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3 years ago

Wht kind of particles contribute to an overall heating effect on climate?

Chemistry

2 answers:

MakcuM [25]3 years ago

6 0

Answer:

Aerosols like sulfur, nitrate, and organic carbon are formed in the atmosphere and cause global cooling. Thereby they contribute to mask parts of the human induced global warming. On the other hand, black carbon absorbs radiation and thereby has a warming effect on the earth's climate

Explanation:

google hope this helps you

Ivanshal [37]3 years ago

3 0

Answer: Aerosols like sulfur, nitrate, and organic carbon are formed in the atmosphere and cause global cooling. Thereby they contribute to mask parts of the human induced global warming. On the other hand, black carbon absorbs radiation and thereby has a warming effect on the earth's climate. -science daily

Explanation: please mark me brainiest. i need one to get a higher rank :/

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Water is poured into a conical container at the rate of 10 cm3/sec. The cone points directly down, and it has a height of 20 cm

8090 [49]

Answer:

\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{2}

Explanation:

Hello,

The suitable differential equation for this case is:

$\frac{dV}{dt}=10\frac{cm^3}{s}$

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

$\frac{dh}{dt} = ?*\frac{dV}{dt}$

Of course, $?=\frac{dh}{dV}$ .

Now, since the volume of a cone is $V=\pi r^2h/3$ and the ratio $r/h=15/20=3/4$ or $r=3/4h$ , the volume becomes:

$V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3$

We proceed to its differentiation:

$\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}$

Then, we compute $\frac{dh}{dt}$

$\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}$

Finally, at h=2:

$\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}$

Best regards.

4 0

4 years ago

You are requested to reduce the size of 50 ton/hr of a given solid. The size of the feed is such 80% passes a 4-in (76.2 mm) scr

defon

Answer:

1) The power needed to process 50 ton/hr is 135.4 HP.

2) The void fraction of the bed is 0.37.

Explanation:

1) For this type of milling operations, we can estimate the power needed for an operation according to the work index (Ei), the passing size of the circuit feed (F80) and the passing size of the product (P80).

We assume the units of Ei are kWh/t.

The equation that relates this parameters and the power is (size of particles in μm):

$W=Ei*(\frac{10}{\sqrt{P80}} -\frac{10}{\sqrt{F80}} )\\\\W=9.45*(\frac{10}{\sqrt{3175\mu m}} -\frac{10}{\sqrt{76200mm}} )\\\\\\W=9.45*(0.1774+0.0362)=2.019 kWh/t$

The power needed to process 50 ton/hor is

$P=2.0194\frac{kWh}{Ton}*\frac{50Ton}{h}*\frac{1.341HP}{1kW}= 135.4 \, HP$

2) The density of the packed bed can be expressed as

$\rho=f_v*\rho_v+f_s*\rho_s$

being f the fraction and ρ the density of every fraction. We know that the density of the void is 0 (ρv=0) and that fv=1-fs (the sum of the fractions ois equal to the total space).

Then we can rearrange

$\rho=f_v*\rho_v+f_s*\rho_s\\\\\rho=f_v*0+(1-f_v)*\rho_s\\\\\rho/\rho_s=1-f_v\\\\f_v=1-\rho/\rho_s=1-990/1570=1-0.63=0.37$

The void fraction of the bed is 0.37.

6 0

3 years ago

Describe where a hydrogen bond can form among water molecules.

stiv31 [10]

Answer:

In water hydrogen bonding formed between partially negative oxygen and partially positive hydrogen atoms.

Explanation:

Hydrogen bond is a partially electrostatic attraction between hydrogen (H), which is bonded to a more electronegative atom such as nitrogen (N), oxygen (O) or fluorine (F), and another adjacent atom bearing a lone pair of electrons.

In water molecule hydrogen and oxygen have difference in electronegativity, due to which dipoles are form. One is partial positive on hydrogen atom and other is partial negative on oxygen atom. The electrostatic attraction occur between two oppositely charged poles which is known as hydrogen bonding.