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3 years ago

6

A material that breaks down quickly in the environment is called??

2 answers:

irinina [24]3 years ago

5 0

A material that breaks down quickly in the environment is called Biodegradable.

Artist 52 [7]3 years ago

3 0

Biodegradable material

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How can groundwater flow, elevation can transport contaminates

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Answer: it's under explanation

Explanation:Groundwater contamination occurs when man-made products such as gasoline, oil, road salts and chemicals get into the groundwater and cause it to become unsafe and unfit for human use. Materials from the land's surface can move through the soil and end up in the groundwater.

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3 years ago

Read 2 more answers

At one point in a pipeline, the water's speed is 3.57 m/s and the gauge pressure is 68.7 kPa. Find the gauge pressure at a secon

ArbitrLikvidat [17]

Answer:

The pressure at point 2 is $P_2 = 254.01 kPa$

Explanation:

From the question we are told that

The speed at point 1 is $v_1 = 3.57 \ m/s$

The gauge pressure at point 1 is $P_1 = 68.7kPa = 68.7*10^{3}\ Pa$

The density of water is $\rho = 1000 \ kg/m^3$

Let the height at point 1 be $h_1$ then the height at point two will be

$h_2 = h_1 - 18.5$

Let the diameter at point 1 be $d_1$ then the diameter at point two will be

$d_2 = 2 * d_1$

Now the continuity equation is mathematically represented as

$A_1 v_1 = A_2 v_2$

Here $A_1 , A_2$ are the area at point 1 and 2

Now given that the are is directly proportional to the square of the diameter [i.e $A= \frac{\pi d^2}{4}$ ]

which can represent as

$A \ \ \alpha \ \ d^2$

=> $A = c d^2$

where c is a constant

so $\frac{A_1}{d_1^2} = \frac{A_2}{d_2^2}$

=> $\frac{A_1}{d_1^2} = \frac{A_2}{4d_1^2}$

=> $A_2 = 4 A_1$

Now from the continuity equation

$A_1 v_1 = 4 A_1 v_2$

=> $v_2 = \frac{v_1}{4}$

=> $v_2 = \frac{3.57}{4}$

$v_2 = 0.893 \ m/s$

Generally the Bernoulli equation is mathematically represented as

$P_1 + \frac{1}{2} \rho v_1^2 + \rho * g * h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho * g * h_2$

So

$P_2 = \rho * g (h_1 -h_2 )+P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

=> $P_2 = \rho * g (h_1 -(h_1 -18.3) + P_1 + \frac{1}{2} * \rho (v_1^2 -v_2 ^2 )$

substituting values

$P_2 = 1000 * 9.8 (18.3) )+ 68.7*10^{3} + \frac{1}{2} * 1000 ((3.57)^2 -0.893 ^2 )$

$P_2 = 254.01 kPa$

8 0

3 years ago

[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.

kumpel [21]

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

v(0) = 0

b) solution to the IVP

v(t) = 40(1 - e^(-t/10))

c) Limiting velocity

Vo = 40 ft/s

Position of the car after 12 hours

X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

4 0

3 years ago