What type of damage is reduced by installing impact resistant glass functional shutters and double door top and bottom latches
Answer: you are reducing bullet damage or the damage of a person getting shattered glass all over them
Explanation: i dont
have one
Luis is installing some 12 gage wire. How much resistance will there be throughout a distance of 400 feet
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]
The practice of honoring those who have been lost in battle dates back to which two ancient
civilizations?
In Illinois, once a person has obtained their boating education certificate what is the minimum age to operate a motorized vessel without adult supervision?
Answer:
I dont really know, I am sorry, but I am going to ask my teacher
What is one way a C47 can be used on set?
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
A refrigerated space is maintained at -15℃， and cooling water is available at 30℃， the refrigerant is ammonia. The refrigeration capacity is 105 kJ/h. If the compressor is operated reversibly:
(1) What is the value of ε for Carnot refrigerator?
(2) Calculate the ε for the vapor-compression cycle;
(3) Calculate the circulation rate for the refrigerant;
(4) Calculate the rating power of the compressor.
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.
what are some preventions and treatments for the listing below?
-Human papillomavirus
-Chlamydia
-Trichomoniasis
-Gonorrhea
-Syphilis
-HIV/AIDS
In Florida, bike lanes are painted
Answer: green
Explanation:
3. Airbags are supplemental protection and are designed to deploy in all crashes.
A. O TRUE
B. O FALSE
Airbags are supplemental protection and are not designed to deploy in all crashes. Therefore, it's false.
What are airbag?It should be noted that airbags are out in vehicles in order to reduce the impact on an individual during accidents.
Therefore, airbags are supplemental protection and are not designed to deploy in all crashes. They're designed to work with seatbelts.
Learn more about airbag on:
brainly.com/question/2607849
#SPJ9
Pure ethanol has an octane rating of about 113. E85, Which contains 35 percent oxygen by weight, has an octane rating of about
Florida's No-Fault requires owners of motor vehicles must cover the
below minimum Personal Injury Liability (PIP) insurance coverage:
o $10,000 of Personal Injury Protection (PIP)
o $30,000 of Personal Injury Protection (PIP)
o $50,000 of Personal Injury Protection (PIP)
o $100,000 of Personal Injury Protection (PIP)
Answer:
The correct option is;
$10,000 of Personal Injury Protection (PIP)
Explanation:
Personal injury protection, PIP, otherwise known as No-Fault Law coverage is a part of auto insurance that takes care of medical cost of treatment of an accident victim regardless of who is at fault
In Florida, the No-Fault Law requires motor vehicle owners to have and retain PIP providing $10,000 in medical care, resulting disability and expenses of a funeral resulting from an accident in which the motorist is involved
Therefore, the correct option is $10,000 of Personal Injury Protection.