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2 years ago

Determine the frequency of light whose wavelength is 4.257 x 10-7 cm

1 answer:

7 0

$v = f (wavelength)$ i don't know what symbol ya'll use for wavelength so i just put the word instead.We use the greek symbol lambda.So just plug in everything you know.
wavelength=4.257×10^-7x10^-2 and
v=speed of light = 3×10^8
So you should get f= 7.04 ×10^15Hz

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3 years ago

If the position of a body is 3 m, 6 m, 9 m, 12 m at instants 2 s, 4 s, 6 s

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a car accelerates from 4 meters/second to 16 meter/second in 4 seconds. The cars acceleration is how many meter/seconds.

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Answer:

=3 metre per second ^2

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2 years ago

What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.

valkas [14]

<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period $T$ of a body (moon, planet, satellite) orbiting a greater body in space with the size $a$ of its orbit.

This Law is originally expressed as follows:

$T^{2}=\frac{4\pi^{2}}{GM}a^{3}$ (1)

Where;

$G$ is the Gravitational Constant and its value is $6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}$

$M=6.39(10)^{23}kg$ is the mass of Mars

$a=3.4(10)^{6}m$ is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

$T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}$ (2)

$T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}$ (3)

$T=\sqrt{11581157.44 s^{2}}$ (4)

Finally:

$T=3403.1099s=56.718min$ This is the orbital period of a spacecraft in a low orbit near the surface of mars

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