Answer:
Part a)
![Q_1 = 2\mu C](https://tex.z-dn.net/?f=Q_1%20%3D%202%5Cmu%20C)
Part b)
![\Delta V = 4.5 \times 10^4 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%204.5%20%5Ctimes%2010%5E4%20V)
Explanation:
As we know that conducting sphere is treated like spherical capacitor
so we can say
![\frac{C_1}{C_2} = \frac{0.4}{1}](https://tex.z-dn.net/?f=%5Cfrac%7BC_1%7D%7BC_2%7D%20%3D%20%5Cfrac%7B0.4%7D%7B1%7D)
also we know that
![Q = CV](https://tex.z-dn.net/?f=Q%20%3D%20CV)
so charge on two spheres will be in the ratio of their capacitance
Part a)
![Q_1 + Q_2 = 7 \mu C](https://tex.z-dn.net/?f=Q_1%20%2B%20Q_2%20%3D%207%20%5Cmu%20C)
![\frac{Q_1}{Q_2}= 0.4](https://tex.z-dn.net/?f=%5Cfrac%7BQ_1%7D%7BQ_2%7D%3D%200.4)
![1.4Q_2 = 7\mu C](https://tex.z-dn.net/?f=1.4Q_2%20%3D%207%5Cmu%20C)
![Q_2 = 5 \mu C](https://tex.z-dn.net/?f=Q_2%20%3D%205%20%5Cmu%20C)
![Q_1 = 2\mu C](https://tex.z-dn.net/?f=Q_1%20%3D%202%5Cmu%20C)
Part b)
Now potential difference between two sphere can be given as
![\Delta V = \frac{Q}{C}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7BQ%7D%7BC%7D)
![\Delta V = \frac{2\mu C}{4\pi \epsilon_0 R_1}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7B2%5Cmu%20C%7D%7B4%5Cpi%20%5Cepsilon_0%20R_1%7D)
![\Delta V = \frac{2\mu C}{4\pi \epsilon_0(0.400)}](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%20%5Cfrac%7B2%5Cmu%20C%7D%7B4%5Cpi%20%5Cepsilon_0%280.400%29%7D)
![\Delta V = 4.5 \times 10^4 V](https://tex.z-dn.net/?f=%5CDelta%20V%20%3D%204.5%20%5Ctimes%2010%5E4%20V)
Answer:
Explanation:
If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.
Answer: liquid
explanation: 1 liter is a measurement of liquids, not solids, or gases.
Liquids also have a set volume, but can flow to take the shape of the bottom of their container.
Answer:
a) 1.67 m/s
b) 23kJ
Explanation:
We need to apply the linear momentum conservation formula, that states:
![m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}](https://tex.z-dn.net/?f=m1%2Av_%7Bo1%7D%2Bm2%2Av_%7Bo2%7D%3Dm1%2Av_%7Bf1%7D%2Bm2%2Av_%7Bf2%7D)
in this case:
![3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s](https://tex.z-dn.net/?f=3.45%2A10%5E4kg%2A2.60m%2Fs%2B2%2A3.45%2A10%5E4kg%2A1.20m%2Fs%3D3%2Am1%2Av_%7Bf%7D%5C%5Cv_f%3D1.67m%2Fs)
the initital kinetic energy is:
![K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ](https://tex.z-dn.net/?f=K_i%3D%5Cfrac%7B1%7D%7B2%7D%2A3.45%2A10%5E4kg%2A%282.60m%2Fs%29%5E2%2B2%28%5Cfrac%7B1%7D%7B2%7D%2A3.45%2A10%5E4kg%2A%281.20m%2Fs%29%5E2%5C%5CK_i%3D167kJ)
and the final:
![K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ](https://tex.z-dn.net/?f=K_f%3D3%2A%5Cfrac%7B1%7D%7B2%7D%2A3.45%2A10%5E4kg%2A%281.67m%2Fs%29%5E2%5C%5CK_f%3D144kJ)
The energy lost is given by:
![E_l=|K_f-K_i|\\E_l=23kJ](https://tex.z-dn.net/?f=E_l%3D%7CK_f-K_i%7C%5C%5CE_l%3D23kJ)