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vlada-n [284]
3 years ago
13

Do you think that government should institute measures to control the human population? Give me your honest opinion please:)!

Biology
2 answers:
postnew [5]3 years ago
5 0

Answer:

Although I think the human population does need to be slowed down, I do not think it is right for any government on earth to try and control it. That is like trying to play God and I don't think that's right.

Explanation:

.

IRISSAK [1]3 years ago
4 0

Answer:

Look below.

Explanation:

Personally I say it is a two way anwer a yes and a no (mostly no). A yes because if their is a carrying capacity for a state or area, it obviously needs to be controlled to avoid problems such as food shortages. A no because the govt is "weird" (my opinion dont come at me). This whole virus suspiciously just appeared. I speculate that this virus was manipulated by the government to control the human population. I could be totally wrong but this could be a possibility to control man kind. So yes government should institute measures to control the human population, if its under understandable circumstances. And a no if the government does it without human kind knowing and hurting the population to decrease vastly.

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Answer:

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Explanation:

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3 years ago
2.86 A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight
Rashid [163]

Answer:

Distance Below the top = 6,02 m

Explanation:

To get the Final velocity of the first ball (that will be the intial velocity of the second) you need to solve the kinematic equation for Velocity:

(1) V_{1f} =V_{1O}  - gt

As the ball is dropped from rest V_{1O} = 0, so:

(2) V_{1f} = - gt

Note that the velocity is going to be negative as the ball is going down. To get the time it would take the ball to reach de base you can use the kinematic equation for position:

(3) h_{1} = h_{1O} + V_{1O} *t - \frac{1}{2} gt^{2}

We need the answer when h=0, and from the initial conditions V_{1O} = 0, h_{1O} = 24m, you get:

(4) 24m = \frac{1}{2} gt_{1f}^{2}

Solving for t, with 9,8\frac{m}{s^{2}} and :

(5) t_{1f} = \sqrt{\frac{24m*2}{g} } = 2,21 s

Replacing this time in (2), the final velocity is:

(6) V_{1f} = -21,66  \frac{m}{s}

So the initial velocity of ball 2 is equial to this but oppossite in direction so: V_{2O}= -V_{1f} = 21,66  \frac{m}{s}

The general position equation for ball 2 is (considering h_{2O} = 0 :

(7) h_{2} = V_{2O} *t - \frac{1}{2} gt^{2}

They cross paths when h_{1}=h_{2} so:

(8) h_{1O} - \frac{1}{2} gt^{2} = V_{2O} *t - \frac{1}{2} gt^{2}

Rearranging:

(9) t_{cross} =\frac{h_{1O}}{V_{2O} }

Replacing values:

t_{cross} = 1,108 s

To get the absolute position replace t_{cross} on equation (3) or (7):

h_{cross} = 17,98 m

To get it below the top of te cliff:

h_{cross, from above} = 24 m - 17,98 m

h_{cross, from above} = 6,02 m

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Tcecarenko [31]

Answer:

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Explanation:

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2 years ago
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