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prisoha [69]
3 years ago
14

A CD-ROM drive in a computer spins the 12-cm-diameter disks at 9500 rpm. Find acceleration in units of g that a speck of dust on

the outside edge of the disk experiences
Physics
1 answer:
NeTakaya3 years ago
4 0
The net force on an object is:
F = ma
The only force acting on the speck of dust as it is lays on the CD is centripetal force given by:
F = (mv²)/r
Equating the two
ma = (mv²)/r
We get:
a = v²/r
v is the linear velocity in this case; however, we can calculate only the angular velocity with the given data. Therefore, we must use:
v = ωr; where
ω = 2πf; f is the rotations per second
f = 9500 / 60
f = 950/6
ω = 2π(950/6)
ω = (950π)/3
v = (950π)/3 × 0.12
v = 38π
a = (38π)²/0.12
a = 1.2 × 10⁵ m/s²
To express this in terms of gravitational acceleration, we divide by the value of gravitational acceleration, 9.81
a = 1.2 × 10⁴ g
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Explanation:

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A 3.30 x 10-2-kg block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring const
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Answer:

0.17547 m

Explanation:

m = Mass of block = 3.3\times 10^{-2}\ kg

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k = Spring constant = 125 N/m

A = Amplitude

The kinetic energy of the system is conserved

\dfrac{1}{2}mv^2=\dfrac{1}{2}kA^2\\\Rightarrow \\\Rightarrow A=\sqrt{\dfrac{mv^2}{k}}\\\Rightarrow A=\sqrt{\dfrac{3.3\times 10^{-2}\times 10.8^2}{125}}\\\Rightarrow A=0.17547\ m

The amplitude of the resulting simple harmonic motion is 0.17547 m

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3 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
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