Ask question

Ask question

All categories

4 years ago

5

Give an example of nonpoint-source pollution

1 answer:

Ket [755]4 years ago

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 :)

You might be interested in

Why does your weight change on the moon

lorasvet [3.4K]

Answer:

the moon has different gravitational force.

5 0

3 years ago

Read 2 more answers

If the observed test value of a hypothesis test is outside of the established critical value(s), a researcher would __________.

sashaice [31]

I just had this question, the awnser is A.

6 0

3 years ago

Read 2 more answers

In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance

maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = $X_{L} = j\omega L = 2\pi fL$

where

R = resistance

$X_{L} = Inductive Reactance$

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

$C = \frac{\epsilon_{o}A}{x}$

where

x = separation between the parallel plates

Thus

C ∝ $\frac{1}{x}$

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

$Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}$

Also,

Z ∝ I

Therefore,

$\frac{Z}{I} = \frac{Z'}{I'}$

$\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}$

${R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})$

${R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})$

Solving the above eqn:

$X_{C} = 4R$

6 0

3 years ago

How do you know there is current in a circuit?

Ksivusya [100]

if there is not a reaction

5 0

3 years ago

You need to design a 60.0-Hz ac generator that has a maximum emf of 5200 V. The generator is to contain a 180-turn coil that has

OlgaM077 [116]

Answer:

0.082 T

Explanation:

Given:

Frequency of the generator (f) = 60.0 Hz

Maximum emf of the generator (E) = 5200 V

Number of turns (N) = 180

Area per turn (A) = 0.94 m²

Magnetic field magnitude (B) = ?

We know that, the angular frequency of the generator is given as:

$\omega=2\pi f$

Plug in the value of 'f' and solve for 'ω'. This gives,

$\omega=2\pi\times 60.0=120\pi\ rad/s$

Now, the maximum emf of the generator is given by the formula:

$E=NAB\omega$

Rewriting in terms of magnetic field, 'B', we get:

$B=\frac{E}{NA \omega}$

Plug in the given values and solve for 'B'. This gives,

$B=\frac{5200\ V}{180\times 0.94\ m^2\times 120\pi\ rad/s}\\\\B=\frac{5200\ V}{63786.897}\\\\B=0.082\ T$

Therefore, the magnitude of the magnetic field in which the coil rotates is 0.082 T.

8 0

4 years ago