Give an example of nonpoint-source pollution

1 answer:

8 0

Nonpoint source pollution is pollution that affects bodies of water, so remembering this can help in the future with this type of question. Some examples of nonpoint source pollution are; Excess fertilizers and herbicides, oil and toxic chemical runoff, as well as sediment from eroding stream banks. I hope this helps :)

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1. What is the acceleration of a car that goes from 20 Km/hr to 100 km/hr in 8 seconds?

san4es73 [151]

Answer:

10 :)

You have to divide the difference of speed and divide it by the time. So 100-20 would be 80, and if you divide that by 8 it would be 10.

Hope this helps.

7 0

3 years ago

Read 2 more answers

1: In a longitudinal wave, the particle displacement is __________ the direction of wave movement

jenyasd209 [6]

The answer is parallel

5 0

3 years ago

PLEASE HURRY 20 PTS

Crank

Answer:

A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.

Explanation:

The electric force exerted on a charge by an electric field is given by:

where

F is the force

q is the charge

E is the electric field

We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.

5 0

3 years ago

A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have

denpristay [2]

1) The total mechanical energy of the rock is:
$E=U+K$
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
$E_i=U=mgh$
where $m=4 kg$ is the mass, $g=9.81 m/s^2$ is the gravitational acceleration and $h=5 m$ is the height.
Putting the numbers in, we find the potential energy
$U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J$

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
$E_f=K= \frac{1}{2}mv^2$
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
$K=U=196.2 J$

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
$W=196.2 J-0 J=196.2 J$

6 0

3 years ago

An open pipe of length 0.39m vibrates in the third harmonic with a frequency of 1400Hz. What is the distance from the center of

Gnoma [55]

Length of the pipe = 0.39 m

Third harmonic frequency = 1400 Hz

For the third harmonic:

Wavelength = $\frac{2L}{3}$