The sample calculation for iron oxide in the IDEAS section of this experiment used known atomic weights to calculate an empirical formula. However, early chemists did not have any references in which they could look up atomic weights. Instead, they guessed at the formulas of compounds and measured the percent compositions of elements in compounds in order to calculate atomic weights. Calculate an atomic weight for iron using the hypothetical formula Fe101 and the composition data given in the example in the IDEAS section. You may assume the atomic weight of oxygen is known from other sources to be 16 amu.

Answer:

Therefore the value of the line currents IR, IY, and IB are

[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]

Explanation:

Apply KVL for loop 1  

[tex]415\angle 120^{o}=\left ( 3+j4+3+j4 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 6+j8 \right )I_{1}-\left ( 3+j4 \right )I_{2}\\415\angle 120^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} \rightarrow \left ( 1 \right )[/tex]

Apply KVL for loop 2

[tex]415\angle 0^{o}=\left ( 3+j4+3+j4 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 6+j8 \right )I_{2}-\left ( 3+j4 \right )I_{1}\\415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}-\left ( 5\angle 53.13^{o}\right )I_{1} \rightarrow \left ( 2 \right )[/tex]

Solving the above equations,

[tex]415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{1}-\left ( 5\angle 53.13^{o}\right )I_{2} +\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{1}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 20\angle 53.13^{o}\right )I_{2}- \left ( 10\angle 53.13^{o} \right )I_{2}\\415\angle 120^{o}+830\angle 0^{o}=\left ( 10\angle 53.13^{o} \right )I_{2}\\I_{2}= 71.88\angle -23.13^{o}A[/tex]&

[tex]415\angle 0^{o}=\left ( 10\angle 53.13^{o} \right ) \times 71.88\angle -23.13^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\415\angle 0^{o}=718.8\angle 30^{o}-\left ( 5\angle 53.13^{o} \right )I_{1}\\\left ( 5\angle 53.13^{o} \right )I_{1}=415\angle 60^{o}\\I_{1}= 83\angle 6.87^{o}A[/tex]

Hence,

[tex]I_{R}=I_{1}=83\angle 6.87^{o}A\\I_{B}=-I_{2}=-71.88\angle-23.13^{o}A\\I_{Y}=I_{2}-I_{1}\\I_{Y}=-I_{2}=41.50\angle113.3^{o}A[/tex]

Answer:

0.2063

Explanation:

Given data:

packing factor = 0.5

percentage of reduction of powders = 70%

Calculate the final porosity

after sintering Bulk specific volume = 0.9 * 0.7 = 0.63

assuming true specific volume = 1

packing factor = 0.5 , bulk specific volume = 2

packing factor after pressing and sintering

= 1 / ( 2 * 0.63 ) = 0.7937

hence : porosity = 1 - packing factor

                            = 1 - 0.7937 = 0.2063

Solution :

Given :

A six order Butterworth high pass filter.

∴ n = 6, [tex]w_c=1 \ rad/s[/tex]

a). The location at poles :

    [tex]$s^6-(w_c)^6=0$[/tex]

   [tex]$s^6=(w_c)^6=1^6$[/tex]

  ∴ [tex]$s^6 = 1$[/tex]

Therefore, it has 6 repeated poles at s = 1.

b). The transfer function H(S) :

    Transfer function H(S) [tex]$=\frac{1}{1+j\left(\frac{w_c}{s}\right)^6}$[/tex]

                                         [tex]$=\frac{1}{1-\left(\frac{w_c}{s}\right)^6}$[/tex]

  ∴    H(S) [tex]$=\frac{s^6}{s^6-(w_c)^6}=\frac{s^6}{s^6-1}$[/tex]

   H(S) [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

c). The corresponding LCCDE description :

  [tex]$=\frac{Y(s)}{X(s)}=\frac{s^6}{s^6-1}$[/tex]

   [tex]$Y(s)(s^6-1) = s^6 \times (s)$[/tex]

   [tex]$Y(s)s^6-y(s).1 = s^6 \times (s)$[/tex]

By taking inverse Laplace transformation on BS

   [tex]$L^{-1}[Y(s)s^6-Y(s)1]=L^{-1}[s^6 \times (s)]$[/tex]

   [tex]$\frac{d^6y(t)}{dt^6}-y(t)=\frac{d^6 \times (t)}{dt^6}$[/tex]

  Hence solved.

Answer:

The correct solution is "5.74%".

Explanation:

The given values are:

Gravity of aggregate,

[tex]G_{agg}=2.7[/tex]

Gravity of asphalt,

[tex]G_{asp}=1.0[/tex]

Asphalt concrete mixture,

[tex]W_{agg}=0.94 \ W_m[/tex]

We know that,

[tex]W_{asp}=W_m-W_{agg}[/tex]

        [tex]=0.06 \ W_m[/tex]

Now,

The theoretical specific gravity of mix,

⇒ [tex]G_t=\frac{W_{agg}+W_{asp}}{\frac{W_{agg}}{G_{agg}} +\frac{W_{asp}}{G_{asp}} }[/tex]

By putting the values, we get

         [tex]=\frac{0.94 \ Wm+0.06 \ Wm}{\frac{0.94 \ Wm}{2.7} +\frac{0.06 \ Wm}{1} }[/tex]

         [tex]=2.45[/tex]

hence,

The percentage of voids will be:

⇒  %V = [tex]\frac{G_t-G_m}{G_t}\times 100[/tex]

           = [tex]\frac{2.45-2.317}{2.45}\times 100[/tex]

           = [tex]\frac{0.133}{2.317}\times 100[/tex]

           = [tex]5.74[/tex] (%)  

This question is incomplete, the complete question is;

Consider a venturi with a throat-to-inlet area ratio of 0.75, mounted on the side of an airplane fuselage. The airplane is in flight at standard sea level. If the static pressure at the throat is 2050 lb/ft2 , calculate the velocity of the airplane.

Note that standard sea level density and pressure are 1.23 kg/m3 (0.002377 slug/ft3) and 1.01 x 105 N/m2 (2116lb/ft3), respectively.

Answer:

the velocity of the airplane is 267.2 ft/s

Explanation:

Given the data in the question;

throat-to-inlet area ratio A₂/A₁ = 0.75

density of air ρ = 0.002377 slug/ft³

the pressure at inlet p₁ = 2116 lb/ft³

the pressure at the throat p₂ = 2050 lb/ft³

Now, for a venturi duct, the velocity of the airplane V is given as;

V = √[ (2( p₁ - p₂ )) / (ρ( [A₁/A₂]² - 1 )) ]

so we substitute in our values

V = √[ (2( 2116 - 2050 )) / (0.002377 ( [1/0.75]² - 1 )) ]

V = √[ ( 2 × 66 ) / (0.002377 ( 1.7778 - 1 )) ]

V = √[ ( 2 × 66 ) / (0.002377 × 0.7778 ) ]

V = √[ 132 / 0.0018488 ]

V = √[ 71397.663349 ]

V = 267.2 ft/s

Therefore, the velocity of the airplane is 267.2 ft/s

78950W the answer

Explanation:

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor

Answer:

The weight will be [tex]\frac{1}{4}[/tex] of its weight on earth surface. A further explanation is provided below.

Explanation:

According to the question,

h = R

The value of gravitational acceleration at height equivalent to radius of earth R.

⇒ [tex]g=\frac{g_0}{(1+\frac{h}{R} )^2}[/tex]

or,

⇒ [tex]g=\frac{g_0}{(1+\frac{R}{R} )^2} =\frac{g_0}{4}[/tex]

here,

[tex]g_0[/tex] = gravitational acceleration earth's surface

then,

⇒ [tex]mg=\frac{mg_0}{4}[/tex]

Thus, the above is the appropriate solution.

Answer:

[tex]n=5.36kg[/tex]

Explanation:

From the question we are told that:

Volume [tex]V=2.5m^3[/tex]

Pressure[tex]\rho=500Kpa[/tex]

Temperature [tex]T=28^o[/tex]

Atmospheric pressure [tex]\rho_{atm} =97 kPa.[/tex]

Generally the equation for an Ideal gas is mathematically given by

 [tex]PV=nRT[/tex]

Therefore

 [tex]n=\frac{500*2.5}{8.314*28}[/tex]

 [tex]n=5.36kg[/tex]

Answer:

5354

Explanation:

Answer:

1.2727 stokes

Explanation:

specific gravity of fluid A = 1.65

Dynamic viscosity = 210 centipoise

Calculate the kinematic viscosity of Fluid A

First step : determine the density of fluid A

Pa = Pw * Specific gravity =  1000 * 1.65 = 1650 kg/m^3

next : convert dynamic viscosity to kg/m-s

210 centipoise = 0.21 kg/m-s

Kinetic viscosity of Fluid A = dynamic viscosity / density of fluid A

                                            = 0.21 / 1650 = 1.2727 * 10^-4 m^2/sec

Convert to stokes = 1.2727 stokes

Answer:

Option C = Only 1 path for current to flow.

Answer: hello attached below is the question properly written

a) 2670 Kw

b) product will be made up of vapor and liquid

c) Product will be a super cooled liquid

Explanation:

mass Flow rate ( m ) = 175 kg/min

pressure = 1 atm

molecular weight of ethylene glycol ( mw ) = 62.07 g/mol

enthalpy of vaporization ( ΔHv ) = 56.9 KJ/mol

Using values from the table 8.1 related to the question

a) Determine the rate at which heat must be transferred from condenser

Using values from the table 8.1 related to the question

ΔH = 2670 Kw

b) If heat is transferred  at a lower temperature the product will be made up of vapor and liquid

c) If heat was transferred at a higher temperature the product will be a super cooled liquid

Answer:

When you come up on a steady yellow traffic light you should always yield to cross traffic if you can yield safely. The flashing yellow light is there to inform drivers to be careful and to slow down.

Explanation:

hope it helped!

Answer:

Many techniques have been developed for the measurement of pressure and vacuum. Instruments used to measure and display pressure in an integral unit are called pressure meters or pressure gauges or vacuum gauges.

Answer:

Đường dây siêu cao áp 500kV: Những chuyện giờ mới kể ... ​Ngày 27/5/1994, hệ thống đường dây điện siêu cao áp 500kV Bắc - Nam chính thức đưa ... Tại thời điểm đó, các nước như Pháp, Úc, Mỹ khi xây dựng đường dây dài nhất ... và chế ra các máy kéo dây theo đặc thù công việc của từng đơn vị.

Explanation:

False, Good housekeeping eliminates accident and fire hazards. It also maintains safe, healthy work conditions; saves time, money, materials, space, and effort; improves productivity and quality; boosts morale; and reflects an image of a well-run, successful organization.

Hope it helps you❤️

A find the ganeral solution to this equation

An object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

Velocity is defined as the speed at which an object's position changes in relation to time and a frame of reference.  Speed is the rate at which an object travels along a path over time, whereas velocity is the speed and direction of an item's motion. In other words, speed is a scalar value, but velocity is a vector.

As given

Initial velocity = 40 m / sec

Velocity after 2 seconds = 30 m / sec

So the velocity after 10 seconds will be = 0 m / sec

That is the object will stop moving.

Thus. an object moving through the water has an initial velocity of 40 m/s. Two seconds later, the velocity has decreased to 30 m/s. The velocity after ten seconds is 0.

To learn more about velocity, refer to the link below:

https://brainly.com/question/28738284

#SPJ2

Your question is incomplete, but probably your complete question was

Solution :

For R-134a, we are given :

[tex]$T_i = 25^\circ C$[/tex]

[tex]$P_i=750 \ kPa$[/tex]

[tex]$P_e=165 \ kPa$[/tex]

Now we have one inlet and one exit flow, no work and no heat transfer. The energy equation is :

[tex]$h_e+\frac{1}{2}.v_e^2= h_i+\frac{1}{2}.v_i^2 $[/tex]

We also know that the gas is throttled and there is no change in the kinetic energy.

So, [tex]$v_e=v_i$[/tex]

Now from the energy equation above, we can see that the inlet and the exit enthalpies are also the same. Therefore,

[tex]$h_i=h_e$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the exit saturation temperature.

[tex]$T_e=-15^\circ C$[/tex]

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific enthalpies :

[tex]$h_f = 180.19 \ kJ/kg$[/tex]

[tex]$h_{fg} = 209 \ kJ/kg$[/tex]

Calculating the exit flow quality factor,

[tex]$x_e=\frac{h_e-h_f}{h_{fg}}$[/tex]

    [tex]$=\frac{234.59-180.19}{209}$[/tex]

   = 0.26

From the saturated R-134a table, corresponding to [tex]P_e = 165 \ kPa[/tex], we can find the specific volumes :

[tex]$v_f = 0.00746 \ m^3/kg$[/tex]

[tex]$v_{fg} = 0.11932 \ m^3/kg$[/tex]

Calculating the exit specific volume :

[tex]$v_e=v_f+x_e(v_{fg})$[/tex]

   = 0.000746 + 0.26 (0.11932)

   = 0.0318 [tex]m^3/kg[/tex]

The mass flow is equal to :

[tex]$\dot{m} = A_i . \frac{v}{v_i}$[/tex]

  [tex]$=A_e . \frac{v}{v_e}$[/tex]

So, [tex]$\frac{A_e}{A_i}=\frac{v_e}{v_i}$[/tex]

Therefore, the ratio of the exit pipe and the inlet pipe diameter is equal to

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{A_e}{A_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{v_e}{v_i}}$[/tex]

[tex]$\frac{D_e}{D_i}=\sqrt{\frac{0.0318}{0.000829}}$[/tex]

[tex]$\frac{D_e}{D_i}=6.19$[/tex]

Answer:

The type of succession is:

Primary succession

Explanation:

This is a type of succession that occurs after a volcanic eruption or earthquake; it involves the breakdown of rocks by lichens to create new, nutrient rich soils.

Primary succession is one of the two types of succession we have. It begins on rock formations, such as volcanoes or mountains, or in a place with no organisms or soil.

Answer:

its A

Explanation:

Answer:

1 NAND gate

Explanation:

The minimum number of 2 input NAND gates that can be used to implement the combinational circuit = 1

The only true combinations conditions that can produce a false result ( i.e. condition/result different from the expected result as stated in the question )

Sensor 2 activated + Emergency switch pressed = False ( Line will keep moving )

Given:

There are seven sessions to be scheduled in seven different consecutive time periods. Only one session can be scheduled for each period. The sessions are abbreviated as M, N, O, P, S, T, U. The restrictions are:

(i) M & O must occupy consecutive periods.

(ii) M must be scheduled for an earlier period than U.

(iii) O must be scheduled for a later period than S.

(iv) If S does not occupy the fourth period, then P must occupy the fourth period.

(v) U & T cannot occupy consecutively numbered periods.

Solution:

We will construct the sequence of sessions based on the given restrictions.

Since M & O must occupy consecutive periods, we can have the sequence as {..., M, O, ...} or {..., O, M, ...}

Since M must be scheduled for an earlier period than U, we can have the sequence as {..., M, O, ..., U, ...} or {..., O, M, ..., U, ...}

Since O must be scheduled for a later period than S, we can have the sequence as {..., S, ..., M, O, ..., U, ...} or {..., S, ..., O, M, ..., U, ...}

We can see that, according to the given restrictions, M cannot be assigned the first period as S has to be assigned before M. Thus option (a) is incorrect.

We can see that, according to the given restrictions, S cannot be assigned to the seventh period as seventh period is the last period and M, O & U has to be assigned after S. Thus option (c) is incorrect.

Now, T can be assigned in the following ways:

(I) After U: In this case, there are at least 4 sessions before T, the last of which is U. Moreover, according to the given restrictions, U & T cannot occupy consecutive periods. Also, since we are assuming that S is the first element, the fourth element has to be P, so that U is assigned to 5th period or after. Thus T has to be assigned to 7th, if we skip the period after U. That is, T cannot be assigned to the 6th period in this case.

(II) Between O, M & U: Even if U is assigned to the last (7th) period, since U & T cannot occupy consecutive periods, T cannot be assigned the 6th period in this case.

(III) Between S & O, M: This would imply that there are at least 3 sessions after T. This would automatically imply that T cannot be assigned to the 6th period in this case.

(IV) Before S: This implies that there are at least 4 sessions after T. Thus, T cannot be assigned to the 6th period in this case either.

Thus, T cannot be assigned to the sixth period in any case. That is, option (d) is incorrect.

Now, following all the given restrictions, one of the arrangements can be,

{1-N, 2-P, 3-T, 4-S, 5-O, 6-M, 7-U}

We can see that S is occupying the 4th period & U and T are not occupying consecutive periods. Thus, all the restrictions are followed. We can see that it is possible for O to be assigned to fifth period by following all the restrictions. Thus option (b) is the correct choice.

Final answer:

Option (b) is the correct choice. That is, based on the given restrictions, O can be assigned to the fifth period.

The option that is true as regards the 7 sessions for the consecutive time periods under the given conditions is;

B; O is assigned to the fifth period.

We are given the seven sessions during the day as;

M, N, O, P, S, T and U.

There are seven consecutive time periods for the sessions with the following conditions;

M, O, U or O, M, U.

- S, M, O, P, U

- T/U

, S

, N

, P

, T/U

, M/O

, M/O

- S

, T/U

, N

, P

, T/U

, M/O

, M/O

- S, O, M, P, U

U, N, T or T, N, U

Read more about Logical reasoning at; https://brainly.com/question/14458200

Answer:

Answer:

The volume for this is 29.7

Explanation:

Trust me on this I'm an expert

Explanation:

sory sorry sorry sorrysorrysorry

Solution :

Given data :

p = 315612 Pa

[tex]$V_1=7.07 \ m/sec$[/tex]

At exit of B,

p = [tex]$P_{atm}$[/tex]

[tex]$V_B = 26.1 \ m/sec$[/tex]

At exit of A,

[tex]p=P_{atm}[/tex]

[tex]$V_{A} = 26.1 \ m/s$[/tex]

We need to determine X component of force ([tex]$R_x$[/tex]) to hold in its place.

From figure,

[tex]$\sum F_x = m_0'V_{0x} - m_iV_{ix} $[/tex]

[tex]$=F_x+P_1A_1\sin 30=-mVA-mV_1 \sin 30$[/tex]

[tex]$=F_x=-pA_1\sin 30-m_AV_AA-m_B \sin30$[/tex]

Substitute all the values,

[tex]$=F_x=[-315612 \times \frac{\pi}{4}(0.3)^2 \sin 30]-[26.1 \times 1000 \times 26.1 \frac{\pi}{4}(0.1)^2]-[7.07 \times 1000\times 0.5 \sin 30]$[/tex][tex]$=F_x = -11154.64-5350.21-1767.28$[/tex]

[tex]$F_x = -18.2733 \ kN$[/tex]

Therefore, the force required to hold the nozzle in its place along horizontal direction.

[tex]$F_x = -18.2733 \ kN$[/tex]

Answer:

a. 67607.9psi

b. 123278.33

Explanation:

to get the yield strength of the material

= load/ cross sectional area

cross sectional area = π * 0.253²/4

= 0.0502927

The yield strenght

= 3400/0.0502927

= 67609.9 psi

b. the uts of the material

= maximum load/cross sectional area

= 6200/0.0502927

= 123278.33

Answer:

The view factor for radiation emitted by surface 2 to surface 1 is 0.1

Explanation:

Given

[tex]F_{12} = 0.4[/tex]

[tex]A_1 = 0.01m^2[/tex]

[tex]A_2 = 0.04m^2[/tex]

Required

Determine [tex]F_{21}[/tex]

To do this, we make use of the following equivalent ratio

[tex]A_1 * F_{12} = A_2 * F_{21}[/tex]

Make [tex]F_{21[/tex] the subject

[tex]F_{21} = \frac{A_1 * F_{12}}{ A_2}[/tex]

Substitute values into the equation

[tex]F_{21} = \frac{0.01m^2 * 0.4}{0.04m^2}[/tex]

[tex]F_{21} = \frac{0.01 * 0.4}{0.04}[/tex]

[tex]F_{21} = \frac{0.004}{0.04}[/tex]

[tex]F_{21} = 0.1[/tex]

Answer:

For  lower disk :   V = e^θ​rω(0)/h  = 0

At the upper disk:  V = e^θ​rω(h)/h  = e^θ​rω

Hence The physical boundary conditions are satisfied

Explanation:

Velocity field ( V ) = e^θ​rωz/h

Upper disk located at  z = h

Determine the dimensions of the velocity field

velocity field is two-dimensional ; V = V( r , z )

applying the no-slip condition

condition : The no-slip condition must be satisfied

For  lower disk Vo = 0 when disk is at rest z = 0

∴  V = e^θ​rω(0)/h  = 0

At the upper disk  V = e^θ​rω  given that a upper disk it rotates at z = h

∴ V = e^θ​rω(h)/h  = e^θ​rω

Hence we can conclude that the velocity field satisfies the appropriate physical boundary conditions.