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mihalych1998 [28]
3 years ago
5

To understand the standard formula for a sinusoidal traveling wave. One formula for a wave with a y displacement (e.g., of a str

ing) traveling in the x direction is y(x,t)=Asin(kx−ωt).
All the questions in this problem refer to this formula and to the wave it describes.
1) What is the phase ϕ(x,t) of the wave? Express the phase in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π
ϕ(x,t)=
2) What is the wavelength λ of the wave? Express the wavelength in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
λ=
3) What is the period T of this wave? Express the period in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
T=
4) What is the speed of propagation v of this wave? Express the speed of propagation in terms of one or more given variables ( A, k, x, t, and ω) and any needed constants like π.
v=
Physics
1 answer:
AysviL [449]3 years ago
3 0

Explanation:

The displacement of the string in the x direction is given by :

y(x,t)=A\ sin(kx-\omega t)

Where

A is the amplitude of wave

k is the propagation constant

1. Here, the phase of the wave is (kx-\omega t)

2. The propagation constant of a wave is given by :

k=\dfrac{2\pi}{\lambda}

\lambda=\dfrac{2\pi}{k}

3. Since, \omega=2\pi f

f=\dfrac{\omega}{2\pi}

Time period, T=\dfrac{1}{f}

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

4. Speed of this wave is given by :

v=f\times \lambda

v=\dfrac{\omega}{2\pi}\times \dfrac{2\pi}{k}

v=\dfrac{\omega}{k}

Hence, this is the required solution.

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A steel piano wire, of length 1.150 m and mass 4.80 g is stretched under a tension of 580.0 N.
kaheart [24]

A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be  372.77 m/s

<h3>What is a sound wave?</h3>

It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.

For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula

V= √F/μ

where v is the wave velocity of the wave travel on the string

F is the tension in the string of piano

μ is the mass per unit length of the string

As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.

The μ is the mass per unit length of the string would be

μ = 4.80/(1.150×1000)

μ = 0.0041739 kg/m

By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity

V= √F/μ

V=√(580/0.0041739)

V =  372.77 m/s

Thus,  the speed of transverse waves on the wire comes out to be  372.77 m/s

Learn more about sound waves from here

brainly.com/question/11797560

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6 0
1 year ago
10 points!! If you help!!!
marta [7]

For Mass

K.E = (1/2*mv^2)

Explanation:

Kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. For example, if a an object with a mass of 10 kg (m = 10 kg) is moving at a velocity of 5 meters per second (v = 5 m/s), the kinetic energy is equal to 125 Joules, or (1/2 * 10 kg) * 5 m/s2.

4 0
3 years ago
What is the distance that a car travels if it was brought to stop in 5 seconds and if it was traveling at 110 Km/h
Triss [41]

Answer:

Suppose that the acceleration is a constant, a.

a(t) = a.

To write the velocity equation, we must integrate over time, and the constant of integration will be equal to the initial velocity, in this case is 110km/h.

v(t) = a*t + 110km/h

And we know that at t = 5s, the car was brought to stop, so the velocity must be zero.

v(5s) = 0 = a*5s + 110km/h.

a = (110km/h)*(1/5s)

now we have that:

1 hour = 3600 seconds.

1km = 1000m

then:

110km/h = (110*1000/3600)m/s = 30.56 m/s

Then we have:

a = (-30.55 m/s)/5s = -6.11 m/s^2

Now the velocity equation is:

v(t) = -6.11m/s^2*t + 30.56m/s

To write the positon equation we must integrate over time again, we can get:

p(t) = (1/2)*(-6.11m/s^2)*t^2 + (30.56m/s)*t + p0

Where p0 is the initial position, here i will assume that is zero, because it does no really mater.

The total displacement of the car will be equal to p(5s)

p(5s) = (1/2)*(-6.11m/s^2)*(5s)^2 + (30.56m/s)*(5s) = 76.425 meters.

6 0
3 years ago
A string is stretched by two equal and opposite forces F newton each then the tension in sting is?
Arlecino [84]

Answer: The tension in the string is zero

Explanation:

3 0
3 years ago
Saturated ethylene glycol at 1 atm is heated by a horizontal chromiumplated surface which has a diameter of 200 mm and is mainta
Paha777 [63]

Here is the full question.

Saturated ethylene glycol at 1 atm is heated by a chromium-plated surface which is circular in shape and has a diameter of 200-mm and is maintained at 480 K.  

At 470 K the properties of the saturated liquid are mu = 0.38 * 10-3 N. s/m^2, Cp = 3280 J/kg. K and Pr = 8.7. The saturated vapour density is p= 1.66 kg/m^3. Take the liquid to surface constants to be Cnb = 0.010 and m=4.1.  

Estimate the heating power requirement and the rate of evaporation  

What fraction is the power requirement of the maximum power associated with the critical heat flux

Answer:

The heating power requirement = 559.2 W

The rate of evaporation = 6.89*10^{-4}kg/s

The fraction of the power requirement  of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

Explanation:

From the thermodynamics tables; we deduced the value for enthalpy at the pressure 1 atm and T_{sat} = 470 K   for the saturated ethylene glycol.

Value for enthalpy of formation h_{fg} = 812 kJ/kg

Density of saturated ethylene glycol \rho___l = 1111 kg/m³

Surface tension \sigma = 32.7*10^{-3}N/m

The heat flux can be calculated by using the formula:

q"s = \mu___l}}}h_{fg}{[\frac{g(\rho{__l}- \rho{__v} }{\sigma} ]^{1/2}  [\frac{C_p*\delta T_c}{C_{sf}*h_{fg}P_r} ]^3

= [0.38*10^{-3}\frac{NS}{m^2} *812*10^3\frac{J}{kg} (\frac{9.81m/s^2*(1111-1.66)kg/m^3}{32.7*10^{-3}N/m} )^{1/2}*(\frac{3280J/kg.K(480-470)K}{0.01*812*10^3\frac{J}{kg}*(8.7)^1 } )]

= 308.56 × 576.6 × 0.1

= 1.78 × 10⁴ W/m²

Now; to find the heating power requirement; we have:

q_{boil} = q__s }*A S

= 1.78*10^4 \frac{W}{m^2}*(\frac{\pi}{4}*(0.2m))^2

Thus, the heating power requirement = 559.2 W

The rate of evaporation is given as:

m= \frac{q_{boil}}{h_[fg}}

= \frac{559.2}{812*10^3}

= 6.89*10^{-4}kg/s

Thus, the rate of evaporation = 6.89*10^{-4}kg/s

To determine to what fraction in the power requirement of the maximum power is associated with the critical total flux ; we needed to first calculate the critical heat flux.

So, the  calculation for the critical heat is given as:q"max = 0.149*h_{fg}}* \rho{___l}}}}[ \frac{\sigma_g (\rho_l - \rho_v}{\rho_v^2} ]^{1/4}

= q"max = 0.149*812810^3* 1.66[ \frac{32.7*10^{-3}*9.8 (1111- 1.66}{1.66^2} ]^{1/4}

= 200840.08 × 3.37

= 6.77 × 10⁵ W/m²

Finally, the fraction of the power requirement  of operating heat flux to the maximum power associated with the critical heat flux is as follows:

= \frac{q''s}{q''max}

= \frac{1.78*10^4}{6.77*10^5}

= 0.026

Thus, the fraction of the power requirement  of operating heat flux to the maximum power associated with the critical heat flux is = 0.026

7 0
3 years ago
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