Answer:
Now e is due to the ring at a
So
We say
1/4πEo(ea/ a²+a²)^3/2
= 1/4πEo ea/2√2a³
So here E is faced towards the ring
Next is E due to a point at the centre
So
E² = 1/4πEo ( e/a²)
Finally we get the total
Et= E²-E
= e/4πEo(2√2-1/2√2)
So the direction here is away from the ring
Answer:

Explanation:
As we know that the orbital speed is given as

here we know that
v = 5500 m/s


now we have


now acceleration due to gravity of planet is given as



now range of the projectile on the surface of planet is given as



Answer:
(a) 45 micro coulomb
(b) 6 micro Coulomb
Explanation:
C = 3 micro Farad = 3 x 10^-6 Farad
V = 15 V
(a) q = C x V
where, q be the charge.
q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb
(b)
V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad
q = C x V
where, q be the charge.
q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb
This would be typical of an elastic collision.
Answer:
32cm³
Explanation:
Given parameters:
Density of substance = 2.7g/cm³
Mass of substance = 86.4g
Unknown:
Volume of substance = ?
Solution:
Density is the mass per unit volume of a substance.
Density = 
Since the unknown is volume we solve for it;
mass = density x volume
86.4 = 2.7 x volume
volume =
= 32cm³