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Gwar [14]
3 years ago
11

Suppose that the process were repeated, except that in step 3 a neutral acrylic rod instead of a finger is used to touch the ele

ctroscope's post, which is then removed in step 4. How would this modification affect the experiment just described? Select all correct statements The net charge on the vane would change following step 3, but revert to its original value following step 4. The net charge on the vane in all steps would be exactly as it was when the experimenter used a finger. The net charge on the vane would not change as a consequence of step 3. The net charge on the vane + post + disk would be zero for all steps. The net charge on the vane following steps 3 through 5 would have a sign opposite what it was when the experimenter used a finger.
Physics
1 answer:
Marta_Voda [28]3 years ago
8 0

Answer:

b) True Only if the finger is isolated from ground

c) True. The total charge does not change since the system is isolated

Explanation:

When the electroscope is touched with an acrylic rod, some charges are transferred from the electroscope to the rod, until the charge in both is equal.

In the case it know when the electroscope is touched with a finger, two things can happen.

- The body is isolated from the ground, the efective charge is redistributed between the two bodies. Case similar to insulating rod

- The body is connected to ground, the charge is transferred to the finger and from here to the ground until the total charge is transferred and the Earth and the final charge of the electroscope is zero.

Let's review the final statements

a) False, when part of the load is touched, it passes to the rod, so when it separates it does not return to the initial load

b) True Only if the finger is isolated from ground

c) True. The total load does not change since the system is isolated

d) False. The value of the load changes =, but its sign does not

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A 60.00-cm guitar string under a tension of 50.000 N has a mass per unit length of 0.100 00 g/cm chegg
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Fundamental frequency,

f=v2l=T/μ−−−−√2l

=(50)/0.1×10−3/10−22×0.6−−−−−−−−−−−−−−−−−−−√

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Let, n th harmonic is the hightest frequency, then

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∴N=339.38

Hence, 339 is the highest frequency.

∴fmax=(339)(58.93)Hz=19977Hz.

<h3>What is frequency?</h3>

In physics, frequency is the number of waves that pass a given point in a unit of time as well as the number of cycles or vibrations that a body in periodic motion experiences in a unit of time. After moving through a sequence of situations or locations and then returning to its initial position, a body in periodic motion is said to have experienced one cycle or one vibration. See also simple harmonic motion and angular velocity.

learn more about frequency refer:

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