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Studentka2010 [4]
4 years ago
5

Light striking a mirror at a 50° angle will be reflected at an angle _____.

Physics
2 answers:
jasenka [17]4 years ago
7 0

Answer:Actually the answer would be correctly less than 50

Explanation:Because Light is always reflected at an angle equal and opposite to the incoming wave. For example, light striking a mirror at a 45° angle will be reflected in the opposite direction at a 45° angle.

Stells [14]4 years ago
5 0
Equal to 50


law of reflection: angle of incidence equals angle of reflection
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Does the coefficient of kinetic friction depend on the weight of the block
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No it does not because this weight is being balanced by a normal source
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4 years ago
Which occurrence demonstrates dispersion?
katrin2010 [14]
A rainbow. Dispersion is the splitting of radiation into it's different wavelengths.
4 0
3 years ago
Read 2 more answers
A single slit is illuminated by light of wavelengths λa and λb, chosen so that the first diffraction minimum of the λa component
agasfer [191]

Answer:

λ_A = 700 nm ,   m_B = m_a 2

Explanation:

The expression that describes the diffraction phenomenon is

         a sin θ = m λ

where a is the width of the slit, lam the wavelength and m an integer that writes the order of diffraction

a) They tell us that now lal_ A m = 1

         a sin θ = λ_A

coincidentally_be m = 2

          a sin  θ = m λ_b

as the two match we can match

         λ _A = 2 λ _B

         λ_A = 2 350 nm

         λ_A = 700 nm

b)

For lam_B

       a sin  λ_A  = m_B  λ_B

For lam_A

        a sin θ_A = m_ λ_ A

to match they must have the same angle, so we can equal

           m_B  λ_B = m_A  λ_A

           m_B = m_A  λ_A / λ_B

           m_b = m_a 700/350

           m_B = m_a 2

8 0
3 years ago
A fillet weld has a cross-sectional area of 25.0 mm2and is 300 mm long. (a) What quantity of heat (in joules) is required to acc
HACTEHA [7]

Answer:

77362.56 J

163730.28571 J

Explanation:

A = Area = 25 mm²

l = Length = 300 mm

K = Constant = 3.33\times 10^{-6}

\eta = Heat transfer factor = 0.75

f_m = Melting factor = 0.63

T = Melting point of low carbon steel = 1760 K

Volume of the fillet would be

V=Al\\\Rightarrow V=25\times 300\\\Rightarrow V=7500\ mm^3=7500\times 10^{-9}\ m^3

The unit energy for melting is given by

U_m=KT^2\\\Rightarrow U_m=3.33\times 10^{-6}\times 1760^2\\\Rightarrow U_m=10.315008\ J/mm^3

Heat would be

Q=U_mV\\\Rightarrow Q=10.315008\times 7500\\\Rightarrow Q=77362.56\ J

Heat required to weld is 77362.56 J

Amount of heat generation is given by

Q_g=\dfrac{Q}{\eta f_m}\\\Rightarrow Q_g=\dfrac{77362.56}{0.75\times 0.63}\\\Rightarrow Q_g=163730.28571\ J

The heat generated at the welding source is 163730.28571 J

7 0
4 years ago
What are the chances?
Sindrei [870]

Answer:

what ar the chances of what.

5 0
3 years ago
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