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3 years ago

Air pollution affects everyone equally. True or false

2 answers:

8 0

Overcourse this is true.

Air pollution causes distortions to the air organisms breathe.
The type of pollution that is caused by particles in the air is called Particulates and it is a type of air pollution.These are highly deadly for all living beings. The minute solid or liquid particles that are present in the air or atmosphere can cause a great amount of damage to the lungs of humans. WHO or World Health Organization has designated Particulates as Group 1 carcinogen. The particulates are considered as the deadliest form of air pollution as it can mix with blood unhindered and create death, heart attack or heart ailment and even go as far as creating DNA mutation.

prohojiy [21]3 years ago

7 0

False. bc those with asthma are affect more than others.

You might be interested in

Which of the following compounds is not likely to have ionic bonds? Question 5 options: LiF NaCl CH4 MgF2

Harrizon [31]

Answer:

CH4

Explanation:

CH4 is joined together by a covalent bond, aka a bond between two non-metals. Non-metals are found on the right side of the periodic table and include Carbon (C) and Hydrogen. Although Hydrogen is technically on the left side of the table, it has the characteristics of a non-metal. Futhermore, Ionic bonds generally are between an element on the right joined with an element on the left. This is because ionic bonds want charges that will cancel out to create a neutral molecule.

example: LiF

Li→ Li+

F→F-

(Li+)+(F-)=charges cancel out.

7 0

3 years ago

Suppose the voltage source for a series RL-circuit were given as V0sin(ωt) instead of V0cos(ωt). Write an expression for the cur

Gala2k [10]

Answer:

Explanation:

This is an RL circuit, therefore:

Impedance; z = $\mathbf{\sqrt{R^2+L^2}}$

$\mathbf{z = \sqrt{R^2+(Lw)^2}}$

Current amplitude

$\mathbf{I_o = \dfrac{V_o}{z}} \\ \\ \mathbf{I_o = \dfrac{V_o}{\sqrt{R^2+L^2\omega ^2}}}$

Given that:

$V_o = 1.9 \ V \\ \\ \omega= 51 \ rad/s\\\\ R = 21 \Omega \\ \\ L = 0.52 H$

∴

$I_o= \dfrac{1.9}{\sqrt{21^2+(0.52\times 51)^2}}$

$\mathbf{I_o= 0.0562} \\ \\ \mathbf{I_o = 56.2 \ mA}$

Phase constant :

$tan \ \phi = \dfrac{L \omega}{R } \\ \\ tan \ \phi = \dfrac{0.52 \times 51}{21} \\ \\ tan \phi = 1.263$

$\text{Phase constant : }\phi = tan^{-1} (1.263) \\ \\ \phi = 51.6^0\\ \\\text{To radians} \phi = 51.6 \times \dfrac{\pi}{180} \\ \\ \phi = 0.287 \pi \\ \\ \mathbf{\phi = 0.9 \ rad}$

3 0

3 years ago

A ray of light traveling in air is incident on the surface of a block of clear ice (of index 1.309) at an angle of 25.5 ◦ with t

sergij07 [2.7K]

Answer: $135.29\º$

The figure attached will be helpful to understand the solution.

Here we see two cases, reflection and refraction of light.

According to the laws of reflection:

1. The incident ray, the reflected ray and the normal are in the same plane.

2. The angle of incidence is equal to the angle of reflection.

*Note the normal is perpendicular to the plane, with a $90\º$ angle with the surface

And this can be visualized in the figure, where the angle ${\theta}_{1}$ with which the incident ray hits the surface is equal to the angle with which this same ray is reflected.

On the other hand we have Refraction, a phenomenon in which the light bends or changes its direction when passing through a medium with a refractive index different from the other medium.

In this context, the Refractive index is a number that describes how fast light propagates through a medium or material.

According to Snell’s Law:

$n_{1}sin(\theta_{1})=n_{2}sin(\theta_{2})$ (1)

Where:

$n_{1}=1$ is the first medium refractive index (the air)

$n_{2}=1.309$ is the second medium refractive index (the ice)

$\theta_{1}=25.5\º$ is the angle of the incident ray

$\theta_{2}$ is the angle of the refracted ray

Now, firstly we have to find $\theta_{2}$ and then, by geometry, find $\alpha$ and $\beta$ which sum the angle between the reflected and the refracted light.

Finding $\theta_{2}$ from (1):

$(1)sin(25.5\º)=(1.309)sin(\theta_{2})$

$sin(\theta_{2})=0.328$

$\theta_{2}=arcsin(0.328)$

$\theta_{2}=19.201\º$ (2)

Remembering that the Normal makes a $90\º$ angle with the surface, we can say:

$90\º=\theta_{1}+\beta$ (3)

Finding $\beta$ :

$\beta=90\º-25.5\º=64.5\º$ (4)

Doing the same with $\theta_{2}$ and $\alpha}$ :

$90\º=\theta_{2}+\alpha$ (5)

Finding $\alpha$ :

$\alpha=90\º-19.201\º=70.79\º$ (6)

Adding both angles (4) and (6):

$\alpha+\beta=70.79\º+64.5\º$ (7)

$\alpha+\beta=135.29\º$ >>>This is the angle between the reflected and the refracted light.

5 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s lat

victus00 [196]

Answer / Explanation

It is worthy to note that the question is incomplete. There is a part of the question that gave us the vale of V₀.

So for proper understanding, the two parts of the question will be highlighted.

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.19s later. You may ignore air resistance.

a) What must the height of the building be for both balls to reach the ground at the same time if (i) V₀ is 6.0 m/s and (ii) V₀ is 9.5 m/s?

b) If Vo is greater than some value Vmax, a value of h does not exist that allows both balls to hit the ground at the same time.

Solve for Vmax

Step Process

a) Where h = 1/2g [ (1/2g - V₀)² ] / [(g - V₀)²]

Where V₀ = 6m/s,

We have,

h = 4.9 [ ( 4.9 - 6)²] / [( 9.8 - 6)²]

= 0.411 m

Where V₀ = 9.5m/s

We have,

h = 4.9 [ ( 4.9 - 9.5)²] / [( 9.8 - 9.5)²]

= 1152 m

b) From the expression above, we got to realise that h is a function of V₀, therefore, the denominator can not be zero.

Consequentially, as V₀ approaches 9.8m/s, h approaches infinity.

Therefore Vₙ = V₀max = 9.8 m/s

4 0

3 years ago

Select the correct answer.

Vesna [10]

Answer:

A. continental-oceanic convergent

Explanation:

I knew it couldn't be B because it's oceanic and continental, not oceanic and oceanic.

Next, I noticed the word convergent, which implies "coming together" to me.

I looked it up and noticed the term convergent referred to a plate boundary where a plate slips under (subducted) another, so I knew it was A.

Hopefully, this helps you understand the question better. Have a great day!

3 0

3 years ago