Frequency = 1/time period = 1/0.05 = 20s^-1.
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
![E = \frac{KQx}{[x^2+ R^2]^{3/2}}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BKQx%7D%7B%5Bx%5E2%2B%20R%5E2%5D%5E%7B3%2F2%7D%7D)

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
There are many porperties. You can use Altitude, Temperature, Pressure and Density, but the best one is temperature. The resaon for that is that based on the temperature changes then the athmosphere can be broken into four major layers. Remember that the layers are the following: <span>the </span>troposphere,the<span> </span>stratosphere, <span>the </span>mesosphere<span>, and the</span>thermosphere<span>.</span>
Each point along the track of one solar mass star represents the star's surface temperature and luminosity at one time.
<h3>What is the one-solar mass star?</h3>
A star having a mass equal to the mass of the Sun is called a one-solar mass star.
Its life track shows the luminous intensity as well as the surface temperature.
Learn more about one-solar mass star.
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