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Dmitry [639]
1 year ago
9

HII!!! I need serious help with this assignment so I'm offering half my points so pLS accurately answer!! SO for this assignment

i need to know the order it goes in conducter, battery, or light bulb, battery conducter, light bulb, (ETC...etc....)
Please tell me which order they go in! Also please tell me why it is correct ??

Physics
1 answer:
ehidna [41]1 year ago
3 0

Answer:

Hey, I'm not sure I understand exactly what your assignment is but hopefully this helps . . .

put the other end against the side of the bubls base to light up the light bulb.

This will act as the bulbs conductor, lighting up bc you are making a complete electrical circuit between positive and negative terminals of the battery.

Ok hopefully this helps you out

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A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t
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Answer:

C=\frac{ab}{k(b-a)}

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]

Hence, the capacitance is

C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}

but R1=a and R2=b

C=\frac{ab}{k(b-a)}

HOPE THIS HELPS!!

3 0
3 years ago
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Select the correct answer. A ball is thrown straight down from the top of a building at a velocity of 16 ft/s. The building is 4
BartSMP [9]

Answer:

The ball takes 5s to reach the ground

Explanation:

in order to solve this problem we use the kinematics equation with gravity as acceleration:

h=v_0t+1/2*gt^2

we replace the values

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Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

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E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

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Answer:

Explanation:

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