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ASHA 777 [7]
3 years ago
14

Your grandmother enjoys creating pottery as a hobby. She uses a potter's wheel, which is a stone disk of radius R-0.520 m and ma

ss M-100 kg. In operation, the wheel rotates at 60.0 rev/min. While the wheel is spinning, your grandmother works clay at the center of the wheel with her hands into a pot-shaped object with circular symmetry. When the correct shape is reached, she wants to stop the wheel in as short a time interval as possible, so that the shape of the pot is not further distorted by the rotation. She pushes continuously with a wet rag as hard as she can radially inward on the edge of the wheel and the wheel stops in 6.00 s
(a) You would like to build a brake to stop the wheel in a shorter time interval, but you must determine the coefficient of friction between the rag and the wheel in order to design a better system. You determine that the maximum pressing force your grandmother can sustain for 6.00 s is 50.0N. k0.544

(b) What If? If your grandmother instead chooses to press down on the upper surface of the wheel a distance r 0.250 m from the axis of rotation, what is the force (in N) needed to stop the wheel in 6.00 s? Assume that the coefficient of kinetic friction between the wet rag and the wheel remains the same as before (Enter the magnitude.) 25.99 Remember that a torque is a product of a force and a distance. N
Physics
1 answer:
Lesechka [4]3 years ago
5 0

Answer:

0.54454

104.00902 N

Explanation:

m = Mass of wheel = 100 kg

r = Radius = 0.52 m

t = Time taken = 6 seconds

\omega_f = Final angular velocity

\omega_i = Initial angular velocity

\alpha = Angular acceleration

Mass of inertia is given by

I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}

Equation of rotational motion

\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454

The coefficient of friction is 0.54454

At r = 0.25 m

\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N

The force needed to stop the wheel is 104.00902 N

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Pleaseee HElpp!!!!!!!
lutik1710 [3]

Coulomb's Law

Given:

F = 3.0 x 10^-3 Newton

d = 6.0 x 10^2 meters

Q1 = 3.3x 10^-8 Coulombs

k = 9.0 x 10^9 Newton*m^2/Coulombs^2

Required:

Q2 =?

Formula:

F = k • Q1 • Q2 / d²

Solution:

So, to solve for Q2

 

Q2 = F • d²/ k • Q1

Q2 = (3.0 x 10^-3 Newton) • (6.0 x 10^2 m)² / (9.0 x 10^9 Newton*m²/Coulombs²) • (3.3x 10^-8 Coulombs)

Q2 = (3.0 x 10^-3 Newton) • (360 000 m²) / (297 Newton*m²/Coulombs)

Q2 = 1080 Newton*m²/ (297 Newton*m²/Coulombs)

Then, take the reciprocal of the denominator and start multiplying

Q2 = 1080 • 1 Coulombs/297

Q2 = 1080 Coulombs / 297

Q2 = 3.63636363636 Coulombs

Q2 = 3.64 Coulumbs

6 0
3 years ago
Read 2 more answers
An 1800 kg helicopter rises with an upward acceleration of 2.0 m/s?. What lifting force is supplied by its rotating blades?
Viktor [21]

Answer:

Lifting force, F = 21240 N

Explanation:

It is given that,

Mass of the helicopter, m = 1800 kg

It rises with an upward acceleration of 2 m/s². We need to find the lifting force  supplied by its rotating blades. It is given by :

F = mg + ma

Where

mg is its weight

and "ma" is an additional acceleration when it is moving upwards.

So, F=1800\ kg(9.8\ m/s^2+2\ m/s^2)

F = 21240 N

So, the lifting force supplied by its rotating blades is 21240 N. Hence, this is the required solution.

4 0
3 years ago
Someone please help me Im confuzled.
fenix001 [56]

its a solid from the pressure they put on a tennis ball would make it full of moving at fast speed matter bites compacted together but still moving.

6 0
2 years ago
Two thermometers are calibrated, one in de
zlopas [31]
-40 c = -40 f but k would be 233.15
6 0
3 years ago
Block A has mass 1.00 kg and block B has mass 3.00 kg. The blocks collide and stick together on a level, frictionless surface. A
Pani-rosa [81]

Answer:

1/2mv²=0

1/2(4kg)(v²)=0

2=-v²

square root -2=v

v=1.414

5 0
2 years ago
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