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dimulka [17.4K]
3 years ago
11

Freeee Poinntssss 100!!!!!! Hi how you doin?

Engineering
2 answers:
Andreas93 [3]3 years ago
6 0
Thank u very much , im doin good wby :)
Zinaida [17]3 years ago
3 0

Answer:

Explanation:

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FAST PLLZZ!! Ideally, the backrest is tilted back slightly, so when you turn the wheel your shoulders are _______ the seat.
exis [7]

Answer:

touching

Explanation:

The backrest of the seat should be tilted back ever so slightly, and when turning the steering wheel your shoulders should remain in contact with the seat – rather than hunched forward.

8 0
3 years ago
Bob and Alice are solving practice problems for CSE 2320. They look at this code: for(i = 1; i <= N; i = (i*2)+17 ) for(k = i
MissTica

Answer:

Alice is correct.

The loop are dependent.

Explanation:

for(i = 1; i <= N; i = (i*2)+17 )

for(k = i+1; k <= i+N; k = k+1) // notice i in i+1 and i+N

printf("B")

This is a nested for-loop.

After the first for-loop opening, there is no block of statement to be executed rather a for-loop is called again. And the second for-loop uses the value of i from the first for-loop. The value of N is both called from outside the loop.

So, the second for-loop depend on the first for loop to get the value of i. For clarity purpose, code indentation or use of curly brace is advised.

8 0
3 years ago
Read 2 more answers
Why would the shear stress be considered as the momentum flux.
oksano4ka [1.4K]

Answer:

A fluid flowing along a flat plate will stick to it at the point of contact

Explanation:

and this is known as the no-slip condition. ... This is the precise reason why shear stress in a fluid can also be interpreted as the flux of momentum.

3 0
2 years ago
What similarities do wind and solar energy share?
Viefleur [7K]

Answer:

Both come from the sun

Both are reusable sources

and both don't cause pollution

Explanation:

3 0
2 years ago
A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R1
ella [17]

Answer:

a)

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b) 3.1 secs

Explanation:

a) Determine the normal times in TMUs for these motion elements

1) R16C ; Tn = 17 TMU

2) G4A ; Tn = 7.3 TMU

3) M10B5 ; Tn = 15.1 TMU

4) RL1 ; Tn = 2 TMU

5) R14B ; Tn = 14.4 TMU

6) G1B ; Tn = 3.5 TMU

7) M8C3 ; Tn = 14.7 TMU

8) P1NSE ; Tn = 10.4 TMU

9) RL1 ; Tn = 2 TMU

b ) Determine the total time for this work element in seconds

first we have to determine the total TMU = ∑ TMU = 86.4 TMU

note ; 1 TMU = 0.036 seconds

hence the total time for the work in seconds = 86.4 * 0.036 = 3.1 seconds

7 0
2 years ago
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