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1 year ago

Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?

1 answer:

6 0

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

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Which describes the transformation of chemical energy into mechanical energy and why?

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B
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3 years ago

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To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass c

olasank [31]

Answer:

the mass should be bring closer to the point about which we are finding torque

Explanation:

τ = Σr × F = rmg

where m is the mass, g is acceleration due to gravity, and r is the distance

Torque is directly proportional to -

1.mass, m , of object

2. distance, r, of the mass from the point about which we are finding the torque.

So if we increase or decrease them then the torque will also increase or decrease.

So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.

Therefore the mass should be bring closer to the point about which we are finding torque.

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3 years ago

We have seen that when you generate a wave on a stretched spring, as long as the medium doesn’t change (that is, the tension and

TiliK225 [7]

Answer:

No, the acceleration is not always zero.

Explanation:

It does not mean that the acceleration of the particle is zero.

The velocity of wave is different from the velocity of particle.

The acceleration of wave is different from the acceleration of particle.

the acceleration of the particle is given by

$a =- w^2 y$

where, w is the angular frequency and y is the displacement from the mean position.

So, the acceleration is zero at mean position only and it varies as the position changes.

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3 years ago

The amplitude of a lightly damped oscillator decreases by 3.42% during each cycle. What percentage of the mechanical energy of t

frozen [14]

Answer:

The percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

Explanation:

Mechanical energy (Potential energy, PE) of the oscillator is calculated as;

PE = ¹/₂KA²

During the first oscillation;

PE₁ = ¹/₂KA₁²

During the second oscillation;

A₂ = A₁ - 0.0342A₁ = 0.9658A₁

PE₂ = ¹/₂KA₂²

PE₂ = ¹/₂K (0.9658A₁)²

PE₂ = (0.9658²)¹/₂KA₁²

PE₂ = (0.9328)¹/₂KA₁²

PE₂ = 0.9328PE₁

Percentage of the mechanical energy of the oscillator lost in each cycle;

$Change \ in \ percent= \frac{PE_2 - PE_1}{PE_1} \\\\Change \ in \ percent= \frac{0.9328PE_1 -PE_1}{PE_1} *100\\\\Change \ in \ percent= \frac{-0.0672PE_1}{PE_1}*100 \\\\Change \ in \ percent= -0.0672*100\\\\Change \ in \ percent= -6.72 \ \% \\\\Loss \ in \ percent= 6.72 \ \%$

Therefore, the percentage of the mechanical energy of the oscillator lost in each cycle is 6.72%

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3 years ago

No links please, links never work, this is science not physics

stiks02 [169]

Answer:

b and c are the answers

A is an opinion, D is a superstitious belief because they haven’t found Jesus obviously

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3 years ago