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3 years ago

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A car accelerates uniformly from rest and reaches a speed of 21.9 m/s in 8.99 s. Assume the diameter of a tire is 57.5 cm. (a) F

ind the number of revolutions the tire makes during this motion, assuming that no slipping occurs.

1 answer:

Marat540 [252]3 years ago

7 0

Answer:

Explanation:

initial velocity, u = 0 m/s

time, t = 8.99 s

final velocity, v = 21.9 m/s

diameter of tyre, D = 57.5 cm

(a) Let a is the acceleration, and n be the number f revolutions made by the tyre.

Use first equation of motion

v = u + at

21.9 = 0 + a x 8.99

a = 2.44 m/s²

Use third equation of motion

v² = u² + 2 a s

21.9 x 21.9 = 0 + 2 x 2.44 x s

s = 98.28 m

Circumference of wheel, S = π x D = 3.14 x 0.575 = 1.8055 m

number of revolutions, n = distance / circumference

n = 98.28 / 1.8055

n = 54.4

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Answer:

F = 7.143 kN

Explanation:

given,

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$F = \dfrac{\Delta P}{t}$

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3 years ago

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Complete Question

The complete question is shown on the first uploaded image

Answer:

Explanation:

From the question we are told

The amplitude of the lateral force is $F = 25 \ N$

The frequency is $f = 1 \ Hz$

The mass of the bridge per unit length is $\mu = 2000 \ kg /m$

The length of the central span is $d = 144 m$

The oscillation amplitude of the section considered at the time considered is $A = 75 \ mm = 0.075 \ m$

The time taken for the undriven oscillation to decay to $\frac{1}{e}$ of its original value is t = 6T

Generally the mass of the section considered is mathematically represented as

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Generally the oscillation amplitude of the section after a time period t is mathematically represented as

$A(t) = A_o e^{-\frac{bt}{2m} }$

Here b is the damping constant and the $A_o$ is the amplitude of the section when it was undriven

So from the question

$\frac{A_o}{e} = A_o e^{-\frac{b6T}{2m} }$

=> $\frac{1}{e} =e^{-\frac{b6T}{2m} }$

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Generally the amplitude of the section considered is mathematically represented as

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4 years ago

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Sphinxa [80]

Answer:

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lisov135 [29]

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