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2 years ago

A driver on a leveltwo-lane highway observes a

Engineering

1 answer:

NARA [144]2 years ago

3 0

Answer:

SSD = 565.54 ft

Explanation:

Given data:

driver speed = 60 mph

clear distance available to stop the truck on emergency can be calculated by using stopping sight distance

SSD = Lag distance + breaking distance

$SSD = vt + \frac{v^2}{2gf}$

let take comfortable acceleration rate as a = 11.2 ft/s^2

g =32.2 ft/s^2

a = fg

$f = \frac{11.2}{32.2} = 0.348$

speed is given as 60 mph

we know that 1 mile 5290 ft

i hr = 3600 sec

therefore v = 88 ft./s

from standard assumption time to reaction is taken as 2.5 sec

so $SSD = 88 \times 2.5 + \frac{88^2}{2\times 32.3\times 0.348}$

SSD = 565.54 ft

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Please write the following code in Python 3. Also please show all output(s) and share your code.

maksim [4K]

Answer:

sum2 = 0

counter = 0

lst = [65, 78, 21, 33]

while counter < len(lst):

sum2 = sum2 + lst[counter]

counter += 1

Explanation:

The counter variable is initialized to control the while loop and access the numbers in lst

While there are numbers in the lst, loop through lst

Add the numbers in lst to the sum2

Increment counter by 1 after each iteration

6 0

2 years ago

Consider an aircraft powered by a turbojet engine that has a pressure ratio of 9. The aircraft is stationary on the ground, held

77julia77 [94]

Answer:

The break force that must be applied to hold the plane stationary is 12597.4 N

Explanation:

p₁ = p₂, T₁ = T₂

$\dfrac{T_{2}}{T_{1}} = \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} }$

${T_{2}}{} = T_{1} \times \left (\dfrac{P_{2}}{P_{1}} \right )^{\frac{K-1}{k} } = 280.15 \times \left (9 \right )^{\frac{1.333-1}{1.333} } = 485.03\ K$

The heat supplied = $\dot {m}_f$ × Heating value of jet fuel

The heat supplied = 0.5 kg/s × 42,700 kJ/kg = 21,350 kJ/s

The heat supplied = $\dot m$ · $c_p(T_3 - T_2)$

$\dot m$ = 20 kg/s

The heat supplied = 20* $c_p(T_3 - T_2)$ = 21,350 kJ/s

$c_p$ = 1.15 kJ/kg

T₃ = 21,350/(1.15*20) + 485.03 = 1413.3 K

p₂ = p₁ × p₂/p₁ = 95×9 = 855 kPa

p₃ = p₂ = 855 kPa

T₃ - T₄ = T₂ - T₁ = 485.03 - 280.15 = 204.88 K

T₄ = 1413.3 - 204.88 = 1208.42 K

$\dfrac{T_5}{T_4} = \dfrac{2}{1.333 + 1}$

T₅ = 1208.42*(2/2.333) = 1035.94 K

$C_j = \sqrt{\gamma \times R \times T_5}$ = √(1.333*287.3*1035.94) = 629.87 m/s

The total thrust = $\dot m$ × $C_j$ = 20*629.87 = 12597.4 N

Therefore;

The break force that must be applied to hold the plane stationary = 12597.4 N.

5 0

2 years ago

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ZanzabumX [31]

Answer is yes.
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2 years ago

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larisa86 [58]

Answer:

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2 years ago

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tankabanditka [31]

Answer:

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