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3 years ago

To be considered acid rain, precipitation would have to

2 answers:

6 0

The two primary sources of acid rain are sulfur dioxide (SO2), and oxides of nitrogen (NOx). Sulfur dioxide is a colourless, prudent gas released as a by-product of combusted fossil fuels containing sulfur. A variety of industrial processes, such as the production of iron and steel, utility factories, and crude oil processing produce this gas. In iron and steel production, the smelting of metal sulfate ore, produces pure metal. This causes the release of sulfur dioxide. Metals such as zinc, nickel, and copper are commonly obtained by this process. Sulfur dioxide can also be emitted into the atmosphere by natural disasters or means. This ten percent of all sulfur dioxide emission comes from volcanoes, sea spray, plankton, and rotting vegetation. Overall, 69.4 percent of sulfur dioxide is produced by industrial combustion. Only 3.7 percent is caused by transportation. Read more at: https://sites.google.com/site/acidrain1project/

irina1246 [14]3 years ago

5 0

Answer: precipitation will have to be deposited with a high level of sulphuric and nitric components

Explanation: precipitation is considered to be acidic when its pH falls below 5.6 (which is 25 times more acidic than pure distilled water). For an acid rain to occur, precipitation will have an acidic component such as sulfuric or nitric acid that fall to the ground from the atmosphere in wet or dry forms.

This can include rain, snow, fog, hail or even dust that is acidic.

You might be interested in

Explain why it took so long for psychology to be recognized and what things helped it to get recognized.

Juli2301 [7.4K]

It took so long because at the time there was no way for people to study the behavior formally. im not sure what helped it get recognized but i know wihelm wundt helped get it recongnized.

sorry i couldnt be much help

4 0

3 years ago

What are two examples of goods and two examples of services

julsineya [31]

The goods and the services make up the basis of every economy. The goods can simply be defined as merchandise or possessions. The services can be defined as the actions through which help is provided, or work is done for someone else. Example of goods are the food and furniture, with the food being crucial for the survival of the people, while the furniture is an essential part of every home and its practicality and decor. Examples of services are teaching and car repairing. The teaching is crucial for the development of the societies, as through it the people get education, while the repairing of cars is very important as lot of people have them, can not afford to buy new ones all the time, and they need for their daily movement over longer distances.

8 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.

kondaur [170]

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

$F_k = \mu_k N$ = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

3 0

3 years ago

A 81 kg person sits on a 3.8 kg chair. Each leg of the chair makes contact with the floor in a circle that is 1.2 cm in diameter

tankabanditka [31]

Answer:

1.9 MPa

Explanation:

Mass of person = 81 kg

Mass of chair = 3.8 kg

Diameter of contact point = 1.2 cm = D

Radius of contact point = 1.2/2 = 0.6 cm

Total mass of chair and person = 81 + 3.8 = 84.8 kg = M

Acceleration due to gravity = 9.81 m/s²

Force acting on the floor

F = Mg

⇒F = 84.8×9.81

⇒F = 831.888 N

Area of the contact point

A = πR²

⇒A = π0.006²

⇒A = π0.000036 m²

Area of the four points is

4A = 0.000144π m²

Pressure

$p=\frac{F}{A}\\\Rightarrow p=\frac{831.888}{0.000144\pi}\\\Rightarrow p=1838876.21\ Pa=1.83887621\times 10^6\ Pa=1.9 MPa$

Pressure exerted on the floor by each leg of the chair is 1.9 MPa

5 0

3 years ago