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Bess
3 months 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:
Keith_Richards [3.2K]3 months ago
7 0

Response:

Clarification:

Provided

weight of disk m=12.5 kg

diameter of disc R=0.23 m

weight 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

Overall moment of inertia

=Disc's moment of inertia +Ring's Moment of Inertia

=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

At this point, Torque is T=F\times R=I\cdot \alpha

9.7\times 0.23=0.7\times \alpha

\alpha =3.18 rad/s^2

Utilizing \omega _f=\omega +\alpha t

\omega _f=0 in this scenario

0=6\pi -3.18\times t

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

t=5.92 s

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Result:

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Explanation:

The electromagnetic attraction between the electron and the proton in the nucleus is equivalent to the centripetal force:

k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}

where

k represents the Coulomb constant

e denotes the charge of the electron

e denotes the charge of the proton in the nucleus

r signifies the distance from the electron to the nucleus

v indicates the velocity of the electron

is the mass of the electron

Rearranging for v, we determine

v=\sqrt{k\frac{e^2}{m_e r}}

Inside a hydrogen atom, the distance separating the electron from the nucleus is roughly

r=5.3\cdot 10^{-11}m

while the mass of the electron is

m_e = 9.11\cdot 10^{-31}kg

and the charge is

e=1.6\cdot 10^{-19} C

By plugging in the values into the formula, we achieve

v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s

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On a foggy day, you are driving at 70 mph and Granny is driving at 50 mph. As you try to pass her, you both see an overturned tr
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3 months ago
An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground
Ostrovityanka [3204]

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = the position of the ladder's center of mass = (0.5) L = (0.5) 8 = 4 m

AC = distance of the climber from the bottom of the ladder = x

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F_{g} = weight of the climber = 710 N

F = force exerted by the wall on the ladder

N = normal force acting on the ladder from the ground =?

By applying force equilibrium in the vertical direction

N = F_{g} + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f = static friction force on the ladder

Static friction force can be expressed as

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f = (0.55) (950)

f = 522.5 N

The equation for force along the horizontal axis reads

F = f

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using torque equilibrium around point A

F Sin50 (AD) = W Cos50 (AB) + (F_{g} Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

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