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3 years ago

What is a common environmental problem caused by mining

2 answers:

7 0

Erosion, formation of sinkholes, loss of biodiversity, and contamination of soil, groundwater and surface water by chemicals from mining processes.

Anarel [89]3 years ago

3 0

Answer:

Among the main impacts that mining can cause to the environment are the destruction of the earth's crust, the contamination of the waters, the impact on the flora and fauna of the environment close to mining and the negative effects on human health of nearby populations to the mine, as well as deforestation, earthworks, chemical contamination and other substances used to extract the ore.

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Which object has the most kinetic energy? a bus a car a plane a bicycle

loris [4]

Answer:I would guess a plane

Assuming they all Thad the same velocity....

4 0

3 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

#SPJ4

8 0

1 year ago

A runner accelerated to 4 m/s^2 for 20 seconds before winning the race.How far did he/she run?

creativ13 [48]

Answer:

s=800 m

Explanation:

Given that,

Acceleration of a runner, a = 4 m/s²

Time, t = 20 seconds

We need to find the distance covered by her. Initially, she was at rest. It means its initial velocity is equal to 0. So, using second equation of motion as follows :

$s=ut+\dfrac{1}{2}at^2$

Herre, u = 0

$s=\dfrac{1}{2}at^2\\\\s=\dfrac{1}{2}\times 4\times (20)^2\\\\s=800\ m$

So, she will cover a distance of 800 m.

8 0

4 years ago

If a certain gas occupies a volume of 12 l when the applied pressure is 6.0 atm , find the pressure when the gas occupies a volu

dolphi86 [110]

From Boyle's law, the volume of a fixed mass of a gas is inversely proportional to its pressure at constant absolute temperature.
Therefore; P1V1 =P2V2; where PV is a constant
hence; 12 × 6 = 3× p2
p2 = 72/3
= 24 atm
Therefore; the new pressure will be 24 atm

5 0

3 years ago

A closely wound, circular coil with a diameter of 4.30 cm has 470 turns and carries a current of 0.460 A .

Nadusha1986 [10]

Hi there!

a)
Let's use Biot-Savart's law to derive an expression for the magnetic field produced by ONE loop.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

dB = Differential Magnetic field element

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

R = radius of loop (2.15 cm = 0.0215 m)

i = Current in loop (0.460 A)

For a circular coil, the radius vector and the differential length vector are ALWAYS perpendicular. So, for their cross-product, since sin(90) = 1, we can disregard it.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{r^2}$

Now, let's write the integral, replacing 'dl' with 'ds' for an arc length:
$B = \int \frac{\mu_0}{4\pi} \frac{ids}{R^2}$

Taking out constants from the integral:
$B =\frac{\mu_0 i}{4\pi R^2} \int ds$

Since we are integrating around an entire circle, we are integrating from 0 to 2π.

$B =\frac{\mu_0 i}{4\pi R^2} \int\limits^{2\pi R}_0 \, ds$

Evaluate:
$B =\frac{\mu_0 i}{4\pi R^2} (2\pi R- 0) = \frac{\mu_0 i}{2R}$

Plugging in our givens to solve for the magnetic field strength of one loop:

$B = \frac{(4\pi *10^{-7}) (0.460)}{2(0.0215)} = 1.3443 \mu T$

Multiply by the number of loops to find the total magnetic field:
$B_T = N B = 0.00631 = \boxed{6.318 mT}$

Now, we have an additional component of the magnetic field. Let's use Biot-Savart's Law again:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} \times \hat{r}}{r^2}$

In this case, we cannot disregard the cross-product. Using the angle between the differential length and radius vector 'θ' (in the diagram), we can represent the cross-product as cosθ. However, this would make integrating difficult. Using a right triangle, we can use the angle formed at the top 'φ', and represent this as sinφ.

$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l} sin\theta}{r^2}$

Using the diagram, if 'z' is the point's height from the center:

$r = \sqrt{z^2 + R^2 }\\\\sin\phi = \frac{R}{\sqrt{z^2 + R^2}}$

Substituting this into our expression:
$dB = \frac{\mu_0}{4\pi} \frac{id\vec{l}}{(\sqrt{z^2 + R^2})^2} }(\frac{R}{\sqrt{z^2 + R^2}})\\\\dB = \frac{\mu_0}{4\pi} \frac{iRd\vec{l}}{(z^2 + R^2)^\frac{3}{2}} }$

Now, the only thing that isn't constant is the differential length (replace with ds). We will integrate along the entire circle again:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} \int\limits^{2\pi R}_0, ds$

Evaluate:
$B = \frac{\mu_0 iR}{4\pi (z^2 + R^2)^\frac{3}{2}}} (2\pi R)\\\\B = \frac{\mu_0 iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Multiplying by the number of loops:
$B_T= \frac{\mu_0 N iR^2}{2 (z^2 + R^2)^\frac{3}{2}}}$

Plug in the given values:
$B_T= \frac{(4\pi *10^{-7}) (470) (0.460)(0.0215)^2}{2 ((0.095)^2 + (0.0215)^2)^\frac{3}{2}}} \\\\ = 0.00006795 = \boxed{67.952 \mu T}$

5 0

2 years ago

Read 2 more answers