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ehidna [41]
2 years ago
13

Imagine an isolated positive point charge Q (many times larger than the charge on a single proton). There is a charged particle

A (whose charge is much smaller than charge Q) at a distance from the point charge Q. On which of the following quantities does the magnitude of the electric force on this charged particle A depend:_____.
1. the size and shape of the point charge Q
2. the specific location of the point charge Q (while the distance between Q and A is fixed)
3. the mass of the charged particle A
4. the size and shape of the charged particle A
5. the distance between the point charge Q and the charged particle A
6. the type of charge on the charged particle A
7. the mass of the point charge Q
8. the amount of the charge on the point charge Q
9. the magnitude of charge on the charged particle A
10. the specific location of the charged particle A (while the distance between Q and A is fixed)
11. the relative orientation between Q and A (while the distance between Q and A is fixed)
Physics
1 answer:
egoroff_w [7]2 years ago
7 0

Answer:

Explanation:

The magnitude of the electric force on this charged particle A depends upon the following

5. the distance between the point charge Q and the charged particle A

8. the amount of the charge on the point charge Q

9. the magnitude of charge on the charged particle A

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An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The curren
Maurinko [17]

i

CHECK COMPLETE QUESTION BELOW

inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.

Part A)What is the resistance RR of the inductor

PART B) what is inductance L of the conductor

Answer:

A)R=1818.18 ohms

B)L=1.0446H

Explanation:

We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.

There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,

.

a)A)What is the resistance RR of the inductor?

The current flowing into RL circuit can be calculated using below expresion

i=ε/R[1-e⁻(R/L)t]

at t=∞ there is maximum current

i(max)= ε/R

Where ε emf of the battery

R is the resistance

R=ε/i(max)

= 12V/(6.60*10⁻³A)

R=1818.18 ohms

Therefore, the resistance R=1818.18 ohms

b)what is inductance L of the conductor?

i(t=0.80ms and 4.96mA

RT/L = ⁻ln[1- 1/t(max)]

Making L subject of formula we have

L=-RT/ln[1-i/i(max)]

If we substitute the values into the above expresion we have

L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]

L=1.0446H

Therefore, the inductor L=1.0446H

3 0
3 years ago
What prominent sea floor fetaure is found in the central altlantic ocean
gayaneshka [121]
The prominent sea floor feature that is found in the central Atlantic Ocean is the Mid-Atlantic Ridge. 
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5 0
3 years ago
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Answer:

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Explanation:

4 0
2 years ago
A rectangular certificate has a perimeter of 32 inches. Its area is 63 square inches. What are the dimensions of the certificate
irina1246 [14]
Perimeter = 2 ( L + W )
32 = 2 ( L + W )
16 = L + W
L = 16 - W

Area = L W
63 = L W
63 = (16-W) W
63 = 16W - W²
-W² + 16 W - 63 = 0
By factorizing W = 9 or W = 7
So the dimensions are 7 and 9
3 0
2 years ago
A light year is approximately 9.5 million km long. 'Barnard's Star' is 6 light years away from Earth. Calculate how many million
hammer [34]

Answer:

6 light years = 57 million km

Explanation:

Given;

A light year = 9.5 million km

To calculate how far is 6 light years;

6 light years = 6 × 1 light year = 6 × 9.5 million km

6 light years = 57 million km

7 0
3 years ago
Read 2 more answers
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