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Anastasy [175]
2 years ago
12

A spin bike has a flywheel in two parts—a 12.5 kg disk with radius 0.23 m, and a 7.0 kg ring with mass concentrated at the outer

edge of the disk. A friction pad exerts a force of 9.7 N on the outside of the disk. A cyclist is pedaling, spinning the disk at a typical 180 rpm. If she stops pedaling, how long will it take for the flywheel to come to a stop?
Physics
1 answer:
Nitella [24]2 years ago
7 0

Answer:

Explanation:

Given

mass of disk m=12.5 kg

radius of disc R=0.23 m

mass of ring m_r=7 kg

Force F=9.7 N

N=180 rpm

\omega =\frac{2\pi N}{60}

\omega =6\pi rad/s

Total moment of inertia

=Moment of inertia of Disc +Moment of Inertia of ring

=0.5\cdot 12.5\times 0.23^2+7\times 0.23^2

=13.25\times 0.23^2=0.7009 kg-m^2

Now Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Now using \omega _f=\omega +\alpha t

\omega _f=0 here

0=6\pi -3.18\times t

t=\frac{6\pi }{3.18}

t=5.92 s                  

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Katen [24]

electric field between mid point of two charges is given by

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here we have

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E_1 = 225 N/c

now similarly for other charge

E_2 = \frac{kq_2}{r^2}

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8 0
2 years ago
Read 2 more answers
When two resistors are wired in series with a 12 V battery, the current through the battery is 0.33 A. When they are wired in pa
MA_775_DIABLO [31]

Answer:

If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

Explanation:

R₁ = Resistance of first resistor

R₂ = Resistance of second resistor

V = Voltage of battery = 12 V

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I = Current = 1.6 A (parallel)

In series

\text{Equivalent resistance}=R_{eq}=R_1+R_2\\\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{0.33}\\\Rightarrow R_1+R_2=36.36\\ Also\ R_1=36.36-R_2

In parallel

\text{Equivalent resistance}=\frac{1}{R_{eq}}=\frac{1}{R_1}+\frac{1}{R_2}\\\Rightarrow {R_{eq}=\frac{R_1R_2}{R_1+R_2}

\text {From Ohm's law}\\V=IR_{eq}\\\Rightarrow R_{eq}=\frac{12}{1.6}\\\Rightarrow \frac{R_1R_2}{R_1+R_2}=7.5\\\Rightarrow \frac{R_1R_2}{36.36}=7.5\\\Rightarrow R_1R_2=272.72\\\Rightarrow(36.36-R_2)R_2=272.72\\\Rightarrow R_2^2-36.36R_2+272.72=0

Solving the above quadratic equation

\Rightarrow R_2=\frac{36.36\pm \sqrt{36.36^2-4\times 272.72}}{2}

\Rightarrow R_2=25.78\ or\ 10.57\\ If\ R_2=25.78\ then\ R_1=36.36-25.78=10.58\ \Omega\\ If\ R_2=10.57\ then\ R_1=36.36-10.57=25.79\Omega

∴ If R₂=25.78 ohm, then R₁=10.58 ohm

If R₂=10.57 then R₁=25.79 ohm

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2 years ago
Ryan and Becca are moving a folding table out of the sunlight. A cup of lemonade, with the message 0.44 kg is on the table. Becc
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Calculate the weight of the table through the equation,

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and Wy = W(cos θ)

x - component:
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  Wy = (4.312 N)(cos 150°) =<em> -3.734 N</em>
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