Roygbv evjnefvvnefv ekfv k kn ke nv
Answer:
f= 4,186 10² Hz
Explanation:
El sistema descrito es un pendulo de torsión que oscila con con velocidad angular, que esta dada por
w = √ k/I
donde ka es constante de torsion de hilo e I es el momento de inercia del disco
El momento de inercia de indican que giran un eje que pasa por enronqueces
I= ½ M R2
reduzcamos las cantidades al sistema SI
R= 1,4 cm = 0,014 m
M= 430 g = 0,430 kg
substituimos
w= √ (2 k/M R2)
calculemos
w = RA ( 2 370 / (0,430 0,014 2)
w = 2,963 103 rad/s
la velocidad angular esta relacionada con la frecuencia por
w =2pi f
f= w/2π
f= 2,963 10³/ (2π)
f= 4,186 10² Hz
The pipe has a cross-sectional area of
<em>π</em> (0.0100 m)² = 0.000100<em>π</em> m² ≈ 0.000314 m²
Each second, one such cross-section of water travels a distance of
(2.33 m/s) • (1 s) = 2.33 m
so that, per second, the pipe carries
(2.33 m) • (0.000100<em>π</em> m²) = 0.000233<em>π</em> m³ ≈ 0.000732 m³
of water into the tank each second. Put another way, 2.33 m of the pipe holds around 0.000732 m³ of water.
Divide the total volume of the tank by this rate:
(1.00 m³) / (0.000233<em>π</em> m³/s) ≈ 1366.14 s ≈ 1370 s
or about 22.8 min, or about 0.379 hr, etc.
Answer:
a) 3.195*10^{-3}N/C
b) 5.11*10^{-4}N/C
Explanation:
a) to find the induced electric field for different distances you use the following formula:
the area of the solenoid is constant, hence:
where you have assumed that the electric field is perpendicular to the vector of the path integral.
r: distance in which the induced electric field is measured
R: radius of the solenoid = 0.12m
dB/dt: change in the magnetic field: -7.10mT/=-7.1*10^{-3}T/s
a) for r=1.60cm=0.016m you obtain:
b) for r=10.0cm=0.10m