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3 years ago

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A ball whose mass is 1.9 kg is suspended from a spring whose stiffness is 8.0 N/m. The ball oscillates up and down with an ampli

tude of 17 cm. Part 1 (a) What is the angular frequency? ω= radians/s the tolerance is +/-2% By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor.

1 answer:

MArishka [77]3 years ago

4 0

Answer:

2.05 radians/s

Explanation:

This is a simple harmonic motion. The angular frequency of a loaded spring is given by

$\omega = \sqrt{\dfrac{k}{m}}$

where $k$ is the spring constant and $m$ is the mass on the spring.

Using the known values,

$\omega = \sqrt{\dfrac{8}{1.9}} = 2.05$

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A "590-W" electric heater is designed to operate from 120-V lines.

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Answer:

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Explanation:

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Answer:

$\displaystyle X_T=66.6\ km$

Explanation:

<u>Accelerated Motion </u>

When a body changes its speed at a constant rate, i.e. same changes take same times, then it has a constant acceleration. The acceleration can be positive or negative. In the first case, the speed increases, and in the second time, the speed lowers until it eventually stops. The equation for the speed vf at any time t is given by

$\displaystyle V_f=V_o+a\ t$

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The train has two different types of motion. It first starts from rest and has a constant acceleration of $0.987 m/s^2$ for 182 seconds. Then it brakes with a constant acceleration of $-0.321 m/s^2$ until it comes to a stop. We need to find the total distance traveled.

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$\displaystyle X=V_o\ t+\frac{a\ t^2}{2}$

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That is the speed the train is at the moment it starts to brake. We need to compute the time needed to stop the train, that is, to make vf=0

$\displaystyle V_o=179.6,a=-0.321\ m/s^2\ ,V_f=0$

$\displaystyle t=\frac{v_f-v_o}{a}=\frac{0-179.6}{-0.321}$

$\displaystyle t=559.5\ sec$

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$\displaystyle X_2=179.6\times559.5\ \frac{-0.321\times 559.5^2}{2}$

$\displaystyle X_2=50,243.2\ m$

The total distance is

$\displaystyle X_t=x_1+x_2=16,346.7+50,243.2$

$\displaystyle X_t=66,589.9\ m$

$\displaystyle \boxed{X_T=66.6\ km}$

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