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
Answer:
<em>155.80rad/s</em>
Explanation:
Using the equation of motion to find the angular acceleration:

is the final angular velocity in rad/s
is the initial angular velocity in rad/s
is the angular acceleration
t is the time taken
Given the following

Time = 4.1secs
Convert the angular velocity to rad/s
1rpm = 0.10472rad/s
6100rpm = x
x = 6100 * 0.10472
x = 638.792rad/s
Get the angular acceleration:
Recall that:

638.792 = 0 + ∝(4.1)
4.1∝ = 638.792
∝ = 638.792/4.1
∝ = 155.80rad/s
<em>Hence the angular acceleration as the blades slow down is 155.80rad/s</em>
A non <span>foliated </span>rock has interlocking grains with no specific pattern.
The answer would be 981 newtons or 220.46 pounds.
Answer:
<u>the autumnal equinox (when day and night are of equal length)</u>
<u>Explanation:</u>
The first event occurs around the 22nd or 23rd in the month of September. What usually happens is that regions around the Northern Hemisphere (which includes all of North America and Europe) experience an equal length of day and night.
While the second event occurs around March 20th or 21st mainly observable in countries found in the Southern Hemisphere.