a. Derive linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R.
b. Compute and compare linear density values for these same two directions for iron (Fe).

Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

Answer:

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Explanation:

A turbine is a device that works usually at steady state. Given that heat losses exists and changes in kinetic energy are not negligible, the following expression allows us to determine the power supplied by the fluid to the turbine by the First Law of Thermodynamics:

[tex]-\dot Q_{loss} - \dot W_{out} + \dot m \cdot \left[(h_{in}-h_{out}) + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right] = 0[/tex]

Output power is cleared:

[tex]\dot W_{out} = -\dot Q_{loss} + \dot m \cdot \left[(h_{in}-h_{out})+\frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) \right][/tex]

If [tex]\dot Q_{loss} = 0.3\,kW[/tex], [tex]\dot m = 0.85\,\frac{kg}{s}[/tex], [tex]h_{in} = 2800\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 1550\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 45\,\frac{m}{s}[/tex] and [tex]v_{out} = 20\,\frac{m}{s}[/tex], then:

[tex]\dot W_{out} = -0.3\,kW + \left(0.85\,\frac{kg}{s} \right)\cdot \left\{\left(2800\,\frac{kJ}{kg}-1550\,\frac{kJ}{kg} \right)+\frac{1}{2}\cdot \left[\left(45\,\frac{m}{s} \right)^{2}-\left(20\,\frac{m}{s} \right)^{2}\right] \right\}[/tex]

[tex]\dot W_{out} = 1752.825\,kW[/tex]

The power that fluid supplies to the turbine is 1752.825 kilowatts.

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

[tex] V = A*t [/tex]

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]    

Now, using the density we can find the mass:

[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]

Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):

[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

Answer:

The viscosities of the oils are 0.967 Pa.s and 1.933 Pa.s

Explanation:

Assuming the two oils are Newtonian fluids.

From Newton's law of viscosity for Newtonian fluids, we know that the shear stress is proportional to the velocity gradient with the viscosity serving as the constant of proportionality.

τ = μ (∂v/∂y)

There are oils above and below the plate, so we can write this expression for the both cases.

τ₁ = μ₁ (∂v/∂y)

τ₂ = μ₂ (∂v/∂y)

dv = 0.3 m/s

dy = (0.06/2) = 0.03 m (the plate is centered in a gap of width 0.06 m)

τ₁ = μ₁ (0.3/0.03) = 10μ₁

τ₂ = μ₂ (0.3/0.03) = 10μ₂

But the shear stress on the plate is given as 29 N per square meter.

τ = 29 N/m²

But this stress is a sum of stress due to both shear stress above and below the plate

τ = τ₁ + τ₂ = 10μ₁ + 10μ₂ = 29

But it is also given that one viscosity is twice the other

μ₁ = 2μ₂

10μ₁ + 10μ₂ = 29

10(2μ₂) + 10μ₂ = 29

30μ₂ = 29

μ₂ = (29/30) = 0.967 Pa.s

μ₁ = 2μ₂ = 2 × 0.967 = 1.933 Pa.s

Hope this Helps!!!

Answer:

amplitudes : 1 , 0.05, 0.05

frequencies : 50/[tex]\pi[/tex],   105/[tex]2\pi[/tex],  95/2[tex]\pi[/tex]

phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]

Explanation:

signal  s(t) = ( 1 + 0.1 cos 5t )cos 100t

signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb

         s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]

          s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t

               =  cos100t + 0.05cos(105)t + 0.05cos 95t

             = cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]

= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex]  + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )

attached is the remaining part of the solution

Answer:

Exit temperature = 32°C

Explanation:

We are given;

Initial Pressure;P1 = 100 KPa

Cp =1000 J/kg.K = 1 KJ/kg.k

R = 500 J/kg.K = 0.5 Kj/Kg.k

Initial temperature;T1 = 27°C = 273 + 27K = 300 K

volume flow rate;V' = 15 m³/s

W = 130 Kw

Q = 80 Kw

Using ideal gas equation,

PV' = m'RT

Where m' is mass flow rate.

Thus;making m' the subject, we have;

m' = PV'/RT

So at inlet,

m' = P1•V1'/(R•T1)

m' = (100 × 15)/(0.5 × 300)

m' = 10 kg/s

From steady flow energy equation, we know that;

m'•h1 + Q = m'h2 + W

Dividing through by m', we have;

h1 + Q/m' = h2 + W/m'

h = Cp•T

Thus,

Cp•T1 + Q/m' = Cp•T2 + W/m'

Plugging in the relevant values, we have;

(1*300) - (80/10) = (1*T2) - (130/10)

Q and M negative because heat is being lost.

300 - 8 + 13 = T2

T2 = 305 K = 305 - 273 °C = 32 °C

Answer:

In consolidated AB Co 300 shares.

Explanation:

Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.

Answer:

Explanation:

The value of a will be zero as it is provided that the particle is on the x-axis.

Calculate the velocity of particles along x-axis.

[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]

Substitute 0 for y.

[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]

Here,

[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]

Calculate the magnitude of vector V .

[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]

Substitute

[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]

[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]

The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]

Calculate the direction of flow.

[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]

Here, θ is the flow from positive x-axis in a counterclockwise direction.

Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.

[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]

The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

[tex]$ C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2) $[/tex]

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

[tex]$ T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$[/tex]

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

[tex]C_p = 520 \:\: J/kg.K[/tex]

So, the inlet temperature of argon is

[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$[/tex]

[tex]$ T_{in} = \frac{1}{2} \times 119 + 300$[/tex]

[tex]$ T_{in} = 360K $[/tex]

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

[tex]C_p = 5193 \:\: J/kg.K[/tex]

So, the inlet temperature of helium is

[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$[/tex]

[tex]$ T_{in} = \frac{1}{2} \times 12 + 300$[/tex]

[tex]$ T_{in} = 306K $[/tex]

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

[tex]C_p = 1039 \:\: J/kg.K[/tex]

So, the inlet temperature of nitrogen is

[tex]$ T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$[/tex]

[tex]$ T_{in} = \frac{1}{2} \times 60 + 300$[/tex]

[tex]$ T_{in} = 330K $[/tex]

Note: Answers are rounded to the nearest whole numbers.

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload)   ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

[tex]T_{min} =[/tex] 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, [tex]d_c = 0.5 mm[/tex]

Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]

minimum time required to reduce the depth of the plate by 20 mm:

[tex]T_{min} =[/tex] number of passes * Time/pass

[tex]T_{min} =[/tex] n * Time/pass

[tex]T_{min} =[/tex] 40 * 40

[tex]T_{min} =[/tex]  1600 = 26 mins 40 secs

Answer:

the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds

Explanation:

From the given information;

Assuming the tool moves 100 mm/sec

The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:

Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]

Number of passes = 20

We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times

However; the time per passes is;

Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]

where;

length L = 400mm

velocity of the feed is assumed as 100

Time/pass  [tex]=\dfrac{20*400}{100}[/tex]

Time/pass = 80 sec

Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:

[tex]T_{min} = Time/pass *number of passes[/tex]

[tex]T_{min} = 20*80[/tex]

[tex]T_{min} = 1600 \ sec[/tex]

[tex]T_{min}[/tex] = 26 minutes 40 seconds

Answer:

A.) 7.13 degree north east

B.) 806.23 km/h

C.) 7.44 hours

D.) 0.06 hours

Explanation:

Assume Washington is due east of San Francisco and Francisco with Washington in the positive x-direction

Also, the cross wind is blowing from north to south of 100 km/hr in y coordinate direction.

A.) Using Cartesian system of coordinates, the direction of flight for the plane to actually arrive in Washington can be calculated by using the formula

Tan Ø = y/x

Substitute y = 100 km/h and x = 800km/h

Tan Ø = 100/800

Tan Ø = 0.125

Ø = Tan^-1(0. 125)

Ø = 7.13 degrees north east.

Therefore, the direction of flight for the plane to actually arrive in Washington is 7.13 degree north east

B.) The speed in the San Francisco to Washington direction can be achieved by using pythagorean theorem

Speed = sqrt ( 800^2 + 100^2)

Speed = sqrt (650000)

Speed = 806.23 km/h

C.) Let us use the speed formula

Speed = distance / time

Substitute the speed and distance into the formula

806.23 = 6000/ time

Make Time the subject of formula

Time = 6000/806.23

Time = 7.44 hours

D.) If there is no cross wind,

Time = 6000/800

Time = 7.5 hour

Time difference = 7.5 - 7.44

Time difference = 0.06 hours

Answer:

Explanation:

Please

Explanation:

Inputs and Outputs:

There are 3 inputs = I₁, I₂, and S

There are 2 outputs = O₁ and O₂

The given problem is solved in three major steps:

Step 1: Construct the Truth Table

Step 2: Obtain the logic equations using Karnaugh map

Step 3: Draw the logic circuit

Step 1: Construct the Truth Table

The given logic is

When S = 0 then O₁ = I₁ and O₂ = I₂

When S = 1 then O₁ = I₂ and O₂ = I₁

I₁     |     I₂     |    S    |    O₁    |    O₂

0     |     0     |    0    |    0    |     0

0     |     0     |    1     |    0    |     0

0     |     1      |    0    |    0    |     1

0     |     1      |    1     |    1     |     0

1      |     0     |    0    |    1     |     0

1      |     0     |    1     |    0    |     1

1      |     1      |    0    |    1     |     1

1      |     1      |    1     |    1     |     1

Step 2: Obtain the logic equations using Karnaugh map

Please refer to the attached diagram where Karnaugh map is set up.

The minimal SOP representation for output O₁

[tex]$ O_1 = I_1 \bar{S} + I_2 S $[/tex]

The minimal SOP representation for output O₂

[tex]$ O_2 = I_2 \bar{S} + I_1 S $[/tex]

Step 3: Draw the logic circuit

Please refer to the attached diagram where the circuit has been drawn.

Answer:

The conversion in the real reactor is = 88%

Explanation:

conversion = 98% = 0.98

process rate = 0.03 m^3/s

length of reactor = 3 m

cross sectional area of reactor = 25 dm^2

pulse tracer test results on the reactor :

mean residence time ( tm) = 10 s and variance (∝2) = 65 s^2

note:  space time (t) =

t = [tex]\frac{A*L}{Vo}[/tex]   Vo = flow metric flow rate , L = length of reactor , A = cross sectional area of the reactor

therefore (t) = [tex]\frac{25*3*10^{-2} }{0.03}[/tex] = 25 s

since the reaction is in first order

X = 1 - [tex]e^{-kt}[/tex]

[tex]e^{-kt}[/tex] = 1 - X

kt = In [tex]\frac{1}{1-X}[/tex]

k = In [tex]\frac{1}{1-X}[/tex] / t  

X = 98% = 0.98 (conversion in PFR ) insert the value into the above equation then  

K = 0.156 [tex]s^{-1}[/tex]

Calculating Da for a closed vessel

; Da = tk

      = 25 * 0.156 = 3.9

calculate Peclet number Per using this equation

0.65 = [tex]\frac{2}{Per} - \frac{2}{Per^2} ( 1 - e^{-per})[/tex]

therefore

[tex]\frac{2}{Per} - \frac{2}{Per^2} (1 - e^{-per}) - 0.65 = 0[/tex]

solving the Non-linear equation above( Per = 1.5 )

Attached is the Remaining part of the solution

Answer:

The answer is option # 2. (Constraints).

The given question is incomplete. The complete question is as follows.

Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine

(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],

(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.

(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?

Explanation:

Using the property tables

  [tex]T_{1} = 500^{o}C[/tex],    [tex]P_{1}[/tex] = 10 bar

  [tex]v_{1} = 0.354 m^{3}/kg[/tex]

(a) During the process, specific volume remains constant.

  [tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]

  T = [tex](150 - 160)^{o}C[/tex]

Using inter-polation we get,

      T = [tex]154.71^{o}C[/tex]

The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].

(b) When the system will reach at state 3 according to the table at 0.5 bar then

  [tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]

  [tex]v_{g} = 3.24 m^{3} kg[/tex]

Let us assume "x" be the gravity if stream

   [tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]

   [tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]

               = [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]

               = 0.109

At state 3, the fraction of total mass condensed is as follows.

  [tex](1 - x_{5})[/tex] = 1 -  0.109

                = 0.891

The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.

(c) Hence, total mass of the system is calculated as follows.

     m = [tex]\frac{v}{v_{1}}[/tex]

         = [tex]\frac{1}{0.354}[/tex]

         = 2.825 kg

Therefore, at final state the total volume occupied by saturated liquid is as follows.

     [tex]v_{ws} = m \times v_{f}[/tex]

                 = [tex]2.825 \times 0.00103[/tex]

                 = [tex]2.9 \times 10^{-3} m^{3}[/tex]

The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].

KINDLY NOTE that there is a picture in the question. Check the picture below for the picture.

==================================

Answer:

(1). 1.2 metres.

(2). There is going to be the same pressure.

Explanation:

From the question above we can take hold of the statement Below because it is going to assist or help us in solving this particular Question or problem;

" Measurements indicate that the material of tank B will fail and the tank will burst if the air pressure in tank B exceeds 18 kPa."

=> Also, the density of oil = 930

That is if Pressure, P in B > 18kpa there will surely be a burst.

The height, h the can waste oil be poured into tank A is;

The maximum pressure  = height × acceleration due to gravity × density) + ( acceleration due to gravity × density × height, j).

18 × 10^3 = (height, h ×  10 × 930) + 10 × (2 - 1.25) × 1000.

When we make height, h the Subject of the formula then;

Approximately, Height, h = 1.2 metres.

(2). If air is accidentally trapped in the manometer line, what will be the error in the calculation of the height we will have the same pressure.

Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.

Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.

Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.

Thus, the correct option is 2.

Learn more about mathematical modelling

https://brainly.com/question/731147

#SPJ2

Answer:

height ≈ 60.60 m

Explanation:

The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.

Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.

Using tangential ratio,

tan 22° = opposite/adjacent

tan 22° = h/150

h = 150 × tan 22°

h = 150 × 0.40402622583

h = 60.6039338753

height ≈ 60.60 m

Answer:

Electricity cost = $9.36

Explanation:

Given:

Electric power = 400 W = 0.4 KW

Unit cost of electricity = $0.13/kWh

Overall time = 1/4 (30 days) (24 hours) = 180 hours

Find:

Electricity cost

Computation:

Electricity cost = Electric power  x Unit cost of electricity x Overall time

Electricity cost = 0.4 x $0.13 x 180

Electricity cost = $9.36

Given:

Electric power = 400 W = 0.4 KW

Over all Time  = 30(1/4) = 7.5 days

Unit cost of electricity = $0.13/kWh

Find:

Electricity cost.

Computation:

Electricity cost = Electric power x Unit cost of electricity x Over all Time

Electricity cost = 0.4 x 0.13 x 7.5

Electricity cost = $

Answer:

a. The buoyant force on the chamber is 220029.6 N

b. The tension in the cable is 7369.6 N

Explanation:

The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg

Given;

diameter of the sphere, d = 3.50 m

radius of the sphere, r = 1.75 mm = 1.75 m

mass of the chamber, m = 21700 kg

density of water, ρ = 1000 kg/m³

(a)

Buoyant force is the weight of water displaced, which is calculated as;

Fb = ρvg

where;

v is the volume of sphere, calculated as;

[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]

Fb = 1000 x 22.452 x 9.8

Fb = 220029.6 N

(b)

The tension in the cable will be calculated as;

T = Fb - mg

T = 220029.6 N - (21700 x 9.8)

T =  220029.6 N - 212660 N

T = 7369.6 N

Answer:

The correct answer to the following question will be "[tex]a_{x}[/tex] or [tex]a_{y}[/tex]".

Explanation:

So that the above is the appropriate choice.

Answer:

In this example, the delay procedure is given below in the explanation section

Explanation:

Solution

The delay procedure is given below:

LDS # $4000 // load initial memory

LDAA #$FF

STAA  DDRA

LDAA #$00 //load address

STAA DDRB

THERE LDAA PORT B

           ANDA   #%00000011// port A and port B

           CMPA   #%0000011

           BNE     THERE

           LDAA   #$FF

           STAA    PORT A

           JSR       DELAY

           LDAA    #$00

           STAA     PORT A

           BACK     BRA BACK

Answer:

(a)Tb = 330.12 K (b)Tc =304.73 K (c)19.81 K/m (d) h =60.65 W/m². K

Explanation:

Solution

Given that:

The mass flow rate of engine oil m = 1.81 x 10^-3 kg/s

Diameter of the tube, D = 1cm =0.01 m

Electrical heat rate, q =76 W/m

Wall Temperature, Ts = 370 K

Now,

From the properties table of engine oil we can deduce as follows:

thermal conductivity, k =0.139 W/m .K

Density, ρ = 854 kg/m³

Specific heat, cp = 2120 J/kg.K

(a) Thus

The wall heat flux is given as follows:

qs = q/πD

=76/π *0.01

= 2419.16 W/m²

Now

The oil mean temperature is given as follows:

Tb =Ts -11/24 (q.R/k) (R =D/2=0.01/2 = 0.005 m)

Tb =370 - 11/24 * (2419.16 * 0.005/0.139)

Tb = 330.12 K

(b) The center line temperature is given below:

Tc =Ts - 3/4 (qs.R/k)= 370 - 3/4 * ( 2419.16 * 0.005/0.139)

Tc =304.73 K

(c) The flow velocity is given as follows:

V = m/ρ (πR²)

Now,

The The axial gradient of the mean temperature is given below:

dTb/dx = 2 *qs/ρ *V*cp * R

=2 *qs/ρ*[m/ρ (πR²) *cp * R

=2 *qs/[m/(πR)*cp

dTb/dx = 2 * 2419.16/[1.81 x 10^-3/(π * 0.005)]* 2120

dTb/dx = 19.81 K/m

(d) The heat transfer coefficient is given below:

h =48/11 (k/D)

=48/11 (0.139/0.01)

h =60.65 W/m². K

Answer:

A.) Time = 13.75 seconds

B.) Total distance = 339 m

C.) V = 11.18 m/s

D.) V = 10.2 m/s

Explanation: Given that the automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone.

Then,

Initial velocity U of the motorist = 15.65m/s

acceleration a = - 3.05 m/s^2

Initial velocity u of the police man = 11.18 m/s

Acceleration a = 1.96 m/s^2

The police will overtake at distance S as the motorist decelerate and come to rest.

Where V = 0 and a = negative

While the police accelerate.

Using 2nd equation of motion for the motorist and the police

S = ut + 1/2at^2

Since the distance S covered will be the same, so

15.65t - 1/2×3.05t^2 = 11.18t +1/2×1.96t^2

Solve for t by collecting the like terms

15.56t - 1.525t^2 = 11.18t + 0.98t^2

15.56t - 11.18t = 0.98t^2 + 1.525t^2

4.38t = 2.505t^2

t = 4.38/2.505

t = 1.75 seconds approximately

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit.

Therefore, the total time required for the police car to overtake the automobile will be:

12 + 1.75 = 13.75 seconds

B.) Using the same formula

S = ut + 1/2at^2

Where S = total distance travelled

Substitutes t into the formula

S = 11.18(13.75) + 1/2 × 1.96 (13.75)^2

S = 153.725 + 185.28

S = 339 m approximately

C.) The speed of the police car at the time it overtakes the automobile will be constant = 11.18 m/s

D.) Using first equation of motion

V = U - at

Since the motorist is decelerating

V = 15.65 - 3.05 × 1.75

V = 15.65 - 5.338

V = 10.22 m/s

Therefore, the speed of the automobile at the time it was overtaken by the police car is 10.2 m/ s approximately

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

Answer:1 common law

Explanation:

It also says that if Cathy wasn’t wearing a hardhat hat she is responsible for her own injury, more than one answer is correct.

In this case, if Cathy's employer completes compliance and general duty requirements then the organization may not be held liable and again, the law can generally hold Cathy responsible for the injuries as she was not wearing the proper kits for such work.

According to OSHA, Cathy’s employer may not be held liable for her injury if it fulfilled compliance and general duty requirements.

You are entitled to a secure workplace. To stop workers from being murdered or suffering other types of harm at work, the Occupational Safety and Health Act of 1970 (OSH Act) was passed. According to the legislation, companies are required to give their workers safe working environments.

Therefore, more than one answer is correct.

Learn more about OSHA, here:

https://brainly.com/question/13127795

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Answer:

I have attached the diagram for this question below. Consult it for better understanding.

Find the cross sectional area AB:

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

Forces is given by:

F = 2.75 × 10³ N

Horizontal Stress can be found by:

σ (x) = F/A

σ (x) = 2.75 × 10³ / 19.2 × 10⁻⁶m

σ (x) = 143.23 × 10⁶ Pa

Horizontal Strain can be found by:

ε (x) = σ (x)/ E

ε (x) = 143.23 × 10⁶ / 200 × 10⁹

ε (x) = 716.15 × 10⁻⁶

Find Vertical Strain:

ε (y) = -v · ε (y)

ε (y) = -(0.3)(716.15 × 10⁻⁶)

ε (y) = -214.84 × 10⁻⁶

For L = 0.05m

Change (x) = L · ε (x)

Change (x) = 35.808 × 10⁻⁶m

For W = 0.012m

Change (y) = W · ε (y)

Change (y) = -2.5781 × 10⁻⁶m

For t= 0.0016m

Change (z) = t · ε (z)

where

ε (z) = ε (y) ,so

Change (z) = t · ε (y)

Change (z) = -343.74 × 10⁻⁹m

A = A(final) - A(initial)

A = -8.25 × 10⁻⁹m²

(Consult second picture given below for understanding how to calculate area)

The resulting change in the 50-mm gauge length; the width of portion AB of the test coupon; the thickness of portion AB; the cross- sectional area of portion AB are respectively; Δx = 35.808 × 10⁻⁶ m; Δy = -2.5781 × 10⁻⁶m; Δ_z = -343.74 × 10⁻⁹m; A = -8.25 × 10⁻⁹m²

Let us first find the cross sectional area of AB from the image attached;

A = (1.6mm)(12mm) = 19.2 mm² = 19.2 × 10⁻⁶m

We are given;

Tensile Load; F = 2.75 kN = 2.75 × 10³ N

Horizontal Stress is calculated from the formula;

σₓ = F/A

σₓ = (2.75 × 10³)/(19.2 × 10⁻⁶)m

σₓ = 143.23 × 10⁶ Pa

Horizontal Strain is calculated from;

εₓ = σₓ/E

We are given E = 200 GPa = 200 × 10⁹ Pa

Thus;

εₓ = (143.23 × 10⁶)/(200 × 10⁹)

εₓ = 716.15 × 10⁻⁶

Formula for Vertical Strain is;

ε_y = -ν * εₓ

We are given ν = 0.30. Thus;

ε_y  = -(0.3) * (716.15 × 10⁻⁶)

ε_y  = -214.84 × 10⁻⁶

A) We are given;

Gauge Length; L = 0.05m

Change in gauge length is gotten from;

Δx = L * εₓ

Δx = 0.05 × 716.15 × 10⁻⁶

Δx = 35.808 × 10⁻⁶ m

B) From the attached diagram, the width is;

W = 0.012m

Change in width is;

Δy = W * ε_y

Δy = 0.012 * -214.84 × 10⁻⁶

Δy = -2.5781 × 10⁻⁶m

C) We are given;

Thickness of plate; t = 1.6 mm = 0.0016m

Change in thickness;

Δ_z = t * ε_z

where;

ε_z = ε_y

Thus;

Δ_z = t * ε_y

Δ_z = 0.0016 * -214.84 × 10⁻⁶

Δ_z = -343.74 × 10⁻⁹m

D) The change in cross sectional area is gotten from;

ΔA = A_final - A_initial

From calculating the areas, we have;

A = -8.25 × 10⁻⁹ m²

Read more about stress and strain in steel plates at; https://brainly.com/question/1591712