Answer:
1.544*10⁹ Linebackers would be required in order to obtain the same density as an alpha particle
Step-by-step explanation:
Assuming that the pea is spherical ( with radius R= 0.5 cm= 0.005 m), then its volume is
V= 4/3π*R³ = 4/3π*R³ = 4/3*π*(0.005 m)³ = 5.236*10⁻⁷ m³
the mass in that volume would be m= N*L (L= mass of linebackers=250Lbs= 113.398 Kg)
The density of an alpha particle is ρa= 3.345*10¹⁷ kg/m³ and the density in the pea ρ will be
ρ= m/V
since both should be equal ρ=ρa , then
ρa= m/V =N*L/V → N =ρa*V/L
replacing values
N =ρa*V/L = 3.345*10¹⁷ kg/m³ * 5.236*10⁻⁷ m³ /113.398 Kg = 1.544*10⁹ Linebackers
N=1.544*10⁹ Linebackers
Answer:
P = 6200 / (1 + 5.2e^(0.0013t))
increases the fastest
Step-by-step explanation:
dP/dt = 0.0013 P (1 − P/6200)
Separate the variables.
dP / [P (1 − P/6200)] = 0.0013 dt
Multiply the left side by 6200 / 6200.
6200 dP / [P (6200 − P)] = 0.0013 dt
Factor P from the denominator.
6200 dP / [P² (6200/P − 1)] = 0.0013 dt
(6200/P²) dP / (6200/P − 1) = 0.0013 dt
Integrate.
ln│6200/P − 1│= 0.0013t + C
Solve for P.
6200/P − 1 = Ce^(0.0013t)
6200/P = 1 + Ce^(0.0013t)
P = 6200 / (1 + Ce^(0.0013t))
At t = 0, P = 1000.
1000 = 6200 / (1 + C)
1 + C = 6.2
C = 5.2
P = 6200 / (1 + 5.2e^(0.0013t))
You need to change the exponent from negative to positive.
The inflection points are where the population increases the fastest.
I want to say it is an acute angel
Answer:triangles 20 12 28 and15 9 21
Step-by-step explanation:
