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laila [671]
3 years ago
6

A controller on an electronic arcade game consists of a variable resistor connected across the plates of a 0.227 μF capacitor. T

he capacitor is charged to 5.03 V, then discharged through the resistor. The time for the potential difference across the plates to decrease to 0.833 V is measured by a clock inside the game. If the range of discharge times that can be handled effectively is from 11.5 μs to 6.33 ms, what should be the (a) lower value and (b) higher value of the resistance range of the resistor?
Engineering
1 answer:
anyanavicka [17]3 years ago
4 0

Answer:

R min = 28.173 ohm

R max = 1.55 × 10^{4}  ohm

Explanation:

given data

capacitor = 0.227 μF

charged to 5.03 V

potential difference across the plates =  0.833 V

handled effectively = 11.5 μs to 6.33 ms

solution

we know that resistance range of the resistor is express as

V(t) = V_o \times e^{t\RC}    ...........1

so R will be

R = \frac{t}{C\times ln(\frac{V_o}{V})}    ....................2

put here value

so for t min 11.5 μs

R = \frac{11.5}{0.227\times ln(\frac{5.03}{0.833})}

R min = 28.173 ohm

and

for t max 6.33 ms

R max = \frac{6.33}{11.5} \times 28.173  

R max = 1.55 × 10^{4}  ohm

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Mashcka [7]

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4 0
3 years ago
A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h
hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

D_{1} = 150 mm

A_{1}= \frac{\pi }{4}\times 150^{2}

              = 17671.45 mm^{2}

D_{2} = 250 mm

A_{2}= \frac{\pi }{4}\times 250^{2}

              = 49087.78 mm^{2}

The centrifugal force acting on the flywheel is fiven by

F = M ( R_{2} - R_{1} ) x w^{2} ------------(1)

Here F = ( -UTS x A_{1} + UCS x A_{2} )

Since density, \rho = \frac{M}{V}

                        \rho = \frac{M}{A\times t}

                        M = \rho \times A\times tM = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t

                        M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37

                        M = 8252963901

∴ R_{2} - R_{1} = 50 mm

∴ F = 763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}

  F = 33618968.38 N --------(2)

Now comparing (1) and (2)

33618968.38 = 8252963901\times 50\times \omega ^{2}

∴ ω = 4036.61

We know

\omega = \frac{2\pi N}{60}

4036.61 = \frac{2\pi N}{60}

∴ N = 38546.82 rpm

7 0
3 years ago
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Fudgin [204]

Answer:

The percentage volume change is -3.0%

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Cubic structure (a0) for beta-Mn = 0.6326nm = 0.6326e-9m = 2.5316e-28cm3

Density of alpha-Mn = 7.47g/cm3

Density of beta-Mn = 7.26g/cm3

Atomic weight of Mn = 54.938g/mol

Atomic radius of Mn = 0.112nm

STEP1: CALCULATE THE ATOM NUMBER PER CELL IN THE ALPHA-Mn;

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight ) × 100000

(7.47× 7.1236e-28 × 6.02e23) ÷ 54.938 = 58.31

Therefore the number of Atom in alpha-Mn is 58.31 atom per cell

STEP2: CALCULATE THE NUMBER OF ATOM PER CELL IN THE BETA-Mn

Atom per cell = (density × cubic structure × Avogadro's constant) ÷ (atomic weight) × 1000000

(7.26 × 2.5316e-28 × 6.02e23) ÷ 54.938 = 20.14

Therefore the number of Atom in beta-Mn is 20.14 atom per cell

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V% = [(volume of atom × number of atom per cell) ÷ volume of unit cell] × 1000

For Alpha-Mn:

[(1.4049e-30 × 58.31) ÷ 7.1236e-28] × 1000 = 114.998%

For Beta-Mn:

[(1.4049e-30 × 20.14) ÷ 2.5316e-28] × 1000 = 111.766%

STEP4: CALCULATE THE CHANGE IN PERCENTAGE VOLUME FOR ALPHA TO TRANSFORM TO BETA

change = final state - initial state

Therefore;

Change = 111.766 - 114.998 = -3.23%

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victus00 [196]

Answer:

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Therefore, Anna reuses greywater to conserve resources.

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3 years ago
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