<span>Continental crust is less dense than oceanic crust </span>
Answer:
The maximum error is 
Explanation:
From the question we are told that
The length is 
The radius is 
The pressure is 
The rate is 
The viscosity is 
The error in the viscosity is mathematically represented as

Where 
and 
and 
So
![\Delta \eta = \frac{\pi}{8} [ |\frac{r^4}{v} | * \Delta P + | \frac{4 * P * r^3}{v} |* \Delta r + |-\frac{P* r^4}{v^2} |* \Delta v]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7Br%5E4%7D%7Bv%7D%20%20%7C%20%2A%20%20%5CDelta%20%20P%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%20P%20%2A%20r%5E3%7D%7Bv%7D%20%20%7C%2A%20%20%5CDelta%20%20r%20%2B%20%20%7C-%5Cfrac%7BP%2A%20r%5E4%7D%7Bv%5E2%7D%20%20%7C%2A%20%20%5CDelta%20%20v%5D)
substituting values
![\Delta \eta = \frac{\pi}{8} [ |\frac{(0.002)^4}{0.5*10^{-9}} | * 1750 + | \frac{4 * 4 *10^{5} * (0.002)^3}{0.5*10^{-9}} |* 0.0002 + |-\frac{ 4*10^{5}* (0.002)^4}{(0.5*10^{-9})^2} |* 0 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B%20%7C%5Cfrac%7B%280.002%29%5E4%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%20%2A%20%201750%20%20%20%2B%20%20%20%20%7C%20%5Cfrac%7B4%20%2A%20%204%20%2A10%5E%7B5%7D%20%2A%20%280.002%29%5E3%7D%7B0.5%2A10%5E%7B-9%7D%7D%20%20%7C%2A%20%200.0002%20%2B%20%20%7C-%5Cfrac%7B%204%2A10%5E%7B5%7D%2A%20%280.002%29%5E4%7D%7B%280.5%2A10%5E%7B-9%7D%29%5E2%7D%20%20%7C%2A%20%200%20%5D)
![\Delta \eta = \frac{\pi}{8} [56 + 5120 ]](https://tex.z-dn.net/?f=%5CDelta%20%20%5Ceta%20%20%3D%20%5Cfrac%7B%5Cpi%7D%7B8%7D%20%5B56%20%20%2B%20%205120%20%5D)


Answer:


Explanation:
The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

Answer:
The kinetic energy of the system after the collision is 9 J.
Explanation:
It is given that,
Mass of object 1, m₁ = 3 kg
Speed of object 1, v₁ = 2 m/s
Mass of object 2, m₂ = 6 kg
Speed of object 2, v₂ = -1 m/s (it is moving in left)
Since, the collision is elastic. The kinetic energy of the system before the collision is equal to the kinetic energy of the system after the collision. Let it is E. So,

E = 9 J
So, the kinetic energy of the system after the collision is 9 J. Hence, this is the required solution.