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dusya [7]
3 years ago
14

A 6.5 l sample of nitrogen at 25◦c and 1.5 atm is allowed to expand to 13.0 l. the temperature remains constant. what is the fin

al pressure? 1. 0.75 atm 2. 3.0 atm 3. 0.063 atm 4. 0.38 atm 5. 0.12 atm
Physics
1 answer:
ollegr [7]3 years ago
8 0
Since the temperature of the gas remains constant in the process, we can use Boyle's law, which states that for a gas transformation at constant temperature, the product between the gas pressure and its volume is constant:
pV=k
which can also be rewritten as
p_1 V_1 = p_2 V_2 (1)
where the labels 1 and 2 mark the initial and final conditions of the gas.

In our problem, p_1 = 1.5 atm, V_1 =6.5 L and V_2 =13.0 L, so the final pressure of the gas can be found by re-arranging eq.(1):
p_2 = p_1  \frac{V_1}{V_2}= (1.5 atm) \frac{6.5 L}{13.0 L}=0.75 atm

Therefore the correct answer is
<span>1. 0.75 atm</span> 
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3 years ago
Ramesh announced in class: ''Yesterday I had fever and my body temperature was 100 degrees.'' Ravi said: ''We learnt in the last
pogonyaev

Answer:

D. Ramesh and Ravi are correct, but they are using different measurement scales.

\Huge{\underline{\textrm{Explanation}}}Explanation

Here, Ravi says that his body temperature is 100 degrees, but does not mention that whether it is 100 degrees Celsius or 100 degrees Fahrenheit. When the temperature of a human body is more than 100.4 degree Fahrenheit (38°C), or near to it, the person is considered to have fever.

The boiling point of water is 100 degrees Celsius and not 100 degrees Fahrenheit.

Thus, they both are using different measurement scales.

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3 years ago
A mass is oscillating with amplitude A at the end of a spring.
Dmitry_Shevchenko [17]

A) x=\pm \frac{A}{2\sqrt{2}}

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

E=\frac{1}{2}kA^2 (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2 (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

U=\frac{1}{3}K

Using (2) we can rewrite this as

U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}

And using (1), we find

U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2

Substituting U=\frac{1}{2}kx^2 into the last equation, we find the value of x:

\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}

B) x=\pm \frac{3}{\sqrt{10}}A

In this case, the kinetic energy is 1/10 of the total energy:

K=\frac{1}{10}E

Since we have

K=E-U

we can write

E-U=\frac{1}{10}E\\U=\frac{9}{10}E

And so we find:

\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A

3 0
3 years ago
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Explanation:

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Gemiola [76]
D is the correct answer..........
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