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lutik1710 [3]
3 years ago
10

Which statement is true of the particles that make up a substance? A.) Particles in a solid have more energy than particles in a

gas. B.) Particles in a gas have more energy than particles in a liquid. C.) Particles in a liquid have more energy than particles in a gas. D.) Particles in a liquid and particles in a solid have the same amount of energy.
Physics
1 answer:
Black_prince [1.1K]3 years ago
5 0

Answer:

i think the answer is "B"

Explanation:

the particles in a gas are more far apart and have more kinetic energy than liquid because they are more close together and have less kinetic energy than   gas.  

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The missing measurement is distance.
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Consider a 2.54-cm-diameter power line for which the potential difference from the ground, 19.6 m below, to the power line is 11
tiny-mole [99]

Answer:

The line charge density is 1.59\times10^{-4}\ C/m

Explanation:

Given that,

Diameter = 2.54 cm

Distance = 19.6 m

Potential difference = 115 kV

We need to calculate the line charge density

Using formula of potential difference

V=EA

V=\dfrac{\lambda}{2\pi\epsilon_{0}r}\times\pi r^2

\lambda=\dfrac{V\times2\epsilon_{0}}{r}

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\lambda=\dfrac{115\times10^{3}\times2\times8.8\times10^{-12}}{1.27\times10^{-2}}

\lambda=1.59\times10^{-4}\ C/m

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3 years ago
Is it possible that a speed of 254 and a speed of 100 could be the same speed?
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However, if the 254 is 'centimeters per time' and the 100 is 'inches per time',
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3 years ago
Argon gas enters steadily an adiabatic turbine at 900 kPa and 450C with a velocity of 80 m/s and leaves at 150 kPa with a veloc
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Answer:

Temperature at the exit = 267.3 C

Explanation:

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To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.

Assuming Argon behaves as an Ideal gas, we have the specific volume v_{1}

as

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m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec

for Ideal gasses, the enthalpy change can be calculated using the formula

h_{2}-h_{1}=C_{p}(T_{2}-T_{1})

hence we have

W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})

250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})

<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>

evaluating the above equation, we have T_{2}=267.3C

Hence, the temperature at the exit = 267.3 C

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Difference between inverted and upright microscope
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