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Romashka [77]
3 years ago
13

At what tension, in newtons, must the D-string must be stretched in order for it to be properly tuned

Physics
1 answer:
Vsevolod [243]3 years ago
4 0

The fundamental frequency, wavelength of the wave, and mass per unit

length of the string, determines the tension in the string.

  • The tension to which the D-string must be tuned is approximately <u>19.718 Newtons</u>

Reasons:

From a similar question, we have;

Fundamental frequency, f₁ = 146.8 Hz

Oscillating length on the D-string, λ₁ = 0.61 m

Mass of the string = 1.5 × 10⁻³ kg.

We have;

\displaystyle f_1 \cdot \lambda _1 =\mathbf{ \sqrt{\frac{T}{m/L} }}

Therefore;

\displaystyle T =  \mathbf{\frac{f_1^2 \cdot \lambda _1^2 \cdot m }{L}}

Which gives;

\displaystyle T =  \frac{146.8^2 \times 0.61^2 \times 1.5 \times 10^{-3} }{0.61 } \approx  \mathbf{19.718}

The tension to which the D-string must be tuned, T ≈ <u>19.718 Newtons</u>

Learn more here:

brainly.com/question/15589287

The parameters given in a similar question obtained online are;

The fundamental frequency of the tone, f₁ = 146.8 Hz

The oscillating length on the D-string, λ₁ = 0.61 m

The mass of the string = 1.5 × 10⁻³ kg.

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As capacitor was discharging, what did you observe about q on its plates and the motion of charges in the external circuit?
liubo4ka [24]

As capacitor was discharging, The charge on the plate got reversed and the motion of charge is opposite to the flow of current.

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The capacitor is completely charged when the voltage of the electricity supply is equal to that at the capacitor terminals. that is referred to as capacitor charging; and the charging segment is over when modern-day stops flowing thru the electrical circuit.

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8 0
2 years ago
You observe two cars traveling in the same direction on a long, straight section of Highway 5. The red car is moving at a consta
Arte-miy333 [17]

Answer:

1)  3.66 s

2) 124.44 m

3) 3.12 s

Explanation:

Let's start by first listing down the information in the question.

Red Car : 34 m/s

Blue Car: 28 m/s

Distance between them : 22 m

The difference in speed between the cars is: 34 - 28 = 6 m/s

This means that the red car is catching up to the blue car at a speed of 6 m/s.

1) We can solve this by just dividing the distance by the difference in speed. This becomes:    \frac{Distance}{Speed}= \frac{22}{6} =   3.66

Thus it takes 3.66 seconds for the red car to catch up to the blue car.

2) We know from (1) that it took 3.66 seconds for the red car to catch up. Since the speed it was travelling at is constant, we only need to multiply it by the time from (1) to get the distance.

This becomes:    Speed * Time = 34 * 3.66 = 124.44

Thus the red car travels 124.44 m before catching up to the blue car.

3) If the red car starts to accelerate the moment we see it, the time taken to get to the blue car will be less than before. We can find this in a simple way.

We can use the motion equation : s = u*t + \frac{1}{2}(a * t^2)

Here s = 22 m

We can take u as the difference in speed. u = 6 m/s

Acceleration a = (2/3) m/s^2

Substituting the these into the equation we get:

22 = 6t + \frac{1}{2}(\frac{2}{3}t^2)

Solving this for the variable 't' using the quadratic formula we get the following two answers:

t1  = 3.12 s

t2 = - 21.12 s

Since t2 is not possible, the answer is t1. This means it takes 3.12 seconds for the red car to catch up to the blue.

4 0
3 years ago
On a velocity time graph, the slope tells you the ______ of the object.
Paraphin [41]
The acceleration is the correct answer
7 0
3 years ago
A circuit contains a 305 ohm resistor, a 1.1 micro Farad capacitor, and a 42 mH inductor. At resonance, the impedance is determi
r-ruslan [8.4K]

Answer:

The resistance of the inductor at resonance is 258.76 ohms.

Explanation:

Given;

resistance of the resistor, R = 305 ohm

capacitance of the capacitor, C = 1.1 μF = 1.1 x 10⁻⁶ F

inductance of the inductor, L = 42 mH = 42 x 10⁻³ H = 0.042 H

At resonance the inductive reactance is equal to capacitive reactance.

\omega L = \frac{1}{\omega C}\\\\2\pi F_0 L =  \frac{1}{2\pi F_0 C}\\\\F_0 = \frac{1}{2\pi\sqrt{LC} }

Where;

F₀ is the resonance frequency

F_0 = \frac{1}{2\pi\sqrt{LC} } \\\\F_0 = \frac{1}{2\pi\sqrt{(0.024)(1.1*10^{-6})} }\\\\F_0 =980.4 \ Hz

The inductive reactance is given by;

X_l = 2\pi F_0 L\\\\X_l = 2\pi (980.4)(0.042) \\\\X_l = 258.76 \ ohms

Therefore, the resistance of the inductor at resonance is 258.76 ohms.

5 0
3 years ago
The area of the effort and load of the Piston of a hydrolic are 0.5 and 5m respectively. If a force of 100 Newton is applied on
Yanka [14]

Answer:

Explanation:

Using Pascal laws, which states that pressure are the input equals the pressure at the output.

Pressure is given as force/area

P1=P2

Then,

F1/A1=F2/A2

Cross multiply

F1A2=F2A1

Given that

Ae=0.5m² area of effort

Al=5m² area of load

Fl=? Force if load

Fe= 100N. Force of effort

Then applying pascal

Fl/Al=Fe/Ae

Fl/5=100/0.5

FL/5=200

Fl=200×5

Fl=1000N

The first safety load is 1000N

6 0
4 years ago
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