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slava [35]
3 years ago
13

What is the equivalent resistance of a circuit that contains two 50.00

Physics
2 answers:
tigry1 [53]3 years ago
5 0

Answer:

100.0

Explanation:

A

P

E

X

Whitepunk [10]3 years ago
4 0

Answer:

A

Explanation:

Resistors in series add. There is only one path the current can take. That's why Christmas Tree lights sometimes give a lot of trouble. If a bulb burns out, it could be any one of them and time is needed to find the burned out bulb.

That being the case R = R1 + R2

R1 = 50 ohms

R2 = 50 ohms

R = 50 + 50

R = 100 ohms

Answer A

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Answer:

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Explanation:

so all but one light could be burned out, and the last one will still function.

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Which is a characteristic of the atom marked A
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Its very dense. Hey, are you homeschooled?
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You are 12 miles north of your base camp when you begin walking north at a speed of 2 miles per hour. What is your location, rel
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Why do atoms form bonds by donating, accepting or sharing electrons with other atoms?
sineoko [7]

Answer:

atoms form bonds by donating, accepting or sharing electrons with other atoms in order to complete their valence shell electrons

hence , C. Bonding gives an atom the same number of protons as a noble gas.

Explanation:

i hope it helped

7 0
3 years ago
An aluminum wire having a cross-sectional area equal to 2.20 10-6 m2 carries a current of 4.50 A. The density of aluminum is 2.7
Kazeer [188]

Answer:

The drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

Explanation:

We can find the drift speed by using the following equation:

v = \frac{I}{nqA}

Where:

I: is the current = 4.50 A

n: is the number of electrons

q: is the modulus of the electron's charge = 1.6x10⁻¹⁹ C

A: is the cross-sectional area = 2.20x10⁻⁶ m²

We need to find the number of electrons:

n = \frac{6.022\cdot 10^{23} atoms}{1 mol}*\frac{1 mol}{26.982 g}*\frac{2.70 g}{1 cm^{3}}*\frac{(100 cm)^{3}}{1 m^{3}} = 6.03 \cdot 10^{28} atom/m^{3}                  

Now, we can find the drift speed:

v = \frac{I}{nqA} = \frac{4.50 A}{6.03 \cdot 10^{28} atom/m^{3}*1.6 \cdot 10^{-19} C*2.20 \cdot 10^{-6} m^{2}} = 2.12 \cdot 10^{-4} m/s              

Therefore, the drift speed of the electrons in the wire is 2.12x10⁻⁴ m/s.

I hope it helps you!      

4 0
2 years ago
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