Una carga eléctrica de 120 Coulomb pasa uniformemente por la sección transversal de un hilo conductor durante un minuto. La inte
1 answer:
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Answer:
Frequency = 1,550Hz
Explanation:
To solve this we can use the equation:
(frequency = velocity/wavelength).
We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m
Now we can substitute these values into the formula and calculate to solve:
Simplify to 3 significant figures:
f = 1,550Hz
(Which I believe is just below a G6 if you were interested)
Hope this helped!
Answer:
q = - 93.334 nC
Explanation:
GIVEN DATA:
Radius of ring 73 cm
charge on ring 610 nC
ELECTRIC FIELD p FROM CENTRE IS AT 70 CM
E = 2000 N/C
Electric field due tor ring is guiven as
E1 = 3714.672 N/C
electric field due to point charge q
now the eelctric charge at point P is
E = E1 + E2
solving for q
q = - 93.334 nC
Answer:
(a) 135 kV
(b) The charge chould be moved to infinity
Explanation:
(a)
The potential at a distance of <em>r</em> from a point charge, <em>Q</em>, is given by
where
Difference in potential between the points is
(b)
If this potential difference is increased by a factor of 2, then the new pd = 135 kV × 2 = 270 kV. Let the distance of the new location be <em>x</em>.
The charge chould be moved to infinity