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3 years ago

the ____ method is the safest turning technique to use as it does not expose your hands to the airbags deployment area

Engineering

1 answer:

Slav-nsk [51]3 years ago

6 0

Answer:

Push/pull Method

Explanation:

Push/pull method is seen as the best method to use when a driver what to steer a wheel to point the car in a particular direction reason been that PUSH/PULL method enables the driver to sit perfectly well in order to firmly hold the wheel which is why this method is often recommended as the safest and most effective way of steering than other steering method.

Therefore the PUSH/PULL method is the safest and the best turning technique to use because it does not expose the driver hands to the airbags area reason been that air bag are sometimes place in the centre of the steering wheel which means that when a driver cross his/her arms when steering will block the air bag in a situation where the air bag needed to explode out because airbag can explodes out with a lot of force which in turn could force the driver arms towards his/ her face which is dangerous.

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A section of highway has a free-flow speed of 55 mph and a capacity of 3300 veh/hr. In a given hour, 2100 vehicles were counted

alexira [117]

Answer:

Space mean speed = 44 mi/h

Explanation:

Using Greenshield's linear model

q = Uf ( D - $D^{2}$ /Dj )

qcap = capacity flow that gives Dcap

Dcap = Dj/2

qcap = Uf. Dj/4

Where

U = space mean speed

Uf = free flow speed

D = density

Dj = jam density

now,

Dj = 4 × 3300/55

= 240v/h

q = Dj ( U - $U^{2}$ /Uf)

2100 = 240 ( U - $U^{2}$ /55)

Solve for U

U = 44m/h

5 0

3 years ago

an object of mass 2kg is released from a top of inclined plane 30° and height 6m. The coefficient of kinetic friction of the sur

mel-nik [20]

Explanation:

1) Work done = force x distance x cos(θ)

= 0.15 x 6 x cos(30)

= 0.779

2) Ek = ½mv²

v = acceleration due to gravity so 9.81

Ek = ½(2)(9.81)²

Ek = 96.2361

3) v = (√(2em)) / m

= (√(2(96.2361)(2)) / 2

= 9.81 so especially with no time given, I can only assume the acceleration due to gravity but take it with a pinch of salt.

5 0

2 years ago

How many kg / day of NaOH must be added to neutralize a waste stream generated by an industry producing 90,800 kg / day of sulfu

hram777 [196]

Answer:

74.12kg/day

Explanation:

Equation of reaction for the neutralization reaction of sodium hydroxide and sulfuric acid: 2NaOH + H2SO4 yields Na2SO4 + 2H2O

Mass of sulfuric acid produced per day = 90,800kg

Percentage of sulfuric acid in wastewater = 0.1%

Mass of sulfuric acid that ends up in wastewater per day = 0.1/100 × 90,800 = 90.8kg

From the equation of reaction, 2 moles of NaOH (80kg of NaOH) is required to neutralize 1 mole of H2SO4 (98kg of H2SO4)

80kg of NaOH is required to neutralize 98kg of sulfuric acid

90.8kg of sulfuric acid would be neutralized by (90.8×80)/98kg of NaOH = 74.14kg/day of NaOH

7 0

3 years ago

Read 2 more answers

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:

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

The irradiation is $G =46.177\ kW/m^2$

The amount of energy absorbed is $E_B = 461.772 KJ$

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

Consider a Carnot refrigeration cycle executed in a closed system in the saturated liquid–vapor mixture region using 1.06 kg of

Alexxandr [17]

Answer:

$P_m_i_n = 442KPA$

Explanation:

We are given:

m = 1.06Kg

$T_H = 1.2T_L$

T = 22kj

Therefore we need to find coefficient performance or the cycle

$COP_R = \frac {1}{(T_R/T_l) -1}$

$= \frac {1 }{1.2-1}$

= 5

For the amount of heat absorbed:

$Q_l = COP_R Wm$

= 5 × 22 = 110KJ

For the amount of heat rejected:

$Q_H = Q_L + W_m$

= 110 + 22 = 132KJ

[tex[ q_H = \frac{Q_L}{m} [/tex];

= $= \frac{132}{1.06}$

= 124.5KJ

Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c

Convert 69.5°c to K we have 342.5K

To find the minimum temperature:

$T_L = \frac{T_H}{1.2}$ ;

$T_L = \frac{342.5}{1.2}$

= 285.4K

Convert to °C we have 12.4°C

From the refrigerant R -134a table at $T_L$ = 12.4°c we have 442KPa

6 0

3 years ago