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sweet [91]
3 years ago
11

If the coefficient of kinetic friction between tires and dry pavement is 0.98, what is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 34.7 m/s
Physics
1 answer:
Ilia_Sergeevich [38]3 years ago
7 0

Answer:

The shortest distance is 62.7 m

Explanation:

Given;

coefficient of kinetic friction, μk = 0.98

initial velocity, u = 34.7 m/s

Frictional force on the tire;

Fk = -μkN

where;

N is normal reaction = mg

ma = -μkN

ma = -μkmg

a = -μkg

a = - 0.98 x 9.8 = -9.604 m/s²

The shortest distance in which you can stop an automobile by locking the brakes:

Apply equation of motion;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration of the  automobile

0 = 34.7² + 2(-9.604)x

0 = 1204.09 - 19.208x

19.208x = 1204.09

x = 1204.09/19.208

x = 62.7 m

Therefore, the shortest distance in which you can stop an automobile by locking the brakes when traveling at 34.7 m/s is 62.7 m

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The velocity of sound in air saturated with water vapour at 30°C
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Explanation:

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Rd and Rv are the specific gas constants of dry air and water vapor,

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At the first condition:

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Pd = 650 mmHg - 31.7 mmHg = 618.3 mmHg = 82433 Pa

Rv = 461.52 J/kg/K

Rd = 287.00 J/kg/K

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ρ = (82433 / 287.00 / 303.15) + (4226.3 / 461.52 / 303.15)

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(340)² = k / 0.97767

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Now finding the velocity of sound at the second condition:

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