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charle [14.2K]
3 years ago
6

Calculate the change in potential energy of a 91.2 kg man when he takes an elevator from the first floor to 37th floor, if the d

istance between floors is 3.99 m.
Physics
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

128379.69 J

Explanation:

Potential energy: This can be defined as the energy of a body to to position in the gravitational field. The S.I unit of potential energy is Joules (J).

The expression for potential energy is given as,

ΔEp  = mgΔH

ΔEp = mg(H₂-H₁)..................... Equation 1

Where ΔEp =change in  Potential Energy, m = mass of the man, g = acceleration due to gravity, H₂ = Height of the 37th floor, H₁ = height of the first floor

Given:m = 91.2 kg, g = 9.8 m/s², H₁ = 3.99 m, H₂ = 3.99×37 = 147.63 m.

Substitute into equation 1

ΔEp = 91.2(9.8)(147.63-3.99)

ΔEp = 893.76(143.64)

ΔEp = 128379.69 J.

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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
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a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

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a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

k = 4.3 is the dielectric constant

\epsilon_0 =8.85\cdot 10^{-12} F/m is the vacuum permittivity

A=5.1\cdot 10^{-9} m^2 is the surface area

d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

Learn more about capacity:

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