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3 years ago

Which undergoes greater acceleration: an airplane that goes from 1000 km/h to 1005 km/h in 10 seconds or a skateboard that

Physics

1 answer:

Alex_Xolod [135]3 years ago

6 0

Answer:

Skateboard

Explanation:

Acceleration is change in velocity over time.

a = Δv / Δt

The airplane's acceleration is:

a = (1005 km/h − 1000 km/h) / 10 s

a = 0.5 km/h/s

The skateboard's acceleration is:

a = (5 km/h − 0 km/h) / 1 s

a = 5 km/h/s

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Why has the model of the atom changed over time?

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3 years ago

Determine the average value of the translational kinetic energy of the molecules of an ideal gas at (a) 27.8°C and (b) 143°C. Wh

Alinara [238K]

Answer:

a) $k_{avg}=6.22\times 10^{-21}$

b) $k_{avg}=8.61\times 10^{-21}$

c) $k_{mol}=3.74\times 10^{3}J/mol$

d) $k_{mol}=5.1\times 10^{3}J/mol$

Explanation:

Average translation kinetic energy ( $k_{avg}$ ) is given as

$k_{avg}=\frac{3}{2}\times kT$ ....................(1)

where,

k = Boltzmann's constant ; 1.38 × 10⁻²³ J/K

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T = 27.8 + 273 = 300.8 K

substituting the value of temperature in the equation (1)

we have

$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 300.8$

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b) at T = 143° C

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

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$k_{avg}=\frac{3}{2}\times 1.38\times 10^{-23}\times 416$

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c ) The translational kinetic energy per mole of an ideal gas is given as:

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$k_{mol}=3.74\times 10^{3}J/mol$

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$k_{mol}=6.02\times 10^{23}\times 8.61\times 10^{-21}$

$k_{mol}=5.1\times 10^{3}J/mol$

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