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Sveta_85 [38]
3 years ago
11

Linear expansivity?​

Physics
1 answer:
shusha [124]3 years ago
7 0

Linear expansivity, area expansivity and volume or cubic expansivity are

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What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then
Sedbober [7]

Answer:

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

\omega =\sqrt{\frac{k}{m}}

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

x=A\cos(\omega t)

Finally, the equation of the motion of the system is:

x=(0.088m)\cos(\omega t)

or

x=(0.088m)\cos(\sqrt{\frac{k}{m} }  t)

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4 years ago
Name the types of muscle movement
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Other movements are involuntary meaning they are not under conscious control such as the contraction of your pupil in bright light muscle tissue is classified into three types according to structure and function skeletal cardiac and smooth

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A vector is<br> In English
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A vector is a quantity or phenomenon that has two independent properties: magnitude and direction.
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Compare the two numbers using &lt;, &gt;, =
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3 years ago
A wave pulse travels down a slinky. The mass of the slinky is m = 0.87 kg and is initially stretched to a length L = 6.8 m. The
Ber [7]

Answer:

1. v=14.2259\ m.s^{-1}

2. F_T=25.8924\ N

3. \lambda=29.6373\ m

Explanation:

Given:

  • mass of slinky, m=0.87\ kg
  • length of slinky, L=6.8\ m
  • amplitude of wave pulse, A=0.23\ m
  • time taken by the wave pulse to travel down the length, t=0.478\ s
  • frequency of wave pulse, f=0.48\ Hz=0.48\ s^{-1}

1.

\rm Speed\ of\ wave\ pulse=Length\ of\ slinky\div time\ taken\ by\ the\ wave\ to\ travel

v=\frac{6.8}{0.478}

v=14.2259\ m.s^{-1}

2.

<em>Now, we find the linear mass density of the slinky.</em>

\mu=\frac{m}{L}

\mu=\frac{0.87}{6.8}\ kg.m^{-1}

We have the relation involving the tension force as:

v=\sqrt{\frac{F_T}{\mu} }

14.2259=\sqrt{\frac{F_T}{\frac{0.87}{6.8}} }

202.3774=F_T\times \frac{6.8}{0.87}

F_T=25.8924\ N

3.

We have the relation for wavelength as:

\lambda=\frac{v}{f}

\lambda=\frac{14.2259}{0.48}

\lambda=29.6373\ m

8 0
3 years ago
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