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Step2247 [10]
3 years ago
10

Consider a taut inextensible string. You shake the end of the string with some frequency, causing a wave to travel down the stri

ng. In the questions below, assume you only change one aspect of the system at a time.
1. If you shake the end of the string twice as rapidly (double the frequency), what will happen to the speed of the wave?
2. If you double the tension in the string, what will happen to the speed of the wave?
3. If you shake the end of the string twice as rapidly (double the frequency), what will happen to the wavelength of the propagating wave?
4. If you double the tension in the string, without changing the rate at which you're shaking it, what will happen to the wavelength of the wave?
For each question, choose from the following choices:
a. It will double.
b. It will remain unchanged.
c. It will increase by a factor of √ 2.
d. It will increase by a factor of 4.
e. It will be half as fast/long.
Physics
1 answer:
masha68 [24]3 years ago
5 0

Answer:

Part 1:

Option B is correct (It will remain unchanged).

Part 2:

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Option E is correct (It will be half as fast/long.)

Part 4:

Option C is correct (It will increase by a factor of √ 2.)

Explanation:

Formula we are going to use:

V=f*λ

Where:

V is the speed of Sound

f is the frequency of wave

λ is the wavelength.

The speed of wave , tension and linear density have following relation:

V=\sqrt{F/\rho}

Where:

V is the speed of Sound (Initial)

F is the tension in string (Initial)

\rho is the linear density of string (Constant)

Terms:

V' is the new speed

f' is the new frequency

λ' is the wavelength

Solution:

Part 1:

From V=\sqrt{F/\rho}:

Speed of Sound is independent of the frequency of shaking so speed well remain unchanged.

Option B is correct (It will remain unchanged)

Part 2:

If F'=2F then

V=\sqrt{F/\rho}

V'=\sqrt{F'/\rho}\\V'=\sqrt{2F/\rho}\\V'=  \sqrt{2} * \sqrt{F/\rho}\\V'=\sqrt{2}V

Option C is correct (It will increase by a factor of √ 2)

Part 3:

Formula we are going to use:

V=f*λ

Given f'=2f,

Even though frequency is doubled we will keep velocities same. V=V' in order to find the changing wavelength.

V'=f'*λ'

f*λ=f'*λ'

f*λ=2f*λ'

Solving above Equation:

λ'=λ/2

Option E is correct (It will be half as fast/long.)

Part 4:

T'=2T means V'=\sqrt{2}V (From Part 1)

f'=f

Now:

V'=f'*λ'

\sqrt{2}f*\lambda=f'*\lambda '\\\sqrt{2}f*\lambda=f*\lambda '\\ \lambda '=\sqrt{2}*\lambda

Option C is correct (It will increase by a factor of √ 2.)

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Explanation:

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Answer:

<em>The coefficient of static friction between the crate and the floor is 0.41</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration and velocity because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N          [1]

Where \mu is the coefficient of static or kinetics friction and N is the normal force.

If no forces other then the weight and the normal are acting upon the y-direction, then the weight and the normal are equal in magnitude:

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W = 20*9.8

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The normal force is also N=196 N

We can find the coefficient of static friction by solving [1] for \mu:

\displaystyle \mu=\frac{Fr}{N}

The friction force is equal to the minimum force required to start moving the object on the floor, thus Fr=80 N and:

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How did the magnet’s density measurement using the Archimedes’ Principle compare to the density measurement using the calculated
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Answer:

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In an experiment, a variable, position-dependent force F(x)F(x) is exerted on a block of mass 1.0kg1.0kg that is moving on a hor
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Answer:

The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(C) is correct option.

Explanation:

Given that,

Mass of block = 1.0 kg

Dependent force = F(x)

Frictional force = F(f)

Suppose, the following information would students need to test the hypothesis,

(A) The function F(x) for 0 < x < 5 and the value of F(f).

(B) The function a(t) for the time interval of travel and the value of F(f).

(C) The function F(x) for 0 < x < 5, the block's initial velocity, and the value of F(f).

(D) The function a(t) for the time interval of travel, the time it takes the block to move 5 m, and the value of F(f).

(E) The block's initial velocity, the time it takes the block to move 5 m, and the value of F(f).

We know that,

The work done by a force is given by,

W=\int_{x_{0}}^{x_{f}}{F(x)\ dx}.....(I)

Where, F(x) = net force

We know, the net force is the sum of forces.

So, \sum{F}=ma

According to question,

We have two forces F(x) and F(f)

So, the sum of these forces are

F(x)+(-F(f))=ma

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Now put the value in equation (I)

W=\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

We need to find the value of \int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}

Using newton's second law

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{ma\ dx}...(II)

We know that,

Acceleration is rate of change of velocity.

a=\dfrac{dv}{dt}

Put the value of a in equation (II)

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{x_{0}}^{x_{f}}{m\dfrac{dv}{dt}dx}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\int_{v_{0}}^{v_{f}}{mv\ dv}

\int_{x_{0}}^{x_{f}}{(F(x)-F(f))dx}=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

Now, the work done by the net force on the block is,

W=\dfrac{mv_{f}^2}{2}+\dfrac{mv_{0}^2}{2}

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Answer:

Explanation:

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v = velocity of the body (in m/s²)

To get the velocity v, we will use the equation of motion v^{2} = u^{2}+2gh

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Since mass = 3.8kg

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3 years ago
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