A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch W, what tension T in the cable will be required?
The 600-kg trunk has a center
of gravity at G. The felling notch at O is sufficiently large so that the resisting moment there is negligible. Enter your answer in (N) - numerical value only

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:

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

Answer:

219kJ

Explanation:

The work done (W) on a gas in an isothermal process is given by;

W = -P₁V₁ ln[tex]\frac{V_{2}}{V_1}[/tex]      -----------------(i)

Where;

P₁ = initial pressure of the gas

V₁ = initial volume of the gas

V₂ = final volume of the gas

From the question;

P₁ = 600kPa = 6 x 10⁵Pa

V₁ = 0.45m³

V₂ = 0.2m³

Substitute these values into equation (i) as follows;

W = -6 x 10⁵ x 0.45 x ln [tex]\frac{0.2}{0.45}[/tex]

W = -6 x 10⁵ x 0.45 x ln (0.444)

W = -6 x 10⁵ x 0.45 x -0.811

W = 2.19 x 10⁵

W = 219 x 10³

W = 219kJ

Therefore, the work done on the gas during the compression process is 219kJ

Answer is given below

Explanation:

Agreement is a official contract. It is written form or even in orally form. The Agreement can be written in formal or informal terms or we can use purely verbal language.

Agreement made between two or more party that allow the court to decide.  

The main 6 elements are:  

1. Offer

2. acceptance

3. consideration

4. intention to create legal relation

5. certainty

6. capacity  

1. The first elements of the contract herein are ABC Sanctions Inc., without the offer, as it is not valid under the Contract Act 1950, Contract Act.

2. Once the offer is made in the contract, acceptance must take place. The agreement must be approved by Northern Inc. When Northern Inc is clear with the offer, it will accept it once the terms and conditions of the agreement are clear.

3. Contrarification is the most important aspect of a contract, when considering a contract, the other person will give something in return. It is considered an exchange between ABC Construction Inc and Northern Inc.

4. It is necessary to have these elements in the contract. Contract law 1950 is one of the requirements of a valid contract, although there is silence about the need for a legal relationship.

5. Another important aspect of the contract is of course. The Contract Agreement sets out the terms and conditions that must be clearly understood by both ABC Contracts Inc and Northern Inc.

6. The ability of a contract to have a legal capacity on either side of the contract is more than eighteen years, since the age of 18 years is specified as the age at which the contract is entered into.

Answer:

Velocity = 4.73 m/s.

Explanation:

Work done by friction is;

W_f = frictional force × displacement

So; W_f = Ff * Δs = (μF_n)*Δs

where; magnitude of the normal force F_n is equal to the component of the weight perpendicular to the ramp i.e; F_n = mg*cos 24

Over the distance ab, Potential Energy change mgΔh transforms into a change in Kinetic energy and the work of friction, so;

mg(3 sin 24) = ΔKE1 + (0.22)*(mg cos 24) *(3).

Similarly, Over the distance bc, potential energy mg(2 sin 24) transforms to;

ΔKE2 + (0.16)(mg cos 24)(2).

Plugging in the relevant values, we have;

1.22mg = ΔKE1 + 0.603mg

ΔKE1 = 1.22mg - 0.603mg

ΔKE1 = 0.617mg

Also,

0.813mg = ΔKE2 + 0.292mg

ΔKE2 = 0.813mg - 0.292mg

ΔKE2 = 0.521mg

Now total increase in Kinetic Energy is ΔKE1 + ΔKE2

Thus,

Total increase in kinetic energy = 0.617mg + 0.521m = 1.138mg

Putting 9.81 for g to give;

Total increase in kinetic energy = 11.164m

Finally, if v = 0 m/s at point a, then at point c, KE = ½mv² = 11.164m

m cancels out to give; ½v² = 11.164

v² = 2 × 11.164

v² = 22.328

v = √22.328

v = 4.73 m/s.

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:

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:

Waterfall model

Explanation:

The waterfall model is amenable to the projects. It focused on the data structure. The software architecture and detail about the procedure. It will interfere with the procedure. It interfaces with the characterization of the objects. The waterfall model is the first model that is introduced first. This model also called a linear sequential life cycle model.

The waterfall model is very easy to use. This is the earliest approach of the SDLC.

There are different phase of the waterfall:

A) The linear density expressions for BCC [110] and [111] directions in terms of the atomic radius R are;

i) LD_110 = √3/(4R√2)

ii) LD_111 = 1/(2R)

B) The linear density values for these same two directions for iron (Fe) are;

i) LD_110 = 2.4 × 10^(9) m^(-1)

ii) LD_111 = 4 × 10^(9) m^(-1)

A) i) To find linear density expression for BCC 110, first of all we will calculate the length of the vector using the length of the unit cell which is 4R/√3 and the cell edge length which is 4R. Thus, the vector length can now be calculated from this expression;

√((4R)² - (4R/√3)²)

This reduces to; 4R√(1 - 1/3) = 4R√(2/3)

Now, the expression for the linear density of this direction is;

LD_110 =

Number of atoms centered on (110) direction/vector length of 110 direction

In this case, there is only one atom centered on the 110 direction. Thus;

LD_110 = 1/(4R√(2/3))

LD_110 = √3/(4R√2)

ii)  The length of the vector for the direction 111 is equal to 4R, since

all of the atoms whose centers the vector passes through touch one another. In addition, the vector passes through an equivalent of 2 complete atoms. Thus, the linear density is;

LD_111 = 2/(4R) = 1/(2R)

B)i) From tables, the atomic radius for iron is 0.124 nm or 0.124 x 10^(-9) m. Therefore, the linear density for the [110] direction is;

LD_110 = √3/(4R√2) = √3/(4*0.124*10^(-9)(√2))

LD_110 = 2.4 × 10^(9) m^(-1)

ii) for the 111 direction, we have;

LD_111 = 1/(2R) = 1/(2*0.124*10^(-9))

LD_111 = 4 × 10^(9) m^(-1)

Read more about Linear Density of Crystalline Structures at; https://brainly.com/question/14831455

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:

a) 2.45 KN

b) 0.0375 m

Explanation:

[tex](a) \quad \sigma_{v}=400 \times 10^{6} \mathrm{Pa} \quad A=\frac{\pi}{4} d^{2}=\frac{\pi}{4}(5)^{2}=19.635 \mathrm{mm}^{2}=19.635 \times 10^{-6} \mathrm{m}^{2}[/tex]

[tex]P_{U}=\sigma_{U} A=\left(400 \times 10^{6}\right)\left(19.635 \times 10^{-6}\right)=7854 \mathrm{N}[/tex]

[tex]P_{\text {al }}=\frac{P_{U}}{F S}=\frac{7854}{3.2}=2454 \mathrm{N}[/tex]

(b) [tex]\quad \delta=\frac{P L}{A E}=\frac{(2454)(60)}{\left(19.635 \times 10^{-6}\right)\left(200 \times 10^{9}\right)}=37.5 \times 10^{-3} \mathrm{m}[/tex]

Answer:

hello the required diagram is missing attached to the answer is the required diagram

7.9954 kip.ft

Explanation:

AB = 1550-Ib ( weight acting on AB )

BCD = 190 - Ib ( weight of cage )

169-Ib = weight of man inside cage

Attached is the free hand diagram of the question

calculate distance [tex]x![/tex]

= cos 75⁰ = [tex]\frac{x^!}{10ft}[/tex]

    [tex]x! = 10 * cos 75^{o}[/tex] = 2.59 ft

calculate distance x

= cos 75⁰ = [tex]\frac{x}{30ft}[/tex]

x = 30 * cos 75⁰ = 7.765 ft

The resultant moment  produced by all the weights about point A

∑ Ma = 0

Ma = 1550 * [tex]x![/tex] + 190 ( x + 2.5 ) + 169 ( x + 2.5 + 1.75 )

Ma = 1550 * 2.59 + 190 ( 7.765 + 2.5 ) + 169 ( 7.765 + 2.5 + 1.75 )

      = 4014.5 + 1950.35 + 2030.535

      = 7995.385 ft. Ib ≈ 7.9954 kip.ft

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:

the  pressure drop  is 0.21159 atm

Explanation:

Given that:

length of the reactor L = 2.5 m

inside diameter of the reactor d= 0.025 m

diameter of alumina sphere [tex]dp[/tex]= 0.003 m

particle density  = 1300 kg/m³

the bed void fraction [tex]\in =[/tex]  0.38

superficial mass flux m = 4684 kg/m²hr

The Feed is  methane with pressure P = 5 bar and temperature T = 400 K

Density of the methane gas [tex]\rho[/tex] = 0.15 mol/dm ⁻³

viscosity of methane gas [tex]\mu[/tex] = 1.429  x 10⁻⁵ Pas

The objective is to determine the pressure drop.

Let first convert the Density of the methane gas from 0.15 mol/dm ⁻³  to kg/m³

SO; we have :

Density =  0.15 mol/dm ⁻³  

Molar mass of methane gas (CH₄) = (12 + (1×4) ) = 16 mol

Density =  [tex]0.1 5 *\dfrac{16}{0.1^3}[/tex]

Density =  2400

Density [tex]\rho_f[/tex] =  2.4 kg/m³

Density = mass /volume

Thus;

Volume = mass/density

Volume of the methane gas =  4684 kg/m²hr / 2.4 kg/m³

Volume of the methane gas = 1951.666 m/hr

To m/sec; we have :

Volume of the methane gas = 1951.666 * 1/3600 m/sec = 0.542130 m/sec

[tex]Re = \dfrac{dV \rho}{\mu}[/tex]

[tex]Re = \dfrac{0.025*0.5421430*2.4}{1.429*10^5}[/tex]

[tex]Re=2276.317705[/tex]

For Re > 1000

[tex]\dfrac{\Delta P}{L}=\dfrac{1.75 \rho_f(1- \in)v_o}{\phi_sdp \in^3}[/tex]

[tex]\dfrac{\Delta P}{2.5}=\dfrac{(1.75 *2.4)(1- 0.38)*0.542130}{1*0.003 (0.38)^3}[/tex]

[tex]\Delta P=8575.755212*2.5[/tex]

[tex]\Delta = 21439.38803 \ Pa[/tex]

To atm ; we have

[tex]\Delta P = \dfrac{21439.38803 }{101325}[/tex]

[tex]\Delta P =0.2115903087 \ atm[/tex]

ΔP  ≅  0.21159 atm

Thus; the  pressure drop  is 0.21159 atm

Answer:

No, water can't be used.

Explanation:

No, water cannot be used as the working fluid in air-conditioning applications.

This is because, if we assume the water is maintained at 10°C in the evaporator, the evaporator pressure will now be the saturation pressure that corresponds to this pressure, which in this case would be 1.2 kPa.

So we can conclude that for the refrigerants in the evaporator the temperature of a saturated pressure would be very low and so it's not practical to maintain it with water

Thus, it's is not practical to design refrigeration or air conditioning devices with water as the working fluid because it will involve extremely low pressures.

Answer:

There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.

Explanation:

The masses of silicon and germanium contained in a cubic centimeter of the germanium-silicon alloy by apply the concepts of mass ([tex]m[/tex]), density ([tex]\rho[/tex]) and volume ([tex]V[/tex]), as well as the mass-mass proportion of Germanium ([tex]x[/tex]):

[tex]m_{Ge} = x \cdot \rho_{Ge}\cdot V_{sample}[/tex]

[tex]m_{Ge} = 0.15\cdot \left(5.32\,\frac{g}{cm^{3}} \right)\cdot (1\,cm^{3})[/tex]

[tex]m_{Ge} = 0.798\,g[/tex]

The amount of moles of Germanium is obtained after dividing previous outcome by its atomic weight. That is to say:

[tex]n = \frac{m_{Ge}}{M_{Ge}}[/tex]

[tex]n = \frac{0.798\,g}{72.64\,\frac{g}{mol} }[/tex]

[tex]n = 0.011\,mol[/tex]

There are 0.011 moles in a cubic centimeter of the germanium-silicon alloy. According to the Law of Avogadro, there are [tex]6.022 \times 10^{23}\,atoms[/tex] in a mole of Germanium. The quantity of atoms in a cubic centimeter is therefore found by simple rule of three:

[tex]y = \frac{0.011\,mol}{1\,mol}\times \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)[/tex]

[tex]y = 6.624 \times 10^{21}\,atoms[/tex]

There are [tex]6.624 \times 10^{21}\,atoms[/tex] of Germanium in a germanium-silicon alloy that contains 15 wt% Ge and 85 wt% Si.

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:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

[tex]dQ = m \times c_p \times (T_2 -T_1)[/tex]

For ideal gas, [tex]c_p[/tex] = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

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

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.

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

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Answer:

Explanation:

Please

Answer:

Explanation:

The formula for critical stress is

[tex]\sigma_c=\frac{K}{Y\sqrt{\pi a} }[/tex]

[tex]\sigma_c =\texttt{critical stress}[/tex]

K is the plane strain fracture toughness

Y is dimensionless parameters

We are to Determine the Critical stress

Now replacing the critical stress with 54.8

a with 0.2mm = 0.2 x 10⁻³

Y with 1

[tex]\sigma_c=\frac{54.8}{1\sqrt{\pi \times 0.2\times10^{-3}} } \\\\=\frac{54.8}{\sqrt{6.283\times10^{-4}} } \\\\=\frac{54.8}{0.025} \\\\=2186.20Mpa[/tex]

The fracture will not occur because this material can handle a stress of 2186.20Mpa  before fracture. it is obvious that is greater than 2023Mpa

Therefore, the specimen does not failure for surface crack of 0.2mm

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:

The answer is option # 2. (Constraints).

Answer is given below

Explanation:

Answer:

Fifty (50) percent. [50%]

Explanation:

Water heater is a home appliance that comprises of an electric or gas heating unit as well as a water-tank where water is heated and stored for use.

When using an alternative method of sizing with two vent connectors for draft hood-equipped water heaters, the effective area of the common vent connector or vent manifold and all junction fittings shall not be less than the area of the larger vent connector plus fifty (50) percent of the areas of smaller flue collar outlets.

A water heater is primarily vented with an approved and standardized plastic or metallic pipe such as flue or chimney, which allows gas to flow out of the water heater into the surrounding environment.

For a draft hood-equipped water heater, both the water heater and the barometric draft regulators must be installed in the same room. Also, the technician should ensure that the vent is through a concealed space such as conduit and should be labeled as Type L or Type B.

The minimum capacity of a water heater should be calculated based on the number of bathrooms, bedrooms and its first hour rating.

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:

The exit temperature of the gas = 32° C

Explanation:

Solution

Given that:

Inlet temperature T₁ = 27°C ≈ 300.15 K

Inlet pressure P₁ = 100 KPa = 100 * 10^3 Pa

Volume flow rate , V = 15 m/s³

Diameter of the deduct, D = 500 mm = 0.5 m

Electric heater power, W heater = 130 kW = 130 * 10^3 W

The heat lost Q = 80 kW =  80 * 10^3 W

Now,

From the ideal gas law, density of the air at the inlet is given as :

ρ₁ = P₁/RT₁ = 100 * 10^3/500 * 300

=0.6667 kg/m³

The mass flow rate through the duct is computed below:

m = ρ₁ V = 0.6667 * 15 = 10 kg/s

Thus

Applying the first law of thermodynamics to the process is shown below:

Q + m (h₁ + V₁²/2 + gz₁) = m (h₂ + V₂²/2 + gz₂) + W (Conservation energy)

So,

If we neglect the potential and kinetic energy changes of the air, the above equation can be written again as:

Q + m (h₁) = m (h₂) + W

or

Q - W heater =m (h₂ - h₁) or Q - W heater =m (T₂ - T₁)

Thus

h₂ - h₁ = Cp T₂ - T₁

Now by method of substitution the known values are:

(- 80 *10^3) - (-130 * 10^3) = 10 * 100 * (T₂ -27)

Note: The heat transfer is  taken as negative because the heat is lost by the gas and work done is also taken as negative because the work is done on the gas

So,

Solving for T₂,

T₂ = 32° C

Therefore the exit temperature of the gas = 32° C