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Vedmedyk [2.9K]
3 years ago
10

You have a lens whose focal length is 6.91 cm. You place an object on the axis of the lens at a distance of 11.1 cm from it. How

far is the object\'s image from the lens?
Physics
1 answer:
stellarik [79]3 years ago
7 0

Answer:

4.2591 cm

Explanation:

We have given focal length of the lens f=6.91 cm

Object distance u =11.1 cm

We have to find the image distance that is v

For the lens we know that \frac{1}{f}=\frac{1}{v}-\frac{1}{u}

So \frac{1}{6.91}=\frac{1}{v}-\frac{1}{11.1}

0.1447=\frac{1}{v}-0.09

\frac{1}{v}=0.2347

v=4.2591cm

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What type of wave has a wavelength?<br><br><br>​
Viktor [21]

Answer:

. Radio waves, infrared rays, visible light, ultraviolet rays, X-rays, and gamma rays are all types of electromagnetic radiation.

Explanation:

Radio waves have the longest wavelength, and gamma rays have the shortest wavelength.

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Which type of wave is more dangerous for humans? Microwaves or x-rays?
Dvinal [7]
X- rays. It can damage living tissue
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Write an expression for a transverse harmonic wave that has a wavelength of 2.5 m and propagates to the right with a speed of 13
lyudmila [28]

Answer:

y = 0.14 Cos\left ( 2.512x-34.66t \right )

Explanation:

wavelength, λ = 2.5 m

speed, v = 13.8 m/s

Amplitude, A = 0.14 m

The general equation of the transverse harmonic wave which is travelling right is given by

y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right  )

where, Ф is phase

At t = 0, x = 0 , y = 0.14 m

0.14 = 0.14 Sin Ф

Ф = π/2

So, the equation is

y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right  )

y = 0.14 Cos\left ( 2.512x-34.66t \right )

3 0
3 years ago
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
2 years ago
A stationary police car emits a sound of frequency 1240 HzHz that bounces off of a car on the highway and returns with a frequen
Tju [1.3M]

Answer

given,

frequency from Police car= 1240 Hz

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Using Doppler's effect formula

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u is the speed of sound = 340 m/s

v is the speed of the car

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now, inserting the value of equation (1)

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   2.02822 v = 340 x 0.028822

   2.02822 v = 9.799

   v = 4.83 m/s

hence, the speed of the car is equal to v = 4.83 m/s

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