Answer:
1 Frequency
2 Wavelength
3 Amplitude
4 Crest
Hope it helps pls mark brainliest
Complete question:
What is the peak emf generated by a 0.250 m radius, 500-turn coil is rotated one-fourth of a revolution in 4.17 ms, originally having its plane perpendicular to a uniform magnetic field 0.425 T. (This is 60 rev/s.)
Answer:
The peak emf generated by the coil is 15.721 kV
Explanation:
Given;
Radius of coil, r = 0.250 m
Number of turns, N = 500-turn
time of revolution, t = 4.17 ms = 4.17 x 10⁻³ s
magnetic field strength, B = 0.425 T
Induced peak emf = NABω
where;
A is the area of the coil
A = πr²
ω is angular velocity
ω = π/2t = (π) /(2 x 4.17 x 10⁻³) = 376.738 rad/s = 60 rev/s
Induced peak emf = NABω
= 500 x (π x 0.25²) x 0.425 x 376.738
= 15721.16 V
= 15.721 kV
Therefore, the peak emf generated by the coil is 15.721 kV
Given that,
Atmospheric Pressure = 14.7 psi
Cooking Pressure = 14.7 +11.1 = 25.8 psi
Take, Atmospheric Temperature = 25 °C
Cooking Temperature = ??
Since, we know that Gas equation is given by:
PV = nRT
or
P ∝ T
P1 / T1 = P2 / T2
14.7/ 25 = 25.8/ T2
T2 = 25*25.8/14.7
T2 = 43.87 °C
The cooking pressure will be 43.87 °C.
Average velocity over a given time interval is the distance traveled divided by the time:
