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3 years ago

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A wave pulse travels along a string at a speed of 230 cm/s . Note that parts a - d are independent and refer to changes made to

the original string.(a) What will be the speed if the string's tension is doubled?(b) What will be the speed if the string's mass is quadrupled (but its length is unchanged)?(c) What will be the speed if the string's length is quadrupled (but its mass is unchanged)?

1 answer:

Oksanka [162]3 years ago

3 0

a) The speed of the wave will increase by a factor $\sqrt{2}$ .

b) The speed of the wave will halve

c) The speed of the wave will double

Explanation:

a)

The speed of a standing wave on a string is given by

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension in the string

m is the mass of the string

L is the length of the string

In this part of the problem, the tension in the string is doubled, so that the new tension is

T' = 2T

Substituting into the equation, we find the new speed of the wave in the string:

$v'=\sqrt{\frac{T'}{m/L}}=\sqrt{\frac{2T}{m/L}}=\sqrt{2}\sqrt{\frac{T}{m/L}}=\sqrt{2}v$

So, the speed will increase by a factor $\sqrt{2}$ .

b)

We can solve also this part by referring to the formula

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the string mass is quadrupled, so the new mass is:

m' = 4m

Substituting into the equation, we find what happens to the speed of the wave:

$v'=\sqrt{\frac{T}{m'/L}}=\sqrt{\frac{T}{4m/L}}=\frac{1}{\sqrt{4}}\sqrt{\frac{T}{m/L}}=\frac{1}{2}v$

So, the speed of the wave will halve.

c)

Again, we can solve this part by referring to the same equation

$v=\sqrt{\frac{T}{m/L}}$

where

T is the tension

m is the mass

L is the length

In this case, the length of the string is quadrupled, so the new length is:

L' = 4L

Substituting into the equation, we find that the new speed is:

$v'=\sqrt{\frac{T}{m/L'}}=\sqrt{\frac{T}{m/(4L)}}=\sqrt{4}\sqrt{\frac{T}{m/L}}=2v$

So, the speed of the wave will double.

Learn more about waves:

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We know the maser radiowave has a frequency $f$ of $1,420,405,751.786 cycles/s$

In addition we know there is an inverse relation between frequency and time $T$ :

$f=\frac{1}{T}$ (1)

Isolating $T$ : $T=\frac{1}{f}$ (2)

$T=\frac{1}{1,420,405,751.786 cycles/s}$ (3)

$T=7.04(10)^{-10} s$ (4) This is the time for 1 cycle

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If $1h=3600s$ and we already know the amount of cycles per second $1,420,405,751.786 cycles/s$ , then:

$1,420,405,751.786 \frac{cycles}{s}(3600s)=5.11(10)^{12} cycles$ This is the number of cycles in an hour

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Firstly, we have to convert this from years to seconds:

$4.6(10)^{9} years \frac{365 days}{1 year} \frac{24 h}{1 day} \frac{3600 s}{1 h}=1.45(10)^{17} s$

Now we have to multiply this value for the frequency of the maser radiowave:

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<h2>d) By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth?</h2>

If we have 1 second out for every 100,000 years, then:

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Answer:

Electric flux, $\phi=6.668\times 10^4\ Nm^2/C$

Explanation:

It is given that,

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We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane. $A=Ak$

The electric flux is given by :

$\phi=E{\cdot}A$

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Since, k.k=i.i=j.j = 1

So,

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Answer: final Displacement = 0 km, total distance covered =7 km

Explanation:

Given the following :

From home to library = 1.3 km East

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Displacement is the net change in distance traveled.

Eastward distance :

To = (1.3 + 0.68)km = 1.98km East

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Δ distance = (1.98 - 1.98) = 0

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Answer:

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Explanation:

The equations for the position and velocity vectors of the cat are as follows:

r = (x0 + v0x · t, y0 + v0y · t + 1/2 · g · t²)

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Where:

r = position vector of the cat at time t.

x0 = initial horizontal position.

v0x = initial horizontal velocity.

t = time.

y0 = initial vertical position.

v0y = initial vertical velocity

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