Answer: 7.53 μC
Explanation: In order to explain this problem we have to use the gaussian law so we have:
∫E.dS=Qinside/εo we consider a gaussian surface inside the conducting spherical shell so E=0
Q inside= 0 = q+ Qinner surface=0
Q inner surface= 1.12μC so in the outer surface the charge is (8.65-1.12)μC=7.53μC
Answer:
1%
Explanation:
Percent error can be found by dividing the absolute error (difference between measure and actual value) by the actual value, then multiplying by 100.
The measured value is 2.02 meters and the actual value is 2.00 meters.
First, evaluate the fraction. Subtract 2.00 from 2.02
Next, divide 0.02 by 2.00
Finally, multiply 0.01 and 100.
The percent error is 1%.
Multiply 6 x 60 to get the frequency.
Answer:
2.62A
Explanation:
Given
V = 0.43 V
I = 3.1 A
Then, V = IR, R = V/I
R = 0.43/3.1
R = 0.14 Ω
The induced emf = dB/dt * A
So that, dB/dt = emf/A
Since dB/dt is constant then Emf/A(circle) = Emf/A square
So Emf (square)/Emf (circle) = A square / A circle
A circle = πr². The perimeter of the square is 2πr which also is the circumference of the square.
Since the perimeter is 2πr, then each side would be πr/2. Thus, the area of the square would be, (πr/2)² = π²r²/4
So A square/Acircle = (π²r²/4) / πr² = π/4 = 0.79
this means that, emf square = emf circle * 0.79
emf square = 0.43*0.79 = 0.34V
I = V/R
I = 0.34/0.13
I = 2.62A
Answer:
It works by testing the ph levels in those areas it is placed in.
Explanation:
