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Elis [28]
2 years ago
7

Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?

provide the solution: __________ x 10-7 m
Physics
1 answer:
postnew [5]2 years ago
7 0

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

            c = speed of light

            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

Learn more about light here;

brainly.com/question/15200315

#SPJ4

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DENIUS [597]

Answer:

Magnitude of induced emf will be 87.5 volt

Direction of induced emf will be clockwise  

Explanation:

We have given number of turns in the coil N = 50

Initial area A=0.140m^2

Time is given dt = 0.1 sec

Magnetic field B = 1.25 T

From Faraday's law of electromagnetic induction

ne=-N\frac{d\Phi }{dt}=-NA\frac{dB}{dt}

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7 0
4 years ago
You have a reservoir held at a constant temperature of –30°C. You add 400 J of heat to the reservoir. If you have another reserv
marysya [2.9K]

Answer:

449.38 J

Explanation:

ΔS = ΔQ/T

Where ΔS = entropy change

Q = quantity of heat

T = temperature

First reservoir :

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Q = 400 J

Second reservoir :

T = 0°C = 273K

Q =?

To have same increase in entropy for both reservoirs :

Q/T of first reservoir = Q/T of second reservoir

400/243 = Q/273

243 * Q = 400 * 273

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Q = 109,200 / 243

Q = 449.38271

Q = 449.38 J

6 0
4 years ago
Which plot correctly shows the velocity of the two balls as a function of time?
yanalaym [24]
The given in this problem is that two balls are thrown at different times, different heights and velocities. A blue ball is thrown upward at a specific velocity at a lower altitude while a red ball is thrown downwards at a specific speed and at a higher height. In this case, we are asked here to describe the graph of the behavior of the balls as a function of time. The x-axis then is time while the y-axis is the velocity of the ball. The blue ball has a quadratic function while the red ball is more or less exponential. See the attached figure for reference.

5 0
4 years ago
A ray diagram without the produced image is shown.
4vir4ik [10]

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<u> The image produced by the lens is (b) inverted and real</u>

Explanation:

A real image occurs where the rays converge.

Real images can be produced both by the  concave mirrors or  converging lenses, but the condition is that the object of consideration is always  placed far  away from the mirror or the lens than the focal point, and thus the  real image produced is inverted.

A car acting as an object in front of a biconvex lens between F and 2 F on the object side of the lens. There is a light ray parallel to the principal axis that is bent through F on the image side of the lens. There is a ray straight through the center of the lens. The rays intersect below the x axis further than 2 F away from the lens and farther from the principal axis than the object is tall.

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8 0
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Both would go 15N. Hoped I helped!

7 0
3 years ago
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