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Alexus [3.1K]
3 years ago
13

A plastic rod that has been charged to -21 nc touches a metal sphere. afterward, the rod's charge is -8 nc. (a) what kind of cha

rged particle was transferred between the rod and the sphere, and in which direction? that is, did it move from the rod to the sphere or from the sphere to the rod?
Physics
1 answer:
otez555 [7]3 years ago
3 0
The rod transferred negatively charged particle to the metal sphere. The rod becomes less negatively charged than it was originally (originally it was charged to -21 nc, now it has -8, and -8>-21), so it must have moved 13 nc of negative charge to the sphere. 
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Law of inertia would be your answer.

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Two bulbs are connected in parallel across a source of EMF = 8.0V with a negligible internal resistance. One bulb has a resistan
Sonja [21]
<h2>Answer:</h2>

(a) 3.18Ω

(b) 3.18Ω

<h2>Explanation:</h2>

Let the two bulbs be A and B

Given;

R_{A} = Resistance in bulb A = 3.0Ω

R_{B} = Resistance in bulb B = 2.5Ω

Since the two bulbs are connected in parallel;

i. their effective resistance (R_{X}) is given by

\frac{1}{R_{X}} = \frac{1}{R_{A} } + \frac{1}{R_{B} }  ---------------(i)

Substitute the values of R_{A} and R_{B} into equation (i)

=> \frac{1}{R_{X}} = \frac{1}{3.0} + \frac{1}{2.5}

Solve for  R_{X}

R_{X} = 1.36Ω

ii. voltage (potential difference), V, across them is the same;

Therefore we can get the total current (I) that will flow through them if the voltage to be supplied is 2.4V.

Use the Ohm's law;

V = I x R    -----------------(ii)

Where;

V = voltage across them = 2.4V

I = total current flowing through them

R = their effective resistance = R_{X} = 1.36Ω

Substitute these values into equation (ii) as follows;

2.4  = I x 1.36

I = 2.4 / 1.36

I = 1.76A

(a) Now get the value of R

Since the voltage across the two bulbs is 2.4V out of the 8.0V supplied by the source, then the remaining (8.0 - 2.4 = 5.6)V will pass across the resistor R.

Also, since the two bulbs make a series connection with the resistor R, the same total current (I = 1.76A) that flows through these bulbs will flow through the resistor R.

Therefore, to get the value of R, we use the relation

V = I x R   ------------------------------(iii)

Where;

V = voltage across the resistor = 5.6V

I = current through the resistor = 1.76A

<em>Substitute these values into equation (iii)</em>

=> 5.6 = 1.76 x R

=> R  = 5.6 / 1.76

=> R = 3.18Ω

Therefore, the value of R to be chosen in order to supply each bulb with a voltage of 2.4V is 3.18Ω

(b) The potential difference and voltage across refer to the same thing. Therefore, the value of R that would make the potential difference across each of the bulbs be 2.4V is the same as the one calculated in (a) which is 3.18Ω

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I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
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If you were to triple the size of the Earth (R = 3R⊕) and double the mass of the Earth (M = 2M⊕), how much would it change the g
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Answer:

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\frac{F_G}{f_g} = \frac{M_2}{R^2}\frac{r^2}{m_2}

\frac{F_G}{f_g} = \frac{M_2}{m_2}(\frac{r}{R})^2

Since M_2 = 2m_2 and r = R/3

\frac{F_G}{f_g} = \frac{2}{3^2} = 2/9 = 1/4.5

So gravity would have been decreased by a factor of 4.5  

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