the second medium is air.

Air is less dense that water, hence light travel faster in air than in water.

Thus we can see that the magnification property of a glass has nothing to do with the wave speed, just the refractive indices of the media.

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What is the time constant of a series circuit where the capacitor is 0.330μF and the resistor is 10Ω ?
PtichkaEL [24]

Answer:

\tau=3.3*10^{-6}s

Explanation:

Take at look to the picture I attached you, using Kirchhoff's current law we get:

C*\frac{dV}{dt}+\frac{V}{R}=0

This is a separable first order differential equation, let's solve it step by step:

Express the equation this way:

\frac{dV}{V}=-\frac{1}{RC}dt

integrate both sides, the left side will be integrated from an initial voltage v to a final voltage V, and the right side from an initial time 0 to a final time t:

\int\limits^V_v {\frac{dV}{V} } =-\int\limits^t_0 {\frac{1}{RC} } \, dt

Evaluating the integrals:

ln(\frac{V}{v})=e^{\frac{-t}{RC} }

natural logarithm to both sides in order to isolate V:

V(t)=ve^{-\frac{t}{RC} }

Where the term RC is called time constant and is given by:

\tau=R*C=10*(0.330*10^{-6})=3.3*10^{-6}s

3 0
2 years ago
Which of these events is an example of resonance?
vodka [1.7K]
An opera singer breaks a thin glass with only the use of her high frequency voice
8 0
3 years ago
Read 2 more answers
Infrared observations of the orbits of stars close to the galactic center indicate a small object at the center with a mass of a
MariettaO [177]

Answer:

(4.31±0.38) million Solar masses.

Explanation:

The galactic center is the center of the milky way around which the galaxy rotates. It is most likely the location of a supermassive black hole which has a mass of (4.31±0.38) million Solar masses. The location is called Sagittarius A*.

As there is interstellar dust in our line of sight from the Earth infrared observations need to be taken.

5 0
3 years ago
The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an
9966 [12]

Answer:

-4.941*10^8J.

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

\Delta U = \Delta_{perogee}-\Delta_{Apogee}

Where,

U= \frac{-GmM_e}{r}

Replacing

\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}

\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})

Our values are given by,

m = 85.5Kg

M_e = 5.97*10^{24}Kg

r_A = 7330Km

r_p = 6610Km

G = 6.67*10^{-11}Nm^2/Kg^2

Replacing at the equation,

\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})

\Delta U = -4.941*10^8J

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was -4.941*10^8J.

4 0
3 years ago
When a wave strikes a solid barrier it behaves like a basketball hitting a backboard this wave behavior years called?
Mama L [17]
This behavior is called reflection.
Reflection is a change of in direction of the wave when it reaches another medium. Imagine a wave colliding with a glass in a tank of water.
During reflection, some of the initial energy of the wave is lost.
Waves always reflect with at same angle at which it approached the obstacle.
5 0
3 years ago
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