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SOVA2 [1]
3 years ago
8

Bandwidth is the: Select one: A. number of frequencies that can be broadcast through a medium. B. number of cycles per second th

at can be sent through a medium. C. difference between the highest and lowest frequencies that can be accommodated on a single channel. D. total number of bytes that can be sent through a medium per second. E. geographical distance spanned by a network.
Physics
1 answer:
natita [175]3 years ago
4 0

Answer:

C. difference between the highest and lowest frequencies that can be accommodated on a single channel.

Explanation:

Bandwidth is the range of the band of frequencies. It is the amount of the data which can be transmitted over a wide range of frequency in the fixed time.

Bandwidth is difference between highest and lowest frequencies in the continuous band of  the frequencies which can be accommodated on the single channel. It is measured in hertz.

<u>Correct answer - C. difference between the highest and lowest frequencies that can be accommodated on a single channel.</u>

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How much heat has to be added to 508 g of copper at 22.3°C to raise the temperature of the copper to 49.8°C? (The specific heat
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Answer:

Q = 5267J

Explanation:

Specific heat capacity of copper (S) = 0.377 J/g·°C.

Q = MSΔT

ΔT = T2 - T1

ΔT=49.8 - 22.3 = 27.5C

Q = change in energy = ?

M = mass of substance =508g

Q = (508g) * (0.377 J/g·°C) * (27.5C)

Q= 5266.69J

Approximately, Q = 5267J

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Light of wavelength 480 nm illuminates a pair of slits separated by 0.27 mm. If a screen is place 1.7 m from the slits, determin
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Answer:

  Δy = 6.05 mm

Explanation:

The double slit phenomenon is described by the expression

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      m = 0,±1, ±2, ...

As they tell us that they measure the dark stripes, we are in a case of destructive interference, let's use trigonometry to find the sins tea

      tan θ = y / x

      y = x tan θ

In the interference experiments the measured angle is very small so we can approximate the tangent

      tan θ = sin θ / cos θ

     cos θ = 1

     tan θ = sin θ

     y = x sin θ

We substitute in the destructive interference equation

     d (y / x) = (m + ½) λ

    y = (m + ½) λ x / d

The first dark strip occurs for m = 0 and the third dark strip for m = 2. Let's find the distance for these and subtract it

 m = 0

      y₀ = (0+ ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

      y₀ = 1.511 10⁻³ m

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     y₂ = (2 + ½) 480 10⁻⁹ 1.7 / 0.27 10⁻³

     y₂ = 7.556 10⁻³ m

The separation between these strips is Δy

     Δy = y₂-y₀

     Δy = (7.556 - 1.511) 10⁻³

     Δy = 6.045 10⁻³ m

     Δy = 6.05 mm

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