3 years ago

Is it possible for a population with a high birth rate to decrease in size.

Physics

1 answer:

matrenka [14]3 years ago

7 0

Sure, if the mortality (death) rate is even higher than the birth rate.

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Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

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A roller coaster car is going over the top of a 18-m-radius circular rise. at the top of the hill, the passengers "feel light,"

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The solution for this problem is:

If they feel 50% of their weight that means that the centripetal force is also 50% of their weight 1g - 0.5g = 0.5g

Then 0.5* 9.8m/s² * 18m = 88.2 would be v²

Then get the square root, the answer would be:
and v = 9.391 m/s is the answer.

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True or false : Scientific theories have withstood the test of time and are accepted as proven fact.

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False:Laws are theories that have not been proven false.

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Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes,

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