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andre [41]
4 years ago
14

A metal such as copper is a(n) _______________ because it provides a pathway for electric charges to move easily. A material suc

h as rubber is a(n) _______________ because it _______________ the flow of electric charges. A material that partially conducts electric current is a(n) _______________. These materials include _______________ elements.
Physics
2 answers:
notka56 [123]4 years ago
6 0

Explanation:

A metal such as copper is a <u>conductor</u> because it provides a pathway for electric charges to move easily. A material such as rubber is an <u>insulator</u> because it <u>resists</u> the flow of electric charges. A material that partially conducts electric current is a <u>semiconductor</u>. These materials include <u>group 3 and group 5</u> elements.

jeka57 [31]4 years ago
4 0

Answer:

conductor

insulator

resists

semiconductor

group 3 and group 5

Explanation:

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A baseball sits motionless near first base on a baseball diamond. What statement best explains why the baseball remains motionle
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Explanation:

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3 years ago
There are 92 naturally occurring elements. How does this relate to the number of compounds?
raketka [301]

C) Elements combine in different ways to create many more than 92 compounds.

Explanation:

The number of elements we have relates to the compound formation in that, elements combine in different ways to create many more than 92.

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3 0
4 years ago
A 4.4-µF capacitor is initially connected to a 5.1-V battery. Once the capacitor is fully charged the battery is removed and a 2
Grace [21]

Question is incomplete. Missing part:

Find the charge on the capacitor at the following times:

1) t = 0 mu S  

2) t = 1 mu S

3) t = 50 mu S

1) 22.4 \mu C

We start by calculating the initial charge on the capacitor. For this, we can use the following relationship:

C=\frac{Q_0}{V_0}

where

C is the capacitance

Q0 is the initial charge stored

V0 is the initial potential difference across the capacitor

When the capacitor is connected to the battery, we have:

C=4.4\mu F = 4.4\cdot 10^{-6}F

V_0 = 5.1 V

Solving for Q_0,

Q_0 = CV_0 = (4.4\cdot 10^{-6})(5.1)=2.24 \cdot 10^{-5} C = 22.4 \mu C

So, when the battery is disconnected, this is the charge on the capacitor at time t = 0.

2) 20.0\mu C

To find the charge on the capacitor at any other time t, we use the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

where

Q_0 = 22.4 \mu C

t is the time

R=2.0 \Omega is the resistance

C=4.4\mu F is the capacitance

Therefore, at time t=1 \mu s, we have:

Q(t) = (22.4) e^{-\frac{1}{(2.0)(4.4)}}=20.0 \mu C

3) 0.08 \mu C

As before, we use again the equation:

Q(t) = Q_0 e^{-\frac{t}{RC}}

However, here the time to consider is

t=50 \mu C

Substituting into the formula,

Q(t) = (22.4) e^{-\frac{50.0}{(2.0)(4.4)}}=0.08 \mu C

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