Complete Question
A thin, horizontal, 12-cm-diameter copper plate is charged to 4.4 nC . Assume that the electrons are uniformly distributed on the surface. What is the strength of the electric field 0.1 mm above the center of the top surface of the plate?
Answer:
The values is 
Explanation:
From the question we are told that
The diameter is 
The charge is 
The distance from the center is 
Generally the radius is mathematically represented as

=> 
=> 
Generally electric field is mathematically represented as
![E = \frac{Q}{ 2\epsilon_o } [1 - \frac{k}{\sqrt{r^2 + k^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7BQ%7D%7B%202%5Cepsilon_o%20%7D%20%5B1%20-%20%5Cfrac%7Bk%7D%7B%5Csqrt%7Br%5E2%20%2B%20%20k%5E2%20%7D%20%7D%20%5D)
substituting values
![E = \frac{4.4 *10^{-9}}{ 2* (8.85*10^{-12}) } [1 - \frac{(1.00 *10^{-4})}{\sqrt{(0.06)^2 + (1.0*10^{-4})^2 } } ]](https://tex.z-dn.net/?f=E%20%3D%20%20%5Cfrac%7B4.4%20%2A10%5E%7B-9%7D%7D%7B%202%2A%20%288.85%2A10%5E%7B-12%7D%29%20%7D%20%5B1%20-%20%5Cfrac%7B%281.00%20%2A10%5E%7B-4%7D%29%7D%7B%5Csqrt%7B%280.06%29%5E2%20%2B%20%20%281.0%2A10%5E%7B-4%7D%29%5E2%20%7D%20%7D%20%5D)

Answer:
Explanation:
The total fluid mass can be obtained by multiplying the mass flow rate by the time flow rate.
Mass flow rate is given as
m = ρAv
Where
m is mass flow rate
ρ is density
A is area and it is given as πr²
v is velocity
Then,
M = mt
Where M is mass and t is time
Them,
M = ρAv × t
M = ρ× πr² × v × t
Given that, .
Radius of pipe is
r = 0.089m
velocity of pipe is
v = 3.3m/s
Time taken is
t = 1 hour = 3600 seconds
Density of water is
ρ = 1000kg/m³
M = ρ× πr² × v × t
M = 1000 × π × 0.089² × 3.3 × 3600
M = 295,628.52 kg
M = 2.96 × 10^5 kg
Answer:
3.19*10^-24
Explanation:
the equation to find momentum is p=mv so you just multiply the mass times velocity
Answer:
The induced emf 1.43 s after the circuit is closed is 4.19 V
Explanation:
The current equation in LR circuit is :
.....(1)
Here I is current, V is source voltage, R is resistance, L is inductance and t is time.
The induced emf is determine by the equation :

Differentiating equation (1) with respect to time and put in above equation.


Substitute 6.05 volts for V, 0.655 Ω for R, 2.55 H for L and 1.43 s for t in the above equation.

