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:
B. people with OCD know their disorder is irrational
Explanation:
Got it right
Answer:
68kg
Explanation:
1 cm^3 is the same as 1 mL and there are 5000mL in 5L
Therefore if the density is 13.6g/mL we multiply 13.6 by 5000 to get the amount of grams required = 68000g which is 68kg
Answer:
KE₂ = 6000 J
Explanation:
Given that
Potential energy at top U₁= 7000 J
Potential energy at bottom U₂= 1000 J
The kinetic energy at top ,KE₁= 0 J
Lets take kinetic energy at bottom level = KE₂
Now from energy conservation
U₁+ KE₁= U₂+ KE₂
Now by putting the values
U₁+ KE₁= U₂+ KE₂
7000+ 0 = 1000+ KE₂
KE₂ = 7000 - 1000 J
KE₂ = 6000 J
Therefore the kinetic energy at bottom is 6000 J.
Answer:
dV/dt = 9 cubic inches per second
Explanation:
Let the height of the cylinder is h
Diameter of cylinder = height of the cylinder = h
Radius of cylinder, r = h/2
dh/dt = 3 inches /s
Volume of cylinder is given by

put r = h/2 so,

Differentiate both sides with respect to t.

Substitute the values, h = 2 inches, dh/dt = 3 inches / s

dV/dt = 9 cubic inches per second
Thus, the volume of cylinder increases by the rate of 9 cubic inches per second.