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meriva
2 years ago
7

What evidence did astronomers use to prove jets travel in opposite directions?

Physics
2 answers:
mr Goodwill [35]2 years ago
7 0

The evidence that astronomers use to prove jets travel in opposite direction is Spectral lines from a very fast moving ionized gases.

<h3>Who are the astronomers?</h3>

Astronomer is a person who study astronomy which is a scientific study that has to do with space, space bodies, comets, planets, stars and so on.

Therefore, The evidence that astronomers use to prove jets travel in opposite direction is Spectral lines from a very fast moving ionized gases.

Learn more about astronomers below.

brainly.com/question/10826203

#SPJ11

klio [65]2 years ago
7 0
Spectral lines from fast-moving red- and blue-shifted ionized gas
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A 1.2 kg block is held at rest against the spring with a force constant k= 730 N/m. Initially, the spring is compressed a distan
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Answer:

Compression distance: d \approx 0.102\,m

Explanation:

According to this statement, we know that system is non-conservative due to the rough patch. By Principle of Energy Conservation and Work-Energy Theorem, we have the following expression that represents the system having a translational kinetic energy (K), in joules, at the expense of elastic potential energy (U), in joules, and overcoming work losses due to friction (W_{l}), in joules:

K + W_{l} = U (1)

By definitions of translational kinetic and elastic potential energies and work losses due to friction, we expand the equation described above:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2} (2)

Where:

m - Mass of the block, in kilograms.

v - Final velocity of the block, in meters per second.

\mu - KInetic coefficient of friction, no unit.

g - Gravitational acceleration, in meters per square second.

s - Width of the rough patch, in meters.

k - Spring constant, in newtons per meter.

d - Compression distance, in meters.

If we know that m = 1.2\,kg, v = 2.3\,\frac{m}{s}, \mu = 0.44, g = 9.807\,\frac{m}{s^{2}}, s = 0.05\,m and k = 730\,\frac{N}{m}, then the compression distance of the spring is:

\frac{1}{2}\cdot m \cdot v^{2} +\mu\cdot m\cdot g \cdot s = \frac{1}{2}  \cdot k \cdot d^{2}

m\cdot v^{2} + 2\cdot m\cdot g \cdot s = k\cdot d^{2}

d = \sqrt{\frac{m\cdot (v^{2}+2\cdot g\cdot s)}{k} }

d \approx 0.102\,m

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