All categories

4 years ago

What would the consequences be if no one was environmentally aware?

2 answers:

8 0

When did everyone stop being enviromentally aware. If it was in the future, the effects would be undetermined. If it was today, it would likely only be a small amount of time before the planet was walloweng in waste and all of humanity was dead.

rodikova [14]4 years ago

6 0

Answer:

some of the consequences are:

The amount of contamination would be incalculable, the wildlife will be highly affected by it, at the point that some species may be extinct in a few years.

The pollution in the water would greatly affect the life of the ones that live in it and the ones that drink that water.

with no recycling, the amount of trash would be uncontrollable, with more glass threw away, the number of fires should increase, which also affects everyone.

You might be interested in

A 1.2 nf parallel-plate capacitor has an air gap between its plates. Its capacitance increases by 3.0 nf when the gap is filled

Damm [24]

Answer:

2.5

Explanation:

The capacitance of a parallel-plate capacitor filled with dielectric is given by

$C=kC_0$

where

k is the dielectric constant

$C_0$ is the capacitance of the capacitor without dielectric

In this problem,

$C_0=1.2 nF$ is the capacitance of the capacitor in air

$C=3.0 nF$ is the capacitance with the dielectric inserted

Solving the equation for k, we find

$k=\frac{C}{C_0}=\frac{3.0 nF}{1.2 nF}=2.5$

3 0

3 years ago

Read 2 more answers

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$