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3 years ago

What i s the value of a capacitor with 250 V applied and has 500 pC of charge? (a) 200 uF (b) 0.5 pF (c) 500 uF (d) 2 pF

Engineering

1 answer:

exis [7]3 years ago

3 0

Answer:

(d) 2 pF

Explanation: the charge on capacitor is given by the expression

Q=CV

where Q=charge

C=capacitance

V=voltage across the plate of the capacitor

here we have given Q=500 pF, V=250 volt

using this formula C= $\frac{Q}{V}$

=500× $10^{-12}$ × $\frac{1}{250}$

=2× $10^{-12}$

=2 pF

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OleMash [197]

Answer:

t = 25.10 sec

Explanation:

we know that Avrami equation

$Y = 1 - e^{-kt^n}$

here Y is percentage of completion of reaction = 50%

t is duration of reaction = 146 sec

so,

$0.50 = 1 - e^{-k^146^2.1}$

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taking natural log on both side

ln(0.5) = -k(306.6)

$k = 2.26\times 10^{-3}$

for 86 % completion

$0.86 = 1 - e^{-2.26\times 10^{-3} \times t^{2.1}}$

$e^{-2.26\times 10^{-3} \times t^{2.1}} = 0.14$

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3 years ago

Heats of Reaction and Hess's Law experiment tells you to look up three values. Write down these values. Be sure to clearly indic

Anna [14]

Answer:

Enthalpy of reaction (kJoules/mole)

Heat of formation of products (kJoules/mole)

Heat of reaction of reactants (kJoules/mole)

Explanation:

The general expression for calculating the overall enthalpy of reaction is given as following:

ΔH = ∑ΔH[producst] - ∑Δ[reactants]

Thus, the heat of reaction is given as the difference between the formation of the products and the formation of the reactants. The units are expressed as kJ/mol of reactants or products.

Thus, the three values are fundamental in the determination of the overall energy of the reaction from Hess' Law.

8 0

3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

applying equation 1 determine the value of n for several combinations

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

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3 0

3 years ago

Find the percent change in cutting speed required to give an 80% reduction in tool life when the value of n is 0.12.

vaieri [72.5K]

Answer:21.3%

Explanation:

Given

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According to Taylor's tool life

$VT^n$ =c

where V is cutting velocity

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$VT^{0.12}=V'\left ( 0.2T\right )^{0.12}$

$V'=\frac{V}{0.2^{0.12}}$

$V'=\frac{V}{0.824}$ =1.213V

Thus a change of 21.3 %(increment) is required to reduce tool life by 80%

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Natali5045456 [20]

Answer:

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