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 1

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.


Related Questions

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%

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

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

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.

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.

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