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matrenka [14]
3 years ago
7

A 5.0-Ω resistor and a 9.0-Ω resistor are connected in parallel. A 4.0-Ω resistor is then connected in series with this parallel

combination. An ideal 6.0-V battery is then connected across the series-parallel combination of the three resistors. What is the current through (a) the 4.0-Ω resistor? (b) the 5.0-Ω resistor? (c) the 9.0-Ω resistor?
Physics
1 answer:
Oliga [24]3 years ago
4 0

Answer:

Explanation:

The current through the  resistor is 0.83 A

.

Part b

The current through  resistor is 0.53 A

.

Part c

The current through  resistor is 0.30 A

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Answer:

A. The cart will move to the left

Explanation:

Right has more force acting on the cart than the left

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State and explain the principle of superposition of electrostatic force​
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Answer:

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8 0
3 years ago
An electric field of 1139 V/m is applied to a section of silver of uniform cross section. Find the resulting current density if
N76 [4]

Answer:

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So, the next equation you need is J=σE where E is your electric field (3026 V/m) and σ is the electrical conductivity which is the inverse of your answer you got in the previous equation. So find the inverse of that answer and multiply it by your electric field and that will give you the current density.  

I hope this helps!

Explanation:

8 0
3 years ago
The planets Hox and Blox are near each other in the Dorgon system. The Dorgons have very advanced technology, and a Dorgon scien
dexar [7]
<h2>Answer: Decreasing the distance between Hox and Blox, increasing the mass of Hox, or increasing the mass of Hox and Blox. </h2>

Explanation:

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G is the universal gravitation constant.  

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In other words:

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7 0
3 years ago
Read 2 more answers
A current of 0.92 a flows in a wire. how many electrons are flowing past any point in the wire per second? the charge on one ele
Fantom [35]
The current is defined as the ratio between the charge Q flowing through a certain point of a wire and the time interval, \Delta t:
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Now we know that each electron carries a charge of e=1.6 \cdot 10^{-19} C, so if we divide the charge Q flowing in the wire by the charge of one electron, we find the number of electron flowing in one second:
N= \frac{Q}{q} = \frac{0.92 C}{1.6 \cdot 10^{-19} C}=5.75 \cdot 10^{18}
3 0
3 years ago
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