Explanation:
The hydraulic system works on the principle of Pascal's law which says that the pressure in an enclosed fluid is uniform in all the directions. ... As the pressure is same in all the direction, the smaller piston feels a smaller force and a large piston feels a large force.
![▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪](https://tex.z-dn.net/?f=%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%20%20%7B%5Chuge%5Cmathfrak%7BAnswer%7D%7D%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA%E2%96%AA)
Solution is in attachment ~
I hope that you got what you were looking for, and if there's different data then go through the same procedure, using same formula with different values and you will get your answer ~
![\mathrm{✌TeeNForeveR✌}](https://tex.z-dn.net/?f=%5Cmathrm%7B%E2%9C%8CTeeNForeveR%E2%9C%8C%7D)
Complete question is;
If the diameter of the black marble is 3.0 cm, and by using the formula for volume, what is a good approximation of its volume?
Answer:
14 cm³
Explanation:
We will assume that this black marble has the shape of a sphere from online sources.
Now, volume of a sphere is given by;
V = (4/3)πr³
We are given diameter = 3 cm
We know that radius = diameter/2
Thus; radius = 3/2 = 1.5 cm
So, volume = (4/3)π(1.5)³
Volume ≈ 14.14 cm³
A good approximation of its volume = 14 cm³
Answer:
86.43
25.72
Explanation:
Location of center of mass from 59 Kg
![r_1=\frac {xm_2}{m1+m2}=\frac {1.1*73}{59+73}= 0.608333m](https://tex.z-dn.net/?f=r_1%3D%5Cfrac%20%7Bxm_2%7D%7Bm1%2Bm2%7D%3D%5Cfrac%20%7B1.1%2A73%7D%7B59%2B73%7D%3D%200.608333m)
Location of center of mass from 73 Kg
![r_2=\frac {xm_1}{m1+m2}=\ frac {1.1*59}{59+73}= 0.491667m](https://tex.z-dn.net/?f=r_2%3D%5Cfrac%20%7Bxm_1%7D%7Bm1%2Bm2%7D%3D%5C%20frac%20%7B1.1%2A59%7D%7B59%2B73%7D%3D%200.491667m)
Momentum of inertia
![I=m1r_1^{2}+m2r_2^{2}=59*0.608333^{2}+73*0.491667^{2}=21.8341 +17.64674=39.48083 Kgm^{2}](https://tex.z-dn.net/?f=I%3Dm1r_1%5E%7B2%7D%2Bm2r_2%5E%7B2%7D%3D59%2A0.608333%5E%7B2%7D%2B73%2A0.491667%5E%7B2%7D%3D21.8341%20%20%2B17.64674%3D39.48083%20%20Kgm%5E%7B2%7D)
Angular moment
![L=I\omega=I(\frac {2\pi}{T})=39.48083(\frac {2*\pi}{2.87 s})=86.43393Js](https://tex.z-dn.net/?f=L%3DI%5Comega%3DI%28%5Cfrac%20%7B2%5Cpi%7D%7BT%7D%29%3D39.48083%28%5Cfrac%20%7B2%2A%5Cpi%7D%7B2.87%20s%7D%29%3D86.43393Js)
(b)
When separation is 0.6 m
Location of centre of mass from 59 Kg
![r_1=\frac {xm_2}{m1+m2}=\frac {0.6*73}{59+73}= 0.331818 m](https://tex.z-dn.net/?f=r_1%3D%5Cfrac%20%7Bxm_2%7D%7Bm1%2Bm2%7D%3D%5Cfrac%20%7B0.6%2A73%7D%7B59%2B73%7D%3D%200.331818%20m)
Location of centre of mass from 73 Kg
![r_2=\frac {xm_1}{m1+m2}=\ frac {0.6*59}{59+73}= 0.268182 m](https://tex.z-dn.net/?f=r_2%3D%5Cfrac%20%7Bxm_1%7D%7Bm1%2Bm2%7D%3D%5C%20frac%20%7B0.6%2A59%7D%7B59%2B73%7D%3D%200.268182%20m)
Momentum of intertia
![I=m1r_1^{2}+m2r_2^{2}=59*0.331818^{2}+73*0.268182^{2}=6.496095 + 5.250269=11.74636 Kgm^{2}](https://tex.z-dn.net/?f=I%3Dm1r_1%5E%7B2%7D%2Bm2r_2%5E%7B2%7D%3D59%2A0.331818%5E%7B2%7D%2B73%2A0.268182%5E%7B2%7D%3D6.496095%20%2B%205.250269%3D11.74636%20Kgm%5E%7B2%7D)
Angular moment
![L=I\omega=I(\frac {2\pi}{T})=11.74636(\frac {2*\pi}{2.87 s})=25.71588 Js](https://tex.z-dn.net/?f=L%3DI%5Comega%3DI%28%5Cfrac%20%7B2%5Cpi%7D%7BT%7D%29%3D11.74636%28%5Cfrac%20%7B2%2A%5Cpi%7D%7B2.87%20s%7D%29%3D25.71588%20Js)
Answer:
Within the realm of Newtonian mechanics, an internal frame of reference or internal reference frame, is one in which newton's law of motions is valid