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ValentinkaMS [17]

2 years ago

7

Consider a cylindrical specimen of some hypothetical metal alloy that has a diameter of 11.0 mm. A tensile force of 1550 N produ

ces an elastic reduction in diameter of 6.9 x 10^-4 mm. Compute the elastic modulus of this alloy, given that Poisson's ratio is 0.35.

1 answer:

Natasha_Volkova [10]2 years ago

6 0

Answer:

See it in the pic.

Explanation:

See it in the pic.

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Select the correct answer.

Lilit [14]

Answer:

D

Explanation:

Confidential data is not supposed to be shared amongst others.

8 0

3 years ago

Read 2 more answers

Refers to the capability to keep moving forward on a specified grade.

Mademuasel [1]

Answer:

maneuverability

Explanation:

needless to say, I took the quiz

6 0

2 years ago

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

2 years ago

2) The switch in the circuit below has been closed a long time. At t=0, it is opened.

saul85 [17]

Answer:

il(t) = e^(-100t)

Explanation:

The current from the source when the switch is closed is the current through an equivalent load of 15 + 50║50 = 15+25 = 40 ohms. That is, it is 80/40 = 2 amperes. That current is split evenly between the two parallel 50-ohm resistors, so the initial inductor current is 2/2 = 1 ampere.

The time constant is L/R = 0.20/20 = 0.01 seconds. Then the decaying current is described by ...

il(t) = e^(-t/.01)

il(t) = e^(-100t) . . . amperes

8 0

2 years ago

When measuring a Brake Drum, the Brake Micrometer is set to a Base Drum Diameter of 10 Inches plus four notches, and the dial re

kozerog [31]

Answer:

10.5

Explanation:

7 0

2 years ago