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2 years ago

8

A total charge of 7.5 mC passes through a cross-sectional area of a wire in 0.9 s. What is the current in the wire

1 answer:

Elan Coil [88]2 years ago

4 0

Answer:

the current in the wire is 0.008333333 A

Explanation:

The computation of the current in the wire is as follows;

Current in the wire is

= Total charge ÷ cross sectional area of the wire

= 7.5 × 10^-3C ÷ 0.9s

= 0.008333333 A

Hence, the current in the wire is 0.008333333 A

We simply applied the above formula so that the correct current of the wire could come

And, the same is considerd and relevant too

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An example of an atom that has no charge is one that has?

Pavlova-9 [17]

Answer: Equal number of protons and electrons. Example: an atom of oxygen atom has 8 electrons and 8 protons and is neutral.

Explanation:

An atom that has no charge is a neutral atom. It contains electrons equal to protons. For example: A neutral atom of oxygen has 8 protons and 8 electrons.

An atom which has charge is said be ionized. It is either positively charged or negatively charged. It is positively charge when the number of electrons is less than the number of protons. For example: $Na^+$ contains 10 electrons and 11 protons.

And when the number of electrons is greater than the number of protons, the atom is negatively charged. For example, $Cl^-$ has 17 protons and 18 electrons. It readily accepts an electron to complete its octet.

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2 years ago

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Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

Nostrana [21]

Answer:

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

$E_1 = E_2$

Let the position where net field is zero will lie at distance "r" from q1

$\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}$

now we will have

$\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}$

now square root both sides

$\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}$

now we have

$\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1$

so we have

$r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})$

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3 years ago

What are the possible units for a spring constant

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Hooke's Law says that F=-kx where k is the spring constant measured in N/m (newtons per meter)

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3 years ago

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Complete the following table with the appropriate units of measurement. Word bank: gram, meter, kelvin, liter, second, grams/cm3

lara31 [8.8K]

mass gram, time sec, temp kelvin, vol liter, dens grams/cm3

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3 years ago

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A straight wire of length 0.53 m carries a conventional current of 0.2 amperes. What is the magnitude of the magnetic field made

olga55 [171]

Explanation:

It is given that,

Length of wire, l = 0.53 m

Current, I = 0.2 A

(1.) Approximate formula:

We need to find the magnitude of the magnetic field made by the current at a location 2.0 cm from the wire, r = 2 cm = 0.02 m

The formula for magnetic field at some distance from the wire is given by :

$B=\dfrac{\mu_oI}{2\pi r}$

$B=\dfrac{4\pi \times 10^{-7}\times 0.2\ A}{2\pi \times 0.02\ m}$

B = 0.000002 T

$B=10^{-5}\ T$

(2) Exact formula:

$B=\dfrac{\mu_oI}{2\pi r}\dfrac{l}{\sqrt{l^2+4r^2} }$

$B=\dfrac{\mu_o\times 0.2\ A}{2\pi \times 0.02\ m}\times \dfrac{0.53\ m}{\sqrt{(0.53\ m)^2+4(0.02\ m)^2} }$

B = 0.00000199 T

or

B = 0.000002 T

Hence, this is the required solution.

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3 years ago