Answer:
Magnitude of the third vector: 57.85 cm
The direction of the third vector: 44.76 N of W
Explanation:
The electric potential V(z) on the z-axis is : V = 
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
<u>Given data :</u>
V(z) =2kQ / a²(v(a² + z²) ) -z
<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV
----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = 
V = ![\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%5Calpha%20%5B%20%28a%5E2%2Bz%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%20-z%20%5D)
= ![\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%20%28%5Cfrac%7BQ%7D%7B%5Cpi%20%5Calpha%20%5E2%7D%29%5B%28a%5E2%20%2Bz%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20-z%20%5D)
Therefore the electric potential V(z) = 
Also
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
Learn more about electric potential : brainly.com/question/25923373
Explanation:
a = (Vf-Vi)/t
= (10m/s - 15m/s)/2s
a = -5 / 2 m/s^2
• so the dog diaccelerate at the rate of 5/2 m/s^2
hope it helps
Answer:
6.13 s
219 N
Explanation:
Newton's law in the x direction:
∑F = ma
150 cos 30° N − 50 N = (30 kg) a
a = 2.66 m/s²
Δx = v₀ t + ½ at²
(50 m) = (0 m/s) t + ½ (2.66 m/s²) t²
t = 6.13 s
Newton's law in the y direction:
∑F = ma
Fn + 150 sin 30° N − (30 kg) (9.8 m/s²) = 0
Fn = 219 N