Answer:
The maximum emf that can be generated around the perimeter of a cell in this field is 
Explanation:
To solve this problem it is necessary to apply the concepts on maximum electromotive force.
For definition we know that

Where,
N= Number of turns of the coil
B = Magnetic field
Angular velocity
A = Cross-sectional Area
Angular velocity according kinematics equations is:



Replacing at the equation our values given we have that




Therefore the maximum emf that can be generated around the perimeter of a cell in this field is 
Answer:
% differ 1.72%
Explanation:
given data:
P_ideal = 13.51 atm
n = 1.074 mol
V = 1.746 L
T = 267.6 K
According to ideal gas law we have


(P+0.514)(1.711) = 23.59
P_v = 13.276 atm
% differ 

= 1.72%
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Answer:
7200 N/m
Explanation:
Metric unit conversion
100g = 0.1 kg
5 cm = 0.05 m
50 cm = 0.5 m
As the block is released from the spring and travelling to height h = 1.5m off the ground, the elastics energy is converted to work of friction force and the potential energy at 1.5 m off the ground
The work by friction force is the product of the force F = 15N itself and the distance s = 0.5 m

Let g = 10 m/s2. The change in potential energy can be calculated as the following:

Therefore, as elastic energy is converted to potential energy and work of friction:


