Answer:
Explanation:
We are given that
Surface area of membrane=
Thickness of membrane=
Assume that membrane behave like a parallel plate capacitor.
Dielectric constant=5.9
Potential difference between surfaces=85.9 mV
We have to find the charge resides on the outer surface of membrane.
Capacitance between parallel plate capacitor is given by
Substitute the values then we get
Capacitance between parallel plate capacitor=
V=
Hence, the charge resides on the outer surface=
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
Answer:
The correct option is;
A. The potential energy between both like charges and like poles increases as they move closer together
Explanation:
Here we have that when we move the like poles of two bar magnets close to each other, there is an increased resistance in the continuing motion, therefore for each extra gap closer achieved, there is an increase in potential energy
Similarly, when two like charges are brought closer together, the potential energy, or the energy available to push the two like charges apart increases charge as the as the charges are brought closer together
Therefore, the correct option is the potential energy between both like charges and like poles increases as they move closer together.
