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AnnZ [28]
3 years ago
14

The temperature of a gas is directly related to the kinetic energy of the gas particles. Determine which actions below will have

NO EFFECT on the kinetic
energy of gas particles.

I. Poking a hole in a tire.
II. Sticking an inflated balloon in a freezer.
III. Adding more air to a balloon.
IV. Lighting a candle.
V. Setting a block of ice on the dining room table.


A) IV and V
B) I and III
C) I, III, and V
D) II, III, and IV
Physics
2 answers:
kicyunya [14]3 years ago
6 0

Answer:The correct answer is option is B.

Explanation:

The kinetic energy of the gas molecule:

K=\frac{3\times R\times T}{2\times N_A}

K\propto T

The temperature of a gas is directly proportional to kinetic energy of the gas particles.

According to question actions which will have no effect on the kinetic  energy of gas particles ; poking a hole in a tire and adding more air to the a balloon will not effect the temperature of the gas molecule.

Where as action will result in change of the temperature of the gas particles which will change the kinetic energy of the gas particles.

So, action (I) and( III) will not the effect the kinetic energy of the gas particles. Hence,the correct answer is option is B.

alexandr402 [8]3 years ago
4 0
The answer to your question is B) I and III
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Sergio039 [100]

Resistance = (voltage) / (current)

Resistance = (12v) / (0.33 A)

Resistance = (12/0.33) ohms

<em>Resistance = 36.4 ohms</em>

8 0
3 years ago
A 65.0-kg woman steps off a 10.0-m diving platform and drops straight down into the water. 1) If she reaches a depth of 3.20 m,
timurjin [86]

Answer:

F=2627.6N

Explanation:

The work done by this resistive force while traveling a distance <em>d</em> underwater would be:

W=F.d=-Fd

where the minus sign appears because the force is upwards and the displacement downwards.

This work is equal to the change of mechanical energy. At the diving plataform and underwater, when she stops moving, the woman has no kinetic energy, so all can be written in terms of her total change of gravitational potential energy:

W=\Delta E=U_f-U_i=mgh_f-mgh_i=mg(h_f-h_i)

Putting all together:

F=-\frac{W}{d}=-\frac{mg(h_f-h_i)}{d}=-\frac{(65kg)(9.8m/s^2)(-3.2m-10m)}{3.2m}=2627.6N

7 0
3 years ago
When a burning stick of increase is moved fast in a circle a circle of red light is seen.​
anzhelika [568]

Answer:

The impression of the image on the retina lasts for about 1/16th of a second after the removal of the object. If a burning stick of incense is revolved at a rate of more than sixteen revolutions per second, we see a circle of red light due to persistence of vision.

Explanation:

7 0
3 years ago
Please help me to solve this and give me the summary of answer
Neko [114]
Using the principle of floatation.

u = w............(a)

Upthrust of fluid is equal to the weight of the object.

Let the volume of the wood be V.

The upthrust u, is related to the volume submerged in water, and that is 1/5 of it volume, that is (1/5)V = 0.2V

Formula for upthrust, u = vdg

where v = volume of fluid displaced
d = density of fluid
g = acceleration due to gravity

weight, w = mg
where m = mass
g = acceleration due to gravity

From (a)

                     u = w

                 vdg =  mg      Cancel out g

                   vd  =  m
 
The v  is equal to 0.2V, which is the submerged volume. Notice that the small letter v is volume of fluid displaced, and capital V is the volume of the solid.

d is density of fluid which is water in this case, 1000 kg/m³

         0.2V * 1000 =  m

           200V =  m

Hence the mass of the object is  200V  kg.

But Density of solid =  Mass of solid / Volume of solid

                                 =    200V / V

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Density of solid = 200 kg/m³     
7 0
3 years ago
Coulomb measured the deflection of sphere A when spheres A and B had equal charges and were a distance d apart. He then made the
sdas [7]

Answer:

The new distance is     d = 0.447 d₀

Explanation:

The electric out is given by Coulomb's Law

         F = k q₁ q₂ / r²

This electric force is in balance with tension.

We reduce the charge of sphere B to 1/5 of its initial value (q_{B}=q₂ = q₂ / 5) than new distance (d = n d₀)

dat

     q₁ = q_{A}

     q₂ = q_{B}

     r = d₀

In order for the deviation to maintain the electric force it should not change, so we apply the Coulomb equation for the two points

         F = k q₁ q₂ / d₀²

         F = k q₁ (q₂ / 5) / (n d₀)²

         .k q₁ q₂ / d₀² = q₁ q₂ / (5 n² d₀²)

          5 n² = 1

          n = √ 1/5

          n = 0.447

The new distance is

         d = 0.447 d₀

6 0
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