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sveticcg [70]
3 years ago
11

A 62.00-cm guitar string under a tension of 70.000 N has a mass per unit length of 0.10000 g/cm. What is the highest resonant fr

equency that can be heard by a person capable of hearing frequencies up to 18,000 Hz? (Enter your answer to at least five significant figures.)
Physics
1 answer:
quester [9]3 years ago
8 0

Answer:

17,947.02 Hz

Explanation:

length (L) = 62 cm = 0.62 m

tension (T) = 70 N

mass per unit length (μ) = 0.10000 g/cm = 0.010000 kg/m

maximum frequency = 18,000 Hz

f = \frac{n}{2L} x \sqrt{\frac{T}{μ}}

f = \frac{n}{2 x 0.62} x \sqrt{\frac{70}{0.01} }

f = n x 67.47

18,000 = n x 67.47

n = 266.8≈ 266

the 267th overtone is the highest overtone that can be heard by this person, and its frequency would be 26 x 67.47 = 17,947.02 Hz

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An electrostatic dust precipitator that is installed in a factory smokestack includes a straight metal wire of length 0.7 m that
ratelena [41]

Answer:

Check the explanation

Explanation:

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7 0
3 years ago
A block of mass 4 kilograms is initially moving at 5m/s on a horizontal surface. There is friction between the block and the sur
Dmitriy789 [7]

Answer:

d = 2.54 [m]

Explanation:

Through the theorem of work and energy conservation, we can find the work that is done. Considering that the energy in the initial state is only kinetic energy, while the energy in the final state is also kinetic, however, this is zero since the body stops.

E_{k1}+W=E_{k2}\\

where:

W = work [J]

Ek1 = kinetic energy at initial state [J]

Ek2 = kinetic energy at the final state = 0.

We must remember that kinetic energy can be calculated by means of the following expression.

\frac{1}{2} *m*v^{2}-W=0\\W= \frac{1}{2} *4*(5)^{2}\\W= 50 [J]

We know that work is defined as the product of force by distance.

W=F*d

where:

F = force [N]

d = distance [m]

But the friction force is equal to the product of the normal force (body weight) by the coefficient of friction.

f=m*g*0.5\\f = 4*9.81*0.5\\f = 19.62 [N]

Now solving the equation for the work.

d=W/F\\d = 50/19.62\\d = 2.54[m]

4 0
3 years ago
When you exert 75 N on a jack to lift a 6000 N car, what is the jack’s actual mechanical advantage? Show your work.
professor190 [17]

Answer:

80

Explanation:

<em>the </em><em>mechanical</em><em> </em><em>advantage</em><em> </em><em>is </em><em>the </em><em>ratio </em><em>of </em><em>the </em><em>load </em><em>to </em><em>the </em><em>effort</em><em> </em><em>so </em><em>it </em><em>doesn't</em><em> </em><em>have </em><em>units.</em><em>t</em><em>o</em><em> </em><em>calculate</em><em> </em><em>it </em><em>you </em><em>use </em><em>the </em><em>formula</em>

<em>mechanical</em><em> advantage</em><em>=</em><em>load/</em><em>effort</em>

<em>in </em><em>this</em><em> case</em><em> </em><em>the </em><em>load </em><em>is </em><em>6</em><em>0</em><em>0</em><em>0</em><em>N</em><em> </em><em>and </em><em>the </em><em>effort</em><em> </em><em>is </em><em>7</em><em>5</em><em>N</em>

<em>Ma=</em><em>6</em><em>0</em><em>0</em><em>0</em><em>/</em><em>7</em><em>5</em>

<em>=</em><em>8</em><em>0</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

3 0
3 years ago
Find the velocity in m/s of a swimmer who swims 110m toward the shore in 72 s
Sladkaya [172]
1.5 m/s toward shore 

8 0
3 years ago
A car travels at a constant speed around a circular track whose radiu is 2.6 km. The goes once arond the track in 360s . What is
AveGali [126]

Answer:

Centripetal acceleration = 0.79 m/s²

Explanation:

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

Radius, r = 2.6 km

Time = 360 seconds

<em><u>Conversion:</u></em>

2.6 km to meters = 2.6 * 1000 = 2600 meters

To find the magnitude of centripetal acceleration;

First of all, we would determine the circular speed of the car using the formula;

Circular \; speed (V) = \frac {2 \pi r}{t}

Where;

  • r represents the radius and t is the time.

Substituting into the formula, we have;

Circular \; speed (V) = \frac {2*3.142*2600}{360}

Circular \; speed (V) = \frac {16338.4}{360}

Circular speed, V = 45.38 m/s

Next, we find the centripetal acceleration;

Mathematically, centripetal acceleration is given by the formula;

Centripetal \; acceleration = \frac {V^{2}}{r}

Where;

  • V is the circular speed (velocity) of an object.
  • r is the radius of circular path.

Substituting into the formula, we have;

Centripetal \; acceleration = \frac {45.38^{2}}{2.6}

Centripetal \; acceleration = \frac {2059.34}{2600}

<em>Centripetal acceleration = 0.79 m/s²</em>

3 0
3 years ago
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