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8090 [49]
3 years ago
15

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

Physics
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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A 95kg clock initially at rest on a horizontal floor requires a 560N horizontal force to set it in motion. After the clock is in
miv72 [106K]

Complete Question

A 95 kg clock initially at rest on a horizontal floor requires a 650 N horizontal force to set it in motion. After the clock is in motion, a horizontal force of 560 N keeps it moving with a constant velocity. Find the coefficient of static friction and the coefficient of kinetic friction.

Answer:

The value for static friction is \mu_s =  0.60

The value for static friction is \mu_k =  0.70

Explanation:

From the question we are told that

The mass of the clock is m  =  95 \  kg

The first horizontal force is F _s  =  560 \  N

    The second horizontal force is    F _k  =  650  \  N

Generally the static frictional force is equal to the first  horizontal force

So

     F _s  =  m  *  g  *  \mu_s

=>   560  =  95  *  9.8  *  \mu_s

=>    \mu_s =  0.60

Generally the kinetic frictional force is equal to the second horizontal force

So

      F _k  =  m  *  g  *  \mu_k

      650 =  95  *  9.8  *  \mu_k

     \mu_k =  0.70

3 0
2 years ago
Which equation represents mass-energy equivalence? E = m2c E = mc2 E = (mc)2 E = mc
motikmotik

Einstein's energy mass equivalence relation say that if the whole given mass is converted to energy then it would be

E = mc^2

where

m = mass in kg

c = speed of light in m/s

this is the origination of quantum physics and by this formula we can relate the dual nature of light and particle

So correct relation above will be

E = mc^2

4 0
3 years ago
Read 2 more answers
A student has a rectangular block. It is 2 cm wide, 2 cm tall, and 25 cm long. It has a mass of 600 g. First, calculate the volu
WINSTONCH [101]

Answer:

6g/cm³

Explanation:

Density is the mass per unit volume of any substance. To solve this problem:

   

   Density  = \frac{mass}{volume}  

 Since mass = 600g

Let us find the volume;

    Volume  = length x width x height

   Volume  = 25cm x 2cm x 2cm  = 100cm³  

Therefore;

         Density  = \frac{600}{100}    = 6g/cm³

4 0
2 years ago
A 2,000 kg rocket is launched 12 km straight up at a constant acceleration into the sky at which point the rocket is travelling
hodyreva [135]

Answer:

797700000 J

Explanation:

From the question,

The work done by the rocket, is given as,

W = Ek+Ep............. Equation 1

Where Ek and Ep are the potential and the kinetic energy of the rocket respectively.

Ep = mgh............ Equation 2

Ek = 1/2mv²............. equation 3

Substitute equation 2 and equation 3 into equation 1

W = mgh+1/2mv².............. Equation 4

Where m = mass of the rocket, h = height, v = velocity of the rocket, g = acceleration due to gravity.

Given: m = 2000 kg, h = 12 km = 12000 m, v = 750 m/s, g = 9.8 m/s²

Substitute into equation 4

W = 2000(12000)(9.8)+1/2(2000)(750²)

W = 235200000+562500000

W = 797700000 J

4 0
3 years ago
The depth of the Pacific Ocean in the Mariana Trench is 36,198 ft. What is the gauge pressure at this depth
FinnZ [79.3K]

Answer:

the pressure at the depth is 1.08 × 10^{8} Pa

Explanation:

The pressure at the depth is given by,

P = h \rho g

Where, P = pressure at the depth

h = depth of the Pacific Ocean in the Mariana Trench = 36,198 ft = 11033.15 meter

\rho = density of water = 1000 \frac{kg}{m^{3} }

g = acceleration due to gravity ≈ 9.8 \frac{m}{s^{2} }

P = 11033.15 × 9.8 × 1000

P = 1.08 × 10^{8} Pa

Thus, the pressure at the depth is 1.08 × 10^{8} Pa

4 0
3 years ago
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