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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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Driving Distraction Brainstorming Session

Leto [7]

texting, phone calls, putting on makeup, brushing hair, movies playing in car, loud music, children, and that's pretty much all I could think of

please give BRAINLIEST ANSWER └[T‸T]┘

5 0

3 years ago

When block C is in position xC = 0.8 m, its speed is 1.5 m/s to the right. Find the velocity of block A at this instant. Note th

Amanda [17]

Answer:

The answer is "2 m/s".

Explanation:

The triangle from of the right angle:

$\to (x_c-0.8)+(1.5+y_4) +\sqrt{x_c^2 + 1.5^2}= constant$

Differentiating the above equation:

$\to V_c +V_A+ \frac{X_cV_c}{\sqrt{x_c^2 +1}}=0\\\\\to 1-V_A+ \frac{0.8 \times 1.5}{\sqrt{ 0.8^2+1.5}}=0\\\\$

$\to V_A= \frac{1.2}{\sqrt{ 0.64+1.5}}+1\\\\$

$= \frac{1.2}{ 1.46}+1\\\\= \frac{1.2+ 1.46}{ 1.46}\\\\ = \frac{2.66}{1.46}\\\\= 1.82 \ \frac{m}{s}\\\\= 2 \ \frac{m}{s}$

3 0

2 years ago

Generally, final design results are rounded to or fixed to three digits because the given data cannot justify a greater display.

creativ13 [48]

Answer:

(a) 1.90 kpsi

(b) 0.40 kpsi

(d) 0.009

(a) 8 MPa

(b) 1.30 cm⁴

(d) 62.2 MPa

Explanation:

(a) σ = M/Z, where M = 1770 lbf·in and Z = 0.943 in³.

1770/0.943 = 1876.988 lbf/in² = 1.90 kpsi

(b) σ = F/A, where F = 9440 lbf and A = 23.8 in².

9440 /23.8 = 396.639 lbf/in² = 0.4 kpsi

F = 270 lbf

l = 31.5 in.

E = 30 Mpsi

I = 0.154 in.⁴

y = 270×31.5³/(3×30×10⁶×0.154) = 0.61 in.

(d) θ = Tl/(GJ), where T = 9740 lbf·in, l = 9.85 in. G = 11.3 Mpsi, and d = 1.00 in.

J = π·d⁴/32 = π/32 in.⁴

∴ θ = 9740 × 9.85 /(11.3 × 10⁶× π/32) = 0.009

(a) σ = F/wt, where F = 1 kN, w = 25 mm, and t = 5 mm

∴ σ = 1000/(0.025 × 0.005) = 8 MPa

(b) I = bh³/12, where b = 10 mm and h = 25 mm.

10×25³/12 = 1.30 cm⁴

I = π × 25.4⁴/64 = 2.04 cm⁴

(d) τ = 16×T/(π×d³), where T = 25 N·m, and d = 12.7 mm.

16×25/(π×0.0127³) = 62.2 MPa.

8 0

3 years ago

technician a uses a current output test to check ac generator output. technician b uses a voltage output test to check output. w

expeople1 [14]

Answer:

both

Explanation:

Both the technician are correct, ac generator output can be tested in both ways. The two ways are current output test to check ac generator output. and voltage output test to check output.

7 0

3 years ago

A spring-mass-damper instrument is employed for acceleration measurements. The spring constant is 12000 N/m. The mass is 5 g. Th

shepuryov [24]

Answer:

a) 246.56 Hz

b) 203.313 Hz

c) Add more springs

Explanation:

Spring constant = 12000 N/m

mass = 5g = 5 * 10^-3 kg

damping ratio = 0.4

a) Calculate Natural frequency

Wn = √k/m = $\sqrt{12000 / 5*10^{-3} }$

= 1549.19 rad/s ≈ 246.56 Hz

b) Bandwidth of instrument

W / Wn = $\sqrt{1-2(0.4)^2}$

W / Wn = 0.8246

therefore Bandwidth ( W ) = Wn * 0.8246 = 246.56 * 0.8246 = 203.313 Hz

C ) To increase the bandwidth we have to add more springs

5 0

3 years ago