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romanna [79]
3 years ago
8

What additional information would make the following problem statement stronger? Select all that apply.

Engineering
1 answer:
Zolol [24]3 years ago
4 0
Idl sjdjdjdkdnsnehusoslowpwmwkw
You might be interested in
25 points!!
adell [148]
Answer 1:

Like most electrical mechanical devices, electric motors have mechanical bearings that eventually wear out.

The graphite brushes and commutator of older DC motors can also wear out over time

Answer2:

Question two is not specific enough!!
Are we talking DC?? Because if so, then yes, nothing will happen. Assuming you are insulted by ground.
Ac is a different story, as the two would light up dangerously and etc.

-your welcome!!
5 0
4 years ago
Two technicians are discussing hand tool use. Technician A says that a 6-point wrench is easier to use in tight places than a 12
Mila [183]

Answer:

<em>Technician B says that a ratchet is used to loosen fasteners that are very tight.</em>

Explanation:

A ratchet is a common wrench device with a fastener component. A ratchet wrench is an essential tool that is used to fasten or loosen nuts and bolts.

8 0
3 years ago
Read 2 more answers
Find the time-domain sinusoid for the following phasors:_________
sattari [20]

<u>Answer</u>:

a.  r(t) = 6.40 cos (ωt + 38.66°) units

b.  r(t) = 6.40 cos (ωt - 38.66°) units

c.  r(t) = 6.40 cos (ωt - 38.66°) units

d.  r(t) = 6.40 cos (ωt + 38.66°) units

<u>Explanation</u>:

To find the time-domain sinusoid for a phasor, given as a + bj, we follow the following steps:

(i) Convert the phasor to polar form. The polar form is written as;

r∠Ф

Where;

r = magnitude of the phasor = \sqrt{a^2 + b^2}

Ф = direction = tan⁻¹ (\frac{b}{a})

(ii) Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid (r(t)) as follows:

r(t) = r cos (ωt + Φ)

Where;

ω = angular frequency of the sinusoid

Φ = phase angle of the sinusoid

(a) 5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

5 + j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

(b) 5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{5^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

5 - j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(c) -5 + j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + 4^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{4}{-5})

Φ = tan⁻¹ (-0.8)

Φ = -38.66°

-5 + j4 = 6.40∠-38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt - 38.66°)

(d) -5 - j4

<em>(i) convert to polar form</em>

r = \sqrt{(-5)^2 + (-4)^2}

r = \sqrt{25 + 16}

r = \sqrt{41}

r = 6.40

Φ = tan⁻¹ (\frac{-4}{-5})

Φ = tan⁻¹ (0.8)

Φ = 38.66°

-5 - j4 = 6.40∠38.66°

(ii) <em>Use the magnitude (r) and direction (Φ) from the polar form to get the general form of the time-domain sinusoid</em>

r(t) = 6.40 cos (ωt + 38.66°)

3 0
3 years ago
Select the correct answer. Why should engineers keep themselves updated on the technological developments in their field? OA. to
Nataly_w [17]
OD is the correct answer
5 0
3 years ago
The permeability coefficient of a type of small gas molecule in a polymer is dependent on absolute temperature according to the
jok3333 [9.3K]

Answer:

Answer for the question is explained in the attachment.

Explanation:

Diffusion flux at 350K for water in polystyrene is equal to

2.8 * 10^-7 .

Download pdf
5 0
3 years ago
Read 2 more answers
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