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DaniilM [7]
4 years ago
6

Consider the following. (Let C1 = 20.80 µF and C2 = 14.80 µF.) A rectangular circuit contains a battery and four capacitors. The

bottom side has a 9.00 V battery with the positive terminal on the left. The left and right sides of the circuit each contain a capacitor labeled C1. The top side splits into two parallel horizontal branches, which recombine before reaching the top right corner. There is a 6.00 µF capacitor on the upper branch and a capacitor labeled C2 on the lower branch.
(a) Find the equivalent capacitance of the capacitors in the figure. µF

(b) Find the charge on each capacitor. on the right 20.80 µF capacitor µC on the left 20.80 µF capacitor µC on the 14.80 µF capacitor µC on the 6.00 µF capacitor µC

(c) Find the potential difference across each capacitor. on the right 20.80 µF capacitor V on the left 20.80 µF capacitor V on the 14.80 µF capacitor V on the 6.00 µF capacitor V

Physics
1 answer:
umka21 [38]4 years ago
7 0

Find the below attachment

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4 0
3 years ago
Two insulated wires, each 2.64 m long, are taped together to form a two-wire unit that is 2.64 m long. One wire carries a curren
nikklg [1K]

Answer:

4.77\ \text{A}

Explanation:

F = Magnetic force = 4.11 N

I_n = Net current

I_2 = Current in one of the wires = 7.68 A

B = Magnetic field = 0.59 T

\theta = Angle between current and magnetic field = 65^{\circ}

l = Length of wires = 2.64 m

I = Current in the other wire

Magnetic force is given by

F=I_nlB\sin\theta\\\Rightarrow I_n=\dfrac{F}{lB\sin\theta}\\\Rightarrow I_n=\dfrac{4.11}{2.64\times 0.59 \sin65^{\circ}}\\\Rightarrow I_n=2.91\ \text{A}

Net current is given by

I_n=I_2-I\\\Rightarrow I=I_2-I_n\\\Rightarrow I=7.68-2.91\\\Rightarrow I=4.77\ \text{A}

The current I is 4.77\ \text{A}.

8 0
3 years ago
An infinite line of charge with linear density λ1 = 6 μC/m is positioned along the axis of a thick insulating shell of inner rad
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Answer: λ2= 2.34 * 10^-6 C/m

Explanation: In order to calculate the value of the  linear charge density of the insulating shell we have to multiply ρ* Volume of the hollow cylinder, so

Volume of cylinder:2*π*b*L *(b-a)  where (b-a) is the thickness, then

λ2=Q/L = 634 *10^-6 C/m^3* 2*π*0.042 m*(0.042-0.26)== 2.34 μ C/m

5 0
4 years ago
Shay reacts solid zinc and aqueous copper sulfate to form aqueous zinc sulfate and solid copper. If he reacts 10.1 grams of zinc
zhenek [66]

Answer:

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

Explanation:

As we know that zinc reacts with copper sulfate

so the reaction is given as

Zn + CuSO_4 --> ZnSO_4 + Cu

so here we have

Zn = 10.1 g

CuSO_4 = 18.6 g

ZnSO_4 = 20 g

Cu = 8.7 g

Now total mass of reactant is given as

M_1 = 10.1 + 18.6 = 28.7 g

Mass of the product is given as

M_2 = 20 + 8.7 = 28.7 g

Now since mass of reactant is equal to mass of the product after the reaction so we can say that mass conservation is applicable here

7 0
4 years ago
The weight of an objects can be calculated by multiplying mass by
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Acceleration due to gravity. 
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3 years ago
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