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tankabanditka [31]
2 years ago
14

A particularly beautiful note reaching your ear from a rare Stradivarius violin has a wavelength of 39.1 cm. The room is slightl

y warm, so the speed of sound is 344 m/s. Part A If the string's linear density is 0.660 g/mg/m and the tension is 160 NN , how long is the vibrating section of the violin string
Physics
1 answer:
USPshnik [31]2 years ago
3 0

Given Information:  

Wavelength =  λ = 39.1 cm = 0.391 m

speed of sound = v = 344 m/s

linear density = μ = 0.660 g/m = 0.00066 kg/m

tension = T = 160 N

Required Information:

Length of the vibrating string = L = ?

Answer:

Length of the vibrating string = 0.28 m

Explanation:

The frequency of beautiful note is

f = v/λ

f = 344/0.391

f = 879.79 Hz

As we know, the speed of the wave is

v = √T/μ

v = √160/0.00066

v = 492.36 m/s

The wavelength of the string is

λ = v/f

λ = 492.36/879.79

λ = 0.5596 m

and finally the length of the vibrating string is

λ = 2L

L = λ/2

L = 0.5596/2

L = 0.28 m

Therefore, the vibrating section of the violin string is 0.28 m long.

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Answer:

v_max = (1/6)e^-1 a

Explanation:

You have the following equation for the instantaneous speed of a particle:

v(t)=ate^{-6t}   (1)

To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:

\frac{dv(t)}{dt}=\frac{d}{dt}[ate^{-6t}]=a[(1)e^{-6t}+t(e^{-6t}(-6))]  (2)

where you have use the derivative of a product.

Next, you equal the expression (2) to zero in order to calculate t:

a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}

For t = 1/6 you obtain the maximum speed.

Then, you replace that value of t in the expression (1):

v_{max}=a(\frac{1}{6})e^{-6(\frac{1}{6})}=\frac{e^{-1}}{6}a

hence, the maximum speed is v_max = ((1/6)e^-1)a

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2 years ago
A 1232 kg car moving north at 25.6 m/s collides with a 2028 kg car moving north at 17.5 m/s . They stick together. In what direc
Citrus2011 [14]

Answer:

I. Angle = 41.7° Northeast.

II. Vr = 7.08m/s

Explanation:

Let the two cars be denoted by A and B

<u>Given the following data;</u>

Mass of car A = 1232 Kg

Velocity of car A = 25.6 m/s

Mass of car B = 2028 Kg

Velocity of car B = 17.5m/s

First of all, we would solve for momentum;

Momentum = mass × velocity

Momentum, M1 = 1232 × 25.6

Momentum, M1 = 31539.2 Kgm/s

Momentum, M2 = 2028 × 17.5

Momentum, M2 = 35490 Kgm/s

Now, let's find the resultant momentum using the Pythagoras theorem;

R² = M1² + M2²

R² = 31539.2² + 35490²

R² = 994721136.6 + 1259540100

R² = 2254261237

Taking the square root of both sides, we have

Resultant momentum, R = 47479.06 Kgm/s

To find the direction;

Angle = tan¯¹(M1/M2)

Angle = tan¯¹(31539.2/35490)

Angle = tan¯¹(0.89)

<em>Angle = 41.7° Northeast.</em>

To find the speed;

R = (M1 + M2)Vr

47479.06 = (31539.2 + 35490)Vr

47479.06 = 67029.2Vr

Vr = 47479.06/67029.2

<em>Vr = 7.08m/s</em>

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2 years ago
A) Define rest and motion?​
7nadin3 [17]

Answer:

Rest - a body is said to be at rest, if it does not change its position with respect to its surrounding with time. Motion - a body is said to be at motion, if it changes its position with time.

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3 years ago
The _____ phrase "answers" the antecedent with a similar rhythm and harmony.
12345 [234]
The correct answer is consequent.

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3 years ago
Using Kepler's 3rd law and Newton's law of universal gravitation, find the period of revolution P of the planet as it moves arou
weeeeeb [17]

Answer:

P =  \pi \sqrt{\frac{(R_1+R_2)^3}{2 \ GMS}}

Explanation:

From the attached diagram below:

AC = a (1 + e) = R₂     -------- equation (1)

CD = a ( 1 - e) = R₁     ---------   equation (2)

⇒ 1 - e = \frac{R_1}{a}

e= 1 - \frac{R_1}{a}

Replacing the value for e into equation (1)

a(1+1- \frac{R_1}{a})= R_2

= 2a - R_1 = R_2

a= \frac{R_1+R_2}{2}

From Kepler's third law;

P = 2 \pi \sqrt{\frac{a^3}{GMS}}

P = 2 \pi \sqrt{\frac{(R_1+R_2)^3}{8GMS}}

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