In how many ways 2450 can be written as a product of 2 factors?
Select one:
O a. 9
O b. 6
C. 18
O d. 10

Answers

Answer 1

Answer:

search up factors of 2450 and divide by two

Answer 2

The number of ways 2450 can be written as a product of 2 factors is 9 ways

How to find factors of numbers?

Factors of 2,450 = 1, 2, 5, 7, 10, 14, 25, 35, 49, 50, 70, 98, 175, 245, 350, 490, 1225, 2450

2450 has 18 factors.

The number of ways 2450 can be written as a product of 2 factors = number of factors / 2

= 18/2

= 9 wats

This means, taking the product of two factors such as;

1 × 2450

2 × 1225

5 × 490

7 × 350

10 × 245

14 × 175

25 × 98

35 × 70

50 × 98

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Related Questions

Which of the following are major components of a Class II BSC: A. Cabinet blower switch B. Foot support C. Drain spillage trough (catch basin) D. Rear grille E. Temperature control

Answers

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 )

Assuming you determine the required section modulus of a wide flange beam is 200 in3, determine the lightest beam possible that will satisfy this condition.

Answers

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

Unit of rate of heat transfer

Answers

Answer:

The units on the rate of heat transfer are Joule/second, also known as a Watt.

Explanation:

Heat flow is calculated using the rock thermal conductivity multiplied by the temperature gradient. The standard units are mW/m2 = milli Watts per meter squared. Thus, think of a flat plane 1 meter by 1 meter and how much energy is transferred through that plane is the amount of heat flow.

hope it helps .

stay safe healthy and happy..

The rate of heat transfer is measured in Joules per second, also known as Watts.

What is heat transfer?

Heat transfer is a thermal engineering discipline that deals with the generation, use, conversion, and exchange of thermal energy between physical systems.

Heat transfer mechanisms include thermal conduction, thermal convection, thermal radiation, and energy transfer via phase changes.

The rate of heat transfer through a unit thickness of material per unit area per unit temperature difference is defined as thermal conductivity. Thermal conductivity varies with temperature and is measured experimentally.

Heat is typically transferred in a combination of these three types and occurs at random. Heat transfer rate is measured in Joules per second, also known as Watts.

Thus, Joules per second or watts is the unit of rate of heat transfer.

For more details regarding heat transfer, visit:

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In a tension test of steel, the ultimate load was 13,100 lb and the elongation was 0.52 in. The original diameter of the specimen was 0.50 in. and the gage length was 2.00 in. Calculate (a) the ultimate tensile stress (b) the ductility of the material in terms of percent elongation

Answers

Answer:

a) the ultimate tensile stress is 66717.8 psi

b) the ductility of the material in terms of percent elongation is 26%

Explanation:

Given the data in the question;

ultimate load P = 13,100 lb

elongation δl = 0.52 in

diameter of specimen d = 0.50 in

gage length l = 2.00 inch

First we determine the cross-sectional area of the specimen

A = [tex]\frac{\pi }{4}[/tex] × d²

we substitute

A = [tex]\frac{\pi }{4}[/tex] × ( 0.50 )²

A = 0.1963495 in²

a) the ultimate tensile stress σ[tex]_u[/tex]

tensile stress σ[tex]_u[/tex] = P / A

we substitute

tensile stress σ[tex]_u[/tex] = 13,100 / 0.1963495

tensile stress σ[tex]_u[/tex] = 66717.766 ≈ 66717.8 psi

Therefore, the ultimate tensile stress is 66717.8 psi

b) ductility of the material in terms of percent elongation;

percentage elongation of specimen = [change in length / original length]100

% = [ δl / l ]100

we substitute

% = [ 0.52 in / 2.00 in ]100

= [ 0.26 ]100

= 26

Therefore, the ductility of the material in terms of percent elongation is 26%

Which option identifies the best way to reduce the environmental impact in the following scenario?
Sheryl has been assessing a factory where cheese is made and packaged. She has determined that the
product's packaging is the component that causes the highest environmental impact.
Sheryl should implement steps to create a cheese that does not need to be packaged.
0 Sheryl should implement steps to educate the factory personnel on the subject of conservation.
O Sheryl should implement steps to optimize the packaging in an eco-friendly manner.
O Sheryl should implement steps to reduce costs in the packaging department.

Answers

Sheryl should implement steps to optimize the packaging in an eco-friendly manner.

The output side of an ideal transformer has 35 turns, and supplies 2.0 A to a 24-W device. Ifthe input is a standard wall outlet, calculate the number of turns on the input side, and the currentdrawn from the outlet.

Answers

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]

Problem 1. Network-Flow Programming (25pt) A given merchandise must be transported at a minimum total cost between two origins (supply) and two destinations (demand). Destination 1 and 2 demand 500 and 700 units of merchandise, respectively. At the origins, the available amounts of merchandise are 600 and 800 units. USPS charges $5 per unit from origin 1 to demand 1, and $7 per unit from origin 1 to demand 2. From origin 2 to demand 1 and 2, USPS charges the same unit cost, $10 per unit, however, after 200 units, the unit cost of transportation increases by 50% (only from origin 2 to demand 1 and 2).
a) Formulate this as a network-flow problem in terms of objective function and constraint(s) and solve using Excel Solver.
b) How many units of merchandise should be shipped on each route and what is total cost?

Answers

Solution :

Cost

Destination           Destination         Destination                     Maximum supply

Origin 1                       5                          7                                           600

Origin 2                     10                         10                                          800

                         15, for > 200            15, for > 200

         Demand          500                       700

Variables

Destination       1          2

Origin 1             [tex]$X_1$[/tex]        [tex]$$X_2[/tex]

Origin 2            [tex]$X_3$[/tex]        [tex]$$X_4[/tex]

Constraints   :   [tex]$X_1$[/tex], [tex]$$X_2[/tex], [tex]$X_3$[/tex], [tex]$$X_4[/tex]  ≥ 0

Supply : [tex]$X_1$[/tex] + [tex]$$X_2[/tex]  ≤ 600

              [tex]$X_3$[/tex] + [tex]$$X_4[/tex] ≤ 800

Demand : [tex]$X_1$[/tex] + [tex]$$X_3[/tex]  ≥ 500

              [tex]$X_2$[/tex] + [tex]$$X_4[/tex] ≥ 700

Objective function :

Min z = [tex]$5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$[/tex]

[tex]$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$[/tex]

Costs :

                  Destination 1       Destination  2

Origin 1         5                             7

Origin 2        10                           10

                     15                            15

Variables :

[tex]$X_1$[/tex]        [tex]$$X_2[/tex]

300    300  

200   400

[tex]$X_3$[/tex]      [tex]$$X_4[/tex]

Objective function : Min z = 10600

Constraints:

Supply    600 ≤ 600

                600 ≤ 800

Demand   500 ≥ 500

                 700 ≥ 500

Therefore, the total cost is 10,600.

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.

Answers

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]

1. A hydro facility operates with an elevation difference of 50 m and a flow rate of 500 m3/s. If the rotational speed is 90 RPM, find the most suitable type of turbine and estimate the power output of the arrangement

Answers

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]

Determine the complex power, apparent power, average power absorbed, reactive power, and power factor (including whether it is leading or lagging) for a load circuit whose voltage and current at its input terminals are given by:

Answers

Answer: hello your question is incomplete attached below is the missing detail  

answer :

Complex power = 2.5 ∠ 50°  VA

apparent power = 2.5 VA

average power = 1.6 Watts

reactive power = 1.915 Var

power factor = 0.64 ( leading )

Explanation:

i) complex power

P = Vrms *  Irms

  = 17.67∠40°  * 0.1414∠-10°

  = 2.5∠50° VA

ii) Apparent power

s = Vrms * Irms

  = 17.67 * 0.1414

  = 2.5 VA

iii) Average power absorbed

Absorbed power ( p )  = Vrms * Irms * cos∅

  = 17.67 * 0.1414  * cos ( 50 )

  = 1.6 watt

iv) Reactive power

P =  Vrms * Irms * sin∅

  = 17.67 * 0.1414  * sin ( 50 )

  = 1.915 VAR

v) power factor

P.F = cos ∅ = p /s

                   = 1.6 watt / 2.5 VA  = 0.64.

dentify the recommended practices when putting a tip on a micropipette. Select one or more: Gently push the micropipette into the tip and tap lightly to load the tip. Hold the micropipette at a 45 degree angle to the tip rack. Use the tip size designed for the micropipette size in use. Remove the tip from the rack and place it on micropipette by hand.

Answers

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.

4 An approach to a pretimed signal has 30 seconds of effective red, and D/D/1 queuing holds. The total delay at the approach is 83.33 veh-s/cycle and the saturation flow rate is 1000 veh/h. If the capacity of the approach equals the number of arrivals per cycle, determine the approach flow rate and cycle length.

Answers

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]

The term variation describes the degree to which an object or idea differs from others of the same type or from a standard.

a. True
b. False

Answers

The answer is true. Thank me later<3
The answer is true mainly because they function

True or false all workers who do class 1 asbestos work must be part of a medical surveillance program

Answers

Answer:

Yes

Explanation:

Answer:

true

Explanation:

hehehe

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7 m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a constant temperature of 100C, find:

Answers

Complete Question

Air at 40C flows over a 2 m long flat plate with a free stream velocity of 7m/s. Assume the width of the plate (into the paper) is 0.5 m. If the plate is at a co temperature of 100C,find:

The total heat transfer rate from the plate to the air

Answer:

[tex]q=1.7845[/tex]

Explanation:

From the question we are told that:

Air Temperature [tex]T_1=40c[/tex]

Length [tex]l=2m[/tex]

Velocity [tex]v=7m/s[/tex]

Width [tex]w=0.5[/tex]

Constant temperature [tex]T_t= 100C[/tex]

Generally the equation for Total heat Transfer is mathematically given by

 [tex]q=hA(T_s-T_\infty)[/tex]

Where

h=Convective heat transfer coefficient

 [tex]h=29.9075w/m^2k[/tex]

Therefore

 [tex]q=h(L*B)(T_s-T_\infty)[/tex]

 [tex]q=29.9075*(2*0.5)(100+273-(40+273))[/tex]

 [tex]q=1794.45w[/tex]

 [tex]q=1.7845[/tex]

Discuss typical advantages and disadvantages of an irrigation system?

Answers

Advantages include low costs and minimal labor.Water stays in the root zone, and foliage stays dry. Drawbacks to surface irrigation include potential overwatering and wasteful runoff.

A cylinder is internally pressurized to a pressure of 100 MPa. This causes tangential and axial stresses in the outer surface of 400 and 200 MPa, respectively. Make a Mohr circle representation of the stresses in the outer surface. What maximum normal and shear stresses are experienced by the outer surface?

Answers

Answer:

[tex]\mu_{max}=200Mpa[/tex]

Explanation:

From the question we are told that:

Internally pressurized [tex]P_i=100MPa[/tex]

Tangential Stress [tex]P_t=400mpa[/tex]

Axial stress [tex]P_a=200mpa[/tex]

Generally the equation for maximum normal and shear stresses are experienced by the outer surface is mathematically given by

 [tex]\mu_{max}=|\frac{P_t-P_a}{2}|,|\frac{P_t}{2}|,|\frac{P_t}{2}|[/tex]

Therefore

 [tex]\mu_{max}=|\frac{400-200}{2}|,|\frac{400}{2}|,|\frac{200}{2}|[/tex]

 [tex]\mu_{max}=200Mpa[/tex]

The National Weather Service has issued an alert for a severe storm that will bring 100 mm of rainfall in one hour. A farmer in the area is trying to decide whether to sand bag the creek that drains the 40 acres of row crops. The soil for the drainage area is a sandy clay loam and has a porosity of 0.398, effective porosity of 0.330, suction pressure of 52.3 cm, a hydraulic conductivity of 0.25 cm/hr and an effective saturation of 90%. Assuming that ponding occurs instantaneously, estimate the total depth of direct runoff in mm from the event using the Green-Ampt infiltration model.

a. 80
b. 89
c. 76
d. 72

Answers

I believe it’s 72 or 76
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