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il63 [147K]
3 years ago
15

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

e note at 110 Hz has two strings at this frequency. If one string slips from its normal tension of 596 N to 538.00 N, what beat frequency is heard when the hammer strikes the two strings simultaneously
Physics
1 answer:
pav-90 [236]3 years ago
3 0

Answer:

5.4893 beats/sec

Explanation:

We have f_1 and f_2 as frequency of two waves

also we know that according to question

f_2=f_1\sqrt{\frac{F'}{F} }

F and F' are forces on the two frequencies.

therefore, beat frequency is

f=f_1-f_2=f_1-f_1\sqrt{\frac{F'}{F} }\\=110(1-\sqrt{\frac{538}{596} })\\=5.4893 \text{ beats/sec}

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A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What
Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

R = \sqrt{(3b)^2+(4b)^2}

Therefore the resultant displacement of the girl is

R = \sqrt{(9b^2+16b^2)}

R = \sqrt{25b^2}

R = 5b

Therefore the girl has displaced around of 5 blocks

4 0
3 years ago
A 6 kg object falls 10 m. The object is attached mechanically to a paddle-wheel which rotates as the object falls. The paddle-wh
maksim [4K]

Answer:16.234^{\circ}C

Explanation:

Given

Mass of object (m)=6 kg

falling height(h)=10 m

mass of water(m_w)=600 gm

temperature of water =15

specific heat of water =4.186 j/g-^{\circ}C

Let T be the Final Temperature of water

Here Object Potential Energy is converted into Heat energy which will be absorbed by water

Potential Energy(P.E.)=mgh=6\times 9.81\times 10=588.6 J

Heat supplied=m_wc(\Delta T)

H.E.=600\times 4.186\times (T-16)

588.6=2511.6\times (T-16)

T-16=0.234

T=16.234^{\circ}C

This is not an efficient way of heating water as there is only0.234^{\circ}Cincrease in temperature.

5 0
3 years ago
A runner begins running at the beginning of 7th period and stops running at the end of the period. She runs at a pace of 10 km/h
MrRissso [65]

Answer:

She run for, t = 0.92 s

Explanation:

Given data,

The velocity of the runner, v = 10 km/h

The distance covered by the runner, d = 9.2 km

The relationship between the velocity, displacement and time is given by the formula,

                           t = d / v

Substituting the given values in the above equation,

                           t = 9.2 / 10

                             = 0.92 s

Hence, she ran for, t = 0.92 s

8 0
3 years ago
A thin rod of length 0.75 m and mass 0.42 kg is suspended
MrRissso [65]

Answer:

a)  K = 0.63 J, b)  h = 0.153 m

Explanation:

a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is

         w² = \frac{m g d}{I}

where d is the distance from the pivot point to the center of mass and I is the moment of inertia.

The rod is a homogeneous body so its center of mass is at the geometric center of the rod.

              d = L / 2

the moment of inertia of the rod is the moment of a rod supported at one end

              I = ⅓ m L²

we substitute

            w = \sqrt{\frac{mgL}{2}  \ \frac{1}{\frac{1}{3} mL^2} }

            w = \sqrt{\frac{3}{2}  \ \frac{g}{L} }

            w = \sqrt{ \frac{3}{2} \ \frac{9.8}{0.75}  }

            w = 4.427 rad / s

an oscillatory system is described by the expression

              θ = θ₀ cos (wt + Φ)

the angular velocity is

             w = dθ /dt

             w = - θ₀ w sin (wt + Ф)

In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1

In the exercise it is indicated that at the lowest point the angular velocity is

           w = 4.0 rad / s

the kinetic energy is

           K = ½ I w²

           K = ½ (⅓ m L²) w²

           K = 1/6 m L² w²

           K = 1/6 0.42 0.75² 4.0²

           K = 0.63 J

b) for this part let's use conservation of energy

starting point. Lowest point

             Em₀ = K = ½ I w²

final point. Highest point

             Em_f = U = m g h

energy is conserved

             Em₀ = Em_f

             ½ I w² = m g h

             ½ (⅓ m L²) w² = m g h

             h = 1/6 L² w² / g

             h = 1/6 0.75² 4.0² / 9.8

             h = 0.153 m

5 0
2 years ago
The message refers to which of the following?
Oliga [24]

The message is the information being communicated from one place to another.

It used to be called the "intelligence".  But as time went on, it became
harder to ignore the obvious fact that that was going too far, and the
label was changed to the more IQ-neutral "message".    

7 0
3 years ago
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