Give several examples of a mechanism, machine or system that has two

g In a certain binary-star system, each star has the same mass which is 8.2 times of that of the Sun, and they revolve about the

Mademuasel [1]

To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.

Kepler's third law tells us that the period is defined as

$T^2 = \frac{4\pi^2 d^3}{2GM}$

The given data are given with respect to known constants, for example the mass of the sun is

$m_s = 1.989*10^{30}$

The radius between the earth and the sun is given by

$r = 149.6*10^9m$

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun

Therefore:

$m = 8.2*1.989*10^{30}$

$d = 6.2*149.6*10^6$

Substituting in Kepler's third law:

$T^2 = \frac{4\pi^2 d^3}{2}$

$T^2 = \frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30 )}$

$T=\sqrt{\frac{4\pi^2(6.2*149.6*10^9)^3}{2(6.674*10^{-11} )(8.2*1.989*10^30)}}$

$T = 120290789.7s$

$T = 120290789.7s(\frac{1year}{31536000s})$

$T \approx 3.8143 years$

Therefore the period of this star is 3.8years

7 0

3 years ago

25 PTs

Nuetrik [128]

Answer:
They are both wrong!

Liquid oxygen really is a pale blue color.
I’ve seen it.

And they cant say that liquid and solid oxygen is blue which makes the sky blue because they’re not and it doesn’t make up for the color of the sky.

The sky is actually blue because It reflects more light than It can absorb
aka Rayleigh Scattering.

-HOPE THAT HELPED

8 0

3 years ago

A 150 kg line backer sacks the 120 kg quarterback. With what force is the quarterback sacked if the line backer has an accelerat

Gekata [30.6K]

Answer:

The force required to move the quarterback with linebacker is <u>1215 N</u>

Explanation:

$\text { Mass of linebacker } \mathrm{m}_{2}=150 \mathrm{kg}$

$\text { Mass of quarterback } \mathrm{m}_{2}=120 \mathrm{kg}$

$\text { Moved at an acceleration }(a)=4.5 \mathrm{m} / \mathrm{s}^{2}$

Using Newton's second law, it is established that F = Ma

Where F is net force acting on the system, a is the acceleration and M is mass of the two object $\left(m_{1}+m_{2}\right)$

Now consider both $\mathrm{m}_{1} \text { and } \mathrm{m}_{2}$ as a system, so net force acting on the system is $\text { Force }=\left(m_{1}+m_{2}\right) a$

Substitute the given values in the above formula,

$\text { Force }=(150+120) \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

$\text { Force }=270 \mathrm{kg} \times 4.5 \mathrm{m} / \mathrm{s}^{2}$

Force = 1215 N

<u>1215 N </u>is the force required to move the quarterback with linebacker.

5 0

3 years ago

Gamma rays have a higher frequency than visible light waves. What can you conclude from this?

NeTakaya

Answer:

i think it d

Explanation:

3 0

3 years ago

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When one person shouts at a football game, the sound intensity level at the center of the field is 58.4 dB. When all the people

Tanzania [10]

Answer:

The number of people at game are approximately 22909

Explanation:

Given data

When one person shout $\beta _{1}=58.4dB$

When n number of person shout together $\beta _{n}=102dB$

The sound intensity level during one person shout is given by:

$\beta _{1}=10log(\frac{I_{1}}{I_{o}} )\\58.4=10log(\frac{I_{1}}{I_{o}} )\\5.84=log(\frac{I_{1}}{I_{o}} )\\\frac{I_{1}}{I_{o}} =10^{5.84}\\I_{1}=10^{5.84}*I_{o}$

The sound intensity level during n number of person shout is given by:

$\beta _{n}=10log(\frac{I_{n}}{I_{o}} )\\102=10log(\frac{I_{n}}{I_{o}} )\\10.2=log(\frac{I_{n}}{I_{o}} )\\\frac{I_{n}}{I_{o}}=10^{10.2}\\I_{n}=10^{10.2}*I_{o}$

Since each person generates same sound intensity and hence total number of persons can be determined as

$=\frac{I_{n}}{I_{1}}\\ =\frac{10^{10.2}I_{o}}{10^{5.84}I_{o}} \\=22909$

Hence

The number of people at game are approximately 22909

3 0

3 years ago

Read 2 more answers