How do Machine Learning and Artificial Intelligence (AI) technologies help businesses use their enterprise data effectively?

Answer:

Following are the responses to the given question:

Explanation:

Effective red duration is applied each cycle r=30 second D/D/1 queuing

In total, its approach delay is 83.33 sec vehicle per cycle

Flow rate(s) of saturated = 1,000 vehicles each hour

Total vehicle delay per cycle[tex]= \frac{v \times 30^2}{2(1-\frac{v}{0.2778})}[/tex]

[tex]\to \frac{v\times 30^2}{2(1-\frac{v}{0.2778})}= 83.33\\\\\to 900v=166.66-599.928v\\\\\to v=0.111 \frac{veh}{sec}\\\\[/tex]

The flow rate for such total approach is 0.111 per second.

The overall flow velocity of the approach is 400 cars per hour

The approach capacity refers to the number of arrivals per cycle.

Environmentally friendly time ratio to cycle length:

[tex],\frac{g}{C} \ is = \frac{400}{1000}=0.4\\\\r= c-g\\\\30\ sec =C - 0.4 C\\\\C=50 \ sec[/tex]

Answer:

It would break I think need to try it out

Explanation:

Answer:

Gently push the micropipette into the tip box and tag tightly to load the tip.

Explanation:

The recommended practice when putting a tip on a micropipette is ;  Gently push the micropipette into the tip box and tag tightly to load the tip.

Given that it is not advisable to remove tip from rack so as not to contaminate it, if we want to put a tip on a micropipette we should gently push the micropipette into the tip box.

Answer:

Opened Push-button Switch (i.e. a PTM Switch)

Explanation:

Tha's just another symbol for a switch, but this one specifies that the switch is a push-button type of switch.

Since it's not touching and completing the line, the state of the switch is initially open.

Answer:

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Answer:

Geography is the science that studies and describes the surface of the Earth in its physical, current and natural aspect, or as a place inhabited by humanity.

Answer:

Resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis.

Explanation:

Pulling the strain gauge in the long axis causes the wires to elongate/thin, the effect of this is that it will lead to an increase in resistance and voltage drop (V = I*R).

As a result of the resultant effect, resistance and voltage drop will still continue to rise, although at a slower pace than on the desired axis, such as the long axis.

Answer:

[tex]Hr=4.2*10^7\ btu/hr[/tex]

Explanation:

From the question we are told that:

Water flow Rate [tex]R=4.5slug/s=144.78ib/sec[/tex]

Initial Temperature [tex]T_1=60 \textdegree F[/tex]

Final Temperature  [tex]T_2=140 \textdegree F[/tex]

Let

Specific heat of water [tex]\gamma= 1[/tex]

And

 [tex]\triangle T= 140-60[/tex]

 [tex]\triangle T= 80\ Deg.F[/tex]

Generally the equation for Heat transfer rate of water  [tex]H_r[/tex] is mathematically given by

Heat transfer rate to water= mass flow rate* specific heat* change in temperature

 [tex]H_r=R* \gamma*\triangle T[/tex]

 [tex]H_r=144.78*80*1[/tex]

 [tex]H_r=11582.4\ btu/sec[/tex]

Therefore

 [tex]H_r=11582.4\ btu/sec*3600[/tex]

 [tex]Hr=4.2*10^7\ btu/hr[/tex]

Answer:

Step On: Your foot forces the clutch pedal down and then causes it to take up the slack. This, in turn, causes the clutch friction disk to slip, creating heat and ultimately wearing your clutch out.

Step Off: When the clutch pedal is released, the springs of the pressure plate push the slave cylinder's pushrod back, which forces the hydraulic fluid back into the master cylinder.

This question is incomplete, the complete question is;

An air-standard Diesel cycle engine operates as follows: The temperatures at the beginning and end of the compression stroke are 30 °C and 700 °C, respectively. The net work per cycle is 590.1 kJ/kg, and the heat transfer input per cycle is 925 kJ/kg. Determine the a) compression ratio, b) maximum temperature of the cycle, and c) the cutoff ratio, v3/v2.

Use the cold air standard assumptions.

Answer:

a) The compression ratio is 18.48

b) The maximum temperature of the cycle is 1893.4 K

c) The cutoff ratio, v₃/v₂ is 1.946

Explanation:

Given the data in the question;

Temperature at the start of a compression T₁ = 30°C = (30 + 273) = 303 K

Temperature at the end of a compression T₂ = 700°C = (700 + 273) = 973 K

Net work per cycle [tex]W_{net[/tex] = 590.1 kJ/kg

Heat transfer input per cycle Qs = 925 kJ/kg

a) compression ratio;

As illustrated in the diagram below, 1 - 2 is adiabatic compression;

so,

Tγ[tex]^{Y-1[/tex] = constant { For Air, γ = 1.4 }

hence;

⇒ V₁ / V₂ = [tex]([/tex] T₂ / T₁ [tex])^{\frac{1}{Y-1}[/tex]

so we substitute

⇒ V₁ / V₂ = [tex]([/tex]  973 K / 303 K  [tex])^{\frac{1}{1.4-1}[/tex]

= [tex]([/tex]  3.21122  [tex])^{\frac{1}{0.4}[/tex]

= 18.4788 ≈ 18.48

Therefore, The compression ratio is 18.48

b) maximum temperature of the cycle

We know that for Air, Cp = 1.005 kJ/kgK

Now,

Heat transfer input per cycle Qs = Cp( T₃ - T₂ )

we substitute

925 = 1.005( T₃ - 700 )

( T₃ - 700 ) = 925 / 1.005

( T₃ - 700 ) = 920.398

T₃ = 920.398 + 700

T₃ = 1620.398 °C

T₃ = ( 1620.398 + 273 ) K

T₃ = 1893.396 K ≈ 1893.4 K

Therefore, The maximum temperature of the cycle is 1893.4 K

c)  the cutoff ratio, v₃/v₂;

Since pressure is constant, V ∝ T

So,

cutoff ratio S = v₃ / v₂  = T₃ / T₂

we substitute

cutoff ratio S = 1893.396 K / 973 K

cutoff ratio S = 1.9459 ≈ 1.946

Therefore, the cutoff ratio, v₃/v₂ is 1.946

Answer:

The current drawn from the outlet is 0.2 A

The number of turns on the input side is 350 turns

Explanation:

Given;

number of turns of the secondary coil, Ns = 35 turns

the output current, [tex]I_s[/tex] = 2 A

power supplied, [tex]P_s[/tex] = 24 W

the standard wall outlet in most homes = 120 V = input voltage

For an ideal transformer; output power = input power

the current drawn from the outlet is calculated;

[tex]I_pV_p = P_s\\\\I_p = \frac{P_s}{V_p} = \frac{24}{120} = 0.2 \ A[/tex]

The number of turns on the input side is calculated as;

[tex]\frac{N_p}{N_s} = \frac{I_s}{I_p} \\\\N_p = \frac{N_sI_s}{I_p} \\\\N_p = \frac{35 \times 2}{0.2} \\\\N_p = 350 \ turns[/tex]

Answer:

a. 6.48 kW b. 1678.34 kg c. 777.92 W d. 201.42 kg

Explanation:

(a) the rate of heat transfer to the iced water in the tank

The rate of heat transfer to the outer surface of the spherical tank is P = P₁ + P₂ where P₁ = rate of heat transfer to the outer surface by radiation and P₂ = rate of heat transfer to the outer surface by convection through air

P₁ = εσAT⁴ where ε = emissivity of outer surface ε= 1, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of outer surface of spherical tank = 4πR² where R = outer radius of spherical tank = inner radius + thickness = inner diameter/2 + 5 cm = 2 m/2 + 0.05 m = 1 m + 0.05 m = 1.05 m and T = temperature of surroundings = 20 °C = 273 + 20 = 293 K.

P₁ = εσAT⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.05 m)² × (293 K)⁴

P₁ = 1 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.1025 m²) × 7370050801 K⁴

P₁ = 184285909263.7647π × 10⁻⁸ W

P₁ = 578951258703.16 × 10⁻⁸ W

P₁ = 5789.51 W

P₂ = hA(T - T₁) where h = coefficient of thermal convection of air = 2.5 W/m²-K, A = outer surface area of spherical tank = 4πR², T = temperature of surroundings = 20 °C = 273 + 20 = 293 K and T₁ = temperature of outer surface of spherical tank = 0 °C = 273 + 0 = 273 K.  

P₂ = hA(T - T₁)

P₂ = 2.5 W/m²-K × 4π(1.05 m)² × (293 K - 273 K)

P₂ = 2.5 W/m²-K × 4π(1.1025 m²) × 20 K

P₂ = 220.5π W

P₂ = 692.72 W

So, P = P₁ + P₂ = 5789.51 W + 692.72 W = 6482.23 W

Since we are neglecting the thermal resistance of the spherical tank, the rate of heat absorption of the outer surface equals the rate of heat absorption in the inner surface. The rate of heat absorption at the inner surface equals the rate of heat transfer to the iced water.

So, rate of heat transfer to the iced water = P = 6482.23 W = 6.48223 kW 6.48 kW

(b) the amount of ice at 0°C that melts during a 24-h period. The heat of fusion of water is 333.7 kJ/kg.

Since the amount of heat, Q = Pt where P = heat transfer rate to iced water = 6482.23 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, Pt = mL

m = Pt/L

= 6482.23 W × 86400 s/333.7 × 10³ J/kg

= 560064672/333.7 × 10³

= 1678.34 kg

(c) the rate of heat transfer to the iced water in the tank for this double-walled tank configuration

Since P is the rate of heat transfer to the outer surface, this is also the rate of heat transfer to the outer 0.5 cm thick wall = P₃ = 6482.23 W

P₃ = kA(T - T₃)/d where k = thermal conductivity of outer wall = 15 W/m²-K

A = surface area of outer wall = 4πR'² where R' = radius of outer wall = radius of inner wall + thickness of inner wall + thickness of vacuum + thickness of outer wall = 2.0 m/2 + 0.5 cm + 1.5 cm + 0.5 cm = 1 m + 2.5 cm = 1 m + 0.025 m = 1.025 m, T = temperature of surroundings = 20 °C = 273 + 20 = 293 K, T₃ = temperature of inner surface of outer wall of spherical tank and d = thickness of outer surface of tank = 0.5 cm = 0.05 m

P₃ = kA(T - T₃)/d

making T₃ subject of the formula, we have

P₃d = kA(T - T₃)

P₃d/kA = (T - T₃)

T₃ = T - P₃d/kA

substituting the values of the variables into the equation, we have

T₃ = 293 K - 6482.23 W × 0.05 m/[15 W/m-K × 4π(1.025 m)²]

T₃ = 293 K - 324.1115 Wm/[15 W/m-K × 4π(1.050625 m²)]

T₃ = 293 K - 324.1115 Wm/[63.0375π W/m-K)]

T₃ = 293 K - 324.1115 Wm/[198.0381 W/m-K)]

T₃ = 293 K - 1.64 K

T₃ = 291.36 K

Since the 1.5 cm thick air space is evacuated, all the heat gets to the inner 0.5 cm thick wall by radiation.

So P = εσAT₃⁴

P₄ = εσAT₃⁴ where ε = emissivity of outer surface ε = 0.15, σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m-K⁴, A = area of inner surface of outer wall of spherical tank = 4πR"² where R" = outer radius of inner thick wall of spherical tank = inner radius + thickness of inner wall = inner diameter/2 + 0.5 cm = 2 m/2 + 0.005 m = 1 m + 0.005 m = 1.005 m and T = temperature of outer wall = 291.36 K.

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.005 m)² × (291.36 K)⁴

P₄ = 0.15 × 5.67 × 10⁻⁸ W/m-K⁴ × 4π(1.010025 m²) × 7206422389.51 K⁴

P₄ = 24762024365.028π × 10⁻⁸ W

P₄ = 77792193833.18 × 10⁻⁸ W

P₄ = 777.92 W

Now P₄ is the heat transfer rate to the inner surface which is at temperature T₄

Since T₄ = 0 °C, P₄ is the rate of heat transfer to the iced water

So, rate of heat transfer to the iced water P₄ = 777.92 W

(d) the amount of ice at 0°C that melts during a 24-h period for this double-walled tank configuration

Since the amount of heat, Q = P₄t where P₄ = heat transfer rate to iced water = 777.92 W and t = time = 24 h = 24 h × 60 min/h × 60 s/min = 86400 s.

Also, Q = the latent heat required to melt the ice at 0 °C = mL where m = mass of ice melted and L = latent heat of fusion of ice = 333.7 kJ/kg

So, P₄t = mL

m = P₄t/L

= 777.92 W × 86400 s/333.7 × 10³ J/kg

= 67212288/333.7 × 10³

= 201.42 kg

Answer:

a) Pelton Turbine

b) [tex]P=2.42*10^{5}KW[/tex]

Explanation:

From the question we are told that:

Height [tex]h=50[/tex]

Flow Rate [tex]R= 500 m^3/s[/tex]

Rotational speed [tex]\omega=\90 RPM[/tex]

Let

Density of water

[tex]\rho=1000[/tex]

Generally the equation for momentum is mathematically given by

[tex]P=\rho gRh[/tex]

[tex]P=1000*9.81*500*50[/tex]

[tex]P=2.42*10^{5}KW[/tex]

Answer:

W18 * 106

Explanation:

Given that the section modulus of the wide flange beam is 200 in^3 the lightest beam possible that can satisfy the section modulus must have a section modulus ≥ 200 in^3. also the value of the section modulus must be approximately closest to 200in^3

From wide flange Beam table ( showing the section modulus )

The beam that can satisfy the condition is W18 × 106  because its section modulus ( s ) = 204 in^3

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Answer:

Cabinet blower switch ( A )

Explanation:

A major component of a class II BSC ( Biological safety cabinet ) is  Cabinet blower switch  because the Cabinet blower is an integral part of a class II BSC hence the switch is also a major component.

Class II BSC provides protection for the user, environment and sample to be manipulated in the laboratory ( mostly ; Pharmaceutical laboratories, Microbiology laboratories )

Answer:

100 kWh

Explanation:

Since the freezer has a rating of 500 watts and runs for 200 hours in a month, the energy consumption can be gotten by getting the product of the rating of the freezer in kilowatts and the amount of time the fridge is on per month.

The rating of the freezer = 500 watts = 0.5 kW, time = 200 h

Energy consumption = rating * time = 0.5 kW * 200 h

Energy consumption = 100 kWh

Therefore the refrigerator uses 100 kWh per month

Answer:

6.33 mm

Explanation:

Stress is an internally resistive force produced by the molecules of the object to resist the deformation when an amount of load acts on the object. The SI unit of stress is measured in Pascal.

Given that force (F) = 13300 N, factor of safety (λ) = 2, yield strength (σy) = 860 MPa

The working stress [tex](\sigma_w)=\frac{\sigma_y}{\lambda}=\frac{860\ MPa}{2}=430\ MPa[/tex]

Let d be the minimum diameter, hence it is calculated using:

[tex]\sigma_w=\frac{F}{A}\\\\430 *10^6=\frac{13300}{A}\\\\A=30.93*10^{-6}\ m^2\\\\A=area=\frac{\pi d^2}{4} \\\\30.93*10^{-6}\ m^2=\frac{\pi d^2}{4} \\\\d=\sqrt{\frac{4*30.93*10^{-6}}{\pi } } \\\\d=0.0063\ m=6.3\ mm[/tex]

Answer:

answer 2

Explanation:

In an ideal transformer, the ratio of input voltage to output voltage is equal to the ratio of the number of turns in primary coil to number of turns in the secondary coil. Therefore, the secondary current in the given case is 1.5 A.

Secondary current refers to the electric current that flows in the secondary winding of a transformer. A transformer is a device that transfers electrical energy from one circuit to another by means of electromagnetic induction.

It consists of two or more coils of insulated wire, called windings, that are wound around a common magnetic core. In a transformer, an alternating current (AC) flows through the primary winding, which produces a magnetic field that induces a voltage in the secondary winding.

The secondary current in the above given case is 1.5 A.

Learn more about secondary current here:

https://brainly.com/question/18763122

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Complete Question

Analyze the boundary work done during the process having a rigid tank contains air at 500 kPa and 150°C. As a result of heat transfer to the surroundings, the temperature and pressure inside the tank drop to 65°C and 400 kPa, respectively.

Determine the boundary work done during this process and heat Lose

Answer:

a)  [tex]W=0[/tex]

b)  [tex]dQ=-61.03KJ/kg[/tex]

Explanation:

From the question we are told that:

Pressure of air [tex]P_1=500kpa[/tex]

Temperature of Air [tex]T_2=150°C[/tex]

Pressure drop [tex]P_2=400kpa[/tex]

Temperature of drop [tex]T_2=65 \textdegree C[/tex]

Generally the Constant Volume Process  is mathematically given by

 [tex]V_1=V_2=V[/tex]

Therefore

a)

Generally the equation for  boundary work w is mathematically given by

 [tex]W=pdv[/tex]

 [tex]W=P(V_2-V_1)[/tex]

 [tex]W=P(V_V)[/tex]

 [tex]W=0KJ[/tex]

b)

Generally the equation for Heat Change is mathematically given by

 [tex]dQ=dU+dW[/tex]

 [tex]dQ=dU[/tex]

 [tex]dQ=C_v(T_2-T_1)[/tex]

Where

   C_v=Specific Heat capacity of Air

  [tex]C_v=0.718 kJ/kg K[/tex]

 [tex]dQ=0.718(338-423)[/tex]

 [tex]dQ=-61.03KJ/kg[/tex]

Answer:

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Explanation:

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Answer:

Dark shadow

Explanation:

Shadow is nothing but space when the light is blocked by an opaque object. It is just that part where light does not reach. When you stand in the sun, you are able to see your shadow behind you. ... This is because our body is opaque and does not allow the light to pass through it

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