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3 years ago

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A floor polisher has a rotating disk that has a 14-cm radius. The disk rotates at a constant angular velocity of 1.3 rev/s and i

s covered with a soft material that does the polishing. An operator holds the polisher in one place for 36 s, in order to buff an especially scuffed area of the floor. How far does a spot on the outer edge of the disk move during this time

2 answers:

luda_lava [24]3 years ago

6 0

Answer:

Distance moved = 41.2m

Explanation:

Detailed explanation and calculation is shown in the image below

erik [133]3 years ago

4 0

Answer:

d = 41.17 m

The outer edge of the disk moves 41.17m during this time

Explanation:

Given;

angular velocity w = 1.3 rev/s

Radius of rotating disc r = 14 cm = 0.14m

Time t = 36s

The linear velocity v can be derived using;

v = 2πwr

And the distance covered can be expressed as;

d = vt = 2πwrt

d = 2πwrt .......1

Where; w,t,r are angular velocity,time and radius respectively.

Substituting the given values into equation 1

d = 2π×1.3×0.14×36

d = 41.17 m

The outer edge of the disk moves 41.17m during this time

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Answer:

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Explanation:

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3 years ago

Which statement describes a scientific law

Olegator [25]

The Earth is a sphere is a statement to describe a scientific law

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3 years ago

A space station, in the form of a wheel 140 m in diameter, rotates to provide an "artificial gravity" of 3.90 m/s2 for persons w

Zigmanuir [339]

Radial acceleration is given by

$a_{rad}= \frac{v^2}{r}$
where

$v=r \omega$
then

$a_{rad}= \frac{r^2 \omega^2}{r}=r\omega^2$

Now

$70\omega^2=3.90 \frac{m}{s^2} \\ \\ \omega= \sqrt{ \frac{3.9}{70} }$

Using the relation

$\omega=2 \pi f$

$2 \pi f= \sqrt{ \frac{3.9}{70} }\\ \\ f= \frac{1}{2 \pi}\sqrt{ \frac{3.9}{70} }Hz$

Putting into rpm

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3 years ago

A rubber ball is dropped and bounces vertically from a horizontal concrete floor. If the ball has a speed of 3 m/s just before s

Ganezh [65]

Answer:

F=12.5N

Explanation:

Net force = rate of change of momentum

$F = m*a$

so find the change of momentum P

Pdown

$P=m*v_1=0.42kg*3m/s$

Pup

$P=m*v_1=0.42kg*6.5m/s$

dP = change in P

$dP= 0.42kg (3- -.6.5)m/s =3.99 kg m/s$

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6 0

3 years ago

The density of mobile electrons in copper metal is 8.4 1028 m-3. Suppose that i = 4.6 1018 electrons/s are drifting through a co

Vesna [10]

Answer:

The time is 106.7 minute.

Explanation:

Given that,

Density $= 8.4\times10^{28}\ m^3$

Current $i = 4.6\times10^{18}\ electron/s$

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Using formula of drift velocity

$v_{d}=\dfrac{I}{neA}$

$v_{d}=\dfrac{Ne}{tne\times\pi r^2}$

Put the value into the formula

$v_{d}=\dfrac{4.6\times10^{18}}{8.4\times10^{28}\times\pi\times(0.6\times10^{-3})^2}$

$v_{d}=4.842\times10^{-5}\ m/s$

We need to calculate the time

Using formula for time

$v_{d}=\dfrac{l}{t}$

$t=\dfrac{l}{v_{d}}$

Where, l = length

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Put the value into the formula

$t=\dfrac{31\times10^{-2}}{4.842\times10^{-5}}$

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$t=106.7\ minute$

Hence, The time is 106.7 minute.

7 0

3 years ago