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quester [9]
3 years ago
8

Scientific notation for 0.000468

Physics
1 answer:
Lubov Fominskaja [6]3 years ago
5 0
I think that’s the correct answer

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masya89 [10]
The type of weather is most likely to be raining

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4 years ago
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An installation consists of a 30-kVA, 3-phase transformer, a 480-volt primary, and a 240-volt secondary. Calculate the largest s
Nikitich [7]

Answer:  45 A

Explanation:

Primary only protection 3-phase

I =  3 phase kVA / ( 1.723 * V)

I = 30000 / ( 1.732 * 480 ) = 36.085 A

Table 450.3(B)

Currents of 9A or more column

primary only protection  = 125%

Max OCPD pri = 125% of I = 1.25 * 36.085 = 45.11 A

 Table 450.3(B) Note 1   does not apply, use next smaller Table 240.6(A)

Next smaller = 45 A

5 0
1 year ago
What is an analogy of two different roads or rivers to compare a series and parallel circuit?
faltersainse [42]

Answer:

In a series circuit, the same amount of current flows through all the components placed in it. On the other hand, in parallel circuits, the components are placed in parallel with each other due to which the circuit splits the current flow.

5 0
2 years ago
A scientific theory states that the universe is expanding. Which of these statements is correct about this theory?
Stolb23 [73]
A scientific theory states that the universe is expanding. the statement is correct about this theory is: d. it is a well supported explanation which cannot become law. a theory is very different from a law. a theory explain why something happen.

hope this help
8 0
3 years ago
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X-rays with an energy of 400 keV undergo Compton scattering with a target. If the scattered X-rays are detected at \theta = 30^{
marissa [1.9K]

Answer:

37.91594 keV

Explanation:

E_i = Incident energy = 400 keV

θ = 30°

h = Planck's constant = 4.135×10⁻¹⁵ eV s = 6.626×10⁻³⁴ J s

Incident photon wavelength

\lambda_i=\frac{hc}{E_i}\\\Rightarrow \lambda_i=\frac{4.135\times 10^{-15}\times 3\times 10^8}{400\times 10^3}\\\Rightarrow \lambda_i=3.101\times 10^{-12}\ m

Difference in wavelength

\Delta \lambda=\frac{h}{m_ec}(1-cos\theta)\\\Rightarrow \Delta \lambda=\frac{6.626\times 10^{-34}}{9.11\times 10^{-31}\times 3\times 10^8}(1-cos30)\\\Rightarrow \Delta \lambda=3.248\times 10^{-13}\ m

\lambda_f=\lambda_i+\Delta \lambda\\\Rightarrow \lambda_f=3.101\times 10^{-12}+3.248\times 10^{-13}\\\Rightarrow \lambda_f=3.426\times 10^{-12}

Final photon wavelength

\lambda_f=\frac{hc}{\lambda_f}\\\Rightarrow E_f=\frac{4.135\times 10^{-15}\times 3\times 10^8}{3.426\times 10^{-12}}\\\Rightarrow E_f=362084.06\ eV = 362.08406\ keV

Energy of the recoiling electron

\Delta E=E_i-E_f\\\Rightarrow \Delta E=400-362.08406=37.91594\ keV

Energy of the recoiling electron is 37.91594 keV

8 0
4 years ago
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