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soldier1979 [14.2K]
3 years ago
12

Carbon dioxide is added to the atmosphere:

Physics
1 answer:
dimulka [17.4K]3 years ago
8 0
Carbon dioxide is added to the atmosphere b) from burning the fossil fuels
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Watching ice melt on a hot surface.
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A bullet is fired straight up from a gun with a
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<span>v(4 seconds)= 300 m/s - 9.8 (m/s^2)(4s) = 260.8 m/s </span>, hope this helps:)
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A closed system’s internal energy changes by 178 J as a result of being heated with 658 J of energy. The energy used to do work
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Answer:

480J

Explanation:

Using the formula:

Delta U = Q - W

Q:Heat (J)

Delta U: Changes in internal Energy (J)

W:Work (J)

We can plug in the give numbers, Q and W.

Delta U = 658J - 178J = 480J

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A proton is accelerated from rest through a potential difference of 2.5 kV and then moves perpendicularly through a uniform 0.60
Bumek [7]

Answer:

1.2cm

Explanation:

V=(2ev/m)^1/2

=(2*1.6*10^19 x2500/ 1.67*10^27)^1/2

=6.2x10^5m/s

Radius of resulting path= MV/qB

= 1.67*10^-27x6.92*10^6/1.6*10^-16 x0.6

=0.012m

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A wire of radius R has a current I uniformly distributed across its cross-sectional area. Ampere's law is used with a concentric
MrMuchimi

Answer:

Please refer to the figure.

Explanation:

The crucial point here is to calculate the enclosed current. If the current I is flowing through the whole cross-sectional area of the wire, the current density is

J = \frac{I}{\pi R^2}

The current density is constant for different parts of the wire. This idea is similar to that of the density of a glass of water is equal to the density of a whole bucket of water.

So,

J = \frac{I}{\pi R^2} = \frac{I_{enc}}{\pi r^2}\\I_{enc} = \frac{Ir^2}{R^2}

This enclosed current is now to be used in Ampere’s Law.

\mu_o I_{enc} = \int {B} \, dl

Here, \int \, dl represents the circular path of radius r. So we can replace the integral with the circumference of the path, 2\pi r.

As a result, the magnetic field is

B = \frac{\mu_0}{2\pi}\frac{Ir}{R^2}

5 0
3 years ago
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