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chubhunter [2.5K]
3 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]3 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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6 0
3 years ago
A 0.272-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and c
Nady [450]

(a) +9.30 kg m/s

The impulse exerted on an object is equal to its change in momentum:

I= \Delta p = m \Delta v = m (v-u)

where

m is the mass of the object

\Delta v is the change in velocity of the object, with

v = final velocity

u = initial velocity

For the volleyball in this problem:

m = 0.272 kg

u = -12.6 m/s

v = +21.6 m/s

So the impulse is

I=(0.272 kg)(21.6 m/s - (-12.6 m/s)=+9.30 kg m/s

(b) 155 N

The impulse can also be rewritten as

I=F \Delta t

where

F is the force exerted on the volleyball (which is equal and opposite to the force exerted by the volleyball on the fist of the player, according to Newton's third law)

\Delta t is the duration of the collision

In this situation, we have

\Delta t = 0.06 s

So we can re-arrange the equation to find the magnitude of the average force:

F=\frac{I}{\Delta t}=\frac{9.30 kg m/s}{0.06 s}=155 N

6 0
3 years ago
ASAP:
goldfiish [28.3K]

Explanation:

...domains are aligned which creates a magnetic field.

... are not always aligned so the resulting element had no apparent magentism

5 0
2 years ago
An insulating cup contains 200 grams of water at 25 ∘C. Some ice cubes at 0 ∘C is placed in the water. The system comes to equil
Nataly [62]

Answer:

The amount of ice added in gram is 32.77g

Explanation:

This problem bothers on the heat capacity of materials

Given data

Mass of water Mw= 200g

Temperature of water θw= 25°c

Temperature of ice θice= 0°c

Equilibrium Temperature θe= 12°c

Mass of ice Mi=???

The specific heat of ice Ci= 2090 J/(kg ∘C)

specific heat of water Cw = 4186 J/(kg ∘C)

latent heat of the ice to water transition Li= 3.33 x10^5 J/kg

heat heat loss by water = heat gained by ice

N/B let us understand something, heat gained by ice is in two phases

Heat require to melt ice at 0°C to water at 0°C

And the heat required to take water from 0°C to equilibrium temperature

Hence

MwCwΔθ=MiLi +MiCiΔθ

Substituting our data we have

200*4186*(25-12)=Mi*3.3x10^5+

Mi*2090(12-0)

837200*13=Mi*3.3x10^5+Mi*2090

10883600=332090Mi

Mi=10883600/332090

Mi= 32.77g

4 0
3 years ago
A girl is sledding down a slope that is inclined at 30º with respect to the horizontal. The wind is aiding the motion by providi
OleMash [197]

Answer:

The sled required 9.96 s to travel down the slope.

Explanation:

Please, see the figure for a description of the problem. In red are the x and y-components of the gravity force (Fg). Since the y-component of Fg (Fgy) is of equal magnitude as Fn but in the opposite direction, both forces get canceled.

Then, the forces that cause the acceleration of the sled are the force of the wind (Fw), the friction force (Ff) and the x-component of the gravity force (Fgx).

The sum of all these forces make the sled move. Finding the resulting force will allow us to find the acceleration of the sled and, with it, we can find the time the sled travel.

The magnitude of the friction force is calculated as follows:

Ff = μ · Fn

where :

μ = coefficient of kinetic friction

Fn =  normal force

The normal force has the same magnitude as the y-component of the gravity force:

Fgy = Fg · cos 30º = m · g · cos 30º

Where

m = mass

g = acceleration due to gravity

Then:

Fgy = m · g · cos 30º = 87.7 kg · 9.8 m/s² · cos 30º

Fgy = 744 N

Then, the magnitude of Fn is also 744 N and the friction force will be:

Ff = μ · Fn = 0.151 · 744 N = 112 N

The x-component of Fg, Fgx, is calculated as follows:

Fgx = Fg · sin 30º = m·g · sin 30º = 87.7 kg · 9.8 m/s² · sin 30º = 430 N

The resulting force, Fr, will be the sum of all these forces:

Fw + Fgx - Ff = Fr

(Notice that forces are vectors and the direction of the friction force is opposite to the other forces, then, it has to be of opposite sign).

Fr = 161 N + 430 N - 112 N = 479 N

With this resulting force, we can calculate the acceleration of the sled:

F = m·a

where:

F = force

m = mass of the object

a = acceleration

Then:

F/m = a

a = 479N/87.7 kg = 5.46 m/s²

The equation for the position of an accelerated object moving in a straight line is as follows:

x = x0 + v0 · t + 1/2 · a · t²

where:

x = position at time t

x0 = initial position

v0 = initial velocity

t = time

a = acceleration

Since the sled starts from rest and the origin of the reference system is located where the sled starts sliding, x0 and v0 = 0.

x = 1/2· a ·t²

Let´s find the time at which the position of the sled is 271 m:

271 m = 1/2 · 5.46 m/s² · t²

2 · 271 m / 5.46 m/s² = t²

<u>t = 9.96 s </u>

The sled required almost 10 s to travel down the slope.

8 0
3 years ago
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