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nata0808 [166]
4 years ago
15

A used car already has 40,000 miles on its odometer. The average owner will drive 12,000 miles per year. Which function best mod

els the linear relationship? y = 12,000x + 40,000 y = 40,000x + 12,000 y = 52,000x + 12,000 y = 12,000x - 40,000
Physics
1 answer:
bekas [8.4K]4 years ago
5 0

Answer:

y = 12,000x + 40,000

Explanation:

A linear relationship that would model the mileage of the car in this example is:

y=mx+q

where

y is the number of miles

x is the number of years

m is the number of miles driven per year

q is the number of miles already in the car at x=0

In this problem, we have

m = 12,000 (number of miles driven per year)

q = 40,000 (number of miles already in the car at x=0)

So substituting into the equation we get

y=12,000 x + 40,000

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The 1.53-kg uniform slender bar rotates freely about a horizontal axis through O. The system is released from rest when it is in
OlgaM077 [116]

Answer:

The spring constant = 104.82 N/m

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Explanation:

From the diagram attached below; we use the conservation of energy to determine the spring constant by using to formula:

T_1+V_1=T_2+V_2

0+0 = \frac{1}{2} k \delta^2 - \frac{mg (a+b) sin \ \theta }{2}  \\ \\ k \delta^2 = mg (a+b) sin \ \theta \\ \\ k = \frac{mg(a+b) sin \ \theta }{\delta^2}

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\delta = \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2}

Thus;

k = \frac{mg(a+b) sin \ \theta }{( \sqrt{h^2 +a^2 +2ah sin \ \theta} - \sqrt{h^2 +a^2})^2}

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b = remaining length in the rod

m = mass of the slender bar

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k = \frac{(1.53*9.8)(0.6+0.2) sin \ 64 }{( \sqrt{0.6^2 +0.6^2 +2*0.6*0.6 sin \ 64} - \sqrt{0.6^2 +0.6^2})^2}

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b

The angular velocity can be calculated by also using the conservation of energy;

T_1+V_1 = T_3 +V_3  \\ \\ 0+0 = \frac{1}{2}I_o \omega_3^2+\frac{1}{2}k \delta^2 - \frac{mg(a+b)sin \theta }{2} \\ \\ \frac{1}{2} \frac{m(a+b)^2}{3}  \omega_3^2 +  \frac{1}{2} k \delta^2 - \frac{mg(a+b)sin \ \theta }{2} =0

\frac{m(a+b)^2}{3} \omega_3^2  + k(\sqrt{h^2+a^2+2ah sin \theta } - \sqrt{h^2+a^2})^2 - mg(a+b)sin \theta = 0

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0.7344 \omega_3^2 = 2.128

\omega _3 = \sqrt{\frac{2.128}{0.7344} }

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