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Oduvanchick [21]
3 years ago
10

The charge that passes a cross-sectional area A=10-4 m2 varies with time according

Physics
1 answer:
Anarel [89]3 years ago
6 0

Answer:

Current = dQ/dt

or I = dQ/dt

Where I represents current.

Which is the rate of flow of charge.

Q=4 + 2t + t²

dQ/dt = 2 + 2t --- This is the relation that gives the instantaneous current.

At time t=2sec

dQ/dt = I = 2 + 2t

= 2 + 2(2)

=2 + 4

= 6A.

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Ocean waves that measure 12m from crest to adjacent crest pass by a fixed point every 2.0s what is the speed of the waves
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The activation energy of a reaction going on its own is 20 kj. If the reaction was treated with a catalyst, which would most lik
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Answer:

15 kJ

Explanation:

We need to know the following;

what is activation energy?

  • Activation energy is the minimum energy required by reactants for a chemical reaction to take place.

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7 0
3 years ago
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When the power in their house unexpectedly went off, Radhika's mother said that she would check and replace the fuse. What exact
denis23 [38]

A) a piece of wire

Explanation:

Radhika's mother is simply replacing piece of wire.

A fuse is made up of a thin strip of wire that melts when there is an over current.

  • A fuse is electrical safety device in circuit that protects appliances from over current.
  • The wire in a fuse has a high resistance and low melting point.
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Learn more:

Current brainly.com/question/3051098

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6 0
3 years ago
A sheet of gold leaf has a thickness of 0.125 micrometer. A gold atom has a radius of 174pm. Approximately how many layers of at
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Answer:

719

Explanation:

Conversion

1 picometer (pm) is equivalent to 1 × 10^{-12} meter

1 micrometer is equivalent to 1 × 10^{-6} meter

To find the number of layers, we divide the overal leaf thickness by the thickness of one atom hence dividing tex]0.125 × 10^{-6}[/tex] meter by 174 × 10^{-12} meter we get that the number of sheets will be as follows

\frac {0.125× 10^{-6}}{174\times 10^{-12}}=718.3908045\approx 719

Therefore, they are approximately 719 sheets

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3 years ago
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