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Natasha2012 [34]
3 years ago
7

What are the inputs, outputs and side effects of a PSP?

Engineering
1 answer:
Kipish [7]3 years ago
5 0

Answer:

Side effects - sudden loss of balance/ repeated falls

Outputs - sever sickness and could me factual

Inputs/corrections of this- medications and experimental treatments to help slow the process of deterioration

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11) (10 points) A large valve is to be used to control water supply in large conduits. Model tests are to be done to determine h
IrinaVladis [17]

Answer:

7.94 ft^3/ s.

Explanation:

So, we are given that the '''model will be 1/6 scale (the modeled valve will be 1/6 the size of the prototype valve)'' and the prototype flow rate is to be 700 ft3 /s. Then, we are asked to look for or calculate or determine the value for the model flow rate.

Note that we are to use Reynolds scaling for the velocity as par the instruction from the question above.

Therefore; kp/ks = 1/6.

Hs= 700 ft3 /s and the formula for the Reynolds scaling => Hp/Hs = (kp/ks)^2.5.

Reynolds scaling==> Hp/ 700 = (1/6)^2.5.

= 7.94 ft^3/ s

7 0
3 years ago
Two previously undeformed rod-shaped specimens of copper are to be plastically deformed by reducing their cross-sectional areas.
mezya [45]
I am not sure I am stuck on this and I have been for 45 min someone please help me and this girl or boy!!
4 0
3 years ago
The liquid-phase reaction:
OLEGan [10]

Answer:

attached below

Explanation:

4 0
3 years ago
Calculate the areas under the stress-strain curve (toughness) for the materials shown in Fig. below, (a) plot them as a
defon

Answer:

Explanation:

Wow

5 0
3 years ago
A sand has a natural water content of 5% and bulk unit weight of 18.0 kN/m3. The void ratios corresponding to the densest and lo
Zinaida [17]

Answer:

Relative density = 0.545

Degree of saturation = 24.77%

Explanation:

Data provided in the question:

Water content, w = 5%

Bulk unit weight = 18.0 kN/m³

Void ratio in the densest state, e_{min} = 0.51

Void ratio in the loosest state, e_{max} = 0.87

Now,

Dry density, \gamma_d=\frac{\gamma_t}{1+w}

=\frac{18}{1+0.05}

= 17.14 kN/m³

Also,

\gamma_d=\frac{G\gamma_w}{1+e}

here, G = Specific gravity = 2.7 for sand

17.14=\frac{2.7\times9.81}{1+e}

or

e = 0.545

Relative density = \frac{e_{max}-e}{e_{max}-e_{min}}

= \frac{0.87-0.545}{0.87-0.51}

= 0.902

Also,

Se = wG

here,

S is the degree of saturation

therefore,

S(0.545) = (0.05)()2.7

or

S = 0.2477

or

S = 0.2477 × 100% = 24.77%

7 0
3 years ago
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