The solution for this problem is:
m = k*(T/2pi)^2
<span>mass of chair is computed by: 606*(0.889/2pi)^2 = 12.13 kg </span>
<span>mass of chair + astronaut is computed by: 606*(2.15/2pi)^2 = 70.96 kg </span>
<span>so mass of astronaut is now 70.96 - 12.13 kg = 58.8 kg (3 s.f.)</span>
<span>The answer is: 30 ft³ ; or, write as: 30 cubic feet .
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Explanation:
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Given:
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Density = 0.4 fish(es) per cubic foot (or, feet);
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or, "Density = 0.4 fish / ft³ ;
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And, given: "12 fish in the tank." ;
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And Density = (value; in this case, "number of fish" / unit volume);
We are given: Density, "D" = 0.4 fish / 1 ft³
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In our case, our volume is in "cubic ft." ; or, "ft³ ;
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We want to find the actual volume of the tank (in "cubic feet"; or, " ft³ ");
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→ (1 ft³ / 0.4 fish) * 12 fish = _____? ft³ ?
→ On the "left-hand side" of the equation; the units of "fish" cancel out to "1"; and we are left with units of "ft³ " ; which is the units for which we wish to solve.
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→ So, so have (1 ft³ * 12) / (0.4) = 12 ft³ / 0.4 = 30 ft³ .
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<span> → The answer is: 30 ft³ ; or, write as: 30 cubic feet .</span>
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Answer:
Explanation:
Given
length of window ![h=2.9\ m](https://tex.z-dn.net/?f=h%3D2.9%5C%20m)
time Frame for which rock can be seen is ![\Delta t=0.134\ s](https://tex.z-dn.net/?f=%5CDelta%20t%3D0.134%5C%20s)
Suppose h is height above which rock is dropped
Time taken to cover ![h+2.9 is t_1](https://tex.z-dn.net/?f=h%2B2.9%20is%20t_1)
so using equation of motion
![y=ut+\frac{1}{2}at^2](https://tex.z-dn.net/?f=y%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2)
where y=displacement
u=initial velocity
a=acceleration
t=time
time taken to travel h is
![h=0+0.5\times g\times (t_2)^2---2](https://tex.z-dn.net/?f=h%3D0%2B0.5%5Ctimes%20g%5Ctimes%20%28t_2%29%5E2---2)
Subtract 1 and 2 we get
![2.9=0.5g(t_1^2-t_2^2)](https://tex.z-dn.net/?f=2.9%3D0.5g%28t_1%5E2-t_2%5E2%29)
![5.8=g(t_1+t_2)(t_1-t_2))](https://tex.z-dn.net/?f=5.8%3Dg%28t_1%2Bt_2%29%28t_1-t_2%29%29)
and from equation ![t_1-t_2=0.134\ s](https://tex.z-dn.net/?f=t_1-t_2%3D0.134%5C%20s)
so ![t_1+t_2=\frac{5.8}{9.8\times 0.134}](https://tex.z-dn.net/?f=t_1%2Bt_2%3D%5Cfrac%7B5.8%7D%7B9.8%5Ctimes%200.134%7D)
![t_1+t_2=4.416\ s](https://tex.z-dn.net/?f=t_1%2Bt_2%3D4.416%5C%20s)
and ![t_1=t_2+\Delta t](https://tex.z-dn.net/?f=t_1%3Dt_2%2B%5CDelta%20t)
so ![t_2+\Delta t+t_2=4.416](https://tex.z-dn.net/?f=t_2%2B%5CDelta%20t%2Bt_2%3D4.416)
![2t_2+0.134=4.416](https://tex.z-dn.net/?f=2t_2%2B0.134%3D4.416)
![t_2=0.5\times 4.282](https://tex.z-dn.net/?f=t_2%3D0.5%5Ctimes%204.282)
![t_2=2.141\ s](https://tex.z-dn.net/?f=t_2%3D2.141%5C%20s)
substitute the value of
in equation 2
![h=0.5\times 9.8\times (2.141)^2](https://tex.z-dn.net/?f=h%3D0.5%5Ctimes%209.8%5Ctimes%20%282.141%29%5E2)
![h=22.46\ m](https://tex.z-dn.net/?f=h%3D22.46%5C%20m)
The wavelength<span> can always be determined by </span>measuring<span> the distance between any two corresponding points on adjacent </span>waves<span>. In the case of a longitudinal </span>wave, awavelength measurement<span> is made by </span>measuring<span> the distance from a compression to the next compression or from a rarefaction to the next rarefaction.
> Please rate 5 stars <</span>
The solar interior includes the core, radiative zone and convective zone. The photosphere is the visible surface of the Sun.
So the answers should be:
<u>A) Photosphere C) Core D) Radioactive zone </u>and<u> F) Convective zone.</u>
I hope this helps. Sorry if I made any mistakes. :)