Answer:
The magnitude of the electric field are
and 
Explanation:
Given that,
Radius of inner shell = 11.0 cm
Radius of outer shell = 14.0 cm
Charge on inner shell 
Charge on outer shell 
Suppose, at r = 11.5 cm and at r = 20.5 cm
We need to calculate the magnitude of the electric field at r = 11.5 cm
Using formula of electric field

Where, q = charge
k = constant
r = distance
Put the value into the formula


The total charge enclosed by a radial distance 20.5 cm
The total charge is

Put the value into the formula


We need to calculate the magnitude of the electric field at r = 20.5 cm
Using formula of electric field

Put the value into the formula


Hence, The magnitude of the electric field are
and 
Answer:
yes
Explanation:
The metal is closer than 20 cm to the magnet which is in the magnetic field.
Answer:
(A) Angular speed 40 rad/sec
Rotation = 50 rad
(b) 37812.5 J
Explanation:
We have given moment of inertia of the wheel 
Initial angular velocity of the wheel 
Angular acceleration 
(a) We know that 
We have given t = 2 sec
So 
Now 
(b) After 3 sec 
We know that kinetic energy is given by 
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
<h3>What is velocity?</h3>
Velocity is a vector quantity that tells the distance an object has traveled over a period of time.
Displacement is a vector quality showing total length of an area traveled by a particular object.
Imagine a time-position graph where the velocity of an object is constant. What will be observed on the graph concerning the slope of the line segment as well as the velocity of the object?
The slope of the line is equal to zero and the object will be stationary.
The position-time graphs show the relationship between the position of an object (shown on the y-axis) and the time (shown on the x-axis) to show velocity.
To learn more about velocity refer to the link
brainly.com/question/18084516
#SPJ2
Answer: 7.38 km
Explanation: The attachment shows the illustration diagram for the question.
The range of the bomb's motion as obtained from the equations of motion,
H = u(y) t + 0.5g(t^2)
U(y) = initial vertical component of velocity = 0 m/s
That means t = √(2H/g)
The horizontal distance covered, R,
R = u(x) t = u(x) √(2H/g)
Where u(x) = the initial horizontal component of the bomb's velocity = 287 m/s, H = vertical height at which the bomb was thrown = 3.24 km = 3240 m, g = acceleration due to gravity = 9.8 m/s2
R = 287 √(2×3240/9.8) = 7380 m = 7.38 km