C. Observations and measurements
Answer:
0.074 V
Explanation:
Parameters given:
Number of turns, N = 121
Radius of coil, r = 2.85 cm = 0.0285 m
Time interval, dt = 0.179 s
Initial magnetic field strength, Bin = 55.1 mT = 0.0551 T
Final magnetic field strength, Bfin = 97.9 mT = 0.0979 T
Change in magnetic field strength,
dB = Bfin - Bin
= 0.0979 - 0.0551
dB = 0.0428 T
The magnitude of the average induced EMF in the coil is given as:
|Eavg| = |-N * A * dB/dt|
Where A is the area of the coil = pi * r² = 3.142 * 0.0285² = 0.00255 m²
Therefore:
|Eavg| = |-121 * 0.00255 * (0.0428/0.179)|
|Eavg| = |-0.074| V
|Eavg| = 0.074 V
Answer:
Explained
Explanation:
Newton would resort to the classical mechanics and say that the momentum of the particle that is moving with a constant velocity will be given by: momentum = mass x velocity
this approach will highlight the particle nature and will not be relativistic.
De-Broglie will say that the momentum of the particle is related to its associated matter wave and the relation between them is given by:
where \lambda = wavelength of the matter wave associated to the particle, h = planck's constant
and
thus, this highlights the wave nature of the particle and is also relativistic.
Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring ?
Data:
For a spring (or an elastic), the elastic potential energy is calculated by the following expression:
Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:
Answer:
The displacement of the spring = 0.8 m
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I Hope this helps, greetings ... Dexteright02! =)
Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point