Answer:
Electric potential, E = 2100 volts
Explanation:
Given that,
Electric field, E = 3000 N/C
We need to find the electric potential at a point 0.7 m above the surface, d = 0.7 m
The electric potential is given by :
V = 2100 volts
So, the electric potential at a point 0.7 m above the surface is 2100 volts. Hence, this is the required solution.
Answer:
I attached an image that should help.
Explanation:
Check it out.
Frequency is the vibration of noise and the vibration determines the pitch, which we depend on to be a pitch or frequency we can hear. If it's too high or too low our ears can't hear it
<span>R = rate of flow = 0.370 L/s
H = height = 2.9 m
T= time = 3.9 s
V = velocity of water when it hits the bucket = sqrt(2gh) = sqrt(2 x 9.8 x 2.9) =7.539 m/s2
G value = 9.8 m/s2
Wb = weight of bucket = 0.690 kg x 9.8 m/s2 = 6.762 N
Wa = weight of accumulated water after 3.9 s
Fi = force of impact of water on the bucket
S = reading on the scale = Wa + Wb + Fi
mass of water accumulated after 3.9 s = R x T = 0.370 x 3.9 = 1.443 L = 1.443 kg
Therefore, Wa = 1.443 x 9.8 = 14.1414 N
Fi = rate of change of momentum at the impact point = R x V (because R = dm/dt)
= 0.37 x 7.539 = 2.78943 N
S = 14.1414 + 6.762 + 2.78943 = 23.692 N</span>
A) Work energy relation;
Work =ΔKE ; work done = Force × distance, while, Kinetic energy = 1/2 MV²
F.s = 1/2mv²
F× 4×10^-2 = 1/2 × 5 ×10^-3 × (600)²
F = 900/0.04
= 22500 N
Therefore, force is 22500 N
b) From newton's second law of motion;
F = Ma
Thus; a = F/m
= 22500/(5×10^-3)
= 4,500,000 m/s²
But v = u-at
0 = 600- 4500,000 t
t = 1.33 × 10^-4 seconds
