The purpose of a block to add a value to the end of an array.
<h3>What is the function that adds an element to the end of an array?</h3>
The push() is known to be often used to add an element to the end of your array.
The end() function is known to be one that transmit the internal pointer to, and also the outputs, the last factor in the array.
A null or zero value is known to be that which marks the end of an array and it is said to be the very equivalent of the null char for any kind of string
Hence, in the above scenario, The purpose of a block to add a value to the end of an array.
Therefore, option D is correct.
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Answer:
Height of oil is 7.06 meters.
Explanation:
The situation is shown in the attached figure
The pressure at the bottom of the tank as calculated by equation of static pressure distribution is given by
![P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(\rho _{water}h_{water}+\rho_{oil}h_{oil})=P_{bottom}-P_{top}](https://tex.z-dn.net/?f=P_%7Bbottom%7D%3DP_%7Btop%7D%2B%5Crho%20_%7Bwater%7Dgh_%7Bwater%7D%2B%5Crho%20_%7Boil%7Dgh_%7Boil%7D%5C%5C%5C%5C%5Ctherefore%20g%28%5Crho%20_%7Bwater%7Dh_%7Bwater%7D%2B%5Crho_%7Boil%7Dh_%7Boil%7D%29%3DP_%7Bbottom%7D-P_%7Btop%7D)
Applying the given values we get
![P_{bottom}=P_{top}+\rho _{water}gh_{water}+\rho _{oil}gh_{oil}\\\\\therefore g(1000\times 2.5+800\times h_{oil})=80\times 10^{3}\\\\\therefore 800\times h_{oil}=\frac{80\times 10^{3}}{9.81}-2500\\\\\therefore 800\times h_{oil}=5654.94\\\\\therefore h_{oil}=7.06m](https://tex.z-dn.net/?f=P_%7Bbottom%7D%3DP_%7Btop%7D%2B%5Crho%20_%7Bwater%7Dgh_%7Bwater%7D%2B%5Crho%20_%7Boil%7Dgh_%7Boil%7D%5C%5C%5C%5C%5Ctherefore%20g%281000%5Ctimes%202.5%2B800%5Ctimes%20h_%7Boil%7D%29%3D80%5Ctimes%2010%5E%7B3%7D%5C%5C%5C%5C%5Ctherefore%20800%5Ctimes%20h_%7Boil%7D%3D%5Cfrac%7B80%5Ctimes%2010%5E%7B3%7D%7D%7B9.81%7D-2500%5C%5C%5C%5C%5Ctherefore%20800%5Ctimes%20h_%7Boil%7D%3D5654.94%5C%5C%5C%5C%5Ctherefore%20h_%7Boil%7D%3D7.06m)
Answer:
a diameter of D₂ = 0.183 inches would be required
Explanation:
appyling pascal's law
P applied to the hydraulic jack = P required to lift the rock
F₁*A₁ = F₂*A₂
since A₁= π*D₁²/4 , A₂= π*D₂²/4
F₁*π*D₁²/4 = F₂* π*D₂²/4
F₁*D₁²=F₂*D₂²
D₂ = D₁ *√(F₁/F₂)
replacing values
D₂ = D₁ *√(F₁/F₂) = 6 in * √(120 lbf/(4000 lbm * 32.174 (lbf/lbm)) = 0.183 inches