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Firlakuza [10]
2 years ago
5

Light of frequency 2.5 x 1015 Hz illuminates a

Physics
1 answer:
Nikitich [7]2 years ago
3 0

Answer:

h f = W + KE

Input energy equals work function plus KE of emitted electron

W = 6.63E-34 * 2.5E15 - 6.3 * 1.6E-19

W = 6.63 * 2.5 * 10^-19 - 10.1 * E-19 ev     (1ev = 1.6E-19 J)

W = (16.6 - 10.1)E-19 = 6.5E-19 J

h f = 6.5E-19 J        for electrons to be emitted with zero KE

f = 6.5E-19 / 6.63E-34 = .98E-15 / sec = 9.8E-14 / sec  (threshold)

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Answer:

Option B

Explanation:

Magnification of Microscope is  

M = M_o \times M_e  

Mo= Magnification of objective lens and  

Me= magnification of the eyepiece.  

Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.  

Magnification,  

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when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.

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The correct answer is Option B

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3 years ago
PLEASE HELP ME WITH THIS PROBLEM
valentinak56 [21]

1) The mass of the continent is 2.13\cdot 10^{21} kg

2) The kinetic energy of the continent is 274.8 J

3) The speed of the jogger must be 2.76 m/s

Explanation:

1)

The continent is a slab of side 5900 km (so the surface is 5900 x 5900, assuming it is a square) and depth 26 km, therefore its volume is:

V=(36)(4600)^2=7.62\cdot 10^8 km^3 = 7.62\cdot 10^{17} m^3

The mass of the continent is given by

m=\rho V

where:

\rho = 2790 kg/m^3 is its density

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Substituting, we find the mass:

m=(2790)(7.62\cdot 10^{17})=2.13\cdot 10^{21} kg

2)

To find the kinetic energy, we need to convert the speed of the continent into m/s first.

The speed is

v = 1.6 cm/year

And we have:

1.6 cm = 0.016 m

1 year = (365)(24)(60)(60)=3.15\cdot 10^7 s

So, the speed is

v=\frac{0.016 m}{3.15 \cdot 10^7 s}=5.08\cdot 10^{-10}m/s

Now we can find the kinetic energy of the continent, which is given by

K=\frac{1}{2}mv^2

where

m=2.13\cdot 10^{21} kg is the mass

v=5.08\cdot 10^{-10}m/s is the speed

Substituting,

K=\frac{1}{2}(2.13\cdot 10^{21})(5.08\cdot 10^{-10})^2=274.8 J

3)

The jogger in this part has the same kinetic energy of the continent, so

K = 274.8 J

And its mass is

m = 72 kg

We can write his kinetic energy as

K=\frac{1}{2}mv^2

where

v is the speed of the man

And solving the equation for v, we find his speed:

v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(274.8)}{72}}=2.76 m/s

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Echo is the property of the sound wave by the virtue of which the sound wave reflects back to the source of the sound after hitting a surface or an object.

Ultrasonic images are obtained from inside organs of our body. This process involves the use of ultrasonic sound waves that have a frequency greater than 20,000 Hz. These sound waves are out of the range of audible sound by the human ear. When these ultrasonic sound waves are sent in form of pulses into the human body by the use of probes, they reflect back from the tissues of different organs to the probe. The probe then records the reflection properties of these sound waves and displays them in form of an image, known as ultrasonic images.

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