Current is defined as the rate of charge flowing a point every second. Having a current of 1 Ampere signifies 1 Coulomb is flowing in a circuit every second. It is measured by the use of an ammeter which is positioned in series to the component to be measured. The current in the problem is calculated as follows:
I = 2.0 x 10^-4 C / 5.0 x 10^-5 s
<span>I = 4 A</span>
When two sides of a membrane are in contact with each other, the distribution of ions will alter as a result of the binding of a signal molecule to a ligand-gated ion channel.
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What is a ligand-gated ion channel?</h3>
Ligand-gated ion channels (LGICs) are membrane proteins that are structurally integral and feature a pore that permits the controlled passage of particular ions across the plasma membrane. The electrochemical gradient for the permeant ions drives the passive ion flux.
When a chemical ligand, such as a neurotransmitter, attaches to the protein, ligand-gated ion channels open. Changes in membrane potential cause voltage channels to open and close. When a receptor physically deforms, as in the case of pressure and touch receptors, mechanically-gated channels open.
Learn more about ligand-gated ion channel here:
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d. the rate at which work is accomplished
Answer:
The BOD concentration 50 km downstream when the velocity of the river is 15 km/day is 63.5 mg/L
Explanation:
Let the initial concentration of the BOD = C₀
Concentration of BOD at any time or point = C
dC/dt = - KC
∫ dC/C = -k ∫ dt
Integrating the left hand side from C₀ to C and the right hand side from 0 to t
In (C/C₀) = -kt + b (b = constant of integration)
At t = 0, C = C₀
In 1 = 0 + b
b = 0
In (C/C₀) = - kt
(C/C₀) = e⁻ᵏᵗ
C = C₀ e⁻ᵏᵗ
C₀ = 75 mg/L
k = 0.05 /day
C = 75 e⁻⁰•⁰⁵ᵗ
So, we need the BOD concentration 50 km downstream when the velocity of the river is 15 km/day
We calculate how many days it takes the river to reach 50 km downstream
Velocity = (displacement/time)
15 = 50/t
t = 50/15 = 3.3333 days
So, we need the C that corresponds to t = 3.3333 days
C = 75 e⁻⁰•⁰⁵ᵗ
0.05 t = 0.05 × 3.333 = 0.167
C = 75 e⁻⁰•¹⁶⁷
C = 63.5 mg/L