Answer:
Induced emf of the wire is 6.36 Volts.
Explanation:
It is given that,
Length of the wire, l = 75 cm = 0.75 m
Magnetic field, B = 0.53 T
Velocity, v = 16 m/s
The wire is moving straight up in the magnetic field. So, an emf is induced in the wire. It is given by :



So, the induced emf of the wire is 6.36 V. Hence, the correct option is (b) "6.36 V".
Answer:
Yes, the energy is not simply the sum of the individual binding energies at each site, it is the product of energy at each binding site of hemoglobin.
Explanation:
Myoglobin and hemoglobin are two different cells. Myoglobin binds only one oxygen while the hemoglobin has the ability to binds four oxygen atoms at its four sides. Myoglobin present in muscle tissue only while hemoglobin is present in the whole body. Oxyhemoglobin is formed when oxygen binds with hemoglobin cell. This oxygen is take to all cells and energy is released due to the breakdown of glucose molecules with this oxygen.
Answer:
f = 3.102 Hz
Explanation:
In this case you have that the required voltage is the maximum induced emf produced by the rotating generator.
In order to calculate the frequency of rotation of the generator that allows one to obtain 12.0V you use the following formula:
(1)
N: turns of the armature = 95
B: magnitude of the magnetic field = 0.800T
A: area of the square armature = (9.0cm)^2 = (0.09m)^2 = 8.1*10^-3 m^2
emf_max = 12.0V
w: angular frequency
you solve the equation (1) for w:

Then, the frequency is:
