The solution for this problem would be:(10 - 500x) / (5 - x)
so start by doing:
x(5*50*2) - xV + 5V = 0.02(5*50*2)
500x - xV + 5V = 10
V(5 - x) = 10 - 500x
V = (10 - 500x) / (5 - x)
(V stands for the volume, but leaves us with the expression for x)
Glucose.... have a nice day
Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:
![F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}](https://tex.z-dn.net/?f=F_%7BAnet%7D%20%3D%20m_A%5Ccdot%20a_A%20%3D%20F%20-%20F_%7Bfr%7D%5C%5CF_%7BBnet%7D%20%3D%20m_B%5Ccdot%20a_B%20%3D%20F_%7Bfr%7D)
where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:
![a_B = a_A-0.5 \frac{m}{s^2}](https://tex.z-dn.net/?f=a_B%20%3D%20a_A-0.5%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:
![30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N](https://tex.z-dn.net/?f=30kg%5Ccdot%20a_A%20%3D%2075N%20-%20F_%7Bfr%7D%5C%5C24kg%5Ccdot%20a_B%20%3D%20F_%7Bfr%7D%5C%5Ca_B%3D%20a_A-0.5%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%5C%5C%5Cimplies%20%5C%5Ca_A%3D%5Cfrac%7B87%7D%7B54%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%2C%5C%2C%5C%2C%5C%2Ca_B%3D%5Cfrac%7B10%7D%7B9%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%5C%5CF_%7Bfr%7D%20%3D%2024kg%20%5Ccdot%20%5Cfrac%7B10%7D%7B9%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%3D%5Cfrac%7B80%7D%7B3%7Dkg%5Cfrac%7Bm%7D%7Bs%5E2%7D%5Capprox%2026.7N)
The force of friction acting on block B is approximately 26.7N.
This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.
Answer:
![1.8\times 105 N/C](https://tex.z-dn.net/?f=1.8%5Ctimes%20105%20N%2FC)
Explanation:
We are given that
![u=2\times 10^7 m/s](https://tex.z-dn.net/?f=u%3D2%5Ctimes%2010%5E7%20m%2Fs)
![v=4\times 10^7 m/s](https://tex.z-dn.net/?f=v%3D4%5Ctimes%2010%5E7%20m%2Fs)
d=1.9 cm=![\frac{1.9}{100}=0.019 m](https://tex.z-dn.net/?f=%5Cfrac%7B1.9%7D%7B100%7D%3D0.019%20m)
Using 1m=100 cm
We have to find the electric field strength.
![v^2-u^2=2as](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2as)
Using the formula
![(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)](https://tex.z-dn.net/?f=%284%5Ctimes%2010%5E7%29%5E2-%282%5Ctimes%2010%5E7%29%5E2%3D2a%280.019%29)
![16\times 10^{14}-4\times 10^{14}=0.038a](https://tex.z-dn.net/?f=16%5Ctimes%2010%5E%7B14%7D-4%5Ctimes%2010%5E%7B14%7D%3D0.038a)
![0.038a=12\times 10^{14}](https://tex.z-dn.net/?f=0.038a%3D12%5Ctimes%2010%5E%7B14%7D)
![a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7B12%7D%7B0.038%7D%5Ctimes%2010%5E%7B14%7D%3D3.16%5Ctimes%2010%5E%7B16%7Dm%2Fs%5E2)
![q=1.6\times 10^{-19} C](https://tex.z-dn.net/?f=q%3D1.6%5Ctimes%2010%5E%7B-19%7D%20C)
Mass of electron,m![=9.1\times 10^{-31} kg](https://tex.z-dn.net/?f=%3D9.1%5Ctimes%2010%5E%7B-31%7D%20kg)
![E=\frac{ma}{q}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7Bma%7D%7Bq%7D)
Substitute the values
![E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}](https://tex.z-dn.net/?f=E%3D%5Cfrac%7B9.1%5Ctimes%2010%5E%7B-31%7D%5Ctimes%203.16%5Ctimes%2010%5E%7B16%7D%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D)
![E=1.8\times 105 N/C](https://tex.z-dn.net/?f=E%3D1.8%5Ctimes%20105%20N%2FC)