Answer:
db / dt = kb
this becomes b(t) = Ce^(kt)
C = 100, the initial population
P(1) = 420 = 100 e^(1k)
4.2 = e^k
ln 4.2 = k
a) thus, b(t) = 100 e^(t ln 4.2)
b) b(3) = 100 e^(3 ln 4.2)
c) growth constant will still be ln 4.2 (constant percentage of populatioin)
d) 10000 = 100 e^(t ln 4.2)
100 = e^(t ln 4.2)
ln 100 = t ln 4.2
t = ln 100 / ln 4.2
Step-by-step explanation:
To determine the answer of Part A draw the equilateral triangle and the to determine the coordinates of of the third charge use that triangle.
To calculate the gravitational field strength in part B from each of the charges use the following equation.
E=kcq/r2
If you would add those values then you can use the symmetry about the y axis to make the vector addition a litter easier.<span />
Answer:32
Step-by-step explanation: manage by hours of work and pay you’ll come to 32
<span>As the age of the U-235 sample is 2.631 billion years, and the half-life of U-235 is 713 million years, the sample has undergone 2.361 X 1,000,000,000 / 713 X 1,000,000 = 3.69 half lives. In each half-life the sample reduces to half its original weight according to the radioactive Half-Life Formula:
ln (Nt /N0) = -kt, where N0 = mass of the original weight of radioactive material, Nt = mass of radioactive material at time t, k = decay constant and t = time interval . We have to put Nt/N0 = 1/2 for time interval = half-life.</span>
The correct answer is B. 30 > 10
Explanation:
In mathematics and related fields, number lines are straight lines that display numbers using intervals this is useful to represent quantities visually, for example in graphs. Additionally, in number lines, numbers are organized in ascending order from left to right, for example, number 1 will be placed to the left of number 2. According to this, number 30 placed to the right of 10 will represent a higher quantity, considering the more you go to the right the higher the number is. Thus, 30 is greater than 10 (30 > 10).
