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Natali5045456 [20]
3 years ago
9

hello:) I don’t really understand this graph. I thought uniform acceleration means the velocity is constant so it’s a constant g

radient?

Physics
1 answer:
Shkiper50 [21]3 years ago
5 0

Answer:

Uniform acceleration means that the object is accelerating at a constant rate. Since there is still acceleration, velocity is not constant. In fact, the velocity is increasing. This means that the object is moving faster and faster at a constant rate.

This graph is a displacement-time graph so its gradient would represent the velocity of the object. The velocity increases hence the gradient should increase (which means that the graph must be steeper and steeper with increasing time).

Constant gradient of a displacement-time graph would mean that there is no acceleration since gradient of the graph does not change so velocity is constant.

Note that it is the<em> </em><em>rate</em><em> </em><em>of</em><em> </em><em>change</em> of velocity that is constant during uniform acceleration, not velocity.

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Leno4ka [110]

Law of inertia would be your answer.

4 0
3 years ago
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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
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Answer:

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

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

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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

T = Temperature in kelvin

a) at T = 27.8° C

or

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

or

T = 143 + 273 = 416 K

substituting the value of temperature in the equation (1)

we have

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

k_{avg}=8.61\times 10^{-21}J

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

       k_{mol}=A_{v}\times k_{avg}

here   A_{v} = Avagadro's number; ( 6.02×10²³ )

now at T = 27.8° C

        k_{mol}=6.02\times 10^{23}\times 6.22\times 10^{-21}

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

d) now at T = 143° C

        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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