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chubhunter [2.5K]
2 years ago
12

Wave motion is characterized by two velocities: the velocity with which the wave moves in the medium (e.g., air or a string) and

the velocity of the medium (the air or the string itself). Consider a transverse wave traveling in a string. The mathematical form of the wave is y(x,t)=Asin(kx−ωt).
a)Find the velocity of propagation v_p of this wave.

Express the velocity of propagation in terms of some or all of the variables A, k, and omega.

b)Find the y velocity v_y(x,t) of a point on the string as a function of x and t.

Express the y velocity in terms of omega, A, k, x, and t.
Physics
1 answer:
Sindrei [870]2 years ago
3 0

Answer: a) v = ω /k, b) v = - ωAcos( kx −ωt)

Explanation:

y(x,t)=Asin(kx−ωt) defines the wave equation.

a)

We are asked to find wave speed (v)

Recall that v = fλ

From the wave equation above,

k = 2π/ λ where k is the wave number and λ is the wavelength, λ = 2π /k

ω = 2πf where f is the frequency and ω is the angular frequency.

f = ω/ 2π.

By substituting for λ and ω into the wave speed formulae, we have that

v =( ω/ 2π) × (2π /k)

v = ω/k

b)

y(x,t)=Asin(kx−ωt)

The first derivative of y with respect to x give the velocity (vy)

By using chain rule, we have that

v = dy/dt = A cos( kx −ωt) × (−ω)

v = - ωAcos( kx −ωt)

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Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.
kramer

Answer:

115 ⁰C

Explanation:

<u>Step 1:</u> The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

q_{1} +q_{2} =-q_{3} -----eqution 1

where,

q_{1} is the heat absorbed by the solid at 0⁰C

q_{2} is the heat absorbed by the liquid at 0⁰C

q_{3} the heat lost by the warmer water sample

Important equations to be used in solving this problem

q=m *c*\delta {T}, where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

\delta {T} is change in temperature

Again,

q=n*\delta {_f_u_s} -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

<u>Step 2:</u> calculate how many moles of water you have in the 100.0-g sample

=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O

<u>Step 3: </u>calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ

This means that equation (1) becomes

79.13 KJ + q_{2} = -q_{3}

<u>Step 4:</u> calculate the final temperature of the water

79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}

Substitute in the values; we will have,

79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})

79.13 kJ + 990.66J* (T_{f}-218}) = -1463J*(T_{f}-100})

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* (T_{f}-218}) = -1.463KJ*(T_{f}-100})

79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3

collect like terms,

2.45366T_{f} = 283.133

∴T_{f} = = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0
2 years ago
What is the total flux φ that now passes through the cylindrical surface? enter a positive number if the net flux leaves the cyl
trasher [3.6K]

Net flux through the cylindrical surface is given as

\phi = \frac{q}{epsilon_0}

here q = enclosed charge in the surface

so here in order to find the value of q

q = \lambda* L

so now we have

\phi = \frac{\lambda * L}{\epsilon_0}

so this is the total flux

now by Gauss's law we can find the electric field

\int E.dA = \phi

\int E.dA = \frac{\lambda * L}{\epsilon_0}

E* 2\pi rL = \frac{\lambda * L}{epsilon_0}

E = \frac{\lambda}{2\pi \epsilon_0 r}

<em>by above expression we can find the electric field at required position</em>

8 0
1 year ago
A boat's capacity plate gives the maximum weight and/or number of people the boat can carry safely in certain weather conditions
erastova [34]
•wind
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5 0
2 years ago
Complete the statements using data from Table A of your Student Guide. The speed of the cart after 8 seconds of Low fan speed is
finlep [7]

Answer:

The speed of the cart after 8 seconds of Low fan speed is  72.0 cm/s

The speed of the cart after 3 seconds of Medium fan speed is   36.0 cm/s

The speed of the cart after 6 seconds of High fan speed is  96.0 cm/s

Explanation:

took the test on edgenuity

4 0
1 year ago
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A 225 kg red bumper car is moving at 3.0 m/s. It hits a stationary 180 kg blue bumper car. The red and blue bumper cars combine
Alex Ar [27]

Given


m1(mass of red bumper): 225 Kg


m2 (mass of blue bumper): 180 Kg


m3(mass of green bumper):150 Kg


v1 (velocity of red bumper): 3.0 m/s


v2 (final velocity of the combined bumpers): ?




The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:


Pa= Pb


Where Pa is the momentum before collision and Pb is the momentum after collision.


Now applying this law for the above problem we get


Momentum before collision= momentum after collision.


Momentum before collision = (m1+m2) x v1 =(225+180)x 3 = 1215 Kgm/s


Momentum after collision = (m1+m2+m3) x v2 =(225+180+150)x v2

=555v2

Now we know that Momentum before collision= momentum after collision.


Hence we get


1215 = 555 v2


v2 = 2.188 m/s


Hence the velocity of the combined bumper cars is 2.188 m/s

4 0
1 year ago
Read 2 more answers
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