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yulyashka [42]
3 years ago
11

PHYSICS HELP PLEASE!! MUST SHOW MATH WORK!!

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
6 0

<u>Answer:</u>

1) Distance traveled by bird = 403 meter

2)Average speed = 1.66 km /hour

3) Zcceleration = 2 m/s^2

<u>Explanation:</u>

1)  Distance traveled = Speed * Time taken = 31 * 13 = 403 meter.

2)  Average speed = Total distance covered / Time taken for that distance to cover.

    Total distance covered = 2+0.5+2.5 = 5 km

    Time taken = 3 hours

     Average speed = 5/3 = 1.66 km /hour

3)    Acceleration is defined as the rate of change of velocity, so acceleration a = change in velocity/time.

  Change in velocity = 14 - 6 = 8 m/s

   Time = 4 seconds

   So acceleration = 8 / 4 = 2 m/s^2

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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}

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

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  

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

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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There are two forces at play:

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If the magnitude of the force your hand exerts on the bucket equals the magnitude of the gravitational force, the bucket is in static equilibrium. That means the bucket is not moving and the forces acting on it balance each other out, making the net force 0.

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