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bezimeni [28]
3 years ago
9

At a retirement party, a coworker described terry as dedicated

Engineering
1 answer:
denis23 [38]3 years ago
8 0

Answer:

At a retirement party, a coworker described Terry as dedicated, hardworking, and dependable. He also said that Terry was a great leader, knew the computer system, and kept the company's finances in order

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Select the creative imaging fields that require knowledge of programming.
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Video game designer for sure
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Can anyone explain how a Halbek Device works
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The Halbek Device can be used effectively on some weapons with practice and certain loadouts. ... This tends to help high-damage weapons or weapons with high multipliers

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List in order first three steps to square a board
LenaWriter [7]

Answer:

STEP1 Cut to Rough Length

STEP2 Cut to Rough Width

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HOPE THAT HELPSSS!!!

5 0
2 years ago
Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces app
stiks02 [169]

Answer:

a) Fb= 275.77 lb   Fc= 142.75 lb

b) M = -779.97 lb.ft (i.e. 779.97 lb.ft in clockwise direction)

c) Fax = 195 lb

   Fay = 337.75 lb

   Fbx = 195 lb

   Fby = 195 lb

Explanation:

Question: Three tugboats are used to turn a barge in a narrow channel. To avoid producing any net translation of the barge, the forces applied should be couples. The tugboat at point A applies a 390 lb force.

(a) Determine FB and FC so that only couples are applied.

(b) Using your answers to Part (a), determine the resultant couple moment that is produced.

(c) Resolve the forces at A and B into x and y components, and identify the pairs of forces that constitute couples.

Solution:

<u>For this problem Right hand side is positive X direction and Upwards is positive Y direction. Couples and moments will be considered positive in counterclockwise direction.</u>

<u />

a) For no translation condition

∑ F_{x} = 0      &     ∑F_{y} = 0

Hence,

F_{A}cos(30) - F_{B}cos(45) - F_{C} = 0

F_{A}sin(30) - F_{B} sin(45) = 0

and

F_{A} = 390 lb

Inserting the value of F_{A} and solving the remaining equations simultaneously yields (magnitudes),

F_{B} = 275.77 lb\\F_{C} = 142.75 lb

b) Summing up moments

M=45 ( -F_{Ay}-F_{Cy}) +5 (-F_{By})+22(-F_{Ax}-F_{Bx})\\ =45(-390cos(30)-142.75)+5(-275.77cos(45))+22 (-390sin(30)-275.77sin(45))

M = -779.97 lb.ft (i.e. 779.97 lb.ft clockwise)

c)

F_{Ax} = 390 sin(30)  = 195 lb

F_{Ay} = 390 cos(30) = 337.75lb\\

F_{Bx} = 275.77 sin(45) = 195lb\\F_{By} = 275.77 cos(45) = 195 lb

8 0
3 years ago
. . . . . . . . . . ..........
Naddika [18.5K]

Answer: Thx for points...

Explanation:

Heres Random Info About Enginnering (because why not): Engineers are involved in all aspects of interactive TV technology, from designing new cables, to creating new film emulsions, to engineering better sound quality. This technology allows viewers to select any program, film, or game from more than 500 channels. Engineers play an instrumental role in the theme park industry?

Hope This Helps Someday!

5 0
2 years ago
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