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larisa86 [58]
4 years ago
6

A spring with spring constant 17.0 N/m hangs from the ceiling. A ball is suspended from the spring and allowed to come to rest.

It is then pulled down 4.00 cm and released. The ball makes 35.0 oscillations in 18.0 seconds.
1) What is its the mass of the ball?
Express your answer using two significant figures.
m = _____ g

2) What is its maximum speed?
Express your answer using two significant figures.

= _____ cm/s
Physics
1 answer:
Sonbull [250]4 years ago
6 0

Explanation:

Given that,

Spring constant of the spring, k = 17 N/m

It is then pulled down 4.00 cm and released, A = 4 cm = 0.04 m

The ball makes 35.0 oscillations in 18.0 seconds.

The time required to complete one oscillation is :

T=\dfrac{18}{35}=0.514\ s

Angular velocity,

\omega=\dfrac{2\pi }{T}\\\\\omega=\dfrac{2\pi }{0.514}\\\\\omega=12.22\ rad/s

(1) The relation between spring constant, angular velocity and mass is given by :

\omega^2=\dfrac{k}{m}\\\\m=\dfrac{k}{\omega^2}\\\\m=\dfrac{17}{(12.22)^2}\\\\m=0.11\ kg

(2) The maximum speed is given by :

v=A\omega\\\\v=0.04\times 12.22\\\\v=0.49\ m/s            

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Answer:

the exposed core of a dead star, supported by electron degeneracy pressure.

Explanation:

A white dwarf is a low luminosity exposed core of a dead star having mass comparable to the sun but volume comparable to the earth . So its density is very high . These stars have lost the capacity to generate energy through the process of fusion . Due to high gravitational energy , it goes on shrinking but ultimately balanced by electron degeneracy pressure. It is not a main sequence star as it has lost the power of fusion .

6 0
3 years ago
The outer layer of cable on a cable reel is 16.2 cm from the center of the reel. The reel is initially stationary and can rotate
ahrayia [7]

Answer:

B. w=12.68rad/s

C. α=3.52rad/s^2

Explanation:

B)

We can solve this problem by taking into account that (as in the uniformly accelerated motion)

\theta=\omega_{0}t+\frac{1}{2}\alpha t^{2}\\\theta = \frac{s}{r}      ( 1 )

where w0 is the initial angular speed, α is the angular acceleration, s is the arc length and r is the radius.

In this case s=3.7m, r=16.2cm=0.162m, t=3.6s and w0=0. Hence, by using the equations (1) we have

\theta=\frac{3.7m}{0.162m}=22.83rad

22.83rad=\frac{1}{2}\alpha (3.6s)^2\\\\\alpha=2\frac{(22.83rad)}{3.6^2s}=3.52\frac{rad}{s^2}

to calculate the angular speed w we can use\alpha=\frac{\omega _{f}-\omega _{i}}{t _{f}-t _{i}}\\\\\omega_{f}=\alpha t_{f}=(3.52\frac{rad}{s^2})(3.6)=12.68\frac{rad}{s}

Thus, wf=12.68rad/s

C) We can use our result in B)

\alpha=3.52\frac{rad}{s^2}

I hope this is useful for you

regards

3 0
3 years ago
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Divide both sides by 48.3

5.22/48.3 = M x 48.3/48.3

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