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VMariaS [17]
3 years ago
14

students may elect to take the Knowledge Test online or at Drivers Office what additional test must be taken at the licensing of

fice
Physics
1 answer:
Nutka1998 [239]3 years ago
4 0
None I mean you have to take a drivers test but you will have to take the knowldege test twice

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To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?
Lena [83]

The volume of the gas at 500^{\circ} \text{C} is \fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}.

Further Explanation:

Consider the pressure of the gas to be constant.  

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

\fbox{\begin\\V\propto T\\\end{minispace}}

The above expression can we written as.

\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

Convert the temperature of the gas into kelvin.

T=273+T^\circ\text{C}}}

Here, T is the temperature in kelvin and T^\circ\text{C}}} is the temperature in centigrade.  

The initial temperature of the gas is 50^\circ\text{C}. The temperature of the gas in kelvin is.  

\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}

The final temperature of the gas is 500^\circ\text{C} . The temperature in kelvin is.  

\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}

Substitute the values of temperature and volume in the expression of the Charles' Law.  

\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}

Thus, the volume of the gas at 500^\circ\text{C}} will be \fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}.

Learn More:  

1. Examples of wind and solar energy brainly.com/question/1062501

2. Stress developed in a wire brainly.com/question/12985068

Answer Details:  

Grade: High school  

Subject: Physics  

Chapter: Gas law  

Keywords:  

Charles law, temperature, volume, initial, final, kelvin, centigrade, 50 C, 500 degree, 500 C, 50 degree, 2,33 L, gas expand, sample, heated.

7 0
3 years ago
Estimate the force required to bind the two protons in the He nucleus together. (Hint: Model the protons as point charges. Assum
scoray [572]

Answer:

 F = 2.30 10⁴ N

Explanation:

The force required to link two gates must be equal to or greater than the electrostatic force of repulsion, because the protons have equal charges.

                 F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁹ N m² / C²

   In this case the proton charge is 1.6 10⁻¹⁹ C and the distance between them is approximately the diameter of the core r = 10⁻¹⁵ m

Let's calculate

              F = 8.99 10⁹ (1.6 10⁻¹⁹)² / (10⁻¹⁵)²

              F = 2.30 10⁴ N

The bond strength must be equal to or greater than this value

3 0
3 years ago
If the units of your answer are kg ∙ m2/s3, which of the following types of quantities could your answer be?
kirza4 [7]

Answer:

  D.  power

Explanation:

  kg represents mass

  (m/v)² represents velocity squared

Then kg·m²/s² represents mass·velocity² = <em>kinetic energy</em> or <em>potential energy</em> or <em>work</em>.

  kg·m²/s³ will be the <em>rate of doing work</em>, which is power

4 0
3 years ago
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Potential energy can be converted into kinetic energy, a good example of this is when a pole-vaulter bends the pole during a lea
lesantik [10]

Answer:

The pole stores elastic potential energy when the pole is bent because its shape is change from its natural shape and it will want to go back to its original form, just like a spring or stretched elastic material.

Question:

Why did you lie about being in college?

6 0
3 years ago
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The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A
love history [14]

Answer:

Part a)

EMF = 14 \times 10^{-3} V

Part b)

EMF = 15.67 \times 10^{-3} V

Explanation:

As we know that magnetic flux through the loop is given as

\phi = B.A

now we have

\phi = B\pi r^2

now rate of change in flux is given as

\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}

now we know that

A = \pi r^2

0.285 = \pi r^2

r = 0.30 m

Now plug in all data

EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)

EMF = 14 \times 10^{-3} V

Part b)

Now the radius of the loop after t = 1 s

r_1 = r_0 + \frac{dr}{dt}

r_1 = 0.30 + 0.037

r_1 = 0.337 m

Now plug in data in above equation

EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)

EMF = 15.67 \times 10^{-3} V

5 0
3 years ago
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