Answer:
The magnitude of rate of change of electric field is
.
Explanation:
Given that,
Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m
Total displacement current through a cross section of the region, 
We need to find the rate of change of electric field. Its is given by the formula as follows :

So, the magnitude of rate of change of electric field is
.
To solve this problem we will use the kinematic equations of angular motion, starting from the definition of angular velocity in terms of frequency, to verify the angular displacement and its respective derivative, let's start:



The angular displacement is given as the form:
In the equlibrium we have to
and in the given position we have to

Derived the expression we will have the equivalent to angular velocity

Replacing,

Finally

Therefore the maximum angular displacement is 9.848°
Answer:
d=0.137 m ⇒13.7 cm
Explanation:
Given data
m (Mass)=3.0 kg
α(incline) =34°
Spring Constant (force constant)=120 N/m
d (distance)=?
Solution
F=mg
F=(3.0)(9.8)
F=29.4 N
As we also know that
Force parallel to the incline=FSinα
F=29.4×Sin(34)
F=16.44 N
d(distance)=F/Spring Constant
d(distance)=16.44/120
d(distance)=0.137 m ⇒13.7 cm
Answer:
Check the explanation
Explanation:
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
The image is virtual
The image is upright
given
R = 1.5 cm
object distance, u = 1.1 cm
focal length of the ball, f = -R/2
= -1.5/2
= -0.75 cm
let v is the image distance
use, 1/u + 1/v = 1/f
1/v = 1/f - 1/u
1/v = 1/(-0.75) - 1/(1.1)
v = -0.446 cm <<<<<---------------Answer
magnification, m = -v/u
= -(-0.446)/1.1
= 0.405 <<<<<<<<<---------------Answer
Kindly check the diagram in the attached image below.