The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.
Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
Please note that your question is incomplete so I gave you a general overview to help you better understand the concept
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Answer:
Explanation:
Given
charge is placed at ![x=0\ cm](https://tex.z-dn.net/?f=x%3D0%5C%20cm)
another charge of
is at ![x=3\ cm](https://tex.z-dn.net/?f=x%3D3%5C%20cm)
We know that Electric field due to positive charge is away from it and Electric field due to negative charge is towards it.
so net electric field is zero somewhere beyond negatively charged particle
Electric Field due to
at some distance r from it
![E_2=\frac{kq_2}{r^2}](https://tex.z-dn.net/?f=E_2%3D%5Cfrac%7Bkq_2%7D%7Br%5E2%7D)
Now Electric Field due to
is
![E_1=\frac{kq_1}{(3+r)^2}](https://tex.z-dn.net/?f=E_1%3D%5Cfrac%7Bkq_1%7D%7B%283%2Br%29%5E2%7D)
Now ![E_1+E_2=0](https://tex.z-dn.net/?f=E_1%2BE_2%3D0)
![\frac{k\times 11.5}{(r+3)^2}\frac{k\times (-1.2)}{r^2}=0](https://tex.z-dn.net/?f=%5Cfrac%7Bk%5Ctimes%2011.5%7D%7B%28r%2B3%29%5E2%7D%5Cfrac%7Bk%5Ctimes%20%28-1.2%29%7D%7Br%5E2%7D%3D0)
![\frac{3+r}{r}=(\frac{11.5}{1.2})^{0.5}](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Br%7D%7Br%7D%3D%28%5Cfrac%7B11.5%7D%7B1.2%7D%29%5E%7B0.5%7D)
![\frac{3+r}{r}=3.095](https://tex.z-dn.net/?f=%5Cfrac%7B3%2Br%7D%7Br%7D%3D3.095)
thus ![r=1.43\ cm](https://tex.z-dn.net/?f=r%3D1.43%5C%20cm)
Thus Electric field is zero at some distance r=1.43 cm right of
One way is manager changes itself and the other one is the same thing i think.
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
Answer:
Final mass of Argon= 2.46 kg
Explanation:
Initial mass of Argon gas ( M1 ) = 4 kg
P1 = 450 kPa
T1 = 30°C = 303 K
P2 = 200 kPa
k ( specific heat ratio of Argon ) = 1.667
assuming a reversible adiabatic process
<u>Calculate the value of the M2 </u>
Applying ideal gas equation ( PV = mRT )
P₁V / P₂V = m₁ RT₁ / m₂ RT₂
hence : m2 = P₂T₁ / P₁T₂ * m₁
= (200 * 303 ) / (450 * 219 ) * 4
= 2.46 kg
<em>Note: Calculation for T2 is attached below</em>