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3 years ago

Which of the following would most likely be considered nonpoint source pollution

1 answer:

8 0

The best answer is b) increased turbidity from erosion.

Nonpoint source pollution generally happens as a result of many systems interacting, and is not directly attributed to one event or pollutant. Generally, natural environmental systems participate in pollution of this kind, regardless of whether or not human activity was a factor. Examples include water runoff, or erosion.

The other pollutants listed have a direct cause and direct effect, the animal waste goes directly from the animals to the ground they live on, the car shop directly sumps the oil on the ground, and the oil tank leaks directly into the earth. Erosion causing turbidity is a less direct form of pollution, and is due to the synthesis of several natural phenomena<span />

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A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

PIT_PIT [208]

Answer:

$9.96\cdot 10^{-10}J$

Explanation:

The capacitance of the parallel-plate capacitor is given by

$C=\epsilon_0 k \frac{A}{d}$

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

$A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2$ is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F$

Now we can calculate the energy of the capacitor, given by:

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

$U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J$

4 0

3 years ago

You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla

lesantik [10]

Answer:

When they are connected in series

The 50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

The power rating of the first bulb is $P_1 = 100 \ W$

The power rating of the second bulb is $P_2 = 50 \ W$

Generally the power rating of the first bulb is mathematically represented as

$P_1 = V^2 R$

Where $V$ is the normal household voltage which is constant for both bulbs

$R_1 = \frac{V^2}{P_1 }$

substituting values

$R_1 = \frac{V^2}{100}$

Thus the resistance of the second bulb would be evaluated as

$R_2 = \frac{V^2}{50}$

From the above calculation we see that

$R_2 > R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 1 = I^2_1 R_1$

This power rating of the first bulb can also be represented mathematically as

$P_ 2 = I^2_2 R_2$

Now given that they are connected in series which implies that the same current flow through them so

$I_1^2 = I_2^2$

This means that

$P \ \alpha \ R$

So when they are connected in series

$P_2 > P_1$

This means that the 50 W bulb glows more than the 100 \ W bulb

3 0

3 years ago

Are the two expressions equvalent 7(8x+5) and 48x + 35

Evgesh-ka [11]

Yes they are equivalent because 7x5=35 and 8x x 5=48x

4 0

3 years ago

A Meteorite Strikes On October 9, 1992, a 27-pound meteorite struck a car in Peekskill, NY, leaving a dent 22 cm deep in the tru

vagabundo [1.1K]

Answer:

Acceleration of the meteorite, $a=-38409.09\ m/s^2$

Explanation:

It is given that,

A Meteorite after striking struck a car, v = 0

Initial speed of the Meteorite, u = 130 m/s

Distance covered by Meteorite, s = 22 cm = 0.22 m

We need to find the magnitude of its deceleration. It can be calculated using the third equation of motion as :

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{-(130)^2}{2\times 0.22}$

$a=-38409.09\ m/s^2$

So, the deceleration of the Meteorite is $-38409.09\ m/s^2$ . Hence, this is the required solution.

7 0

3 years ago

The gravitational self potential energy of a solid ball of mass density ρ and radius R is E. What is the gravitational self pote

Alenkasestr [34]

It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..

8 0

3 years ago