Question 1/5

An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel

Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0

2 years ago

Consider the time domain waveforms below on the left. Match waveforms (a) - (e) to their respective frequency spectrum represent

lozanna [386]

Answer:

(a) ------(3). (b)------(1) (c)-----(5) (d)------(2) ------ (e) -----4

Note: Kindly find an attached copy of the diagram associated with the solution to the question below.

Sources: the diagram to this question was researched from Quizlet

Explanation:

Solution

(1) Part (a)a waveform has a high frequency components compared to another waveform. the corresponding frequency components should be high.

So for the wave form a the corresponding frequency spectrum is (3)

(2) For part (b), waveform has three harmonics, the corresponding frequency spectrum is (1)

(3) The time domain waveform plot (c) is a sine wave but there exists a dc component.

Thus x[0] ≠0

For (c) the corresponding frequency spectrum is (5)

(4) For part (d) the corresponding frequency spectrum is (2)

(5) A sine wave is made of a single frequency only and its spectrum is a single point

For (e) the corresponding frequency spectrum is (4)

3 0

2 years ago

A mass of 12 kg saturated refrigerant-134a vapor is contained in a piston-cylinder device at 240 kPa. Now 300 kJ of heat is tran

Ket [755]

Answer:

I = 12.706 Amps

Explanation:

Given:

- The mass of saturated R-134a m = 12 kg

- The initial Conditions

P_1 = 240 KPa

Saturated Vapor

- The final Conditions

P_2 = 240 KPa

T_2 = 70° C

- The amount of Heat transferred Q_in = 300 KJ

- The voltage of current source V = 110 V

- The current supplied by the source = I

- The time duration Δt = 6 min

Find:

Determine the current supplied I.

Solution:

- Look-up enthalpies h_1 and h_2 at both states using Tables A-11 and A-13.

P_1 = 240 KPa

Saturated Vapor ----------> h_1 = h_g = 247.32 KJ/kg

P_2 = 240 KPa

T_2 = 70° C ----------> h_2 = 314.53 KJ/kg

- Using First Thermodynamic Law, set up an energy balance:

E_in - E_out = ΔE_system

Q_in + W_electric,in - W_out = Δ U

Q_in + V*I*Δt - W_out = Δ U

Q_in + V*I*Δt = Δ H

Q_in + V*I*Δt = m*( h_2 - h_1 )

- Make the current I the subject of the expression above:

V*I*Δt = m*( h_2 - h_1 ) - Q_in

I = [ m*( h_2 - h_1 ) - Q_in ] / V*Δt

- Plug in the values in the expression derived above and evaluate current of source I:

I = [ 12*( 314.53 - 247.32 ) - 300 ]*1000 / 110*6*60

I = 12.706 Amps

- The required source of current is I = 12.706 Amps.

4 0

2 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

a) Determine the yielding factor of safety

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

b) Determine the overload factor of safety

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

C) Determine the factor of safety based on joint separation

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

D) Determine the fatigue factor of safety using the Goodman criterion.

nf = 0.849

attached below is the detailed solution .

4 0

1 year ago

Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia

lukranit [14]

Answer:

α = 7.848 rad/s^2 ... Without disk inertia

α = 6.278 rad/s^2 .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B ) are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

ma*g - T = ma*a

T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

g* ( ma - mb ) = ( ma + mb )*a

a = g* ( ma - mb ) / ( ma + mb )

a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

a = 1.962 m/s^2

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

a = at = 1.962 m/s^2

at = r*α

Where,

α: The angular acceleration of the object ( disk )

α = at / r

α = 1.962 / 0.25

α = 7.848 rad/s^2

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

Sum of moments ∑M = Iα

( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

Ta: The force in string due to mass A

Tb: The force in string due to mass B

I: The moment of inertia of disk = 0.5*M*r^2

( ma*a - mb*a )*r = 0.5*M*r^2*α

α = ( ma*a - mb*a ) / ( 0.5*M*r )

α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

α = ( 3.924 ) / ( 0.625 )

α = 6.278 rad/s^2

6 0

2 years ago