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torisob [31]
3 years ago
13

WILL MARK BRAINLIEST. Look at the circuit below. What is the voltage between points C and D?

Physics
2 answers:
lianna [129]3 years ago
7 0

The hard way:

-- The total resistance in the series circuit is 120 ohms.

-- The current in the circuit is (voltage)/(resistance) = 3/60 = 0.025 A

-- The resistance between C and D is 60 ohms.

-- The voltage across a resistance is (current) x (resistance) = (0.025 x 60) = <em>1.5 volts. </em>(C)

<em>=================================</em>

The easy way:

-- In a series circuit, the total voltage of the power source divides among the circuit's components in proportion to their resistances.

-- Since the bulbs have equal resistances, each one will have 1/2 of the battery voltage across it.

-- (1/2) of (3 volts) = <em>1.5 volts</em> (C)

elena-s [515]3 years ago
6 0

Since two identical bulbs are connected in this circuit in series combination

So here same current will flow into the circuit

So here we can say

V = V_1 + V_2

V = 3 volts

also we know that

V_1 = iR_1

V_2 = iR_2

also it is given that bulbs are identical so we have

V_1 = V_2

as we have

R_1 = R_2

so now we can say

V = 2V_{CD}

3 = 2V_{CD}

V_{CD} = 1.5 Volts

so here voltage across C and D is 1.5 volts

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As per Newton's 2nd law

we know that

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here we know that

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also we know that

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F = 1670 * 6

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3 years ago
An object in motion will have a speed which is a scaler , or ( blank ) which is a vector .
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Velocity

Explanation:

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3 0
3 years ago
How was the Periodic Table of Elements developed and how are the elements arranged on it?
victus00 [196]

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Explanation:

4 0
3 years ago
Calculate the length of wire.<br>​
galben [10]

Answer:

L = 169.5 m

Explanation:

Using Ohm's Law:

V = IR

where,

V = Voltage = 1.5 V

I = Current = 10 mA = 0.01 A

R = Resistance = ?

Therefore,

1.5 V = (0.01 A)R

R = 150 Ω

But the resistance of a wire is given by the following formula:

R = \frac{\rho L}{A}

where,

ρ = resistivity = 1 x 10⁻⁶ Ω.m

L = length of wire = ?

A = cross-sectional area of wire = πr² = π(0.6 mm)² = π(0.6 x 10⁻³ m)²

A = 1.13 x 10⁻⁶ m²

Therefore,

150\ \Omega = \frac{(1\ x\ 10^{-6}\ \Omega .m)L}{1.13\ x\ 10^{-6}\ m^2}\\\\L = \frac{150\ \Omega(1.13\ x\ 10^{-6}\ m^2)}{1\ x\ 10^{-6}\ \Omega .m}\\\\

<u>L = 169.5 m</u>

7 0
3 years ago
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