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Ksivusya [100]
3 years ago
6

A car’s 30.0-kg front tire is suspended by a spring with spring constant k=1.00x10^5 N/m. At what speed is the car moving if was

hboard bumps on the road every 0.750 m drive the tire into a resonant oscillation?
Physics
1 answer:
Taya2010 [7]3 years ago
6 0

we know the equation for the period of oscillation in SHM is as follows:

T = 2 * pi * sqrt(mass/k)

we know f = 1/T, so f = 1/(2 * pi) * sqrt(k/m).

since d = v*T, we can say v = d/t = d * f

the final equation, after combining everything, is as follows:

v = d/(2 * pi) * sqrt(k/m)

by plugging everything in

v = .75/(2 * pi) * sqrt((1 * 10^5)/(30))

We find our velocity to be:

v = 6.89 m/s

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The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

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$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

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