The current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.
Ohm's Law states that the voltage across an electric conductor is directly proportional to the current(I) passing through it provided the resistant is constant.
So;
V ∝ I
V = IR
where
The objective of this question want us to determine: How did the current change for each test provided that Avery uses a 1.5-volt battery, then she uses a 3-volt battery and lastly she uses a 9-volt battery, given that the resistance is constant through out the whole process.
In the first experiment;
In the second experiment;
In the third experiment;
Therefore, we can conclude that the current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.
Learn more about Ohm's Law here:
brainly.com/question/14296509
Answer:
1.551×10^-8 Ωm
Explanation:
Resistivity of a material is expressed as shown;.
Resistivity = RA/l
R is the resistance of the material
A is the cross sectional area
l is the length of the wire.
Given;
R = 0.0310 Ω
A = πd²/4
A = π(2.05×10^-3)²/4
A = 0.000013204255/4
A = 0.00000330106375
A = 3.30×10^-6m
l = 6.60m
Substituting this values into the formula for calculating resistivity.
rho = 0.0310× 3.30×10^-6/6.60
rho = 1.023×10^-7/6.60
rho = 1.551×10^-8 Ωm
Hence the resistivity of the material is 1.551×10^-8 Ωm
It would take about 2 thirds of a second or .66666666 repeating of a second. please give brainliest?
<span>Pice=920kg/m^3
deltaP=PgH=920kg/m^3 X 9.80665m/s^2 X 1000m = 9022118 Pa
P=Po + deltaP=101.325 + 9022 = 9123kPa</span>
