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podryga [215]
3 years ago
9

A) Find the wavelength of an electromagnetic wave with frequency 9.35 GHz = 9.35 109 Hz (G = giga = 109), which is in the microw

ave range. λ = m
(b) Find the speed of a sound wave in an unknown fluid medium if a frequency of 532 Hz has a wavelength of 2.00 m.
Physics
1 answer:
miss Akunina [59]3 years ago
4 0

Answer:

(a) \lambda=0.032\ m

(b) v = 1064 m/s                                                                

Explanation:

(a) Frequency of electromagnetic wave, f=9.35\ GHz=9.35\times 10^9\ Hz

Let \lambda is the wavelength of electromagnetic wave. It can be calculated as :

c=f\lambda

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8\ m/s}{9.35\times 10^9\ Hz}

\lambda=0.032\ m

(b) Frequency of medium, f = 532 Hz

Wavelength, \lambda=2\ m

Let v is the speed of sound wave in an unknown fluid. Using the relation as :

v=f\lambda

v=532\times 2

v = 1064 m/s

Hence, this is the required solution.

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A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
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          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

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         y₁ = 800 m

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        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

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The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

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