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4 years ago

How are temperature, pressure, and volume related dealing with the behavior of a gas

Physics

1 answer:

fredd [130]4 years ago

7 0

Pressure and volume of a gas are inversely related. As one goes up, the other goes down, and vice-versa.

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What statement is true about the part of the electromagnetic spectrum that is visible to the human eye?

Alja [10]

Based on the options above, I see how the last option would actually be most fitting to how "the part of the electromagnetic spectrum that is visible to the human eye". This would be in separate parts, which would be divided into a certain number, which would actually be only 7 ranges of wavelengths. And therefore, this last statement would actually go very well and it would actually be the correct answer. And this would all relate to how the part of the electromagnetic spectrum that is visible to the human eye, it's based on it's own 7 division parts that it has.

Your answer: $\textsf{It is divided into seven ranges of wavelengths}$

7 0

4 years ago

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The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at

denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m _____ eqn (1)

FOR HYDROGEN:

v = √(3KT)/m = 2000 m/s _____ eqn (2)

FOR OXYGEN:

velocity of oxygen = √(3KT)/(mass of oxygen)

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

velocity of oxygen = (1/4)(2000 m/s) = 500 m/s

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

K.E of Oxygen = 4000 J

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

v = 2149.24 m/s

8 0

4 years ago

An engine has an energy input of 125 J, and 35 J of that energy is transformed into useful energy. What is the efficiency of the

xeze [42]

The efficiency of the engine at the given useful energy and input energy is 28%.

The given parameters:

Input energy of the engine, = 125 J
Useful energy of the engine, 35 J

The efficiency of the engine is defined as the ratio of the output energy (useful energy) to the input energy and it is calculated as follows;

$efficiency\ = \frac{0ut put \ energy}{1nput \ energy} \times \ 100\%\\\\efficiency\ = \frac{35}{125} \times 100\%\\\\efficiency\ = 28 \ \%$

Thus, the efficiency of the engine at the given useful energy and input energy is 28%.

Learn more about efficiency of engine here: brainly.com/question/6751595

7 0

2 years ago

PLEASE HELP

Vesnalui [34]

Answer:

Observation 1 represents a solar wind because solar winds are a continuous stream of particles given off by the sun.

Explanation:

7 0

3 years ago

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2 Points How did J. J. Thomson change the atomic model based on his experimental results?

tester [92]

Explanation:

In 1897, J.J. ... Thomson realized that the accepted model of an atom did not account for negatively or positively charged particles. Therefore, he proposed a model of the atom which he likened to plum pudding. The negative electrons represented the raisins in the pudding and the dough contained the positive charge.

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4 years ago