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2 years ago

Testable question: does the water need to be hot or cold to make a cloud.

Chemistry

1 answer:

KengaRu [80]2 years ago

6 0

Water can consist of hot, cold, or both to form a cloud.

With hot water, clouds will form when hot water evaporates then cools when it rises into the air. After, the warm air will expand into the atmosphere. Finally, the warm air will create water vapor that will form water droplets and create the basis of a cloud.
The temperature is very low if the water is cold. When it's cold outside, it results in a lower humidity which evaporates the water faster then it would in warm weather since the humidity would be higher. Water droplets in clouds start to form into bits of ice that fall to the earth because of the mass the ice bit has.

Best of Luck!

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A thermometer having first-order dynamics with a time constant of 1 min is placed in a temperature bath at 100oF. After the ther

sveticcg [70]

Answer:

(a) See below

(b) 103.935 °F; 102.235 °F

Explanation:

The equation relating the temperature to time is

$T = T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )$

1. Calculate the thermometer readings after 0.5 min and 1 min

(a) After 0.5 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-0.5/1} \right )\\ & = & 100 + 10\left (1 - e^{-0.5} \right )\\ & = & 100 + 10 (1 - 0.6065)\\ & = & 100 + 10(0.3935)\\ & = & 100 + 3.935\\ & = & 103.935\,^{\circ}F\\\end{array}$

(b) After 1 min

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-t/\tau} \right )\\ & = & 100 + 10\left (1 - e^{-1/1} \right )\\ & = & 100 + 10\left (1 - e^{-1} \right )\\ & = & 100 + 10 (1 - 0.3679)\\ & = & 100 + 10(0.6321)\\ & = & 100 + 6.321\\ & = & 106.321\,^{\circ}F\\\end{array}$

2. Calculate the thermometer reading after 2.0 min

T₀ =106.321 °F

ΔT = 100 - 106.321 °F = -6.321 °F

t = t - 1, because the cooling starts 1 min late

$\begin{array}{rcl}T & = & T_{0} + \Delta T\left (1 - e^{-(t - 1)/\tau} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-(2 - 1)/1} \right )\\ & = & 106.321 - 6.321\left (1 - e^{-1} \right )\\ & = & 106.321 - 6.321 (1 - 0.3679)\\ & = & 106.321 - 6.321 (0.6321)\\ & = & 106.321 - 3.996\\ & = & 102.325\,^{\circ}F\\\end{array}$

3. Plot the temperature readings as a function of time.

The graphs are shown below.

6 0

2 years ago

Air contains nitrogen, oxygen, argon, and trace gases. Ifthe partial pressure of nitrogen is 592 mm Hg, oxygen is160 mm Hg, argo

Katena32 [7]

Answer:

760 mmHg

Explanation:

Step 1: Given data

Partial pressure of nitrogen (pN₂): 592 mmHg

Partial pressure of oxygen (pO₂): 160 mmHg

Partial pressure of argon (pAr): 7 mmHg

Partial pressure of the trace gas (pt): 1 mmHg

Step 2: Calculate the atmospheric pressure

Since air is a gaseous mixture, the atmospheric pressure is equal to the sum of the gases that compose it.

P = pN₂ + pO₂ + pAr + pt = 592 mmHg + 160 mmHg + 7 mmHg + 1 mmHg = 760 mmHg

6 0

2 years ago

Please help. 1.600x10^8 / 8.000x10^11

sweet-ann [11.9K]

Answer:

2x10^-4 = 2E-4

Explanation:

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1 year ago

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A potted plant is placed under a grow lamp, which provides 6,400 J of energy to the plant and the soil over the course of an hou

STatiana [176]

Answer:

823.7g

Explanation:

Using the formula as follows:

Q = m × c × ∆T

Where;

Q = amount of heat (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

Using the information given in this question as follows:

Q = 6,400 J

m = ?

c of soil = 0.840 J/g°C

∆T = 9.25°C

Using Q = mc∆T

m = Q ÷ c∆T

m = 6,400 ÷ (0.840 × 9.25)

m = 6400 ÷ 7.77

m = 823.7g

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2 years ago

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Mendeleev created this table as he noticed that a

Nimfa-mama [501]

Answer:

Mendeleev had left the noble gases out of his periodic table.

Explanation:

Mendeleev's periodic table is pictured in the image attached to the question.

Mendeleev's table obviously lacked the noble gases. The reason for this grave omission is simple; the noble gases were not known as at the time when he formulated his periodic table. There weren't any known elements whose properties were similar to the properties of the noble gases. This would have lead him to suspect their existence.

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