Ask question

Ask question

All categories

2 years ago

10

Which letter correctly identifies the part of the hydrologic cycle that is most directly affected by impervious building materia

ls, such as concrete and asphalt?

1 answer:

masha68 [24]2 years ago

3 0

Infiltration

Explanation:

The component of the hydrologic cycle affected by impervious building such as concrete and asphalt is infiltration.

Water infiltration is a major component of the hydrologic cycle.
Concretes and other materials can prevent water from going down into the earth.
This affects the ground water system in place.
It leads to increase in surface run off and might cause inundation of an area.
Infiltration is a very important component of water cycle.
It takes water to plant root and recharges groundwater systems.
Impervious structures takes this capability away.

learn more:

Biogeochemical cycle brainly.com/question/3509510

#learnwithBrainly

You might be interested in

Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their

trapecia [35]

The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2

= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer

6 0

2 years ago

3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

5 0

1 year ago

Read 2 more answers

A spring stretches by 15cm when a mass of 300g hangs down from it,if the spring is then stretched an additional 10cm and release

bearhunter [10]

Answer:

0.1 m

Explanation:

It is given that,

Mass of the object, m = 350 g = 0.35 kg

Spring constant of the spring, k = 5.2 N/m

Amplitude of the oscillation, A = 10 cm = 0.1 m

Frequency of a spring mass system is given by :

Time period:

4 0

2 years ago

A tennis player serves a tennis ball such that it is moving horizontally when it leaves the racquet. When the ball travels a hor

nalin [4]

Answer:

$u_x=38.13\ m/s$

Explanation:

Given that initially ball moves in the horizontal direction ,it means that the velocity in the vertical direction is zero.

Horizontal distance = 13 m

Vertical distance = 57 cm

Lets take time to cover 57 cm distance in vertical direction is t.

We know that g is the constant acceleration in the vertical direction so we can apply the equation of motion in the vertical direction.

$S=u_yt+\dfrac{1}{2}gt^2$

Here $u_y=0$

S= 57 cm

$0.57=0\times t+\dfrac{1}{2}\times 9.81\times t^2$

t=0.34 s

Now in the horizontal direction

$x=u_xt$

Here x=13 m

t= 0.34 s

So

$13=u_x\times 0.34$

$u_x=38.13\ m/s$

So the initial speed of ball is 38.13 m/s.

7 0

2 years ago

A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the

hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written as $x(t)=Asin(\omega t+\phi)$

'A' is the amplitude = 6 inches (given)

$\omega =\sqrt{\frac{k}{m}}$ is the natural frequency of the system

At equilibrium we have

$mg=kx\\\\k=\frac{mg}{x}$

Applying values we get

$k=40 lb/ft$

thus natural frequency equals

$\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}$

Thus the equation of motion becomes

$x(t)=6sin(8t+\phi)$

At time t=0 since mass is at it's maximum position thus we have

$A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})$

Thus the position of mass at the given times is as follows

1) at $\frac{\pi}{12}$ $x(t)=5.99inches$

2) at $\frac{\pi}{8}$ $x(t)=5.9909inches$

3) at $\frac{\pi}{6}$ $x(t)=5.98397inches$

4) at $\frac{\pi}{4}$ $x(t)=5.9639inches$

5) at $\frac{9\pi}{32}$ $x(t)=5.954inches$

4 0

1 year ago