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Vera_Pavlovna [14]
3 years ago
7

If the total dissipated power is to be reduced by 10%, how much should the voltage be reduced to maintain the same leakage curre

nt? note: power is defined as the product of voltage and current.
Physics
1 answer:
harkovskaia [24]3 years ago
6 0
<span>Since P = V x I, a 10% reduction of power would lead to a 10% reduction in the product of voltage and current. What is left is 90% of the original power: .9P = .9(V x I). If we assume that current must be the same, then we can regroup the terms on the right-hand side as follows: .9P = (.9V) x I In this case, voltage is also reduced by 10% (100% - 90% = 10%).</span>
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A diffraction grating is to be used to find the wavelength of the emission spectrum of a gas. The grating spacing is not known,
Bumek [7]

Answer:

528.9 nm

Explanation:

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First, we find the grating space d = mλ/sinθ where m = 2 for second order, λ = 632.8 nm = 632.8 × 10⁻⁹ m, θ = 43.2°

d = mλ/sinθ = 2 × 632.8 × 10⁻⁹ m ÷ sin43.2° = 1.849 × 10⁻⁶ m = 1.849 μm

We now find the wavelength of the light to be measured from λ = dsinθ/m

Here, θ = 34.9° and m = 2 for second order. So, we have

λ = dsinθ/m = 1.849 × 10⁻⁶ m × sin34.9° ÷ 2 = 0.5289 × 10⁻⁶ m = 528.9 nm

7 0
3 years ago
What is the wavelength of a radio photon from an "am" radio station that broadcasts at 1120 kilohertz?
Alex73 [517]

The wavelength of the radio photon  is 268 m.

Electromagnetic waves travel with the speed of 3×10⁸m/s in vacuum. The speed of a wave <em>c i</em>s related to its frequency<em> f</em> and wavelengthλ as follows:

c=f\lambda

Thus the wavelength is given by,

\lambda =\frac{c}{f}

Substitute 3×10⁸m/s for <em>c</em> and 1120×10³Hz for <em>f</em>.

\lambda =\frac{c}{f}\\ =\frac{3*10^8m/s}{1120*10^3Hz} \\ =267.8 m\\ =268 m

Thus the wavelength of the radio photon from an AM station is 268 m

6 0
3 years ago
Is the formula for velocity the same as speed or different?
a_sh-v [17]

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2 years ago
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xxTIMURxx [149]

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my mind

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A tuba may be treated like a tube closed at one end. If a tuba has a fundamental frequency of 88.4 Hz, determine the first three
yulyashka [42]

Answer with Explanation:

We are given that

Fundamental frequency,f=88.4 Hz

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We have to find the first three overtones.

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Therefore, the overtone of tuba

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For second overtone

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For third overtone

f'=7\times 88.4=618.8Hz

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3 years ago
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