A baton twirler has a baton of length 0.36 m with masses of 0.48 kg at each end. Assume the rod itself is massless. The rod is f

<span>  </span><span>When the net torque and moment of inertia are given, calculating becomes straightforward.

Using the equation torque = I * alpha, where I represents the moment of inertia and alpha is the angular acceleration.

Consequently, 0.098 / 0.000075 results in 1306.666... rad / s^2

While the angular acceleration stays the same, you can also determine the angular velocity at that instance, which is 21.36 rad / s.</span>

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

Pressure at point (1) P1 = 10 bar = P2

Volume at point (1) V1 = 1 m^3

Volume at point (2) V2 =4 m^3

For Process 2 ---> 3, where V = constant

Pressure at point (3) P3 = 10 bar

Volume at point (3) V3 = 4 m^3

Process 3 ---> 1 defined as PV = constant.

Required

Sketch the processes on the PV coordinates To calculate the work in kJ

Work is calculated by W=

a=V2

b=V1

x=Pdx=dV

For Process 1 ---> 2 where P3 = P4 = 5 bar  

W=

a=V3

b=V2

x=4dx=dV

substituting the values here into the integral gives

W=400 kJFor Process 2 ---> 3

As V = constant in this case, the volume remains unchanged, resulting in W = 0 kJ  

For Process 3 ---> 1  By applying point (1) --> 5 x.2 = C ---> C = 1 P = 5V^-1  

 W=

a=V1

b=V3

x=1V^-1dx=dVsubstituting values into this integral results in

W=| ln V | limits a and b

  = -160.944 KJ

0.000003782 m 0.000001891 m 0.000001197125 m Explanation: Wavelength = 248 nm D = Beam diameter = 1 cm f = Focal length = 0.625 cm. The angle can be expressed as The width is defined as The required width is determined to be 0.000003782 m. The minimum resolvable line separation stands at 0.000001891 m when adjustments are made. The new minimum resolvable line separation, assuming a wavelength reduction to 157 nm, is to be determined.