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Zarrin [17]
4 years ago
7

The time that is required for a vibrating object to complete one full cycle is called : Group of answer choices

Physics
1 answer:
disa [49]4 years ago
4 0

Answer:

Correct option : (c) period.                        

Explanation:

The time that is required for a vibrating object to complete one full cycle is called the time period. If f is the frequency of a wave, then the relation between the frequency and the time period is given by :

T=\dfrac{1}{f}

These are the characteristics of a wave. Some other characteristics are wavelength, amplitude, intensity etc. So, the correct option is (c) "period".

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Answer:

I think it is H2O . Formula of water

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!!!URGENT!!! Worth 100 points!!
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The snail would travel 100mm

Each minutes is 25mm and it took 4 minutes

4 x 25 = 100

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A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w
Zigmanuir [339]
Radial acceleration is given by

a_{rad}= \frac{v^2}{r}
where 

v=r \omega
then

a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2

Now

70\omega^2=3.90 \frac{m}{s^2}  \\  \\ \omega= \sqrt{ \frac{3.9}{70} }

Using the relation

\omega=2 \pi f

2 \pi f= \sqrt{ \frac{3.9}{70} }\\  \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz

Putting into rpm

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8 0
4 years ago
The surface temperature of a planet depends on both the distance of the planet from the Sun and on how the panet's atmosphere di
BaLLatris [955]

Answer:

Aphelion: 6404 W/m2

Perihelion: 14978 W/m2

Explanation:

The solar energy flux depends on the solar power output divided by the surface of a sphere with a radius equal to the distance to the Sun.

\Phi sol = \frac{Psol}{4 * \pi * d^2}

The distances we need are the aphelion and perihelion of Mercury.

Planetary orbits are ellipses. In an ellipse the eccentricity is related to linear eccentricity and the length of the semi major axis:

e = \frac{c}{a}

Where

e: eccentricity

c: linear eccentricity

a: semi major axis

The linear eccentricity is equal to the distance of the focus of the center of the ellipse.

c = a * c =

a = 0.39 AU = 5.83e10 m

c = 5.83e10e * 0.21 = 1.22e10 m

In planetary orbits the Sun is in one of the fucuses. With this we can calculate the prihelion and aphelion as:

Ap = a + c = 5.83e10 + 1.22e10 = 7.05e10 m

Pe = a - c = 5.83e10 - 1.22e10 = 4.61e10 m

And the solar energy fluxes will be:

\Phi Ap = \frac{4e26}{4 * \pi * 7.05e10^2} = 6404 W/m2

\Phi Pe = \frac{4e26}{4 * \pi * 4.61e10^2} = 14978 W/m2

4 0
3 years ago
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