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3 years ago

Which of the following sentences uses the active voice

Engineering

1 answer:

MrRa [10]3 years ago

7 0

Answer: Last week, Nate and I counted all the inventory.

Explanation: all other choices are passive voices

this sentence follows a clear subject + verb + object construct that's why it is an active voice. In fact, sentences constructed in the active voice add impact to your writing. but on the other hand With passive voice, the subject is acted upon by the verb.

Ape x

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The water in a large lake is to be used to generate electricity by the installation of a hydraulic turbine-generator at a locati

Lynna [10]

Answer:

a) 0.76

b) 0.80

c) 1964 kW

Explanation:

GIVEN DATA:

$\dot m = 5000 kg/s$

Assume Mechanical energy at exist is negligible

A) Take lake bottom as reference, and then kinetic and potential energy are taken as zero.

change in mechanical energy is givrn as

$e_{in} - e_{out} = \frac{P}{\rho} - 0 = gh = 9.81 \times 50( \frac{1 kJ/kg}{1000 m^2/s^2}$

= 0.491 kJ/kg

$\Delta \dot E_{mec} = \dot m (e_{in} - e_{out}) = 5000 \times 0.491 = 2455 kW$

$\eta_{OVERALL} = \frac{\dot W}{\Delta \dot E_{mec}} = \frac{1862}{2455} = 0.76$

B) $\eta -{gen} = \frac{\eta_{overall}}{\eta_{gen}} = \frac{0.76}{0.95} = 0.80$

c) $\dot W_{shaft} = \eta_{overall} \left | \Delta \dot E_{mec} \right | = 0.80(2455)$

$\dot W_{shaft} = 1964 kW$

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4 years ago

A BOD test is run using 30 mL of wastewater and 270 mL of dilution water. The initial DO of the mixture is 9.0 mg/L. After 5 day

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3 years ago

11) 6x^2-49x - 45<br> What is the factored answer to this trinomial

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Factores is (6x+5)(x-9)

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3 years ago

A steel cylindrical sample was subjected to a tensile test. The yield load was 2100N. The maximum load was 3400N and the failure

Misha Larkins [42]

Answer:

initial diameter of the sample is 2.95 mm

Explanation:

given data

yield load = 2100 N

maximum load = 3400 N

failure load = 2350 N

ultimate engineering stress = 497.4 MPa = 497 × $10^{6}$ N/m²

to find out

What was the initial diameter of the sample in mm

solution

we will apply here ultimate engineering stress formula that is express as

ultimate engineering stress = $\frac{Pmax}{A}$ ...............1

here A is area and P max is maximum load applied

so area = $\frac{\pi }{4} d^2$

here d is initial diameter

so put all value in equation 1

497 × $10^{6}$ = $\frac{3400}{\frac{\pi }{4} d^2}$

solve it we get d

d = 2.95 × $10^{-3}$ m

so initial diameter of the sample is 2.95 mm

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4 years ago

The 40-ft-long A-36 steel rails on a train track are laid with a small gap between them to allow for thermal expansion. Determin

vfiekz [6]

Answer:

Ф = 0.02838 ft

F = 1,032 N

Explanation:

To find out gap delta,

As it is case of free thermal expansion,

First we start with, some assumptions we have to made to solve this problem.

1. Thermal Expansion Coefficient of Steel is ∝= 6.45 ×10^(-6)

2. Modulas of elasticity for A-36 steel is E= 200 GPa

3. Area of rail is assumed to be unit area.

The gape required can be given by,

Ф = ∝ × ΔT × L ... where Ф= Gap Delta in ft

ΔT= Temperature rise in F

= 90- (-20)

= 110 F

Ф = 6.45 ×10^(-6) × 110 × 40

Ф = 28,380 × 10^(-6) ft

Ф = 0.02838 ft .... total gape required for expansion of steel rails

Stress induced in rails is given by,

σ = ∝ × ΔT × E

= 6.45 ×10^(-6) × 110 × 200

σ = 1,41,900 Pa

Now, let's find axial force in rails,

Here,we have to consider ΔT= 20 F.

As due to temperature change, axial force generated in rails can be find by,

F = A × ∝ × ΔT× E × L

F = 1 × 6.45 × 10^(-6) × 20 × 200 × 10^(-9) × 40

F = 25,800 × 40 × 10^(-3)

F = 10,32,000 × 10^(-3)

F= 1,032 N

Finally, due to temperature change, rail is subjected to axial force, axial stress.

8 0

3 years ago