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coldgirl [10]
2 years ago
12

On a guitar, the lowest toned string is usually strung to the E note, which produces sound at 82.4 82.4 Hz. The diameter of E gu

itar strings is typically 0.0500 0.0500 in and the scale length between the bridge and nut (the effective length of the string) is 25.5 25.5 in. Various musical acts tune their E strings down to produce a "heavier" sound or to better fit the vocal range of the singer. As a guitarist you want to detune the E on your guitar to A# ( 58.3 58.3 Hz). If you were to maintain the same tension in the string as with the E string, what diameter of string would you need to purchase to produce the desired note? Assume all strings available to you are made of the same material.

Physics
1 answer:
Vsevolod [243]2 years ago
8 0

The complete and comprehensive solution is attached.

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The gravitational field strength at a distance R from the center of moon is gR. The satellite is moved to a new circular orbit t
3241004551 [841]

Answer:

g'=\frac{g__R}{4}

Explanation:

Given:

  • gravitational field strength of moon at a distance R from its center, g__R
  • Distance of the satellite from the center of the moon, h=2R

<u>Now as we know that the value of gravity of any heavenly body is at height h is given as:</u>

g'=g__{R}} \times \frac{R^2}{(2R)^2}

g'=\frac{g__R}{4}

∴The gravitational field strength will become one-fourth of what it is at the surface of the moon.

6 0
2 years ago
According to a rule-of-thumb. every five seconds between a lightning flash and the following thunder gives the distance to the f
Bond [772]

Answer:

S_{s}=300 m/s

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

Explanation:

In order to use the rule of thumb to find the speed of sound in meters per second, we need to use some conversion ratios. We know there is 1 mile per every 5 seconds after the lightning is seen. We also know that there are 5280ft in 1 mile and we also know that there are 0.3048m in 1ft. This is enough information to solve this problem. We set our conversion ratios like this:

\frac{1mi}{5s}*\frac{5280ft}{1mi}*\frac{0.3048m}{1ft}=321.87m/s

notice how the ratios were written in such a way that the units got cancelled when calculating them. Notice that in one ratio the miles were on the numerator of the fraction while on the other they were on the denominator, which allows us to cancel them. The same happened with the feet.

The problem asks us to express the answer to one significant figure so the speed of sound rounds to 300m/s.

For the second part of the problem we need to use conversions again. This time we will write our ratios backwards and take into account that there are 1000m to 1 km, so we get:

\frac{5s}{1mi}*\frac{1mi}{5280ft}*\frac{1ft}{0.3048m}*\frac{1000m}{1km}=3.11s/km

This means that for every 3.11s there will be a distance of 1km from the place where the lightning stroke. Since this is a rule of thumb, we round to the nearest integer for the calculations to be made easily, so the rule goes like this:

The rule for kilometers is that every three seconds between a lightning flash and the following thunder gives the distance to the flash in kilometers.

3 0
2 years ago
three neutral metal cans mounted on isulating stands are touching a negatively charge ballon is brought near can a can b is then
Nimfa-mama [501]
Charge on can A is positive. 
Charge on can C is negative.  
Punctuation and capitalization are very useful things to pay attention to and this question would be a lot easier to understand if you had actually used both capitalization and punctuation. If I'm understanding the question, you have 3 metal can that are insulated from the environment and initially touching each other in a straight line. Then a negatively charged balloon is brought near, but not touching one of the cans in that line of cans. While the balloon is near, the middle can is removed. Then you want to know the charge on the can that was nearest the balloon and the charge on the can that was furthermost from the balloon. 
 As the balloon is brought near to can a, the negative charge on the balloon repels some of the electrons from can a (like charges repel). Some of those electrons will flow to can b and in turn flow to can c. Basically you'll have a charge gradient that's most positive on that part of the can that's closest to the balloon, and most negative on the part of the cans that's furthest from the balloon. You then remove can B which causes cans A and C to be electrically isolated from each other and prevents the flow of elections to equalize the charges on cans A and C when the balloon is removed. So you're left with a deficiency of electrons on can A, so can A will have a positive overall charge, and an excess of electrons on can C, so can C will have a negative overall charge.
7 0
2 years ago
Which of the following forms of radiation can penetrate up to a 2-cm layer of skin tissue?
Dmitry_Shevchenko [17]

Answer:

d). X-rays

Explanation:

X -rays are also called photons. They are a packet of electro magnetic radiation. X rays originate from the shell of the electron.  X rays are highly penetrating, and have a shorter wavelength than alpha particles and the beta particles. They are similar to the gamma rays which are also has a high penetrating power and easily pass through the human body.

 Thus X rays can penetrate a skin tissue upto 2 cm thickness.

4 0
2 years ago
Read 2 more answers
Air at 27 ºC and 5 atm is throttled by a valve to 1 atm. If the valve is adiabatic and the change in kinetic energy is negligibl
Citrus2011 [14]

Answer:

27 °C

Explanation:

BY the statement of the question it is clear that it is about an ideal gas - and hence if change in KE is about zero - then there will be no change of temperature.

So, answer is 27 °C

4 0
2 years ago
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