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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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One of your classmates, in a fit of unrestrained ego, jumps onto a lab table:

Anettt [7]

The equilibrium condition allows finding the results for the forces of the system are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Newton's second law indicates that the net force is proportional to the product of the mass and the acceleration of the bodies, in the special case that the acceleration is zero, it is called the equilibrium condition

∑ F = 0

Where the bold letters indicate vectors, F is the external forces

a) A free body diagram is a scheme of the forces without the details of the bodies, in the attachmentt we see a free body diagram of the system.

b) The reaction force of the ground is applied in each of the legs of the table, in general this force has the same magnitude in each leg, therefore in Newton's second law we can place it as a single force

N = N₁ + N₂ + N₃ + N₄₄

Let's apply the equilibrium condition

N - $W_m -w_{table}$ = 0

N = $W_m +w_{table}$

N = $M_m g + w_{table}$

They indicate the pose of the boy is 65 kg, for the weight of the table of a laboratory table is approximately 100 N

N = 65 9.8 + 100

N = 737 N

c) To calculate the mass of the table we use the relation

W = $m_{table}$ g

$m_{table} = \frac{w_{table}}{g}$

$m_{tabble}= \frac{100}{9.8}$

$m_{table}$ e = 10.2 kg

In conclusion using the equilibrium condition we can find the results for the forces are

a) The free body diagram is in the attachment

b) The normal force is N = 737 N

c) The mass of the table is 10.2 kg

Learn more here: brainly.com/question/19860811

7 0

2 years ago

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

How much force is needed to stop a 4000 kg truck moving at 8 m/s in 0 2 seconds?

maksim [4K]

Answer:

32000 N

Explanation:

Force Calculater

8 0

3 years ago

A steel bar of rectangular cross section (1.5 in. 2 3 .0 in.) carries a tensile load P (see fig- ure). The allowable stresses in

lukranit [14]

Explanation:

Value of the cross-sectional area is as follows.

A = $1.5 \times 2.30$

= 3.45 $in^{2}$

The given data is as follows.

Allowable stress = 14,500 psi

Shear stress = 7100 psi

Now, we will calculate maximum load from allowable stress as follows.

$P_{max} = \sigma_{a}A$

= $14500 \times 3.45$

= 50025 lb

Now, maximum load from shear stress is as follows.

$P_{max} = 2 \times \tau_{a} \times A$

= $2 \times 7100 \times 3.45$

= 48990 lb

Hence, $P_{max}$ will be calculated as follows.

$P_{max} = min((P_{max})_{\sigma}, (P_{max})_{\tau})$

= 48990 lb

Thus, we can conclude that the maximum permissible load $P_{max}$ is 48990 lb.

4 0

3 years ago

A block of mass m = 2.0 kg slides head on into a spring of spring constant k = 260 N/m. When the block stops, it has compressed

nasty-shy [4]

Answer:

a) Ws = 2.548 J

b) Wf = 1.153 J

c) v = 1.923 m / s

Explanation:

a) The work done by the spring force

Ws = ½ * k * x²

Ws = ½ * 260 N/m *0.14² m

Ws = 2.548J

b) The increase in thermal energy can by find using

Et = Wf

Wf = µ * m *g * x

Wf = 0.42 * 2.0 kg *9.8 m/s² * 0.14m

Wf = 1.153 J

c) The speed just as the block reaches can by fin using

EK = Ws + Et

Ek = ( 2.548 + 1.153 ) J = 3.7 J

Ek = ½ * m * v²

v² = 2* Ek / m

v = √[2 * 3.7 J / 2.0 kg]

v = 1.923 m / s

8 0

3 years ago