Answer:
The current on the water layer = 1.64×10^-3A
Explanation:
Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:
I = V × pi[(Rwater + Rstring)^2 - (Rstring)^2/ ( Resitivity × L)
I =[ 166×10^6 ×3.142[(0.519×10^-4) + (2.15×10^-3])^2 - ( 2.15×10^-3)^2] / ( 183 × 831)
I =[ 521572000(4.848×10^6)- 4.623×10^-6]/ 154566
I = 252.83 -(4.623×10^-6)/ 154566
I = 252.83/154566
I = 1.64× 10^-3A
Answer:
the final velocity of the wagon is 6 m/s.
Explanation:
Given;
initial velocity of the wagon, u = 4 m/s
mass of the wagon, m = 35 kg
energy applied to the wagon, E = 350 J
The final velocity of the wagon is calculated as;
E = ¹/₂m(v² - u²)
Therefore, the final velocity of the wagon is 6 m/s.
It's a subsection of Geology and Biology
Answer:
8.067 ohms
Explanation:
The relationship between voltage, resistance, and power is ...
P = V²/R
so ...
R = V²/P = 110²/1500 = 8 1/15
R ≈ 8.067 . . . ohms
