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Komok [63]
2 years ago
7

Not all metals have the same current for a given electric potential

Physics
1 answer:
Neporo4naja [7]2 years ago
5 0

The statement 'Not all metals have the same current for a given electric potential' is TRUE. It is conductivity.

<h3>What is conductivity?</h3>

All the metals are able to conduct electric currents, but some metals have a higher conductivity.

Metals are able to conduct electric currents due to the free movement of negatively charged particles i.e., electrons, across the conductor.

The band theory of metals states that metals can conduct electrons (e-) in an electric current by means of the help of the e- valence.

The density of free electrons in metals is around 10^28 m-3, which indicates the number of states at a particular energy level that negative e- can occupy.

Learn more about metal conductivity here:

brainly.com/question/13584414

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What is the intensity of a traveling plane electromagnetic wave?
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S = (c/u0) * B^2

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7 0
4 years ago
6. Velocity is a ____________. a) vector b) value c) arrow d) none of these
Olenka [21]
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4 0
4 years ago
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A block, whose mass is 0.500 kg, is attached to a spring with a force constant of 126 N/m. The block rests upon a frictionless,
erastova [34]

Answer:

(a) F = 15.12 N

(b) a = 30.24 m/s²

(c) To Left

Explanation:

(a)

The magnitude of the spring force is given by Hooke's Law as follows:

F = kx

where,

F = Spring Force = ?

k = Spring Constant = 126 N/m

x = Displacement = A = 0.12 m

Therefore,

F = (126 N/m)(0.12 m)

<u>F = 15.12 N</u>

(b)

The magnitude of acceleration can be found by comparing the spring force with the unbalanced force formula of Newton's Second Law:

F = ma

where,

F = Spring Force = 15.12 N

m = mass of block = 0.5 kg

a = magnitude of acceleration = ?

15.12 N = 0.5 kg (a)

a = 15.12 N/0.5 kg

<u>a = 30.24 m/s²</u>

<u></u>

(c)

Since, the acceleration is always directed towards mean (equilibrium) position in periodic motion. Therefore, the direction of the acceleration at the time of release will be <u>to left.</u>

8 0
3 years ago
The air in a room with volume 180 m3 contains 0.25% carbon dioxide initially. fresher air with only 0.05% carbon dioxide flows i
Butoxors [25]

Answer:

p(t)=\frac{1}{20}(1+2e^{-t/90})

Explanation:

Let p(t) be % of CO_2 at time t(time in minutes).

Amount of CO_2 in room=Volume of room \times p

Rate of change of CO_2 in room=Volume of room x \frac{dp}{dt}

=180\times \frac{dp}{dt}

Rate of inflow of CO_2=(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Inflow\ of \ air)

=(0.05\times \frac{1}{100})\times 2=\frac{1}{100}

Rate of CO_2 outflow

(\% CO_2 \ in\  Fresh\  Air \times \frac{1}{100})\times(Rate \ of  Outflow\ of \ air)\\=(p\times\frac{1}{100})\times2=\frac{p}{50}

Therefore rate of change of CO_2 is \frac{1}{100}-\frac{p}{50}

% rate of change is100\times \ Rate of \ Change=\frac{1}{10}-2p

Therefore we have:

180\frac{dp}{dt}=\frac{1}{10}-2p\\\frac{dp}{dt}=\frac{1-20p}{1800}\\\\\frac{dp}{20p-1}=-\frac{dt}{1800}

Integrate both:

\int\limits \frac{dp}{20p-1}=\int\limits-\frac{dt}{1800}

\frac{1}{20}In|20p-1|=-\frac{t}{1800}+In \ C

In|20p-1|=-\frac{t}{90}+InK

+20In \ C=+In \ K

Raise both sides to base e,

20p-1=e^{-\frac{t}{90}+In \ K}\\\\p=\frac{1}{20}(1+Ke^{-t/90})

Initially, there's only 0.25% of CO_2,

Now substitute p=0.25 and t=0

0.25=\frac{1}{20}(1+K)\\K=4

p=\frac{1}{20}(1+2e^{-t/90})

8 0
3 years ago
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