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Lapatulllka [165]
3 years ago
12

How does charging by conduction occur?

Physics
1 answer:
lozanna [386]3 years ago
4 0

Hello~

The answer to your question is C.

A charged object touches a neutral object.

Explanation:

Even though it would have to be a (negative) charge, that point is that it is charge, so weather it is negative or positive, they are both charged what so ever.

~Hope this Helps!~

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Is the following sentence true or false? Two atoms of the same element can have different numbers of protons.
pickupchik [31]

Two atoms of the same element can have different numbers of protons

false

diff nos neutrons ... isotopes,

diff protons ... diff elements


4 0
3 years ago
Explain why the moon is always half illuminated and half dark no matter where it is in the lunar cycle.
BlackZzzverrR [31]
The easiest way I know to explain it is this:

-- Take a flashlight and a ball into a dark room.

-- Turn on the flashlight and point it at the ball.

-- Half of the ball is lighted up by the flashlight, and the other half is dark.

-- There is no way you can turn or twist the ball to make more or less
than 50% of it lighted up and more or less than 50% of it dark.

<em>Everything</em> in the solar system ... as long as it's shaped like a ball ... is
half illuminated by the sun and half dark.
7 0
3 years ago
Một chất điểm đang chuyển động với vận tốc 10(m/s) thì tăng tốc sau khi đi được 20(s) thì vật có vận tốc 20 (m/S)
Makovka662 [10]

Answer:8huhyuj

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5 0
2 years ago
You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int
uysha [10]
In a RC-circuit, with the capacitor initially uncharged,  when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
Q(t) = Q_0 (1-e^{-t/\tau})
where t is the time, Q_0 = CV is the maximum charge on the capacitor at voltage V, and \tau = RC is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using R=39.1 \Omega and C= 65.7 mF=65.7\cdot 10^{-3}F, the time constant of the circuit is:
\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s

2) To find the charge on the capacitor at time t=1.95 \tau, we must find before the maximum charge on the capacitor, which is
Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C
And then, the charge at time t=1.95 \tau is equal to
Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be Q_0 = 0.59 C. We can see this also from the previous formule, by using t=\infty:
Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C

4 0
3 years ago
How are energy and temperature related?
d1i1m1o1n [39]
Heat is in energy and temperature is measure of energy
5 0
3 years ago
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