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laiz [17]
3 years ago
9

A disk-shaped grindstone of mass 1.7 kg and radius 8 cm is spinning at 730 rev/min. After the power is shut off, a woman continu

es to sharpen her ax by holding it against the grindstone for 9 s until the grindstone stops rotating.
(a)What is the angular acceleration of the grindstone?
(b) What is the torque exerted by the ax on the grindstone? (Assume constant angular acceleration and a lack of other frictional torques.)
Physics
1 answer:
Anastasy [175]3 years ago
8 0

Answer:

0.186 N-m        

Explanation:

mass of the grindstone, m=1.7 kg

radius, r=8 cm

Frequency, f=730 rev/min = 12.16 rev/s

time, t=9s

final angular velocity, \omega=0

Initial angular velocity,

\omega_o=2\pi f\\=2\pi (12.16) rad/s\\= 76.36 rad/s

Angular acceleration of the grind stone is:

\alpha=\frac{\omega-\omega_o}{t}\\\Rightarrow \alpha =\frac{0-76.36}{9} = -8.48 rad/s^2

Moment of inertia:

I=mr^2+mr^2=2mr^2

I=2\times 1.7 kg\times (0.08m)^2= 0.022kg-m^2

Torque exerted by the ax on the grind stone is:

\tau=I\alpha\\\tau=0.022\times (-8.48) \\\tau=0.186N-m

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A one-kilogram mass is still a one-kilogram(as mass is an intrinsic property of the object) but the downward force due to gravity, and therefore it's weight, is only one-sixth of what the object would have on the Earth. So man of mass 180 pounds weights only about 30 pounds-force when visiting the moon

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Solar panels convert light energy from sunlight into electrical energy. What material is most likely used in solar panels, and w
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A metalloid is used because it is a semiconductor and can become more conductive when more light shines on it

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The material used in a solar panel is a metalloid. It can often become conductive when more light shines on it.

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3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
ivanzaharov [21]

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Diamagnetic

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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
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Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

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3 years ago
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