Answer: Increase in the population of lower organisms.
Explanation: Raccoons are mesopredators found in the middle levels of food webs and have critical impacts on the dynamics of many other species. Thus, they greatly contribute to the functioning of the ecosystem.
They are naturally equipped to checkmate the population size of other organism that are below them (i.e., their preys) in the food web. They are; predators, pathogen carriers (such as rabies), and competitors, as they compete with some specialist in the food web.
Hence, the decrease in the population of raccoons will lead to the increase in the population size of lower organisms they prey on in the food web or chain.
Answer:
Each one should decide we is its formula is. Depending on what options one has in live. There is not a general formula for happiness and success. So there is not a general formula to change. Everyone must find its rigth formula.
Answer:
1709.07 ft^3/s
Explanation:
Annual peak streamflow = Log10(Q [ft^3/s] )
mean = 1.835
standard deviation = 0.65
Probability of levee been overtopped in the next 15 years = 1/5
<u>Determine the design flow ins ft^3/s </u>
P₁₅ = 1 - ( q )^15 = 1 - ( 1 - 1/T )^15 = 0.2
∴ T = 67.72 years
Q₁₅ = 1 - 0.2 = 0.8
Applying Lognormal distribution : Zt = mean + ( K₂ * std ) --- ( 1 )
K₂ = 2.054 + ( 67.72 - 50 ) / ( 100 - 50 ) * ( 2.326 - 2.054 )
= 2.1504
back to equation 1
Zt = 1.835 + ( 2.1504 * 0.65 ) = 3.23276
hence:
Log₁₀ ( Qt(ft^3/s) ) = Zt = 3.23276
hence ; Qt = 10^3.23276
= 1709.07 ft^3/s
Answer:
8 μC
Explanation:
By definition, current is the rate of change of charge, so we can write the following equation for current I:
I = ΔQ / Δt
As charge must be conserved, all the charge carried by the current must add to the charge on the plates of the capacitor, so we can finf this incremental charge as follows:
ΔQ = I* Δt (assuming that current remains constant during the charging process)
⇒ ΔQ = 3 A* 2 μsec = 3 coul/sec*2 μsec = 6 μC
As the initial charge must be conserved also, the magnitude of the net electric charge of the capacitor must be as follows:
Qnet = Q₀+ ΔQ = 2 μC + 6 μC = 8 μC
Answer:
About 1.84 amperes
Explanation:
Since calculating amperes is dividing w/v, we can divided 220 by 120.
220/120 = 1.83333333333
Which can be rounded to:
1.84
