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2 years ago

Can inserting a resistor in a circuit produce an effect similar to a short circuit?

Physics

1 answer:

tresset_1 [31]2 years ago

6 0

Answer:

By ‘inserting’ you means to putting a resistor in series. In this case, no, there is no resistance that would produce the same effect as a short circuit.

If you adding a resistor in parallel with the circuit, then if it had a low value It might be similar to a short circuit. I

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A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-

Snezhnost [94]

Answer: -5×10-3

Explanation:

E=kq/r

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2 years ago

To provide the pulse of energy needed for an intense bass, some car stereo systems add capacitors. One system uses a 2.4F capaci

Natasha_Volkova [10]

Answer:

Explanation:

Energy stored in a capacitor

= 1/2 CV²

C is capacitance and V is potential of the capacitor .

When capacitor is charged to 24 V ,

E₁ = 1/2 x 2.4 x 24 x24 = 691.2 J

When it is charged to 12 volt

E₂ = 1/2 CV²

.5 X 2.4 X 12 X12

= 172.8 J

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3 years ago

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

4 0

3 years ago

Convert mm3 into m3.

bixtya [17]

Answer:

$1 \times 10 { - }^{9}$

cubic metre or 1e-9

Explanation:

•By division. Number of cubic millimetre divided(/) by 1000000000, equal(=): Number of cubic metre.

•By multiplication. 83 mm3(s) * 1.0E-9 = 8.3E-8 m3(s)

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3 years ago

Question 8

FinnZ [79.3K]

Answer:

A hope this helps

Explanation:

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3 years ago