Answer:
Uncorrected values for
For circuit P
R = 2.4 ohm
For circuit Q
R = 2.4 ohm
Corrected values
for circuit P
R = 12 OHM
For circuit Q
R = 2.3 ohm
Explanation:
Given data:
Ammeter resistance 0.10 ohms
Resister resistance 3.0 ohms
Voltmeter read 6 volts
ammeter reads 2.5 amp
UNCORRECTED VALUES FOR
1) circuit P
we know that IR =V

2) circuit Q
R = 2.4 ohm as no potential drop across ammeter
CORRECTED VALUES FOR
1) circuit p
IR = V

R= 12 ohm
2) circuit Q


R = 2.3 ohm
Part a
Answer: 17.58 km/h

Total Distance =10 km
Total time =0.5689 h

Part b
Answer: 17.626 km/h

Total Distance =42.195 km
Total time =2.3939 h

A
method of procedure that has characterized natural science since the
17th century, consisting in systematic observation, measurement, and
experiment, and the formulation, testing, and modification of
hypotheses
One is their traits and their characterists that they have in common
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as

We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.

The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is

where ‘h’ is the length of the imaginary Gaussian surface.

2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,

3. At the boundary where r = R:

As can be seen from above, two E-field values are equal as predicted.