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erica [24]
3 years ago
5

When a proton encounters an antiproton, the two particles annihilate each other, producing two gamma rays

Physics
1 answer:
Delvig [45]3 years ago
3 0
When a proton indicators a antiproton the two particles
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A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l
ivolga24 [154]

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

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Help would be greatly appreciated:) thank you! a pendulum clock is brought to mars. How does the bob move on Mars as compared to
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At the _______________ point, the particles in an object have not kinetic energy
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Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each
AURORKA [14]

Answer: A (\sqrt{61},309.8°)

              B (2\sqrt{2}, 315°)

             C (3\sqrt{5}, 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

r=\sqrt{x^{2}+y^{2}} and \theta=tan^{-1}(\frac{y}{x})

For point A:

r=\sqrt{(-5)^{2}+6^{2}}

r=\sqrt{61}

\theta=tan^{-1}(\frac{6}{-5})

\theta=tan^{-1}(-1.2)

\theta=-50.2°

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

\theta=360-50.2

\theta= 309.8°

Polar coordinates for point A is (\sqrt{61}, 309.8°)

For point B:

r=\sqrt{2^{2}+(-2)^{2}}

r=\sqrt{8}

r=2\sqrt{2}

\theta=tan^{-1}(\frac{-2}{2} )

\theta=tan^{-1}(1)

\theta=-45°

Point B is in IV quadrant, so:

\theta=360-45

\theta= 315°

Polar coordinates for point B is (2\sqrt{2}, 315°)

For point C:

r=\sqrt{(-6)^{2}+(-3)^{2}}

r=\sqrt{45}

r=3\sqrt{5}

\theta=tan^{-1}(\frac{-3}{-6} )

\theta=tan^{-1}(0.5)

\theta= 26.56°

Polar coordinates for point C is (3\sqrt{5}, 26.56°)

3 0
3 years ago
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