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yawa3891 [41]
3 years ago
11

Do the math for 30%÷75

Physics
1 answer:
Dafna1 [17]3 years ago
3 0

" 30% " means " 0.30 "

30% ÷ 75  means  (0.30 / 75) = <em>0.004</em>

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If you lift the front wheel of a poorly maintained bicycle off the ground and then start it spinning at 0.69 rev/s , friction in
Sindrei [870]

Answer:

magnitude of the frictional torque is 0.11 Nm

Explanation:

Moment of inertia I = 0.33 kg⋅m2

Initial angular velocity w° = 0.69 rev/s = 2 x 3.142 x 0.69 = 4.34 rad/s

Final angular velocity w = 0 (since it stops)

Time t = 13 secs

Using w = w° + §t

Where § is angular acceleration

O = 4.34 + 13§

§ = -4.34/13 = -0.33 rad/s2

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A 49.0 kg wheel, essentially a thin hoop with radius 0.730 m, is rotating at 114 rev/min. It must be brought to a stop in 22.0 s
belka [17]

Explanation:

Mass of the wheel, m = 49 kg

Radius of the hoop, r = 0.73 m

Initial angular speed of the wheel, \omega_i=114\ rev/min = 11.93\ rad/s

Final angular speed of the wheel, \omega_f=0

Time, t = 22 s

(a) If I is the moment of inertia of the hoop. It is equal to,

I=mr^2

I=49\times (0.73)^2

I=26.11\ kg-m^2

We know that the work done is equal to change in kinetic energy.

W=\Delta E

W=\dfrac{1}{2}I(\omega_f^2-\omega_i^2)

W=-\dfrac{1}{2}\times 26.11\times (11.93^2)

W = -1858.05 Joules

(b) Let P is the average power. It is given by :

P=\dfrac{W}{t}

P=\dfrac{1858.05\ J}{22\ s}

P =84.45 watts

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3 years ago
A contestants spin a wheel when it is their turn in a game show. One contestant gives the wheel an initial angular speed of 3.40
guajiro [1.7K]

Answer:

4.62 s

Explanation:

We are given that

Initial angular speed,\omega=3.4 rad/s

\theta=1\frac{1}{4} rev=\frac{5}{4}\times 2\pi=2.5\pi rad

\omega'=0

\omega'^2-\omega^2=2\alpha \theta

Substitute the values

0-(3.4)^2=2\times 2.5\pi \alpha

\alpha=\frac{-(3.4)^2}{2\times 2.5\pi}=-0.736 rad/s^2

\omega'=\omega+\alpha t

0=3.4-0.736 t

-0.736t=-3.4

t=\frac{-3.4}{-0.736}=4.62 s

Hence, the wheel takes 4.62 s to come to rest.

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4 years ago
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