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3 years ago

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A 9230-kg truck collides head on with a 1250-kg parked car. The vehicles entangle together and slide a linear distance of 10.6 m

eters before coming to rest. Assuming a uniform coefficient of friction of 0.820 between the road surface and the vehicles, determine the pre-collision speed of the truck.

1 answer:

Helen [10]3 years ago

3 0

Answer:

=>By the work energy relation:-

=>W = ∆KE

=>Ff.s = 1/2(M+m)v^2

=>µk x N x s = 1/2(M+m)v^2

=>µk x (M+m) x g x s = 1/2(M+m)v^2

=>v = √[2µkgs]

=>v = √[2 x 0.82 x 9.8 x 10.6]

=>v = 13.05 m/s

By the law of momentum conservation:-

=>Mu = (M+m)v

=>9230u = (9230+1250) x 13.05

=>u = 14.82 m/s

Step-by-step explanation:

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Substituting y=6x we have:

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

3. Find the Taylor series for $sin^2(3x)$

$sin^{2}(3x)=\frac{1}{2}*(1-cos(6x))$ (1)

$cos(6x) = \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$ (2)

Substituting (2) in (1) we have:

$\frac{1}{2} (1-\sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!})\\ \frac{1}{2}-\frac{1}{2} \sum_{n=0}^{\infty}\frac{-1^{n}6^{2n}x^{2n}}{(2n)!}$

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$\frac{1}{2}-\sum_{n=0}^{\infty}\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}$

Extract the term for n=0 from the sum:

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For the power series centered at x=a

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If R=0, the the series converges only for x=a

So we need to evaluate this limit:

$\lim_{n \to \infty} |\frac{\frac{-9^{n}2^{2n-1}x^{2n}}{(2n)!}}{\frac{-9^{n+1}2^{2*(n+1)-1}x^{2*(n+1)}}{(2*(2n+1))!}} |$

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Because this limit is ∞ the radius of converge is ∞ and the interval of converge is $( -\infty, \infty )$ .

6 0

3 years ago