Answer:
7.1934 x 10^12 V/m.s
Explanation:
In order to do this exercise, you need to use the correct formula. Besides that, we need to identify our data.
First we have the radius of the plates which are circular, and it's 0.1 m. The current of the loop (I) is 2.0 A, and the radius of the loop is 0.2 m.
Now with this data, we use the next formula:
I = dE/dt Eo A
Where:
dE/dt = rate of electric field
Eo = constant of permittivity of free space
A = Area of circle
Solving for dE/dT:
dE/dt = I / Eo*A
Now, the area of the circle is A = πr²
A = 3.1416 * (0.1)² = 0.031416 m²
Now solving the electric field:
dE/dt = 2 / (8.85x10^-12 * 0.031416)
dE/dt = 7.1934 x 10^12 V/m.s
Answer:
Electric flux 
Explanation:
Given that,
Electric field acting on the circular area, 
We need to find the electric flux through a circular area of radius 1.83 m that lies in the xy-plane. It lies in xy plane, such that the area vector point in z direction. The electric flux is given by :
Using dot product properties, we get the value of electric flux as :
So, the electric flux through a circular area is . Hence, this is the required solution.
Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
The formula for accelerational displacement is at^2/2, so we know that 3.9t^2/2 = 200, or 3.9t^2 = 400. t =
, at = v, so
Answer:
The correct Answer to the question is : e) 4.1 m/s^2, 52 degrees north of east
Explanation:
F1= 68 N < 24º = 62.12 i + 27.65 j
F2= 32N < 132º = -21.41 i + 23.78 j
R= F1+F2= 40.71 i +51.43 j = 65.59 N < 51.63 º
By 2nd law of newton:
F= m * a
R= m*a
a= R/m
a= 4.1 m/s² < 52º (52 degrees north of east)
I consideer 0º at the EAST axis.
