Answer:
Force of attraction is 2,46*10^-8N![F= G*(m1*m2/r^2)\\where \\\\G=6,67*10^-11 (N*m^2/kg^2)\\m1=42kg\\m2=55kg\\r=2,5m\\therefore\\F=6,67*10^-11*(42*55/2.5^2)\\ F=2,46*10^-8 N](https://tex.z-dn.net/?f=F%3D%20G%2A%28m1%2Am2%2Fr%5E2%29%5C%5Cwhere%20%5C%5C%5C%5CG%3D6%2C67%2A10%5E-11%20%28N%2Am%5E2%2Fkg%5E2%29%5C%5Cm1%3D42kg%5C%5Cm2%3D55kg%5C%5Cr%3D2%2C5m%5C%5Ctherefore%5C%5CF%3D6%2C67%2A10%5E-11%2A%2842%2A55%2F2.5%5E2%29%5C%5C%20%3C%2Fp%3E%3Cp%3EF%3D2%2C46%2A10%5E-8%20N)
Explanation:
Using the Law of Universal Gravitation, proposed by Newton.
Answer:
![v_2=29.13\ m/s](https://tex.z-dn.net/?f=v_2%3D29.13%5C%20m%2Fs)
Explanation:
It is given that,
Dimension of the rectangular roof, (6.17 m × 5.92 m)
The maximum net outward force, ![F=2\times 10^4\ N](https://tex.z-dn.net/?f=F%3D2%5Ctimes%2010%5E4%5C%20N)
The density of air, ![\rho=1.29\ kg/m^3](https://tex.z-dn.net/?f=%5Crho%3D1.29%5C%20kg%2Fm%5E3)
The Bernoulli equation is used to find wind speed of this roof blow outward. It is given by :
![P_1+\dfrac{1}{2}\rho v_1^2=P_2+\dfrac{1}{2}\rho v_2^2](https://tex.z-dn.net/?f=P_1%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_1%5E2%3DP_2%2B%5Cdfrac%7B1%7D%7B2%7D%5Crho%20v_2%5E2)
Here,
(since air inside the roof is not moving)
![v_2=\sqrt{\dfrac{2(P_1-P_2)}{\rho}}](https://tex.z-dn.net/?f=v_2%3D%5Csqrt%7B%5Cdfrac%7B2%28P_1-P_2%29%7D%7B%5Crho%7D%7D)
Since, ![F=(P_1-P_2)A](https://tex.z-dn.net/?f=F%3D%28P_1-P_2%29A)
![v_2=\sqrt{\dfrac{2F}{\rho A}}](https://tex.z-dn.net/?f=v_2%3D%5Csqrt%7B%5Cdfrac%7B2F%7D%7B%5Crho%20A%7D%7D)
![v_2=\sqrt{\dfrac{2\times 2\times 10^4}{1.29\times 6.17\times 5.92 }}](https://tex.z-dn.net/?f=v_2%3D%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%202%5Ctimes%2010%5E4%7D%7B1.29%5Ctimes%206.17%5Ctimes%205.92%20%7D%7D)
![v_2=29.13\ m/s](https://tex.z-dn.net/?f=v_2%3D29.13%5C%20m%2Fs)
So, the wind speed of this roof blow outward is 29.13 m/s. Hence, this is the required solution.
Wavelength = (speed) / (frequency)
Wavelength = (330 m/s) / (1320/s)
Wavelength = (330/1320) m
Wavelength = 0.25 m
<em>Wavelength = 25 cm</em>
Answer:
Ws=8.75 Watts
Explanation:
As per fig. of prob 02.061, it is clear that R5 and R4 are in parallel, its equivalent Resistance will be:
![\frac{1}{Req45}=\frac{1}{R4}+\frac{1}{R5}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BReq45%7D%3D%5Cfrac%7B1%7D%7BR4%7D%2B%5Cfrac%7B1%7D%7BR5%7D)
![\frac{1}{Req45}=\frac{1}{5}+\frac{1}{4}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BReq45%7D%3D%5Cfrac%7B1%7D%7B5%7D%2B%5Cfrac%7B1%7D%7B4%7D)
![\frac{1}{Req45}=0.2+0.25=0.45\\ Req45=2.22](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BReq45%7D%3D0.2%2B0.25%3D0.45%5C%5C%20Req45%3D2.22)
Now, this equivalent Req45 is in series with R3, therefore:
![Req345=R3+Req45\\Req345=4+2.22\\Req345=6.22](https://tex.z-dn.net/?f=Req345%3DR3%2BReq45%5C%5CReq345%3D4%2B2.22%5C%5CReq345%3D6.22)
This Req345 is in parallel with R2, i.e
![Req2345=(R2^{-1}+Req345^{-1} )^{-1}\\ Req2345=(4^{-1}+6.22^{-1} )^{-1} \\Req2345=2.43](https://tex.z-dn.net/?f=Req2345%3D%28R2%5E%7B-1%7D%2BReq345%5E%7B-1%7D%20%20%29%5E%7B-1%7D%5C%5C%20Req2345%3D%284%5E%7B-1%7D%2B6.22%5E%7B-1%7D%20%20%29%5E%7B-1%7D%20%5C%5CReq2345%3D2.43)
Now this gets in series with R1:
![Req12345=R1+Req2345\\Req12345=9+2.43\\Req12345=11.43](https://tex.z-dn.net/?f=Req12345%3DR1%2BReq2345%5C%5CReq12345%3D9%2B2.43%5C%5CReq12345%3D11.43)
Now, the power delivered Ws is:
![Ws=Vs*I=\frac{Vs^{2}}{Req} \\Ws=\frac{10^{2} }{11.43} \\Ws=8.75 Watts](https://tex.z-dn.net/?f=Ws%3DVs%2AI%3D%5Cfrac%7BVs%5E%7B2%7D%7D%7BReq%7D%20%20%5C%5CWs%3D%5Cfrac%7B10%5E%7B2%7D%20%7D%7B11.43%7D%20%5C%5CWs%3D8.75%20Watts)
(A) It will 100 times larger than the original force.