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denpristay [2]
3 years ago
7

A wheel, starting from rest, rotates with a constant angular acceleration of 1.80 rad/s^2. During a certain 7.00 s interval, it

turns through 53.2 rad.
(a) How long had the wheel been turning before the start of the 7.00 s interval?
(b) What was the angular velocity of the wheel at the start of the 7.00 s interval?
Physics
1 answer:
lora16 [44]3 years ago
5 0

Answer:

a) 1.3 rad/s

b) 0.722 s

Explanation:

Given

Initial velocity, ω = 0 rad/s

Angular acceleration of the wheel, α = 1.8 rad/s²

using equations of angular motion, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

where

θ2 - θ1 = 53.2 rad

t2 - t1 = 7s

substituting these in the equation, we have

θ2 - θ1 = ω(0)[t2 - t1] + 1/2α(t2 - t1)²

53.2 =ω(0) * 7 + 1/2 * 1.8 * 7²

53.2 = 7.ω(0) + 1/2 * 1.8 * 49

53.2 = 7.ω(0) + 44.1

7.ω(0) = 53.2 - 44.1

ω(0) = 9.1 / 7

ω(0) = 1.3 rad/s

Using another of the equations of angular motion, we have

ω(0) = ω(i) + α*t1

1.3 = 0 + 1.8 * t1

1.3 = 1.8 * t1

t1 = 1.3/1.8

t1 = 0.722 s

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vovangra [49]

Answer:

(a) 83.06Ω

(b) 0.81A

(c) 25.0W

Explanation:

Comparing Δv = 95 sin 275t with Δv = Vmaxsinωt  

                        ω = 275

Inductive reactance Χ = ωL

                        = 275 × 0.240

                        = 66 Ω

Capacitive reactance Χ = 1/ωc

                        = 1/ (275 × 26 x 10^-6)

                        = 139.86Ω

Impedance Z = \sqrt{R^{2} + (wL - \frac{1}{wc} )^{2}

                      = \sqrt{R^{2} + (X_{l} - X_{c} )^{2}

                      = \sqrt{38^{2} + (66 - 139.86)^{2}

                      = 83.06Ω  

(b) I_{rms} = \frac{V_{rms}}{Z}

solving for V_{rms},

                     V_{rms} = \frac{V_{max}}{\sqrt{2}}

                     V_{rms} = \frac{95}{\sqrt{2}}  

                     V_{rms} = 67.2V  

substituting the value of V_{rms} and Z into I_{rms} equation, we have;

                    I_{rms} = \frac{67.2}{83.06}  

                    I_{rms} = 0.81A  

(c) Average power P = I_{rms}[\tex][tex]V_{rms}cos∅

To get the average power, we first solve for ∅ since it was not given.

                   ∅ = tan^{-1}\frac{X_{l} - X_{c}}{R}  

                   ∅ = tan^{-1}\frac{66 - 139.86}{38}  

                   ∅ = tan^{-1}\frac{-73.86}{38}  

                   ∅ = tan^{-1} -1.9437  

                   ∅ = -62.77°  

Average power P = 0.81 × 67.2 × cos-62.77

                          P = 0.81 × 67.2 × 0.46  

                          P = 25.03872W              

                          P = 25.0W  

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