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Bond [772]
3 years ago
5

When a ray of light passes at an angle to the normal from one material into another material in which its speed is higher?

Physics
1 answer:
Gnesinka [82]3 years ago
4 0
I think your talking about refraction and if so the first medium it existed in before entering the second usually has the higher speed especially it the first medium is less denser than the second
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Conduction can only happen when objects are in:
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Answer:

When it’s gravitational

Explanation:

I’m pretty sure that’s the answer

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n ultraviolet light beam having a wavelength of 130 nm is incident on a molybdenum surface with a work function of 4.2 eV. How f
pashok25 [27]

Answer:

The speed of the electron is 1.371 x 10⁶ m/s.

Explanation:

Given;

wavelength of the ultraviolet light beam, λ = 130 nm = 130 x 10⁻⁹ m

the work function of the molybdenum surface, W₀ = 4.2 eV = 6.728 x 10⁻¹⁹ J

The energy of the incident light is given by;

E = hf

where;

h is Planck's constant = 6.626 x 10⁻³⁴ J/s

f = c / λ

E = \frac{hc}{\lambda} \\\\E = \frac{6.626*10^{-34} *3*10^{8}}{130*10^{-9}} \\\\E = 15.291*10^{-19} \ J

Photo electric effect equation is given by;

E = W₀ + K.E

Where;

K.E is the kinetic energy of the emitted electron

K.E = E - W₀

K.E = 15.291 x 10⁻¹⁹ J - 6.728 x 10⁻¹⁹ J

K.E = 8.563 x 10⁻¹⁹ J

Kinetic energy of the emitted electron is given by;

K.E = ¹/₂mv²

where;

m is mass of the electron = 9.11 x 10⁻³¹ kg

v is the speed of the electron

v = \sqrt{\frac{2K.E}{m} } \\\\v =  \sqrt{\frac{2*8.563*10^{-19}}{9.11*10^{-31}}}\\\\v = 1.371 *10^{6} \ m/s

Therefore, the speed of the electron is 1.371 x 10⁶ m/s.

8 0
3 years ago
In a series circuit, one of the resistors is replaced with a resistor having a lower resistance value. how does this affect the
Assoli18 [71]
Current will increase. 
8 0
3 years ago
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Which diagram represents a closed circuit with the least resistance?
solniwko [45]

Answer:Circuit B is the answer

Explanation:

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3 years ago
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Consider a sealed 20 cm high electronic box whose base dimensions are 40cm x 40cm placed in a vacuum chamber. The emissivity of
Genrish500 [490]

Answer:

T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

Explanation:

The formula we are going to use is:

\dot Q_{rad}=\epsilon\sigma A_s(T_s^4-T_{surr}^4)

Where:

ε is the emissivity

σ is the Stefan constant

T_s is the final temperature of surrounding surfaces

T_{surr} is the required temperature

A_s is the are of surrounding surface

Calculating The area:

A_s=(0.4)(0.4)+4(0.4)(0.2)\\A_s=0.48\ m^2

σ= 5.67*10^{-8}\ W/m^2.K^4

ε =0.95

T_s=55+273

T_s=328 K

\dot Q_{rad=100 W

100=0.95(5.67*10^{-8})(0.48)(328^4-T_{surr}^4)\\3867693926=(328^4-T_{surr}^4)\\T_{surr}^4=7706623130\\T_{surr}=296.289\ K

In Celsius:

T_{surr}=296.289-273\\T_{surr}=23.289^oC

8 0
3 years ago
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