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anastassius [24]
3 years ago
13

HAPPINESS DISCUSSION

Engineering
2 answers:
RideAnS [48]3 years ago
8 0

Answer:

uh because life sucks o_<

KengaRu [80]3 years ago
7 0
I personally have more bad days then good days because I don’t cherish the good days, on the good days I’m happy that I can feel something like sadness, or anger on my bad days, because without my bad days there would be no good days. But on my bad days I don’t care about anything except for what is making my day bad.
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A well penetrates an unconfined aquifer. Prior to pumping, the water level (head) is 25 meters. After a long period of pumping a
GarryVolchara [31]

Answer:

Explanation:

Find attach the solution

3 0
3 years ago
Velocity and temperature profiles for laminar flow in a tube of radius ro = 10 mm have the form: u(r) = 0.15[1 − (r/ro ) 2 ] T(r
antoniya [11.8K]

Answer:

Tm = 366.66k

Explanation:

check for the step by step explanation in the attachment

8 0
3 years ago
“We’re late for homeroom,” said Bonnie, surprised to hear herself say “we.” “EARL is a tool, Bonnie’s mother kept reminding her,
bogdanovich [222]

Answer:

Hi there

Your answer is:

B.

Explanation:

The "metal contraption", as the text goes on to say, is treated like a friend to Bonnie. Her mom comments on this by contrasting the metal contraption to a puppy.

Hope this helps

8 0
3 years ago
Read 2 more answers
Three bars each made of different materials are connected together and placed between two walls when the temperature is 12 oC. D
slega [8]

Answer:

F = 9.11 x 10³ N = 9.11 KN

Explanation:

The areas, lengths, young's modulus, and coefficient of linear thermal expansion are given in the diagram. First we find the equivalent change in length due to temperature change:

ΔL = (ΔL)steel + (ΔL)brass + (ΔL)Copper

ΔL = (∝s)(Ls)(ΔT) + (∝b)(Lb)(ΔT) + (∝c)(Lc)(ΔT)

where,

ΔL = Equivalent Change in Length = ?

ΔT = Change in Temperature = 25°C - 12°C = 13°C

Ls = Length of Steel Segment = 300 mm = 0.3 m

Lb = Length of Brass Segment = 200 mm = 0.2 m

Lc = Length of Copper Segment = 100 mm = 0.1 m

Therefore,

ΔL = (12 x 10⁻⁶ °C⁻¹)(0.3 m)(13 °C) + (21 x 10⁻⁶ °C⁻¹)(0.2 m)(13 °C) + (17 x 10⁻⁶ °C⁻¹)(0.1 m)(13 °C)

ΔL = 46.8 x 10⁻⁶ m + 54.6 x 10⁻⁶ m + 22.1 x 10⁻⁶ m

ΔL = 123.5 x 10⁻⁶ m   ----------------------- equation (1)

Now, we calculate this deflection in terms of an applied force (F):

ΔL = (F)(Ls)/(Es)(As) + (F)(Lb)/(Eb)(Ab) + (F)(Lc)/(Ec)(Ac)

ΔL = (F)(0.3 m)/(200 x 10⁹ Pa)(200 x 10⁻⁶ m²) + (F)(0.2 m)/(100 x 10⁹ Pa)(450 x 10⁻⁶ m²) + (F)(0.1 m)/(120 x 10⁹ Pa)(515 x 10⁻⁶ m²)

ΔL = F(7.5 x 10⁻⁹ m/N + 4.44 x 10⁻⁹ m/N + 1.61 x 10⁻⁹ m/N)

ΔL = F(13.55 x 10⁻⁹ m/N)   --------------------- equation (1)

Comparing equation (1) and equation (2):

123.5 x 10⁻⁶ m = F(13.55 x 10⁻⁹ m/N)

F = (123.5 x 10⁻⁶ m)/(13.55 x 10⁻⁹ m/N)

<u>F = 9.11 x 10³ N = 9.11 KN</u>

6 0
3 years ago
Use the map to answer the question.
djverab [1.8K]

Answer:

Captain falcon

Explanation:

7 0
2 years ago
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