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OLga [1]
3 years ago
14

PLEASE HELP ME WITH THIS ONE QUESTION

Physics
1 answer:
melamori03 [73]3 years ago
3 0

Explanation:

First, we convert the energy from eV to Joules:

2.90\:\text{eV}×\left(\dfrac{1.6×10{-19}\:J}{1\:\text{eV}} \right)

= 4.64×10^{-19}\:\text{eV}

We know from definition that

E=h\nu = \dfrac{hc}{\lambda}

so the wavelength of the photon is

\lambda = \dfrac{hc}{E} = 4.28×10^8\:\text{m}

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The inductance of a solenoid with 450 turns and a length of 24 cm is 7.3 mH. (a) What is the cross-sectional area of the solenoi
Novay_Z [31]

Answer: 0.43 V

Explanation:

L = [μ(0) * N² * A] / l

Where

L = Inductance of the solenoid

N = the number of turns in the solenoid

A = cross sectional area of the solenoid

l = length of the solenoid

7.3*10^-3 = [4π*10^-7 * 450² * A] / 0.24

1.752*10^-3 = 4π*10^-7 * 202500 * A

1.752*10^-3 = 0.255 * A

A = 1.752*10^-3 / 0.255

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emf = -N(ΔΦ/Δt).........1

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6 0
3 years ago
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