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2 years ago

A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). find the period

Physics

2 answers:

Art [367]2 years ago

6 0

The time period of the transverse wave in a rope will be $\boxed{{\text{0}}{\text{.025 s}}}$ .

Further Explanation:

The waves are the disturbance created in the medium particles. A transverse wave in a rope is the disturbance produced in the particles of the rope in the vertical direction.

A transverse wave is a wave in which the disturbance in the medium is perpendicular to the propagation of the wave. The time period of the wave is the time taken in completing one oscillation about the mean position.

The expression of the transverse wave in the rope is given.

$y\left( {x,t} \right) = \left( {0.750\,{\text{cm}}} \right)\cos \left( {\pi \left[ {\left( {0.400\,{\text{cm}} - {\text{1}}} \right)x + \left( {250\,{{\text{s}}^{ - 1}}} \right)t} \right]} \right)$

The standard equation for a wave present in a rope is given as.

$\boxed{y\left( {x,t} \right) = A\cos \left( {kx + \omega t} \right)}$

Here, $A$ is the amplitude of the wave, $y$ is the position of the wave, $k$ is the wave number and $\omega$ is the angular frequency of the wave.

From the above expression, the angular frequency of the wave is obtained as.

$\omega = 250\pi$

We know that the time period is obtained from the angular frequency as.

$T = \dfrac{{2\pi }}{\omega }$

Substitute the value of $\omega$ in above expression.

$\begin{aligned}T &= \frac{{2\pi }}{{250\pi }} \\&= \frac{1}{{125}}\,{\text{s}} \\&\approx {\text{0}}{\text{.008}}\,{\text{s}} \\&\approx {\text{8}}\,{\text{ms}} \\\end{aligned}$

Thus, the time period of the transverse wave in the rope is $\boxed{8\,{\text{ms}}}$ .

Learn More:

Wavelength of electron brainly.com/question/1979815
Charles’ Law and its explanation brainly.com/question/7316997
Magnitude of acceleration of crate while slipping brainly.com/question/10470406

Answer Details:

Grade: High School

Chapter: Waves

Subject: Physics

Keywords: Transverse, wave, amplitude, angular frequency, perpendicular, stationary wave, rope, velocity, position, propagation of wave, wave number.

Angelina_Jolie [31]2 years ago

5 0

Using a graphical tool to obtain the form of the wave we can see that

It is periodic in X with period = 5 cm

It is periodic in t with period = 1/125 s = 8 ms

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a box weighing 155 N is pushed horizontally down the hall at constant velocity. the applied force is 83 n what is the coefficien

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Answer:

μ = 0.535

Explanation:

On a level floor, normal force = weight.

N = W

Friction force = normal force × coefficient of friction.

F = Nμ

Substitute:

F = Wμ

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2 years ago

1. A 9.4×1021 kg moon orbits a distant planet in a circular orbit of radius 1.5×108 m. It experiences a 1.1×1019 N gravitational

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Answer:

26 days

Explanation:

m = 9.4×1021 kg

r= 1.5×108 m

F = 1.1×10^ 19 N

We know Fc = $\frac{m v^{2} }{r}$

==> 1.1 × $10^{19}$ = (9.4 × $10^{21}$ × $v^{2}$ ) ÷ 1.5 × $10^{8}$

==> 1.1 × $10^{19}$ = $v^{2}$ × 6.26× $10^{13}$

==> $v^{2}$ = 1.1 × $10^{19}$ ÷ 6.26× $10^{13}$

==> $v^{2}$ = 0.17571885 × $10^{6}$

==> v= 0.419188323 × $10^{3}$ m/sec

==> v= 419.188322834 m/s

Putting value of r and v from above in ;

T= 2πr ÷ v

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3 0

2 years ago

Read 2 more answers

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$