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d1i1m1o1n [39]
3 years ago
10

wheel rotates from rest with constant angular acceleration. Part A If it rotates through 8.00 revolutions in the first 2.50 s, h

ow many more revolutions will it rotate through in the next 5.00 s?
Physics
1 answer:
Alex73 [517]3 years ago
5 0

Answer:

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

Explanation:

Given;

wheel rotates from rest with constant angular acceleration.

Initial angular speed v = 0

Time t = 2.50

Distance x = 8 rev

Applying equation of motion;

x = vt +0.5at^2 ........1

Since v = 0

x = 0.5at^2

making a the subject of formula;

a = x/0.5t^2 = 2x/t^2

a = angular acceleration

t = time taken

x = angular distance

Substituting the values;

a = 2(8)/2.5^2

a = 2.56 rev/s^2

velocity at t = 2.50

v1 = a×t = 2.56×2.50 = 6.4 rev/s

Through the next 5 second;

t2 = 5 seconds

a2 = 2.56 rev/s^2

v2 = 6.4 rev/s

From equation 1;

x = vt +0.5at^2

Substituting the values;

x2 = 6.4(5) + 0.5×2.56×5^2

x2 = 64 revolutions.

it rotate through 64 revolutions in the next 5.00 s

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BARSIC [14]

Answer:

The axial force is  P =  15.93 k N

Explanation:

From the question we are told that

     The diameter of the shaft steel is  d =  50mm

      The length of the cylindrical bushing  L =100mm

     The outer diameter of the cylindrical bushing  is  D =  70 \ mm

       The diametral interference is \delta _d = 0.005 mm

       The coefficient of friction is  \mu = 0.2

       The Young modulus of  steel is  207 *10^{3} MPa (N/mm^2)

The diametral interference is mathematically represented as

           \delta_d = \frac{2 *d * P_B * D^2}{E (D^2 -d^2)}

Where P_B is the pressure (stress) on the two object held together  

     So making P_B the subject

            P_B = \frac{\delta _d E (D^2 - d^2)}{2 * d* D^2}

Substituting values

                P_B = \frac{(0.005) (207 *10^{3} ) * (70^2 - 50^2))}{2 * (50) (70) ^2 }

                 P_B = 5.069 MPa

Now he axial force required is

             P =  \mu * P_B * A

Where A is the area which is mathematically evaluated as

               \pi d l

So   P  =  \mu P_B \pi d l

Substituting values

      P =  0.2 * 5.069 * 3.142 * 50 * 100

       P =  15.93 k N

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Answer:

The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.

Explanation:

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  • Where ε, is the dielectric constant of the material that fills the space between plates.
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