Which of the following will help sober up and lower the BAL of an intoxicated person? Exercise Black coffee Time A cold shower

Answer: you are reducing bullet damage or the damage of a person getting shattered glass all over them

Explanation: i dont

have one

Answer:

0.635 m

Explanation:

When calculating the resistance R of a wire, we need  its length(l), its cross-sectional area (A) and the resistivity of the material(ρ). The resistance of a wire  is given by the equation:

Resistance (R) = Resistivity(ρ) × length (l) / cross-sectional area (A)

For a 12 guage wire,

Resistivity (ρ) = 1.724 × 10 ⁻⁸ ohm m, length (l) = 400 ft = 121.92 m,

Diameter (d) = 0.00205232, cross-sectional area (A) = πd²/4 = π(0.00205232)²/4 = 3.31 × 10 ⁻⁶ m²

[tex]R=\frac{\rho l}{A}=\frac{1.724*10^{-8}*121.92}{3.31*10^{-6}} =0.635m[/tex]

Answer:

C47 may seem like a fancy word, but in the film world, it is the name of one of the simplest, most useful and versatile tools, which is commonly knows as a clothespin.

Following are major uses of C47 on a film set.

1) Lightning Purposes

C47 are used for attaching materials like gel and diffusion to the adjustable flaps, that control the direction of light

2) Adjusting Wardrobes

C47 are used to shorten, tighten or redesign a material of clothing for a specific purpose. It avoids permanent change and the hassle of sewing.

3) Modifications

These can be used for temporary repairs or to modify the set and shooting locations, e.g if a curtain is not hanging properly, C47 can be used

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, [tex]T_C[/tex]  = -15°C = 258.15 K

Temperature of the cooling water, [tex]T_H[/tex] = 30°C = 303.15 K

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

[tex]\gamma_{Max} = \dfrac{T_C}{T_H - T_C} = \dfrac{258.15}{303.15 - 258.15} = 5.74[/tex]

(2) For ammonia refrigerant, we have;

[tex]h_2 = h_g = 1466.3 \ kJ/kg[/tex]

[tex]h_3 = h_f = 322.42 \ kJ/kg[/tex]

[tex]h_4 = h_3 = h_f = 322.42 \ kJ/kg[/tex]

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

[tex]h_1 = h_{f1} + x_1 \times h_{gf}[/tex]

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

[tex]\gamma = \dfrac{h_1 - h_4}{h_2 - h_1}[/tex]

[tex]\gamma = \dfrac{1278.5 - 322.42}{1466.3 - 1278.5} = 5.09[/tex]

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

[tex]\dot m = \dfrac{Refrigeration \ capacity}{Refrigeration \ effect \ per \ unit \ mass}[/tex]

The refrigeration capacity = 105 kJ/h

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

Therefore;

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

[tex]\dot m[/tex] = 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×[tex]\dot m[/tex]

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