What is refractive index?

Answers

Answer 1

Answer: The refractive index of a substance is the ratio . . .

(speed of light in vacuum) divided by (speed of light in the substance) .

So the speed of light in a substance is

(299,792,458 meters per sec) divided by (the refractive index of the substance).

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walking converts what type of energy into mechanical energy? A) chemical B) electromagnetic C)nuclear D)thermal

Answers

it converts chemical energy into kinetic/mechanical energy

Answer:

A-Chemical Energy

Explanation:

Regan has an empty glass cup. it has a mass of 0.3 kg. she drops the cup on the ground and it shatters into several pieces. if she collected all of the pieces of the cup, how much mass would all of the pieces have combined?

Answers

0.3, because its the same thing, just in pieces.

An article in a magazine denies that Earth is warming and climate change is occurring. The writer reports that this season was the snowiest winter on record for the Northeast, with lower than average temperatures for the region. Because of the low temperatures, the writer concludes that climate change cannot possibly be occurring. What question should you ask the writer to clarify whether his stance on climate change is scientifically sound?(A) Did the snowstorms cause extensive damage?
(B) Was this the coldest month on record for this region?
(C) Were the snowfall totals of each storm larger or smaller than last year?
(D) Were long-term temperature changes investigated in addition to short-term change?

Answers

D.
This is likely to challenge the writer's stance because one piece of data cannot accurately depict a long-term trend.`

Answer:

Explanation:

The excitation of an electron on the surface of a photocell required 5.0 x 10 –27 J of energy. Calculate the wavelength of light that was needed to excite the electron in an atom on the photocell

Answers

c = speed of light in vacuum = about 3 x 10⁸ meters/second
h = Planck's Konstant = 6.63 x 10⁻ ³⁴ joule-second

Energy = (h x frequency) = (h c / wavelength)
Wavelength = (h c) / (energy)

Wavelength = (6.63 x 10^-34 joule-sec x 3 x 10^8 meter/sec) / (5 x 10^-27 joule)

= 19.89 x 10^-26 / 5 x 10^-27 = 39.78 meters

This is an astonishing result ! Simply amazing. That wavelength corresponds
to a frequency of about 7.54 MHz, in one of the short-wave radio bands used by
a lot of foreign-broadcast stations.

If the number in the problem is correct, it means that this 'photocell' responds
to any electromagnetic signal at 7.54 MHz or above ... short-wave radio,
commercial FM or TV signals, FRS walkie-talkies, garage-door openers,
Bluetooth thingies, home WiFi boxes, WiFi from a laptop, microwave ovens,
cellphones, any signal from a satellite, any microwave dish, any heat lamp,
flashlight, LED, black light, or X-ray machine. Some "photocell" !

I'm thinking the number given in the problem for the energy of a photon
at the detection threshold of this device must be wrong by several orders
of magnitude.

(But my math is still bullet-proof.)

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v sin θ / g 1. The same object is then raised again to the same height h but this time is thrown downward with velocity υ1 It now reaches the ground with a new velocity υ2. How is v2 related to v1?

Answers

Answer:

$v^(2)_(2)=v^(2)_(1)+(v^(2)Sin^(2)\theta )/(g^(2))$

Explanation:

Case I:

initial velocity, u = 0 m/s

Final velocity, v' = v Sinθ /g

Height = h

acceleration = g

Use third equation of motion, we get

$v'^(2)=u^(2)+2as$

$\left ( (vSin\theta )/(g) \right )^(2)=0^(2)+2gh$

$h = (v^(2)Sin^(2)\theta )/(2g^(3))$ . ... (1)

Case II:

initial velocity, u = v1

Final velocity, v = v2

height = h

acceleration due to gravity = g

Use third equation of motion, we get

$v^(2)=u^(2)+2as$

$v^(2)_(2)=v^(2)_(1)+2gh$

Substitute the value of h from equation (1) ,we get

$v^(2)_(2)=v^(2)_(1)+2g(v^(2)Sin^(2)\theta )/(2g^(3))$

$v^(2)_(2)=v^(2)_(1)+(v^(2)Sin^(2)\theta )/(g^(2))$

A 76.5 kg cross-country skier skiing on unwaxed skis along dry snow at a constant speed of 4.00 m/s experiences a force of friction of -60.0 N. What is the coefficient of friction between unwaxed skis and dry snow?a. 0.08
b. 0.78
c. 1.28
d. 12.5

Answers

The force of friction = (weight) x (coefficient of friction)

Skier's weight = (mass) x (gravity) = (76.5) x (9.8) = 749.7 N

Force of friction = (749.7) x (coefficient of friction) = 60.0 N

Coefficient of friction = 60 / 749.7 = 0.08 (rounded)

Choice-'A' is the closest choice offered.

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