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babunello [35]
4 years ago
10

A softball player swings a bat, accelerating it from rest to 2.6 rev/srev/s in a time of 0.20 ss . Approximate the bat as a 0.90

-kgkg uniform rod of length 0.95 mm, and compute the torque the player applies to one end of it.
Physics
1 answer:
svetoff [14.1K]4 years ago
5 0

Answer:

\tau=22.13Nm

Explanation:

information we have:

mass: m=0.9kg

lenght: L=0.95m

frequency: f=2.6rev/s

time: t=0.2s

and from the information we have we can calculate the angular velocity \omega. which is defined as

\omega=2\pi f

\omega=2\pi (2.6rev/s)\\\omega=16.336 rev/s

----------------------------

Now, to calculate the torque

We use the formula

\tau=I \alpha

where I  is the moment of inertia and \alpha is the angular acceleration

moment of inertia of a uniform rod about the end of it:

I=\frac{1}{3}mL^2

substituting known values:

I=\frac{1}{3} (0.9kg)(0.95m)^2\\I=0.271kg/m^2

for the torque we also need the acceleration \alpha which is defined as:

\alpha=\frac{\omega}{t}

susbtituting known values:

\alpha=\frac{16.336rev/s}{0.2s} \\\alpha=81.68rev/s^2

and finally we substitute I and  \alpha  into the torque equation \tau=I \alpha:\tau=(0.271kg/m^2)(81.68rev(s^2)\\\tau=22.13Nm

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slamgirl [31]

Answer:

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Explanation:

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<u>The greater the bond energy, the greater is the amount of energy required to break the bond, the more stable is the bond.</u>

Thus, among the following bonds:

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The Tevatron acceleator at the Fermi National Accelerator Laboratory (Fermilab) outside Chicago boosts protons to 1 TeV (1000 Ge
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Answer:

a) v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

c) v = c \cdot 0.994 = 2.97\cdot 10^{8} m/s

d) v = c \cdot 0.999 = 2.997\cdot 10^{8} m/s

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Explanation:

At that energies, the speed of proton is in the relativistic theory field, so we need to use the relativistic kinetic energy equation.

KE=mc^{2}(\gamma -1) = mc^{2}(\frac{1}{\sqrt{1-\beta^{2}}} -1)           (1)

Here β = v/c, when v is the speed of the particle and c is the speed of light in vacuum.

Let's solve (1) for β.

\beta = \sqrt{1-\frac{1}{\left (\frac{KE}{mc^{2}}+1 \right )^{2}}}

We can write the mass of a proton in MeV/c².

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Now we can calculate the speed in each stage.

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v = c \cdot 0.04 = 1.2\cdot 10^{7} m/s

b) Linac (400 MeV)

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\beta = 0.71

v = c \cdot 0.71 = 2.1\cdot 10^{8} m/s

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d) Main ring or injector (150 Gev)

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e) Tevatron (1 TeV)

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Have a nice day!

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EastWind [94]

Answer:

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In this case, k = 2T/L, so the frequency is:

ω = √((2T/L) / m)

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