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
I’m lost at this question, sorry but I would’ve help !
Answer: I don't know how to do this
Explanation: sorry I am not sure.
Surface tension can change with the change in a medium that is just above the layer of the liquid's surface.
Explanation:
Pouring any oil or oily compounds (such as kerosene) on the free surface of the water will reduce the surface tension.
in the atmosphere directly affects the surface tension of the liquid.
If we increase the temperature of the water, then there is a high possibility of the surface tension of the water getting reduced, due to the fact that the net force of attraction is decreased.
Mixing surfactants or emulsifiers into the water will decrease the surface tension.
If the water is subjected to electrification, then the surface tension will be reduced.
Answer:
The acceleration is 
Explanation:
Given the velocity function:

you can obtain the instantaneous acceleration "a" as its first derivative:

To determine the value of "a" when the velocity was 12m/s, you need to figure out the value for "t" when this happens. At what time t is the velocity 12m/s?

This value of t is less than the 5 seconds mentioned in the text - so that is a good sign that the formula is valid for this value. And so you can use t=3.47s in the derivative (acceleration) above: