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RSB [31]
3 years ago
13

A sports car accelerates uniformly from rest to a speed of 87 mi/hr in 8s. Determine: a.The acceleration of the car

Physics
1 answer:
Furkat [3]3 years ago
3 0

Answer:

Part a)

a = 4.86 m/s^2

Part b)

d = 155.52 m

Part c)

v_f = 48.6 m/s

Explanation:

As we know that car start from rest and reach to final speed of 87 mph

so we have

v_f = 87 mph = 38.88 m/s

now we have

Part a)

acceleration is rate of change in velocity

a = \frac{v_f - v_i}{t}

a = \frac{38.88 - 0}{8}

a = 4.86 m/s^2

Part b)

distance moved by car with uniform acceleration is given as

d = \frac{v_f + v_i}{2} t

d = \frac{38.88 + 0}{2} 8

d = 155.52 m

Part c)

As we know that the car start from rest

so final speed after t = 10 s

v_f = v_i + at

v_f = 0 + (4.86)10

v_f = 48.6 m/s

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Answer:

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2) Diagram of both processes on a single PV has been uploaded below

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4) the final volume of container B is 923.36 cm³

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ideal gas constant R = 8.3 J/K·mol

1)

Final Temperature of the gas in A will b GREATER than the temperature in B

2)

Diagram of both processes on a single PV has been uploaded below,

3)

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time t = 15s

the final volume of container B = ?

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( V₂ - 0.8 ) = 375 / 3039.87

V₂ - 0.8 = 0.12336

V₂ = 0.12336 + 0.8

V₂ = 0.92336 Litres

V₂ = 923.36 cm³

Therefore, the final volume of container B is 923.36 cm³

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