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Mrac [35]
4 years ago
9

An unpolarized light shines on an object that is placed below a crystal. When an observer looks at the object through the crysta

l there are two images present. Explain how that can happen using the information learned in optics and e-m radiation
Physics
1 answer:
slavikrds [6]4 years ago
8 0

Answer:

See explanation

Explanation:

Solution:-

- light reflected through calcite crystals difference occurs due to the polarity of the light passing through the crystal. Daylight is composed of light vibrating in all planes, whereas reflected light is often restricted to a single plane that is parallel to the surface from which the light is reflected.

- Light that is reflected from the flat surface of a dielectric (or insulating) material is often partially polarized, with the electric vectors of the reflected light vibrating in a plane that is parallel to the surface of the material.

- light waves that have the electric field vectors parallel to the surface are reflected to a greater degree than those with different orientations.

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It is 6.00 km from your home to the physics lab. as part of your physical fitness program, you could run that distance at 10.0 k
kirill115 [55]
<span>1.) It is 6.00km from your home to the physics lab. As part of your physical fitness program, you could run that distance at 10.0km/hr (which uses up energy at the rate of 700W ), or you could walk it leisurely at 3.00km/hr (which uses energy at 290 W). A.)Which choice would burn up more energy? running or walking? b.)How much energy (in joules) would it burn? c.)Why is it that the more intense exercise actually burns up less energy than the less intense one? Follow 2 answers Report Abuse Answers billrussell42 Best Answer: running, at 10 km/hour for 6 km is 6 km / 10 km/hour = 0.6 hour or 36 min energy used is 700 watts or 700 joules/s x 36 min x 60s/min = 1.512e6 joules or 1.5 MJ walking, at 3 km/hour for 6 km 6 km / 3 km/hour = 2 hour or 120 min energy used is 290 watts or 290 joules/s x 120 min x 60s/min = 1.872e6 joules or 1.8 MJ C) should be obvious PS, this has nothing to do with potential energy. billrussell42 · 5 years ago 0 Thumbs up 1 Thumbs down Report Abuse Comment Simon van Dijk I assume the watt consumption is per hour. Then running 6km at 10.0 km/h results in 700*6/10 = 420 w.h and walking in 290*6/3 = 580 w.h So walking would burn up more energy (kwh) b) 1 kilowatt hour = 3 600 000 joules so 420 wh = 0.42 kwh = 1.51.10^6 joule c) when you put more effort in making the distance your energy is used more efficient. Simon van Dijk · 5 years ago 0 Thumbs up 2 Thumbs down Report Abuse Comment</span>
7 0
3 years ago
A 13.5 μF capacitor is connected to a power supply that keeps a constant potential difference of 22.0 V across the plates. A pie
-BARSIC- [3]

a) 3.27\cdot 10^{-3} J

b) 11.60\cdot 10^{-3} J

c) 8.33\cdot 10^{-3} J

Explanation:

a)

The energy stored in a capacitor is given by

U=\frac{1}{2}CV^2

where

C is the capacitance of the capacitor

V is the potential difference across the plates of the capacitor

For the capacitor in this problem, before insering the dielectric, we have:

C=13.5 \mu F = 13.5\cdot 10^{-6}F is its capacitance

V = 22.0 V is the potential difference across it

Therefore, the initial energy stored in the capacitor is:

U=\frac{1}{2}(13.5\cdot 10^{-6})(22.0)^2=3.27\cdot 10^{-3} J

b)

After the dielectric is inserted into the plates, the capacitance of the capacitor changes according to:

C'=kC

where

k = 3.55 is the dielectric constant of the material

C is the initial capacitance of the capacitor

Therefore, the energy stored now in the capacitor is:

U'=\frac{1}{2}C'V^2=\frac{1}{2}kCV^2

where:

C=13.5\cdot 10^{-6}F is the initial capacitance

V = 22.0 V is the potential difference across the plate

Substituting, we find:

U'=\frac{1}{2}(3.55)(13.5\cdot 10^{-6})(22.0)^2=11.60\cdot 10^{-3} J

C)

The initial energy stored in the capacitor, before the dielectric is inserted, is

U=3.27\cdot 10^{-3} J

The final energy stored in the capacitor, after the dielectric is inserted, is

U'=11.60\cdot 10^{-3} J

Therefore, the change in energy of the capacitor during the insertion is:

\Delta U=11.60\cdot 10^{-3}-3.27\cdot 10^{-3}=8.33\cdot 10^{-3} J

So, the energy of the capacitor has increased by 8.33\cdot 10^{-3} J

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3 years ago
How is an object’s position in the solar system affect its motion and temperature?
SCORPION-xisa [38]

Answer:

it will be base on the climate change.

Explanation: so let just say tomorrow it will be cold so the atmosphere will circle on the coolest atmosphere away from the sun.

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3 years ago
Why might political parties prefer closed primaries to open primaries
sweet [91]

In a closed primary, only voters registered with a given party can vote in that party's primary.

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3 years ago
1The density of an of an object with a volume of 60.0 and mass of 400.0g is______
Kobotan [32]
D is the answer
To get the density you divide mass by volume
So the equation is 400/60=d
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3 years ago
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