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SIZIF [17.4K]
4 years ago
15

A small button placed on a horizontal rotating platform with diameter 0.326 mm will revolve with the platform when it is brought

up to a rotational speed of 42.0 rev/minrev/min , provided the button is a distance no more than 0.149 mm from the axis.1) What is the coefficient of static friction between the button and the platform? 2) How far from the axis can the button be placed, without slipping, if the platform rotates at 57.0 rev/min?
Physics
1 answer:
Tresset [83]4 years ago
3 0

Answer:

1) The coefficient of static friction between the button and the platform is   0. 294

2) The radius is 0.897

Explanation:

knowing that Centipetal force is provided by friction, so

mω²r =  μmg

then, the coefficient of friction μ is

μ = ω<em>²</em>r / g

   = (2π 42/60)² x 0.149 / 9.8

μ = 0. 294

mω²r = μmg

⇒ r = μg / ω²

    = (0.326 x 9.8) / (2π 57/60)²

  r = 3.1948 / 35.6293

    = 0.897

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the half-life of carbon-14 is 5,730 years. After 11,460 year, how much of original carbon-14 remains?
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6 0
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One object (m1 = 0.220 kg) is moving to the right with a speed of 2.10 m/s when it is struck from behind by another object (m2 =
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Answer:

vf₁  = 6.86 m/s , to the right

vf₂ =  2.96 m/s, to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:  

p=m*v  

where  

p:Linear momentum  

m: mass  

v:velocity  

There are 3 cases of collisions : elastic, inelastic and plastic.  

For the three cases the total linear momentum quantity is conserved:  

P₀ = Pf Formula (1)  

P₀ :Initial linear momentum quantity  

Pf : Final linear momentum quantity  

Data

m₁= 0.220 kg : mass of  object₁

m₂= 0.345 kg : mass of  object₂

v₀₁ =  2.1 m/s ₁ , to the right : initial velocity of m₁

v₀₂=   6 m/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

(0.22)*(2.1) + (0.345)*(6) = (0.22)*vf₁ +(0.345)*vf₂

2.532 = (0.22)*vf₁ +(0.345)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

e= \frac{v_{f2}-v_{f1} }{v_{o1} -v_{o2} }

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(2.1 - 6 )  = (vf₂ -vf₁)

-3.9 =  (vf₂ -vf₁)

vf₂ = vf₁ - 3.9

vf₂ = vf₁ - 3.9 Equation (2)

We replace Equation (2) in the Equation (1)

2.532 = (0.22)*vf₁ +(0.345)*( vf₁ - 3.9)

2.532 = (0.22)*vf₁ +(0.345)* (vf₁) -(0.345)( 3.9)

2.532 + 1.3455 = (0.565)*vf₁

3.8775 = (0.565)*vf₁

vf₁  = (3.8775) / (0.565)

vf₁  = 6.86 m/s, to the right

We replace vf₁  = 6.86 m/s in the Equation (2)

vf₂ =  6.86 - 3.9

vf₂ =  2.96 m/s, to the right

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