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siniylev [52]
3 years ago
9

A sinusoidal transverse wave of amplitude ym = 8.4 cm and wavelength = 5.3 cm travels on a stretched cord. Find the ratio of the

maximum particle speed (the speed with which a single particle in the cord moves transverse to the wave) to the wave speed.
Physics
1 answer:
Scilla [17]3 years ago
8 0

Answer:

The ratio is 9.95

Solution:

As per the question:

Amplitude, y_{m} = 8.4\ cm

Wavelength, \lambda = 5.3\ cm

Now,

To calculate the ratio of the maximum particle speed to the speed of the wave:

For the maximum speed of the particle:

v_{m} = y_{m}\times \omega

where

\omega = 2\pi f = angular speed of the particle

Thus

v_{m} = 2\pi fy_{m}

Now,

The wave speed is given by:

v = f\lambda

Now,

The ratio is given by:

\frac{v_{m}}{v} = \frac{2\pi fy_{m}}{f\lambda}

\frac{v_{m}}{v} = \frac{2\pi \times 8.4}{5.3} = 9.95

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Answer: I think is -5

Explanation:

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Is baking soda less dense or more dense than water?
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It’s more dense than air and less dense than liquid!
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An expensive vacuum system can achieve a pressure as low as 1.00×10^{-7} N/m^2 at 20ºC . How many atoms are there in a cubic cen
marta [7]

Answer:

24.70818432141\times 10^7\ atoms

Explanation:

P = Pressure = 1\times 10^{-7}\ N/m^2

V = Volume = 1 cm³

n = Amount of substance

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From the ideal gas law

PV=nRT\\\Rightarrow n=\frac{PV}{RT}

n=\frac{N}{N_A}

\frac{N}{N_A}=\frac{PV}{RT}\\\Rightarrow N=N_A\times \frac{PV}{RT}\\\Rightarrow N=\frac{1\times 10^{-7}\times 1\times 10^{-6}}{8.314\times 293.15}\times 6.022\times 10^{23}\\\Rightarrow N=24708184.32141\ atoms=24.70818432141\times 10^7\ atoms

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8 0
3 years ago
A motion sensor emits sound, and detects an echo 0.0115 s after. A short time later, it again emits a sound, and hears an echo a
Mekhanik [1.2K]

Answer:

1.17 m

Explanation:

From the question,

s₁ = vt₁/2................ Equation 1

Where s₁ = distance of the reflecting object for the first echo, v = speed of the sound in air, t₁ = time to dectect the first echo.

Given: v = 343 m/s, t = 0.0115 s

Substitute into equation 1

s₁ = (343×0.0115)/2

s₁ = 1.97 m.

Similarly,

s₂ = vt₂/2.................. Equation 2

Where s₂ = distance of the reflecting object for the second echo, t₂ = Time taken to detect the second echo

Given: v = 343 m/s, t₂ = 0.0183 s

Substitute into equation 2

s₂ = (343×0.0183)/2

s₂ = 3.14 m

The distance moved by the reflecting object from s₁ to s₂ = s₂-s₁

s₂-s₁ =  (3.14-1.97) m = 1.17 m

7 0
3 years ago
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47. A car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. What
Otrada [13]

Answer: 114 km/h

Explanation:

The formula for determining average speed is expressed as

Average speed = total distance/total time

The car travels 85 km in the first half hour of a trip. The car continues to travel for 2 more hours and travels 200 km. It means that the total distance that the car travels is

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The total time spent by the car is

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Therefore,

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3 0
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