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NeTakaya
3 years ago
10

A steel, rigid container is filled with an ideal gas. The gas inside is heated such that its temperature, as measured in kelvin,

doubles. Choose the following:A. The pressure inside will rise by a factor of 4.A. The pressure inside will rise by a factor of 2. B. The pressure inside will drop by a factor of 4.C. The pressure inside will drop by a factor of 2.D. Nothing will happen to the pressure because this is an isobaric process.
Physics
1 answer:
nikitadnepr [17]3 years ago
8 0

Answer:

If the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then <u><em>the internal pressure will increase by a factor of 2</em></u>.

Explanation:

Gay-Lussac's law can be expressed mathematically as follows:

\frac{P}{T} =k

where V = volume, T = temperature, K = Constant

This law indicates that the ratio between pressure and temperature is constant.

This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, <em>the gas pressure is directly proportional to its temperature.</em> This must be fulfilled since the relationship must remain constant

In short, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.

It is desired to study two different states, an initial state and an final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. When the temperature varies to a new T2 value, then the pressure will change to P2, and the following will be true:

\frac{P1}{T1} =\frac{P2}{T2}

Given the above, it is possible to say that <u><em>if the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2 (doubles too) </em></u>

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You obtain a 100-W light bulb and a 50-W light bulb. Instead of connecting them in the normal way, you devise a circuit that pla
lesantik [10]

Answer:

When they are connected in series

     The  50 W bulb glow more than the 100 W bulb

Explanation:

From the question we are told that

     The power rating  of the first bulb is P_1  = 100 \ W

      The power rating of the second bulb is  P_2  =  50 \ W

     

Generally the power rating of the first bulb is mathematically represented as

      P_1  =  V^2 R

Where  V is the normal household voltage which is constant for both bulbs

  So  

        R_1  =  \frac{V^2}{P_1 }

substituting values

        R_1  =  \frac{V^2}{100}

Thus the resistance of the second bulb would be evaluated as

       R_2  =  \frac{V^2}{50}

From the above calculation we see that

        R_2  >  R_1

This power rating of the first bulb can also be represented mathematically as  

        P_  1  =  I^2_1  R_1

This power rating of the first bulb can also be represented mathematically as    

       P_  2  =  I^2_2 R_2

Now given that they are connected in series which implies that the same current flow through them so

       I_1^2 =  I_2^2

This means  that

       P \ \alpha  \  R

So  when they are connected in series

     P_2  >  P_1

This means that the 50 W bulb glows more than the 100 \ W bulb

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3 years ago
Using the equation zeff=z−s and assuming that core electrons contribute 1.00 and valence electrons contribute 0.00 to the screen
DIA [1.3K]
Thank you for posting your question here at brainly. Below is the solution. I hope the answer will help. 

<span>Cl^- 1s^2 2s^2p^6 3s^2 3p^6 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 17- 10 =7 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6; 1s^2 2s^2p^6 S = 10; 3s^2 3p^6 S = 0 </span>
<span>Zeff = Z-S = 19- 10 = 9 
</span>
S = 2 + 6.8 + 2.45 = 11.25 
<span>Zeff(Cl^-) = 17 – 11.25 = 5.75 </span>
<span>K^+ 1s^2 2s^2p^6 3s^2 3p^6 same S as for Cl^- but Z increases by 2 hence </span>
<span>Zeff(K^+) = 19 - 11.25 = 7.75</span>
5 0
3 years ago
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A student practicing for a cross country meet runs 250 m in 30 s. What is the average speed
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Answer:8.3m/sec 30 sec,

Explanation:

A student practicing for a track meet, ran 250 m in 30 sec. a. What was her average speed? 250 m = 8.3 m/sec 30 sec.

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2 years ago
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A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and t
Readme [11.4K]

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis  ( tan^-1 (13.5 / 16.4)

5 0
3 years ago
Two vehicles A and B accelerate uniformly from rest.
spayn [35]

Answer:

(i) Please find attached the required velocity time graphs plotted with MS Excel

(ii) The velocity of vehicle A at the 18th second = 20 m/s

The velocity of vehicle B at the 18th second = 0 m/s

(iii) The distance between the two vehicles at the moment in (ii) above is 60 meters

Explanation:

The given parameters of the motion of vehicles A and B are;

The acceleration of vehicles A and B = Uniform acceleration starting from rest

The maximum velocity attained by vehicle A = 30 m/s

The time it takes vehicle A to attain maximum velocity = 10 s

The maximum velocity attained by vehicle B = 30 m/s

The time it takes vehicle B to attain maximum velocity = The time it takes vehicle A to attain maximum velocity = 10 s

The time duration vehicle A maintains its maximum velocity = 6 s

The time duration vehicle B maintains its maximum velocity = 4 s

(i) From the question, we get the following table;

\begin{array}{ccc}Time &V_A&V_B\\0&0&0\\10&30&40\\14&30&40\\16&30&20\\18&20&0\\22&0&\end{array}

From the above table the velocity time graphs of vehicles A and B is created with MS Excel and can included here

(ii) The velocity of vehicle A at the start = 0 m/s

After accelerating for 10 seconds, the velocity of vehicle A = The maximum velocity of vehicle A = 30 m/s

The maximum velocity is maintained for 6 seconds which gives;

At 10 s + 6 s = 16 s, the velocity of vehicle A = 30 m/s

The time it takes vehicle A to decelerate to rest = 6 s

The deceleration of vehicle A, a_A = (30 m/s - 0 m/s)/(6 s) = 5 m/s²

Therefore, we get;

v = u - a_A·t

At the 18th second, the deceleration time, t = 18 s - 16 s = 2 s

u = 30 m/s

∴ v₁₈ = 30 - 5 × 2 = 20

The velocity of vehicle A at the 18th second, V_{18A} = 20 m/s

For vehicle B, we have;

At the 14th second, the velocity of vehicle B = 40 m/s

Vehicle B decelerates to rest in, t = 4 s

The deceleration of vehicle B, a_B = (40 m/s - 0 m/s)/(4 s) = 10 m/s²

For vehicle B, at the 18th second, t = 18 s - 14 s = 4 s

∴ v_{18B} = 40 m/s - 10 m/s² × 4 s = 0 m/s

The velocity of the vehicle B at 18th second, v_{18B} = 0 m/s

(iii) The distance covered by vehicle A up to the 18th second is given by the area under the velocity-time graph as follows;

The area triangle A₁ = (1/2) × 10 × 30 = 150

Area of rectangle, A₂ = 6 × 30 = 180

Area of trapezoid, A₃ = (1/2) × (30 + 20) × 2 = 50

The distance covered in the 18th second by vehicle S_A = A₁ + A₂ + A₃

∴ S_A = 150 + 180 + 50 = 380

The distance covered in the 18th second by vehicle S_A = 380 m

The distance covered by the vehicle B in the 18th second is given by the area under the velocity time graph of vehicle B as follows;

Area of trapezoid, A₅ = (1/2) × (18 + 4) × 40 = 440

The distance covered by the trapezoid, S_B = 440 m

The distance of the two vehicles apart at the 18t second, S_{AB} = S_B - S_A

∴ S_{AB} = 440 m - 380 m = 60 m

The distance of the two vehicles from one another at the 18th second, S_{AB} = 60 m.

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3 years ago
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