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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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A woman can row a boat at 5.60 km/h in still water. (a) If she is crossing a river where the current is 2.80 km/h, in what direc
katrin2010 [14]

Answer:

a) θ=210°, b) t=1.155hr, c) t=1.333hr, d) t=1.333hr, e) θ=180° (straight across), f) t=1hr.

Explanation:

So, the very first thing we nee to do when solving this problem is draw a diagram that represents it. In the attached picture I show a diagram for each part of this problem.

part a)

So, for her to move in a direction directly opposite her starting point, the x-component of her velocity must be de same as the velocity of the river in the opposite direction. We can use this fact to find the angle we need. If we analize the triangle I drew in the diagram, we can ses that:

cos \theta = \frac {V_{river}}{V_{boat}}

When solving for theta, we get that:

\theta =cos^{-1} ( \frac {V_{river}}{V_{boat}})

so now we can substitute the corresponding values:

\theta =cos^{-1} ( \frac {2.80km/hr}{5.60km/hr}})

Which yields:

\theta = 60^{o}

but we are measuring the angle relative to the line perpendicular to the river, positive if down the river. So we need to subtract the angle from 270° so we get:

θ=270°-60°=210°

part b)

for part b, we need to find what the y-component for the velocity of the boat is for an angle of 210° as shown in the problem, so we get that:

V_{y}=5.60km/hr*cos(210^{o})

V_{y}=-4.85km/hr

The woman will head in a negative 5.60km distance from one side to the other, so we get that the time it takes her to go to the other side of the river is:

t=\frac{y}{V_{y}}

t=\frac{5.60km}{4.85km/hr}=1.155hr

part c)

In order to find the time it takes her to travel 2.80km down and up the river, we need to find the velocities she will have in both directions. First, down stream:

V_{ds}=V_{river}+V{boat}

V_{ds}=2.80km/hr+5.60km/hr=8.40km/hr

and now up stream:

V_{us}=V_{boat}-V{river}

V_{us}=5.60km/hr-2.80km/hr=2.80km/hr

Once we got these two velocities we will now need to find the time to take each trip:

time down stream:

t_{ds}=\frac{x}{v_{ds}}

t_{ds}=\frac{2.80km}{8.40km/hr}=0.333hr

and the time up stream:

t_{us}=\frac{x}{v_{us}}

t_{us}=\frac{2.80km}{2,80km/hr}=1hr

so the total time will be:

t_{ds}+t_{us}=0.333hr+1hr=1.333hr

d) the time it takes the boat to go upstream and then downstream for the same distance is the same as the time we got on part c, since both times will be the same but they will come in different order, but their sum will be just the same:

t=1.333hr

e) For her to cross the river faster, she must row in a 180° direction (this is in a direction straight accross the river) that way she will use all her velocity to move across the river. (Even though she will move a certain distance horizontally and will not reach a point opposite to the starting point.)

f) In order to find the time it takes her to get to the other side, we need to divide the distance into the velocity of the boat.

t=\frac{d}{v_{boat}}

t=\frac{5.60km}{5.60km/hr}

so

t= 1hr

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4 years ago
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alexgriva [62]
The <span>physical properties of non-metals are as below:

1) Poor conductors - generally non-metals are poor conductors of heat and electricity.
2)Low melting points - all metals have low melting and boiling points, with graphite being an exception.
3) Brittle - all nonmetals are brittle in nature.</span>
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3 years ago
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Si un movil parte del reposo logrando una aceleracion de 5 metros por segundo al cuadrado durante 8 segundos calcular la velocid
Komok [63]

Answer:

Velocidad final, V = 40 m/s

Explanation:

Dados los siguientes datos;

Aceleración = 5 m/s²

Velocidad inicial = 0 m/s (ya que comienza desde el reposo)

Tiempo = 8 segundos

Para encontrar la velocidad final, usaríamos la primera ecuación de movimiento;

V = U + at

Dónde;

  • V es la velocidad final.
  • U es la velocidad inicial.
  • a es la aceleración.
  • t es el tiempo medido en segundos.

Sustituyendo en la fórmula, tenemos;

V = 0 + 5*8

V = 0 + 40

V = 40

<em>Velocidad final, V = 40 m/s</em>

6 0
3 years ago
Calculate the average speed of an airplane that traveled 1,100 miles in 3.5 hours
Ivan

Answer:

314.29 m/hrs

Explanation:

v = d / t

v = 1100 m / 3.5 hrs

v = 314.29 m/hrs

Note that I around the answer to the nearest hundredth. Hope this helps, thank you :) !!

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3 years ago
If earth did not rotate how would air at the equator move?
Dmitry_Shevchenko [17]
Heat rises, and it is warmer at the equator, so I think warm air would rise at the equator and move towards the cooler poles.
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4 years ago
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