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muminat
3 years ago
14

A sound has 13 crests and 15 troughs in 3 seconds. When the second crest is produced the first is 2cm away from the source? Calc

ulate
        a. The wavelength

        b. The frequency

        c. The wave speed

​
Physics
1 answer:
Yuki888 [10]3 years ago
5 0

Answer:

The wavelength will be 4 cm, frequency will be 4.66 Hz and wave speed is 18.6 cm/sec

Explanation:

Given:

No. of crest = 13

No. of trough = 15

Time = 3 seconds

Hence, 1 crest or 1 trough = \frac{1}{2}\lambda

therefore,

13 C + 15 T = 28(\frac{1}{2}\lambda)

=14\lambda

Time given 3 seconds

  = \frac{3}{14}s

\nu= \frac{14}{3}

\nu= 4.6 Hz \approx 5 Hz

2 cm distance is travelled is time period

\lambda = 4 cm

Again wave will travel in 1 T = 4 cm

wave speed v =\lambda\times\nu

= 4\times\frac{14}{3}

= 18.6 cm/s

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Answer:

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Explanation:

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8 0
3 years ago
An 1100kg car reached 30m/s in 9s from rest. What force did the engine provide?
malfutka [58]

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≅3666.67 N

Explanation:

Use Newton's 2nd law, F = ma where F=force applied, m = mass of the object,

a = acceleration acquired by the object.

a= (v-u)/t where v = final velocity, u = initial velocity and t = time taken

calculate a = (30-0)/9 ≅ 3.33 m/s2

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3 years ago
How much work is required to compress 5.05 mol of air at 19.5°C and 1.00 atm to one-eleventh of the original volume by an isothe
Rus_ich [418]

Explanation:

(a)  For an isothermal process, work done is represented as follows.

             W = -nRT ln(\frac{V_{2}}{V_{1}})

Putting the given values into the above formula as follows.

        W = -nRT ln(\frac{V_{2}}{V_{1}})

             = - 5.05 mol \times 8.314 J/mol K \times (19.5 + 273) K \times ln (\frac{\frac{V_{1}}{11}}{V_{1}})

             = -12280.82 \times ln (0.09)

             = -12280.82 \times -2.41

             = 29596.78 J

or,         = 29.596 kJ       (as 1 kJ = 1000 J)

Therefore, the required work is 29.596 kJ.

(b) For an adiabatic process, work done is as follows.

         W = \frac{P_{1}V^{\gamma}_{1}(V^{1-\gamma}_{2} - V(1-\gamma)_{1})}{(1 - \gamma)}

              = \frac{-nRT_{1}(11^{\gamma - 1} - 1)}{1 - \gamma}

              = \frac{-5.05 \times 8.314 J/mol K \times 292.5 (11^{1.4 - 1} - 1)}{1 - 1.4}

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Therefore, work required to produce the same compression in an adiabatic process is 49.41 kJ.

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               P_{1}V_{1} = P_{2}V_{2}

or,       P_{2} = \frac{P_{1}V_{1}}{V_{2}}

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          P_{1}V^{\gamma}_{1} = P_{2}V^{\gamma}_{2}

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Answer:

approximately 5.8 seconds

Explanation:

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