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Artemon [7]
2 years ago
10

Which components of static electricity does a photocopier use?

Physics
1 answer:
Salsk061 [2.6K]2 years ago
8 0

Answer:

For a photocopier to work, a field of positive charges must be generated on the surface of both the drum and the copy paper. These tasks are accomplished by the corona wires. These wires are subjected to a high voltage, which they subsequently transfer to the drum and paper in the form of static electricity.

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Priya lifts a box of apples from the ground to above her head. How was energy transferred?
Vanyuwa [196]

Answer:

Energy is transferred from Priya to the box.

Explanation:

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3 years ago
Pls tell the answer asap
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The maximum force acts between B and C as the graph is steepest showing maximum deceleration
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A solenoid 50-cm long with a radius of 5.0 cm has 800 turns. You find that it carries a current of 10 A. The magnetic flux throu
ioda

Answer:

126 mWb

Explanation:

Given that:

length (L) = 50 cm = 0.5 m, radius (r) = 5 cm = 0.05 m, current (I) = 10 A, number of turns (N) = 800 turns.

We assume that the magnetic field in the solenoid is constant.

The magnetic flux is given as:

\phi_m=NBAcos(\theta)\\Where\ B\ is\ the\ magnetic\ field\ density,A\ is \ the\ area.\\But\ B =\mu_onI.\ n \ is\ the\ number\ of\ turns\ per\ unit \ length=N/L\\Therefore,B=\frac{\mu_oNI}{L} \\substituting\ the\ value\ of\ B\ in\ the\ equation: \\\phi_m=\frac{NAcos(\theta)*\mu_oNI}{L} .\ But \ \theta=0,cos(\theta)=1\ and\ A=\pi r^2\\ \phi_m=\frac{N^2\pi r^2\mu_oI}{L} \\Substituting\ values:\\\phi_m=\frac{800^2*(\pi*0.05^2)*(4\pi*10^{-7})*10}{0.5}=0.126\ Wb=126\ mWb

8 0
3 years ago
hat are the wavelengths of peak intensity and the corresponding spectral regions for radiating objects at (a) normal human body
bagirrra123 [75]

Answer:

(a)

\lambda _{m}=9.332 \times 10^{-6}m

(b)

\lambda _{m}=1.632 \times 10^{-6}m

(c) \lambda _{m}=4.988 \times 10^{-7}m

 

Explanation:

According to the Wein's displacement law

\lambda _{m}\times T = b

Where, T be the absolute temperature and b is the Wein's displacement constant.

b = 2.898 x 10^-3 m-K

(a) T = 37°C = 37 + 273 = 310 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{310}

\lambda _{m}=9.332 \times 10^{-6}m

(b) T = 1500°C = 1500 + 273 = 1773 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{1773}

\lambda _{m}=1.632 \times 10^{-6}m

(c) T = 5800 K

\lambda _{m}=\frac{b}{T}

\lambda _{m}=\frac{2.893\times 10^{-3}}{5800}

\lambda _{m}=4.988 \times 10^{-7}m

5 0
3 years ago
2. Willingness to take turns is one way we can express our attitudes through
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B because willingness shows confidence in your answer
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