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kenny6666 [7]
3 years ago
13

Vince is trying to figure out the volume of two mystery matters. The volume of one of the substances needs to be measured by sub

merging it in water and the other needs to be measured using a graduated cylinder. Based on the properties of two mystery matters, what are they? (2 points)
Group of answer choices

A rock and orange juice

Helium and a golf ball

Lemonade and milk

Orange juice and helium
Engineering
1 answer:
pshichka [43]3 years ago
8 0

Answer:

Rock and orange juice

Explanation:

The mystery matter to be submerged in water must be a solid, therefore we can eliminate the Lemonade and Milk, and Orange juice and Helium, as these pairs do not contain solids. The graduated cylinder is used to measure the volume of a liquid, therefore the only remaining option is Rock and Orange Juice.

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A refrigerated space is maintained at -15℃, and cooling water is available at 30℃, the refrigerant is ammonia. The refrigeration
Illusion [34]

Answer:

(1) 5.74

(2) 5.09

(3) 3.05×10⁻⁵ kg/s

(4) 0.00573 kW

Explanation:

The parameters given are;

Working temperature, T_C  = -15°C = 258.15 K

Temperature of the cooling water, T_H = 30°C = 303.15 K

(1) The Carnot coefficient of performance is given as follows;

\gamma_{Max} = \dfrac{T_C}{T_H - T_C}  =  \dfrac{258.15}{303.15 - 258.15}   = 5.74

(2) For ammonia refrigerant, we have;

h_2 = h_g = 1466.3 \ kJ/kg

h_3 = h_f = 322.42 \ kJ/kg

h_4 = h_3 = h_f = 322.42 \ kJ/kg

s₂ = s₁ = 4.9738 kJ/(kg·K)

0.4538 + x₁ × (5.5397 - 0.4538) = 4.9738

∴ x₁ = (4.9738 - 0.4538)/(5.5397 - 0.4538) = 0.89

h_1 = h_{f1} + x_1 \times h_{gf}

h₁ = 111.66 + 0.89 × (1424.6 - 111.66) = 1278.5 kJ/kg

\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}

\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09

(3) The circulation rate is given by the mass flow rate, \dot m as follows

\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}

The refrigeration capacity = 105 kJ/h

The refrigeration effect, Q = (h₁ - h₄) = (1278.5 - 322.42) = 956.08 kJ/kg

Therefore;

\dot m = \dfrac{105}{956.08}  = 0.1098 \ kg/h

\dot m = 0.1098 kg/h = 0.1098/(60*60) = 3.05×10⁻⁵ kg/s

(4) The work done, W = (h₂ - h₁) = (1466.3 - 1278.5) = 187.8 kJ/kg

The rating power = Work done per second = W×\dot m

∴ The rating power = 187.8 × 3.05×10⁻⁵ = 0.00573 kW.

6 0
3 years ago
Select the correct answer. Sean is a recent engineering graduate who has joined a new company. Read the profiles of his colleagu
PSYCHO15rus [73]

Answer:

giving advertising to the governor with the state

4 0
3 years ago
If safeguarding is not possible what must be used instead
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Answer: since safeguarding isn't possible distance and location must be used instead

5 0
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Wind blows on the side of a fully enclosed hospital located on open flat terrain where V= 120 mi/h. Determine the external press
ss7ja [257]

Answer:

The external pressure is p = -21.9 psf or p = -8.85 psf

Explanation:

Given :

Velocity of wind, v = 120 mi / hr

$k_d = k_c =1 $   (wind direction factor)

$k _{zt} = 1 $  = topographical factor (for flat terrain)

$ q_n$ = velocity pressure at height h

$ \therefore q_n = 0.00256 k_z k_{zt} k_d v^2 $

$  q_n = 0.00256 \times  k_z (1)(1)(120)^2 $

    $ = 36.86 k_z$

But for height h = 30 ft, $ k_z$ = 0.98 (from table)

$ \therefore q_n = 36.86 \times 0.98 $

        = 36.16

Now, $ \frac{L}{B}= \frac{200}{200} =1$ ,   so $C_p=-0.5 $ (from table)

$p = q(G)(C_p)-q_n(GC_{pi})$

where, p = external pressure

            G = 0.85 = gust factor (for typical rigid building)

            $GC_{pi} = \pm 0.18 $   (internal pressure co efficient)

Therefore putting the values,

$p = (36.13)(0.85)(-0.5)-(36.13)(\pm 0.18)$

p = -21.9 psf or p = -8.85 psf

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Calculate the unit length of a roof using common rafters if the unit run is 7 inches. Round your answer to two decimal places.
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Answer:

Explanation:

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