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4 years ago

How does tightening a string on a guitar affect its natural frequency?

2 answers:

6 0

Well, this is the answer in which, will be selected as a correct answer, on a answer choice selection: the natural frequency increases

iragen [17]4 years ago

3 0

By tightening a string you are actually putting more stress on the string you are giving it a new frequency that isn't natural.

Hope this helps
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You might be interested in

at what pressure range does the technician valve off the vacuum pump when evacuating to a deep vacuum?

choli [55]

When transferring a vacuum furnace chamber from atmospheric pressure to a supposedly "rough" vacuum, the roughing valve is utilized (typically in the range of 10 -2 to 10 -3 Torr).

This degree of vacuum is sometimes sufficient to carry out the targeted process. Another common name for vacuum is negative pressure (or soft vacuum). This happens when an application needs to keep track of both pressure increases and pressure drops in both directions above and below the atmosphere. The difference in pressure between the exhaust exit and input provides the basis for a vacuum.

We all know that water and air travel from high altitude to low altitude, and that a vacuum is produced by a difference in pressure between the two.

To know more about vacuum furnace

brainly.com/question/28965352

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8 0

1 year ago

A ball with a mass of 160 g which contains 4.40×108 excess electrons is dropped into a vertical shaft with a height of 115 m . A

faltersainse [42]

Answer:

$F= 6.69*10^{-10}N$

Explanation:

To find the force we proceed by defining the variables we have,

$m= 160g\\n= 4.40*10^8\\h=115m\\B=0.2T\\e= 1.602*10^{-19}c$

The charge on one of the balls is defined under the equation,

$q=ne$

$q= 4.40*10^8(1.602*10^{-19}c)$

$q= 7.0488*10^{-11}$

Due to the height we need to calculate the potential energy at the height of 115m,

$PE=mgh$

The kinetic energy would be given by

$KE= \frac{1}{2}mv^2$

From the law of conservation we equate the two equations

$\frac{1}{2}mv^2=mgh$

$v=\sqrt{2gh}$

$v=\sqrt{2(9.8)(115)}$

$v= 47.47m/s$

In this way we now calculate the strength of the particle

$F=qVB$

$F= (7.0488*10^{-11})(47.47)(0.2)$

$F= 6.69*10^{-10}N$

8 0

3 years ago

An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries

jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

$V_{plant} = 1500$ KV

L = 143 km

I = 500 A

$\rho = 2.4$ $\Omega / km$

So, we will find the voltage potential provided for the city as:

$V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8$ kV

$V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2$ kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

$P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7$ W

Since the charge of plant is not given for electric energy, let's assume it randomly as $x = \frac{\dollar 0.081}{kW.hr}$

Then, we will find the price of energy transmitted to the city as:

$Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8$ $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

$I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25$ $/hr

The amount of money saved per hour = $6949.8 - 6949.8/1.44 = 2123.55$ $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0

3 years ago

QUICKK I NEED HELP!!<br> How is a mineral different from an element rock

Irina18 [472]

Answer:here you goo

Explanation:

A mineral is a naturally occurring inorganic element or compound having an orderly internal structure and characteristic chemical composition, crystal form, and physical properties. ... A rock is an aggregate of one or more minerals, or a body of undifferentiated mineral matter.

7 0

3 years ago

Read 2 more answers

Which term describes what a british thermal unit measures?.

Yakvenalex [24]

Answer:

it is a measure of the heat content of fuels or energy sources

Explanation:

8 0

2 years ago