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3 years ago

14

You plan a car trip for which you want to average 90 km/h. You cover the first half of the distance at an average speed of only

48 km/h. What must your average speed be in the second half of the trip to meet your goal?

1 answer:

xz_007 [3.2K]3 years ago

7 0

T1=S/V1T2=Ttot-T1=(2S/Vtot)-S/V1V2=S/T2 = S / (2S/Vtot-S/V1)Simplify in V2 = V1Vtot/(2V1-Vtot) = 48 * 90 / 6 = 720 km/hr

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Answer:

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2 years ago

A force acts on a 9.90 kg mobile object that moves from an initial position of to a final position of in 5.40 s. Find (a) the wo

horrorfan [7]

Given that,

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Time =5.40 s

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Using formula of displacement

$s=r_{2}-r_{1}$

Where, $r_{1}$ = initial position

$r_{2}$ = final position

Put the value into the formula

$s= (-4.10i + 3.30j + 5.40k)-(2.70i - 2.90j + 5.50k)$

$s= -6.80i+6.20j-0.1k$

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Using formula of work done

$W=F\cdot s$

Put the value into the formula

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Using formula of power

$P=\dfrac{W}{t}$

Where, P = power

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Put the value into the formula

$P=\dfrac{41.67}{5.40}$

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Using formula of angle

$\theta=\cos^{-1}(\dfrac{r_{1}r_{2}}{|r_{1}||r_{2}|})$

Put the value into the formula

$\theta=\cos^{-1}\dfrac{(-4.10i + 3.30j + 5.40k)\cdot(2.70i - 2.90j + 5.50k)}{7.54\times6.778})$

$\theta=79.7^{\circ}$

Hence, (a). The work done on the object by the force in the 5.40 s interval is 41.67 J.

(b). The average power due to the force during that interval is 7.71 Watt.

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MariettaO [177]

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