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3 years ago

Which one is correct and why?

Physics

1 answer:

Gelneren [198K]3 years ago

6 0

'A' and 'C' are exactly the same circuit, except the voltmeter's terminals are flipped.

'A' is the correct way to hook everything up.

If you start at the positive terminal of the battery, and follow the flow of current through the circuit and around to the negative terminal, you're following the path where the voltage gets lower and lower and lower all the way.

So each time you come to any device in the circuit ... whether it's a resistor or a meter ... you would be hitting the positive side of it first, and then the voltage where you come out on the other side of it would be lower.

So the left side of the resistor is more positive, and the right side is more negative. The voltmeter is connected correctly in 'A', but it's backwards in 'C'. If you connect the voltmeter like in 'C' and turn things on, the voltmeter will try to go down from zero. You can't read the number on it, and It's possible that the voltmeter might be damaged.

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abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

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3 years ago

Using the equation zeff=z−s and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screen

DIA [1.3K]

Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help.

Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 
Zeff = Z-S = 17- 10 =7 
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For an object starting from rest and accelerating with constant acceleration, distance traveled is proportional to the square of

natali 33 [55]

The problem states that the distance travelled (d) is directly proportional to the square of time (t^2), therefore we can write this in the form of:

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3 years ago

Which one is correct and why?​

Which one is correct and why?