Ask question

Ask question

All categories

zhannawk [14.2K]

4 years ago

7

Discuss any two current processes that can be used to ensure that there is enough clean water

1 answer:

Tanzania [10]4 years ago

3 0

Two current processes that is used in ensuring that there is enough clean water is maintaining environmental sanitation in communities in which will help promote cleanliness and orderliness and another thing, having people to be educated with cleanliness and proper waste disposal as this will contribute to making the water clean and maintain it that way.

You might be interested in

A cart starts from rest and accelerates uniformly at 4.0 m/s2 for 5.0 s. It next maintains the velocity it has reached for 10 s.

wlad13 [49]

Answer:

12m/s

Explanation:

$v_f=v_o+at$

Let's call the velocity that the car maintains for 10 seconds $v_f_1$ , and the final velocity $v_f_2$ .

$v_f_1=0+(4)(5)=20m/s \\\\v_f_2=20+(-2)(4)=12m/s$

Hope this helps!

5 0

3 years ago

The amount of friction divided by the weight of an object forms a unit less number called the

Romashka [77]

Answer:

Coefficient of friction.

Explanation:

The amount of friction divided by the weight of an object is equal to the coefficient of friction. It is a dimensional less number. It can be given by :

$F=\mu N$

N is normal force.

$\mu$ = coefficient of friction

$\mu=\dfrac{F}{N}$

3 0

3 years ago

Some folded mountains result from the collision of continental plates. <br> a. True<br> b. False

alexgriva [62]

it true be cause when one goes under another the ground rise and makes what we call a mountain so therefore it is treu<span />

3 0

3 years ago

A drag racing car with a weight of 1600 lbf attains a speed of 270 mph in a quarter-mile race. Immediately after passing the tim

Kaylis [27]

Answer:

15.065ft

Explanation:

To solve this problem it is necessary to consider the aerodynamic concepts related to the Drag Force.

By definition the drag force is expressed as:

$F_D = -\frac{1}{2}\rho V^2 C_d A$

Where

$\rho$ is the density of the flow

V = Velocity

$C_d$ = Drag coefficient

A = Area

For a Car is defined the drag coefficient as 0.3, while the density of air in normal conditions is 1.21kg/m^3

For second Newton's Law the Force is also defined as,

$F=ma=m\frac{dV}{dt}$

Equating both equations we have:

$m\frac{dV}{dt}=-\frac{1}{2}\rho V^2 C_d A$

$m(dV)=-\frac{1}{2}\rho C_d A (dt)$

$\frac{1}{V^2 }(dV)=-\frac{1}{2m}\rho C_d A (dt)$

Integrating

$\int \frac{1}{V^2 }(dV)= - \int\frac{1}{2m}\rho C_d A (dt)$

$-\frac{1}{V}\big|^{V_f}_{V_i}=\frac{1}{2m}(\rho)C_d (\pi r^2) \Delta t$

Here,

$V_f = 60mph = 26.82m/s$

$V_i = 120.7m/s$

$m= 1600lbf = 725.747Kg$

$\rho = 1.21 kg/m^3$

$C_d = 0.3$

$\Delta t=7s$

Replacing:

$\frac{-1}{26.82}+\frac{1}{120.7} = \frac{1}{2(725.747)}(1.21)(0.3)(\pi r^2) (7)$

$-0.029 = -5.4997r^2$

$r = 2.2963m$

$d= r*2 = 4.592m \approx 15.065ft$

4 0

4 years ago

The gravitational attraction between two objects will what is the object move further apart

finlep [7]

The farther apart the two objects, the weaker the gravitational attraction between them.

3 0

3 years ago