1. (10 p) a) Find out the specific weight of the liquid. b) Determine the required force to hold the tablet in the same liquid. 8 cm 5 cm 2 cm 10 cm $\gamma = 7 kN/m^3$ Liquid $\gamma = ? kN/m^3$ F = ? N 10 cm 5 cm $\gamma = 7 kN/m^3$
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The dimensions of the rectangular block are 8 cm, 5 cm, and 2 cm. Volume = length × width × height = 8 cm × 5 cm × 2 cm = 80 $cm^3$ = $80 \times 10^{-6} m^3$ The submerged height of the block is 2 cm = 0.02 m. Show more…
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(a) What is the weight density of a liquid that exerts a total force of $433 \mathrm{N}$ on the sides of a $3.00-\mathrm{m}$ -tall cylindrical tank? The radius of the tank is $0.913 \mathrm{m},$ and it is filled to $1.75 \mathrm{m}$. (b) What liquid might this be?
We must first calculate the pressure difference inside the film from that outside. This is Here $2 r_{1}|\cos \theta|=h$ and $r_{2} \sim-R$ the radius of the tablet and can be neglected. Thus the total force exerted by mercury drop on the upper glass plate is We should put $h / n$ for $h$ because the tablet is compresed $n$ times. Then since $H g$ is nearly, incompressible, $\pi R^{2} h=$ constants so $R \rightarrow R \sqrt{n}$. Thus, total force $=\frac{2 \pi R^{2} \alpha|\cos \theta|}{h} n^{2}$ Part of the force is needed to keep the $\mathrm{Hg}$ in the shape of a table rather than in the shape of infinitely thin sheet. This part can be calculated being putting $n=1$ above. Thus $m g+\frac{2 \pi R^{2} \alpha|\cos \theta|}{h}=\frac{2 \pi R^{2} \alpha|\cos \theta|}{h} n^{2}$ or $m=\frac{2 \pi R^{2} \alpha|\cos \theta|}{h g}\left(n^{2}-1\right)=0 \cdot 7 \mathrm{~kg}$
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make sure show the work and make sure to provide me an answer for the Note please
Ankur S.
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