Answer:
Distance of the point where electric filed is 2.45 N/C is 1.06 m
Explanation:
We have given charge per unit length, that is liner charge density 
Electric field E = 2.45 N/C
We have to find the distance at which electric field is 2.45 N/C
We know that electric field due to linear charge is equal to
, here
is linear charge density and r is distance of the point where we have to find the electric field
So 
r = 1.06 m
So distance of the point where electric filed is 2.45 N/C is 1.06 m
(a) 5.66 m/s
The flow rate of the water in the pipe is given by

where
Q is the flow rate
A is the cross-sectional area of the pipe
v is the speed of the water
Here we have

the radius of the pipe is
r = 0.260 m
So the cross-sectional area is

So we can re-arrange the equation to find the speed of the water:

(b) 0.326 m
The flow rate along the pipe is conserved, so we can write:

where we have

and where
is the cross-sectional area of the pipe at the second point.
Solving for A2,

And finally we can find the radius of the pipe at that point:

Speed = distance / time
S= 40 000m / 5400s
S= 7.41m/s
By working with percentages, we want to see how many inches is the center of gravity out of the limits. We will find that the CG is 1.45 inches out of limits.
<h3>What are the limits?</h3>
First, we need to find the limits.
We know that the MAC is 58 inches, and the limits are from 26% to 43% MAC.
So if 58 in is the 100%, the 26% and 43% of that are:
- 26% → (26%/100%)*58in = 0.26*58 in = 15.08 in
- 43% → (43%/100%)*58in = 0.43*58 in = 24.94 in.
But we know that the CG is found to be 45.5% MAC, then it measures:
(45.5%/100%)*58in = 0.455*58in = 26.39 in
We need to compare it with the largest limit, so we get:
26.39 in - 24.94 in = 1.45 in
This means that the CG is 1.45 inches out of limits.
If you want to learn more about percentages, you can read:
brainly.com/question/14345924