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Studentka2010 [4]
4 years ago
10

You stretch a spring by a distance of 0.3 m. The spring has a spring constant of 440 N/m. When you release the spring, it snaps

back. What is the kinetic energy of the spring as it reaches its natural length?
A.5.94 J
B.8.22 J
C.21.5 J
D.19.8 J
Physics
1 answer:
QveST [7]4 years ago
3 0
The kinetic Energy has to be the same as the Energy with which the spring was stratched:
1/2×m×v^2=1/2×D×s^2
Ekin=1/2×440×0.3^2= 1/2×440×0.09=19.8. ( Of course, the kinetic Energy is normally Not calculated Like this , but in this case both Formulas have the Same solution so you can Wright it Like This)
By the way, if the spring Had Not reached its Natural length yet, you wouldnt be right if you calculate it Like above, because a Part of the Energy is still in the spring : Espring=Ekin+Espring2
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A mass-spring system has k = 56.8 N/m and m = 0.46 kg.
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Answer:

A. \omega=11.1121\ rad.s^{-1}

B. f=1.7685\ Hz

C. T=0.5654\ s

Explanation:

Given:

  • spring constant, k=56.8\ N.m^{-1}
  • mass attached, m=0.46\ kg

A)

for a spring-mass system the frequency is given as:

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

\omega=\sqrt{\frac{56.8}{0.46}}

\omega=11.1121\ rad.s^{-1}

B)

frequency is given as:

f=\frac{\omega}{2\pi}

f=\frac{11.1121}{2\pi}

f=1.7685\ Hz

C)

Time period of a simple harmonic motion is given as:

T=\frac{1}{f}

T=0.5654\ s

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A rotating space station is said to create "artificial gravity" - a loosely-defined term used for an acceleration that would be
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Answer:

\omega=0.31\frac{rad}{s}

Explanation:

The artificial gravity generated by the rotating space station is the same centripetal acceleration due to the rotational motion of the station, which is given by:

a_c=\frac{v^2}{r}(1)

Here, r is the radius and v is the tangential speed, which is given by:

v=\omega r(2)

Here \omega is the angular velocity, we replace (2) in (1):

a_c=\frac{(\omega r)^2}{r}\\\\a_c=\omega^2r

Recall that r=\frac{d}{2}=\frac{200m}{2}=100m.

Solving for \omega:

\omega=\sqrt{\frac{a_c}{r}}\\\omega=\sqrt{\frac{9.8\frac{m}{s^2}}{100m}}\\\omega=0.31\frac{rad}{s}

3 0
4 years ago
On a popular amusement park ride, the riders stand along the inside wall of a large cylinder with a diameter of 8.27 m. The cyli
____ [38]

Answer:

Explanation:

D = 8.27 m   ⇒  R = D / 2 = 8.27 m / 2 = 4.135 m

ω = 0.66 rev/sec = (0.66 rev/sec)*(2π rad/1 rev) = 4.1469 rad/s

We can apply the equation

Ff = W  ⇒  μ*N = m*g   <em>(I)</em>

then we have

N = Fc = m*ac = m*(ω²*R)

Returning to the equation <em>I</em>

<em />

μ*N = m*g    ⇒    μ*m*ω²*R = m*g   ⇒ μ = g / (ω²*R)

Finally

μ = (9.81 m/s²) / ((4.1469 rad/s)²*4.135 m) = 0.1379

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