Question 5
1 pts
Lumber which has been put through a planer is known as surfaced or
]

Answer:

Tech A is correct.          

Explanation:

A front-wheel-drive pulling to the left can result from several factors. One of them is definitely a faulty break.

A correct diagnosis linking the problem to the brakes is when there is an internal restriction and the pull is constant to one side and gets worse when the brakes are applied.

To confirm this, one would need to lift the vehicle and rotate each wheel by hand to check for excessive friction.

So the restriction may be caused by:

Cheers

Answer:

835,175.68W

Explanation:

Calculation to determine the required power input to the pump

First step is to calculate the power needed

Using this formula

P=V*p*g*h

Where,

P represent power

V represent Volume flow rate =0.3 m³/s

p represent brine density=1050 kg/m³

g represent gravity=9.81m/s²

h represent height=200m

Let plug in the formula

P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m

P=618,030 W

Now let calculate the required power input to the pump

Using this formula

Required power input=P/μ

Where,

P represent power=618,030 W

μ represent pump efficiency=74%

Let plug in the formula

Required power input=618,030W/0.74

Required power input=835,175.68W

Therefore the required power input to the pump will be 835,175.68W

Hey there ..

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Answer:

During starts and stops

Explanation:

"Many of the energy inefficiencies in vehicles occur during starts and stops."

found it in the text.

Answer:

a) the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b) length of the specimen is 66.97 mm

Explanation:

Given the data in the question;

a) Determine the maximum stress that can be applied without plastic deformation

when know that; maximum stress σ[tex]_{max}[/tex]  = F / A

where F is the force in the rod ( 39872 N )

A is the cross-sectional area of the rod ( 100 mm² )

so we substitute;

σ[tex]_{max}[/tex]  = 39872 N / 100 mm²

σ[tex]_{max}[/tex]  = 398.72 N/mm²

Therefore, the maximum stress that can be applied without plastic deformation is 398.72 N/mm²  

b)  

strain in the members can be calculated using the expression

ε = σ / E

where σ is the stress in the rod

E is the module of elasticity (  110 GPa = 110000 N/mm² )

(Sl-L) / L = σ/E

where Sl-L is the change in length of the member

L is the original length of the specimen

so we substitute

(67.21 - L) / L = 398.72 / 110000

110000( 67.21 - L) = 398.72L

7393100 - 110000L = 398.72L

7393100 = 398.72L+ 110000L

7393100 = 110398.72L

L = 7393100 / 110398.72

L = 66.97 mm

Therefore; length of the specimen is 66.97 mm

Answer:

- Vernier thrust is 5.37 kN

- mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Explanation:  

Given that;

For individual thrust chambers (vacuum);

Fc = 73.14 kN ,  Cc = 3279 m/sec

For Overall engine with Vernier (vacuum);

Foa = 297.93 kN = ,  Coa = 3197 m/sec.

- determine the Vernier thrust

Vernier thrust Fv =  Foa - ( 4 × Fc )

Vernier thrust Fv  = 297.93 - ( 4 × 73.14)

Vernier thrust Fv  = 297.93 - 292.56

Vernier thrust Fv  = 5.37 kN

Therefore, Vernier thrust is 5.37 kN

-

Vernier mass flow rate;

we know that

[tex]Co_{a}[/tex] = Fc + Fv  / mc + mv

mv = Foa/Coa - Fc/Cc

we convert kilonewton to kilograms

1 kn = 102 kg

Fc = 73.14 kN = 73.14 × 102 = 7460.28 kg

Foa = 297.93 kN = 297.93 × 102 = 30388.86 kg

we substitute

mv = (30388.86 / 3197) - (( 4 × 7460.28) / 3279)

mv = 9.5054 - 9.1006

mv = 0.4048 kg/s

Therefore, mass flow rate of the four Vernier gas nozzles is 0.4048 kg/s

Answer: Why is even here then.

Explanation:

Answer:

False

Explanation:

I got it wrong picking true

Answer:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

Explanation:

Given

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38,\ 0.40,\ 0.40,\ 0.40,\ 0.41,[/tex]

[tex]0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.43,\ 0.44,\ 0.45,\ 0.46,[/tex]

[tex]0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49,\ 0.51,\ 0.54,\ 0.54,\ 0.55,[/tex]

[tex]0.58,\ 0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68,\ 0.78.[/tex]

Required

Plot a steam and leaf display for the given data

Start by categorizing the data by their tenth values:

[tex]0.32,\ 0.35,\ 0.36,\ 0.36,\ 0.37,\ 0.38.[/tex]

[tex]0.40,\ 0.40,\ 0.40,\ 0.41,\ 0.41,\ 0.42,\ 0.42,\ 0.42,\ 0.42,\ 0.42,[/tex]

[tex]0.43,\ 0.44,\ 0.45,\ 0.46,\ 0.46,\ 0.47,\ 0.48,\ 0.48,\ 0.49.[/tex]

[tex]0.51,\ 0.54,\ 0.54,\ 0.55,\ 0.58.[/tex]

[tex]0.63,\ 0.66,\ 0.66,\ 0.67,\ 0.68.[/tex]

[tex]0.78.[/tex]

The 0.3's is will be plotted as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

The 0.4's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \ \end{array}[/tex]

The 0.5's is as follows:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \ \end{array}[/tex]

The 0.6's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \ \end{array}[/tex]

Lastly, the 0.7's is as thus:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]

The combined steam and leaf plot is:

[tex]\begin{array}{ccc}{Steam} & {\vert} & {Leaf} \ \\ \\ {0.3} & {\vert} & {2\ 5\ 6\ 6\ 7\ 8} \ \\ \\{0.4} & {\vert} & {0\ 0\ 0\ 1\ 1\ 2\ 2\ 2\ 2\ 2\ 3\ 4\ 5\ 6\ 6\ 7\ 8\ 8\ 9} \ \\ \ \\ {0.5} & {\vert} & {1\ 4\ 4\ 5\ 8} \ \\ \ \\ {0.6} & {\vert} & {3\ 6\ 6\ 7\ 8} \ \\ \ \\ {0.7} & {\vert} & {8} \ \ \end{array}[/tex]