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
true if you are refering to the desing of the experimnt as it does identify the variable
Air pressure is the weight of air on an area. The weight of air
is due to the gravitational forces between the Earth and the
molecules of its atmosphere.
To solve this problem it is necessary to apply the kinematic equations of angular motion.
Torque from the rotational movement is defined as

where
I = Moment of inertia
For a disk
Angular acceleration
The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

Where
Final and Initial Angular velocity
Angular acceleration
Angular displacement
Our values are given as






Using the expression of angular acceleration we can find the to then find the torque, that is,




With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so




Therefore the torque exerted on it is 
Answer:
(A) 667.5 N/m
(B)
Explanation:
(A) Let the spring constant be k.
Using the formula F = kx
k = 251 / 0.376
K = 667.5 N/m
(B)
Work done
W = 0.5 × kx^2
W = 0.5 × 667.5 × 0.376 × 0.376
W = 47.2 J