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3 years ago

Which of the following processes are spontaneous? Check all that apply.

Physics

1 answer:

-Dominant- [34]3 years ago

5 0

Answer:

(1) A hot drink cooling to room temperature.

(2) The combustion of natural gas.

Explanation:

The spontaneous process is the process in which there is a release of energy and moves towards lower energy and a more thermodynamically stable energy state. All the natural processes are spontaneous.

There are two processes which are spontaneous in the given question are:

(1) A hot drink cooling to room temperature: In this, there is a decrease in energy and also it is a natural process and we know that all the natural processes are spontaneous.

(2) The combustion of natural gas: The fire is an example of an exothermic reaction. The combustion is a combination of a decrease in energy and an increase in entropy. So, this process occurs spontaneously.

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4 years ago

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A force which resist the relative motion of two bodies in direct contact is called _______

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Frictional Force is the answer..

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4 years ago

A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it

Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

$I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}$

The required initial angular speed is obtained herein:

$g= \omega_{o}^{2}\cdot R_{ss}$

$\omega_{o}=\sqrt{\frac{g}{R_{ss}} }$

$\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }$

$\omega_{o} \approx 0.253\,\frac{rad}{s}$

The initial moment of inertia is:

$I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}$

The final moment of inertia is:

$I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}$

Now, the final angular speed is obtained:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )$

$\omega_{f} = 0.331\,\frac{rad}{s^}$

The apparent acceleration is:

$a_{f} = \omega_{f}^{2}\cdot R_{ss}$

$a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)$

$a_{f} = 16.763\,\frac{m}{s^{2}}$

This is approximately 1.709g.

4 0

4 years ago

A beam of light has a wavelength of 620 nm in vacuum.

crimeas [40]

Explanation:

It is given that,

Wavelength of a beam of light in vacuum, $\lambda_o=620\ nm$

(a) Let v is the speed of light in a liquid whose index of refraction at this wavelength is 1.56.

We know that,

Refractive index,

$n=\dfrac{c}{v}\\\\v=\dfrac{c}{n}\\\\v=\dfrac{3\times 10^8}{1.56}\\\\v=1.92\times 10^8\ m/s$

(b) The wavelength of light in material,

$\lambda=\dfrac{\lambda_o}{n}$

Let $\lambda_1$ is the wavelength of these waves in the liquid. So,

$\lambda_1=\dfrac{\lambda_o}{n}\\\\\lambda_1=\dfrac{620\times 10^{-9}}{1.56}\\\\\lambda_1=3.97\times 10^{-7}\ m$

Hence, this is the required solution.

7 0

3 years ago

Refrigerators are usually kept at about 5∘C, while room temperature is about 20∘C. If you were to take an "empty" sealed 2-liter

expeople1 [14]

Answer:

The volume will not contract to one fourth of the original.

Explanation:

Applying Charles's Law as:

$\frac {V_1}{T_1}=\frac {V_2}{T_2}$

Given ,

V₁ = ?

V₂ = 2 L

T₁ = 5 °C

T₂ = 20 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (5 + 273.15) K = 278.15 K

T₂ = (20 + 273.15) K = 293.15 K

Using Charles law as:

$V_1=\frac {V_2}{T_2} \times {T_1}$

$V_1=\frac {2\ L}{293.15 K} \times {278.15 K}$

$V_1=1.8977 L$

Thus, the volume will not contract to one fourth of the original. (1/4 of 2 L is 0.5 L).

8 0

4 years ago