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NISA [10]
3 years ago
11

What is non-point source pollution, and why is this kind of pollution hard to monitor and control?

Physics
1 answer:
antoniya [11.8K]3 years ago
7 0

The use of pesticides in farming land is example of non-point source pollution. The area of emitting of pollution is very large. That is why, it is difficult to monitor and control this type of pollution.

Americans get their water for everyday use through aquifer. Aquifers can be conserve by avoiding high amount of chemical uses and reducing household waste. Also the sustainable use of water cause conservation of natural resource.

Aquifer are basically underground water present in ground pores. The structure of aquifer just look like sponge. The water moves through the porous surfaces of the ground.

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The difference in electronegativity values between H and O is 1.4. What type of bond will these elements form?
Gnoma [55]
Who what where when why how ?
6 0
3 years ago
Spitting cobras can defend themselves by squeezing muscles around their venom glands to squirt venom at an attacker. Suppose a s
Alenkasestr [34]

Answer: 1.289 m

Explanation:

The path the cobra's venom follows since it is spitted until it hits the ground, is described by a parabola. Hence, the equations for parabolic motion (which has two components) can be applied to solve this problem:

<u>x-component: </u>

x=V_{o}cos\theta t  (1)

Where:

x is the horizontal distance traveled by the venom

V_{o}=3.10 m/s is the venom's initial speed

\theta=47\° is the angle

t is the time since the venom is spitted until it hits the ground

<u>y-component: </u>

y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}   (2)

Where:

y_{o}=0.44 m  is the initial height of the venom

y=0  is the final height of the venom (when it finally hits the ground)

g=-9.8m/s^{2}  is the acceleration due gravity

Let's begin with (2) to find the time it takes the complete path:

0=0.44 m+3.10 m/s sin\theta(47\°)+\frac{-9.8m/s^{2} t^{2}}{2}   (3)

Rewritting (3):

-4.9 m/s^{2} t^{2} + 2.267 m/s t + 0.44 m=0   (4)

This is a quadratic equation (also called equation of the second degree) of the form at^{2}+bt+c=0, which can be solved with the following formula:

t=\frac{-b \pm \sqrt{b^{2}-4ac}}{2a} (5)

Where:

a=-4.9 m/s^{2

b=2.267 m/s

c=0.44 m

Substituting the known values:

t=\frac{-2.267 \pm \sqrt{2.267^{2}-4(-4.9)(0.44)}}{2(-4.9)} (6)

Solving (6) we find the positive result is:

t=0.609 s (7)

Substituting (7) in (1):

x=(3.10 m/s)cos(47\°)(0.609 s)  (8)

We finally find the horizontal distance traveled by the venom:

x=1.289 m  

7 0
3 years ago
WORD BANK:
Mkey [24]

Answer:

1. Vector, base

2. Vector, derived

3. Vector, ?

4. Scalar, derived

5. scalar, base

4 0
2 years ago
Determine the acceleration due to gravity for low Earth orbit (LEO) given: MEarth = 6.00 x 1024 kg, rEarth = 6.40 x 106 m, G = 6
Nana76 [90]

Answer:

The answer to the question is as follows

The  acceleration due to gravity for low for orbit is  9.231 m/s²

Explanation:

The gravitational force is given as

F_{G}= \frac{Gm_{1} m_{2}}{r^{2} }

Where F_{G} = Gravitational force

G = Gravitational constant = 6.67×10⁻¹¹\frac{Nm^{2} }{kg^{2} }

m₁ = mEarth = mass of Earth = 6×10²⁴ kg

m₂ = The other mass which is acted upon by  F_{G} and = 1 kg

rEarth = The distance between the two masses = 6.40 x 10⁶ m

therefore at a height of 400 km above the erth we have

r = 400 + rEarth = 400 + 6.40 x 10⁶ m = 6.80 x 10⁶ m

and  F_{G} = \frac{6.67*10^{-11} *6.40*10^{24} *1}{(6.8*10^{6})^{2} } = 9.231 N

Therefore the acceleration due to gravity =  F_{G} /mass  

9.231/1 or 9.231 m/s²

Therefore the acceleration due to gravity at 400 kn above the Earth's surface is  9.231 m/s²

4 0
3 years ago
Read 2 more answers
An archer defending a castle is on an 15.5 m high wall. He shoots an arrow straight down at 22.8 m/s. How much time does it take
kodGreya [7K]

Answer:

  about 602 milliseconds

Explanation:

The motion can be approximated by the equation ...

  y = -4.9t^2 -22.8t +15.5

where t is the time since the arrow was released, and y is the distance above the ground.

When y=0, the arrow has hit the ground.

Using the quadratic formula, we find ...

  t = (-(-22.8) ± √((-22.8)^2 -4(-4.9)(15.5)))/(2(-4.9))

  = (22.8 ± √823.64)/(-9.8)

The positive solution is ...

  t ≈ 0.60195193

It takes about 602 milliseconds for the arrow to reach the ground.

8 0
3 years ago
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