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schepotkina [342]

3 years ago

15

Problem 7.16 (GO Tutorial) A cylindrical bar of steel 10.4 mm (0.4094 in.) in diameter is to be deformed elastically by applicat

ion of a force along the bar axis. Determine the force that will produce an elastic reduction of 3.2 ×10-3 mm (1.260 ×10-4 in.) in the diameter. For steel, values for the elastic modulus (E) and Poisson's ratio (ν) are, respectively, 207 GPa and 0.30.

1 answer:

shepuryov [24]3 years ago

4 0

Answer:

P = 18035.25 N

Explanation:

Given

D = 10.4 mm

ΔD = 3.2 ×10⁻³ mm

E = 207 GPa

ν = 0.30

If

σ = P/A

A = 0.25*π*D²

σ = E*εx

ν = - εz / εx

εz = ΔD / D

We can get εx as follows

εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴

Now we find εx

ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³

then

σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa

we have to obtain A:

A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²

Finally we apply the following equation in order o get P

σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N

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a. Determine R for a series RC high-pass filter with a cutoff frequency (fc) of 8 kHz. Use a 100 nF capacitor. b. Draw the schem

Readme [11.4K]

Answer:

a) 199.04 ohms

b) attached in image

c) -0.696dB

Explanation:

We are given:

Fc = 8Khz = 8000hz

$C = 100nF = 100*10^-^9F$

a)Using the formula:

$F_c = \frac{1}{2pie*Rc}$

$8000= \frac{1}{2*3.14*R*100*10^-^9}$

$R =\frac{1}{2*3.14*100*10^-^9*8000}$

R = 199.04 ohms

b) diagram is attached

c) $H(w) = \frac{V_out(w)}{Vin(w)} = \frac{1}{1-j\frac{wc}{w}}$

$H(F) = \frac{1}{1-j\frac{fc}{f}}$

At F = 20KHz and Fc= 8KHz we have:

$H(F)= \frac{1}{1-j\frac{8}{20}} = \frac{1}{1-j(0.4)}$

$|H(F)|= \frac{1}{\sqrt{1^2+(0.4)^2}}$

=0.923

|H(F)| in dB = 20log |H(F)|

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= -0.696dB

5 0

3 years ago

Un material determinado tiene un espesor de 30 cm y una conductividad térmica (K) de 0,04 w/m°C. En un instante dado la distribu

aksik [14]

Answer:

Para x=0:

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-2.4 W/m^{2}$

Explanation

Podemos utilizar la ley de Fourier par determinar el flujo de calor:

$\phi=-k\frac{dT}{dx}$ (1)

Por lo tanto debemos encontrar la derivada de T(x) con respecto a x primero.

Usando la ley de potencia para la derivda, tenemos:

$\frac{dT(x)}{dx}=300x-30$

Remplezando esta derivada en (1):

$\phi=-0.04(300x-30)$

Para x=0:

$\phi=0.04(30)$

$\phi=1.2 W/m^{2}$

Para x=30 cm:

$\phi=-0.04(300*0.3-30)$

$\phi=-2.4 W/m^{2}$

Espero que te haya ayudado!

4 0

3 years ago

Colonial Adventure Tours calculates the total price of a trip by adding the trip price plus other fees and multiplying the resul

Vilka [71]

Answer:

The following query is used to display TripName. Reservationld, FirstName. LastName and TotalCost of trip by adding the trip price plus other fees and multiplying the result by the number of persons which have number of persons >4.

Query:

SELECT ReservationlD, Trip.TripName. Customer.LastName. Customer.FirstName. (TripPrice+OtherFees) 'NumPersons as TotalCost FROM Reservation, Trip, Customer WHERE NumPersons>4 AND Reservation.TriplD=Trip.TriplD AND

Customer. CustomerNum=Reservation.CustomerNum:

Explanation:

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ReservationID. Trip.TripName. Customer.LastName. Customer.FirstName are the column name of table.
(TripPrice+OtherFees) 'NumPersons will calculate the total cost of the Trip and stored it into TotalCost column.
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FROM clause specifies one or more table from where records to be retrieved. o Reservation. Trip and Customer are the table name.
WHERE clause is used in SQL query to retrieve only those records that satisfy the specified condition

3 0

3 years ago

5. A typical paper clip weighs 0.59 g and consists of BCC iron. Calculate (a) the number of

marta [7]

Answer:

(a) $3.185*10^{21} cells$

(b) $6.37*10^{21} atoms$

Explanation:

(a)

Volume, V of unit cell

$V=(2.866*10^{-8})^{3}=2.354*10^{-23}$

Number of unit cells, N

$N=\frac {W_{mat}}{V\rho_{mat}}$ Where $W_{mat}$ is weight of material and $\rho_{mat}$ is density of material

$N=\frac{0.59}{7.87*(2.354*10^{-23}}=3.185*10^{21} cells$

(b)

Number of atoms in paper clip

This is a product of number of unit cells and number of atoms per cell

Since iron has 2 atoms per cell

Number of atoms of iron= $3.185*10^{21} cells*2 atoms/cell=6.37*10^{21} atoms$

8 0

3 years ago

(a) A non-cold-worked 1040 steel cylindrical rod has an initial length of 100 mm and initial diameter of 7.50 mm. is to be defor

serg [7]

Answer:

A) 1040 steel is not a possible candidate for this application

B) 35.94%

Explanation:

Initial length = 100 mm = 0.1 m

Initial diameter ( d ) = 7.5 mm = 0.0075 m

Tensile load ( p ) = 18,000 N

Condition : The 1040 steel must not experience plastic deformation or a diameter reduction of more than 1.5 * 10^-5 m

<u>A) would the 1040 steel be a possible candidate for this application</u>

<em>Yield strength of 1040 steel < stress ( in order to be a possible candidate )</em>

stress = p / A0 = ( 18000 ) / ( $\frac{\pi }{4}$ ) * 0.0075^2

= 18,000 / (4.418 * 10^-5 ) = 407.424 MPa

Yield strength of 1040 steel = 450 MPa

stress = 407.424 MPa

∴ Yield strength ( 450 MPa ) > stress ( 407.424 MPa )

Therefore 1040 steel is not a possible candidate for this application

<u>B) Determine How much cold work would be required to reduce the diameter of the steel to 6.0 mm</u>

Area1 = ( $\frac{\pi }{4}$ ) ( 0.006 )^2 = 2.83 * 10^-5 m^2

therefore % of cold work done = ( A0 - A1 ) / A0 * 100 = 35.94%

6 0

2 years ago