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uysha [10]
4 years ago
13

20 POINTS PLZ HELP- Describe how two type of land use within your watershed can influence the water quality in Rivers in your ar

ea?
Physics
1 answer:
JulsSmile [24]4 years ago
7 0

Answer:

Pollution Sources & Land Use. Imperviousness & Water Quality ... protect streams, rivers, and ground water, land use practices can be implemented ... In built-up areas with pavement and buildings, little rainfall soaks into the soil Ground water flows slowly into streams, usually over a period of months,

Explanation:

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A _____________ is defined as a push or pull that acts on objects and, in some instances, changes their position.
Shkiper50 [21]
Force !!!!!!!!!!!!!!!!!!!!!!!!!!!
6 0
3 years ago
Read 2 more answers
A drag racer starts from the rest and accelerates at 7.4m/s^2.how far will he travel in 2.0 seconds?
Vesna [10]

Answer:

14.8 m

Explanation:

S= ut + \frac{1}{2}at^{2}

where u = initial velocity

S= (0 \frac{m}{s})(2s) + \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=  \frac{1}{2}(7.4\frac{m}{s^{2} })(2s^{2})

S=14.8 m

8 0
3 years ago
How an application of atmospheric device work?example siphon​
Vika [28.1K]

Answer:

A practical siphon, operating at typical atmospheric pressures and tube heights, works because gravity pulling down on the taller column of liquid leaves reduced pressure at the top of the siphon (formally, hydrostatic pressure when the liquid is not moving).

I hope it's helpful!

6 0
3 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
4 years ago
(16 POINTS)
natta225 [31]
Halflife is the time taken by a radioactive substance by half its original mass. In this case the half life of the substance is 3 hours. 
Therefore; New mass = Original mass × (1/2)^n where n is the number of half lives. In this case, the half life is 3 hours and therefore; the number of half lives in 6 hours will be two.
 = if the original mass is 100%
new mass = 100 × (1/2)^2
                 = 100 × 1/4 = 25%
The remaining mass will be 1/4 of the original mass, meaning  a fraction of 3/4 of the  initial mass will have decayed.
Thus; the answer is 3/4 












7 0
3 years ago
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