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AlladinOne [14]
3 years ago
13

Which statement is true regarding copper's ability to conduct electricity? Select one: A. Copper is a good conductor of electric

ity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms. B. Copper is a poor conductor of electricity because its atoms are held in positions that cannot move. C. Copper is a good conductor of electricity because its atoms have electrons that are tightly bound to their shells and will resist movement. D. Copper is a poor conductor of electricity because it has a free electron in its outer shell that will flow to other copper atoms.
Physics
2 answers:
Elan Coil [88]3 years ago
5 0
<span>A.  Copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms.</span>
lidiya [134]3 years ago
5 0

Answer: Option (A) is the correct answer.

Explanation:

Copper is a transition metal and its atomic number is 29. The electrons in its shell are distributed as 2, 8, 18, 1.

As there is only one electron in the outer most shell and as it is far away from the nucleus, therefore, it is a loosely held electron. Also being a metal, copper is a good conductor of heat and electricity.

Thus, we can conclude that out of the given options the statement copper is a good conductor of electricity because its atoms have a loosely held electron in their outer shell that is able to move freely to other atoms is true regarding copper's ability to conduct electricity.

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An energy source forces a constant current of 2A to flow through a light bulbfilament for twenty seconds. If 4.6 kJ is given off
QveST [7]

Answer:

The voltage drop across the bulb is 115 V

Explanation:

The voltage drop equation is given by:

V=\frac{\Delta W}{\Delta q}

Where:

ΔW is the total work done (4.6kJ)

Δq is the total charge

We need to use the definition of electric current to find Δq

I=\frac{\Delta q}{\Delta t}

Where:

I is the current (2 A)

Δt is the time (20 s)

2=\frac{\Delta q}{20}

q=40 C

Then, we can put this value of charge in the voltage equation.

V=\frac{4600}{40}=115 V

Therefore, the voltage drop across the bulb is 115 V.

I hope it helps you!

6 0
3 years ago
Can someone help me ​
Julli [10]
The answer is the first one
6 0
3 years ago
Read 2 more answers
A circuit contains two 1.5 volt battery and a bulb with a resistance of 3 ohms. Calculate the current
horrorfan [7]

Answer:

<em>The current is 1 A</em>

Explanation:

<u>Current in a Series Connection </u>

When two or more elements are connected in series, all of them have the same current, and the sum of their individual voltages is the total voltage applied to the circuit.

According to Ohm's law:

V=R.I

Where V is the voltage, R is the resistance and I is the current of a circuit.

We have a voltage of V=1.5 V + 1.5 V = 3 V and a resistance of R=3 ohms.

We can calculate the current by solving for I:

\displaystyle I=\frac{V}{R}=\frac{3}{3}=1\ A

The current is 1 A

3 0
2 years ago
A raindrop of radius r falls from a certain height h above the ground. The work done by
lisabon 2012 [21]

Answer:

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Explanation:

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5 0
3 years ago
A 15 kg box is sliding down an incline of 35 degrees. The incline has a coefficient of friction of 0.25. If the box starts at re
valina [46]

The box has 3 forces acting on it:

• its own weight (magnitude <em>w</em>, pointing downward)

• the normal force of the incline on the box (mag. <em>n</em>, pointing upward perpendicular to the incline)

• friction (mag. <em>f</em>, opposing the box's slide down the incline and parallel to the incline)

Decompose each force into components acting parallel or perpendicular to the incline. (Consult the attached free body diagram.) The normal and friction forces are ready to be used, so that just leaves the weight. If we take the direction in which the box is sliding to be the positive parallel direction, then by Newton's second law, we have

• net parallel force:

∑ <em>F</em> = -<em>f</em> + <em>w</em> sin(35°) = <em>m a</em>

• net perpendicular force:

∑<em> F</em> = <em>n</em> - <em>w</em> cos(35°) = 0

Solve the net perpendicular force equation for the normal force:

<em>n</em> = <em>w</em> cos(35°)

<em>n</em> = (15 kg) (9.8 m/s²) cos(35°)

<em>n</em> ≈ 120 N

Solve for the mag. of friction:

<em>f</em> = <em>µ</em> <em>n</em>

<em>f</em> = 0.25 (120 N)

<em>f</em> ≈ 30 N

Solve the net parallel force equation for the acceleration:

-30 N + (15 kg) (9.8 m/s²) sin(35°) = (15 kg) <em>a</em>

<em>a</em> ≈ (54.3157 N) / (15 kg)

<em>a</em> ≈ 3.6 m/s²

Now solve for the block's speed <em>v</em> given that it starts at rest, with <em>v</em>₀ = 0, and slides down the incline a distance of ∆<em>x</em> = 3 m:

<em>v</em>² - <em>v</em>₀² = 2 <em>a</em> ∆<em>x</em>

<em>v</em>² = 2 (3.6 m/s²) (3 m)

<em>v</em> = √(21.7263 m²/s²)

<em>v</em> ≈ 4.7 m/s

4 0
3 years ago
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