Answer:
E=
Explanation:
We are given that
Charge on ring= Q
Radius of ring=a
We have to find the magnitude of electric filed on the axis at distance a from the ring's center.
We know that the electric field at distance x from the center of ring of radius R is given by

Substitute x=a and R=a
Then, we get




Where K=
Hence, the magnitude of the electric filed due to charged ring on the axis of ring at distance a from the ring's center=
The correct answer as the first one above !
The runner has initial velocity vector

and acceleration vector

so that her velocity at time
is

She runs directly east when the vertical component of
is 0:

It's not clear what you're supposed to find at this particular time... possibly her position vector? In that case, assuming she starts at the origin, her position at time
would be

so that after 10.4 s, her position would be

which is 19.9 m away from her starting position.
Answer:
7.39 m/s
Explanation:
Applying
K.E = 1/2mv²..................... Equation 1
Where K.E = Kinetic Energy, m = mass of the ball, v = velocity of the ball.
Make v the subject of the equation
v = √(2K.E/m)................. Euqation 2
From the question,
Given: K.E = 30 J, m = 1.1kg
Substitute these values into equation 2
v = √(2×30/1.1)
v = √54.54
v = 7.39 m/s
Answer:
-1.65
Explanation:
First of all, we find the position of the image by using the lens equation:

where:
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
For the lens in this problem:
f = 21.0 cm (the focal length of a convex lens is positive)
p = 33.7 cm
Solving for q, we find the position of the image:

Then, the magnification of the image is given by:

And substituting,

Which means that the image is inverted (negative sign) and enlarged (because M is larger than 1).