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ANTONII [103]
3 years ago
14

Under which set of circumstances will the pressure of a gas DEFINITELY increase? A) Increasing the moles and increasing the volu

me of the gas. B) Raising the temperature and reducing the volume of the gas. C) Lowering the temperature and reducing the volume of the gas. D) Raising the temperature and increasing the volume of the gas.
Physics
2 answers:
Alika [10]3 years ago
6 0

Answer: Option (B) is the correct answer.

Explanation:

According to the ideal gas equation, pressure times volume equals n times R times temperature.

Mathematically,        PV = nRT

where       P = pressure

                 V = volume

                n = number of moles

                R = gas constant

                T = temperature

Therefore, we can see that pressure is directly proportional to temperature and inversely proportional to volume.

Thus, we can conclude that by raising the temperature and reducing the volume of the gas, the pressure of a gas DEFINITELY increase.

Dominik [7]3 years ago
5 0
Hi Gamingclan4444

Raising the temperature and reducing the volume of the gas will make the pressure of a gas DEFINITELY increase.

Answer: B) or the second option.

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AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
Walking across a carpet is an example of charge being transferred by
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static electricity and friction

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3 years ago
Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du
Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time t is

\theta=\dfrac\alpha2t^2

It rotates 44.5 rad in this time, so we have

44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}

b. Since acceleration is constant, the average angular velocity is

\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2

where \omega_f is the angular velocity achieved after 6.00 s. The velocity of the disk at time t is

\omega=\alpha t

so we have

\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}

making the average velocity

\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}

Another way to find the average velocity is to compute it directly via

\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}

c. We already found this using the first method in part (b),

\omega=14.8\dfrac{\rm rad}{\rm s}

d. We already know

\theta=\dfrac\alpha2t^2

so this is just a matter of plugging in t=12.0\,\mathrm s. We get

\theta=179\,\mathrm{rad}

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2

Then for t=6.00\,\rm s we would get the same \theta=179\,\rm rad.

7 0
3 years ago
What is the tangential velocity of a record player which makes 11 revolutions in 20 seconds?
slava [35]

Answer:

The tangential velocity of a rotating object is:

v = r*w

where r is the radius, and w is the angular velocity.

w = 2*pi*f

where f is the frequency.

We know that the record plater does 11 revolutions in 20 seconds, then it does:

11 rev/20s = 0.55 rev/s = f

then we have:

w = 2*pi*0.55 s^-1 = 2*3.14*0.55 s^-1 = 3.454 s^-1

The radius of a record player is really variable, it is around 10 inches, so i will use r = 10in, which is the rotating part of the record player.

then the tangential velocity is:

v = 10in*3.454 s^-1 = 34.54 in/s

7 0
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local AM radio station broadcasts at a frequency of 685.9 kHz. Calculate the wavelength at which they are broadcasting. Waveleng
natta225 [31]

Answer:

λ = 437 m.

Explanation:

  • Since a radio wave is a electromagnetic type of wave, it propagates approximately at the same speed of light in vacuum, 3*10⁸ m/s.
  • As in any wave, there exist a fixed relationship between the propagation speed, the wavelength and the frequency of the radio wave, as follows:

       c = \lambda* f (1)

  • Replacing by the values of the speed and the frequency, and solving for the wavelength λ, we get:

       \lambda = \frac{c}{f} =\frac{3e8 m/s}{685.9e31/s} =437 m  (2)

8 0
2 years ago
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