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djverab [1.8K]
3 years ago
11

When driving during the daytime in reduced visibility, such as rain, smoke, or fog,

Engineering
1 answer:
riadik2000 [5.3K]3 years ago
3 0

Answer:A

Explanation:

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Technician A says that crush zones on a vehicle are designed to collapse in a controlled manner. Technician B says that reinforc
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3 years ago
The explosion of a hydrogen bomb can be approximated by a fireball with a temperature of 7200 K, according to a report published
Iteru [2.4K]

Answer:

a

The rate of radiation of the energy is  E_r = 1.523747635*10^9 W/m^2

b

The irradiation is  G =46.177\ kW/m^2

c

The amount of energy absorbed is E_B = 461.772 KJ

d

The oak Tree would catch fire because the temperature of the blast(7200 K) is higher than the flammability limit (650 K) of the oak tree and secondly the thickness is very small

Explanation:

  From the question we are told that

        The  temperature is  T =  7200K

        The diameter of the ball is  d = 1.5 km = 1.5 *1000 = 1500m

       Hence the radius  == \frac{1500}{2} = 750m

 The total energy radiated can be mathematically represented as

                         E = \sigma A T^4

Where \sigma is the Stefan-Boltzmann constant \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 5.67*10^{-8} W \ \cdot m^{-2} K^{-4}

            A is the area of a sphere  = \pi d^2  = 3.142 * 1500^2 = 7.069500 *10 ^6\ m^2

 Substituting values we have

                    E = 5,67*10^{-8} * 7.069500*10^6 * 7200^4

                        =1.077*10^{15} W

Now the state of the energy is mathematically represented as

                           Rate  \ of \ energy \ radiation (E_r)= \frac{E}{A} = \sigma T^4

                                                            = 5.67*10^{-8} * 7200^2

                                                            = 1.523747635*10^9 W/m^2

A sketch illustrating the b part of the question is shown on the first uploaded image

     looking at the height at which the blast occurs(16km) as compared to the height of the wall we notice that the height of the wall is negligibly small

      from the diagram x can be calculated as follows

                      x = \sqrt{40^2 + 16^2}

                        = 43.0813 Km

This value of x represents the radius of the blast(assuming it is spherical ) when it is at that wall

Now the irradiation G is mathematically represented as

                              G = \frac{E}{4 \pi r^2}

Here r = 43.0813 Km = 43.0813 × 1000 = 43081.3 m

                            G= \frac{1.077*10^15}{4 \pi (431081.3^2)}

                                G =46.177\ kW/m^2

Generally the amount of energy absorbed can be mathematically represented as

                            Amount \ of \ energy \ absorbed \ (E_B ) = G * t

Where t is the time taken

       Therefore     E_B = 46.177 *10 = 461.77 KJ

       

                         

                       

             

   

6 0
3 years ago
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