Answer:
the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow.
Explanation:
We can answer this exercise using Gauss's law
Ф = ∫ e . dA = 
/ ε₀
field flow is directly proportionate to the charge found inside it, therefore if we place a Gaussian surface outside the plastic spherical shell. the flow must be zero since the charge of the sphere is equal induced in the shell, for which the net charge is zero. we see with this analysis that this shell meets the requirement to block the elective field
From the same Gaussian law it follows that if the sphere is not in the center, the only effect it has is to create more induced charge at the closest points, but the net face remains zero, so it has no effect on the flow , so no matter where the sphere is, the total induced charge is always equal to the charge on the sphere.
W = 1/2k*x^2.
k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).
W = 1/2(2500)(0.04)^2 = 2J.
Answer: 35*10^3 N/m
Explanation: In order to explain this problem we know that the potential energy for spring is given by:
Up=1/2*k*x^2 where k is the spring constant and x is the streching or compresion position from the equilibrium point for the spring.
We also know that with additional streching of 2 cm of teh spring, the potential energy is 18J. Then it applied another additional streching of 2 cm and the energy is 25J.
Then the difference of energy for both cases is 7 J so:
ΔUp= 1/2*k* (0.02)^2 then
k=2*7/(0.02)^2=35000 N/m
Answer:
T = 693.147 minutes
Explanation:
The tank is being continuously stirred. So let the salt concentration of the tank at some time t be x in units of kg/L.
Therefore, the total salt in the tank at time t = 1000x kg
Brine water flows into the tank at a rate of 6 L/min which has a concentration of 0.1 kg/L
Hence, the amount of salt that is added to the tank per minute = 
Also, there is a continuous outflow from the tank at a rate of 6 L/min.
Hence, amount of salt subtracted from the tank per minute = 6x kg/min
Now, the rate of change of salt concentration in the tank = 
So, the rate of change of salt in the tank can be given by the following equation,
or, 
or, T = 693.147 min (time taken for the tank to reach a salt concentration
of 0.05 kg/L)
Hello!
The independent variable is the variable deliberately changed.
The dependent variable is the variable that responds to change. So the answer is A.
Hope this helps. Any questions please just ask! Thank you!
