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Natasha2012 [34]
3 years ago
11

The guitarist shortens the oscillating length of the properly tuned D-string by 0.15 m by pressing on the string with a finger.

What is the fundamental frequency, in hertz, of the new tone that is produced when the string is plucked?
Physics
1 answer:
Greeley [361]3 years ago
3 0

Answer:

The fundamental frequency is "190.52 Hz".

Explanation:

The given question is incomplete, please find attachment of the complete problem.

The given values are:

Frequency,

f = 146.8 Hz

Length of D-string,

L = 0.61

As we know,

⇒ f = \frac{v}{2L}

On putting the given values, we get

⇒ 146.8 = \frac{v}{(2\times 0.61)}

⇒       v=146.8\times 1.22

⇒       v=179 \ m/s

So,

The new length will be:

⇒  f = \frac{179}{(2\times 0.47)}

⇒     =\frac{179}{0.94}

⇒     =190.52 \ Hz

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algol13

Answer:

None

Explanation:

Force, F is given by ma where m is the mass of an object and a is acceleration

Acceleration is the rate of change in velocity per unit time. Since skaters with mass of 75 kg and 50 kg are moving at a constant speed, there is no acceleration hence F=50*0=0 and F=75*0=0

For skater of 45 kg, he is at rest to mean the initial and final velocitu of the skater is zero hence no acceleration, the force will be 45*0=0

Therefore, none of the skaters will experience a greater net force.

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Explain the process of why the balloon is attracted to the wall, and why electrons are not transferred in this process. Is the w
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Answer:

The process by which the balloon is attracted and possibly sticks to the wall is known as static electricity which is the attraction or repulsion between electric charges which are not free to move.

The wall is an insulator.

Explanation:

When a balloon is blown and tied off, and then the balloon is rubbed on the woolly object once in one direction, and the side that was rubbed against the wool is brought near a wall and then released, it is observed that the balloon is attracted to and sticks to the wall. The above observation is due to static electricity.

Static electricity refers to electric charges that are not free to move or that are static. One of the means of generating such charges is by friction. When the balloon is rubbed on the woollen material, electrons are given away to the balloon's surface. Since the balloon is an insulator (materials which do not allow electricity to pass through them easily), the electrons are not free to move. When the balloon is brought near to a wall, there is a rearrangement of the charges present on the wall. Negative charges on the wall move farther away while the positive charges on the wall are attracted to the electrons on the balloon's surface. Because the wall is also an insulator, the charges are not discharged immediately. Therefore, this attraction between opposite charges as well as the static nature of the charges results in the balloon sticking to the wall.

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Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect
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Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

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We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b)  r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 \pi r^{2}  = \frac{Q1}{E_{0} }

So,

Rearranging the above equation to get Electric field, we will get:

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   }

Multiply and divide by r1^{2}

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } x \frac{r1^{2} }{r1^{2} }

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

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E x 4 \pi r^{2}  = \frac{Q1 + Q2}{E_{0} }

Rearranging the above equation for E:

E = \frac{Q1+Q2}{E_{0} . 4 \pi. r^{2}   }

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

As we know from above, that:

\frac{Q1}{E_{0} . 4 \pi. r^{2}   } =  (σ1 x r1^{2}) /(E_{0} x r^{2})

Then, Similarly,

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So,

E = \frac{Q1}{E_{0} . 4 \pi. r^{2}   } + \frac{Q2}{E_{0} . 4 \pi. r^{2}   }

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d) Under what conditions,  E = 0, for r > r2?

For r > r2, E =0 if

σ1 x r1^{2} = - σ2 x r2^{2}

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