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zhannawk [14.2K]
4 years ago
7

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

Physics
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.
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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
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