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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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Slav-nsk [51]

Answer:

Yes, the errors are likely to be relevant

Explanation:

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3 0
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Answer:

The gauge pressure is  P_g  =  2058 \ P_a

Explanation:

From the question we are told that

       The height of the water contained is  h_w  =  30 \ cm  =  0.3 \ m

        The height of liquid in the cylinder is  h_t  =  40 \ cm  = 0.4 \ m

       

At the bottom of the cylinder the gauge pressure is  mathematically represented as

        P_g  =  P_w + P_o

Where  P_w is the pressure of water which is mathematically represented as

      P_w  =  \rho_w  *  g * h_w

Now  \rho_w is the density of water with a constant values of  \rho_w  = 1000 \ kg /m^3

   substituting values

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     P_w  =  2940 \  Pa

While P_o is the pressure of oil which is mathematically represented as

          P_o  =  \rho_o *  g *  (h_t -h_w )

Where \rho _o is the density of oil with a constant value

         \rho _o  = 900 \ kg / m^3

substituting values

       P_o  =  900 *  9.8 * (0.4 - 0.3)

       P_o  =  882 \ Pa

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3rd order polynomial

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Given that the increase in the order of the polynomial the error between the curve fit and measured data will decreases hence  :

The polynomial order that is best to use is the 3rd order polynomial, this is because using a 3rd order polynomial will produce a less variance and a low Bias

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