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3 years ago

WILL AWARD BRAINLIEST

Engineering

1 answer:

ValentinkaMS [17]3 years ago

3 0

I think it is C because if have to send it to someone you would want them to be able to open it

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The speed of sound is 1150 ft/s convert to mile/h

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Answer:

784.090909mph

Explanation:

1ft/s=0.681818 mph

1150ft/s=0.681818 x 1150=784.090909

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3 years ago

A Carnot power cycle is executed in a piston-cylinder system using 1 kg of water. The cycle consists of an isobaric expansion of

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b) Determine the heat transfer into the cycle and the net work for the cycle, in kJ.

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3 years ago

A motor vehicle has a mass of 1.8 tonnes and its wheelbase is 3 m. The centre of gravity of the vehicle is situated in the centr

seropon [69]

Answer:

1) The normal reactions at the front wheel is 9909.375 N

The normal reactions at the rear wheel is 8090.625 N

2) The least coefficient of friction required between the tyres and the road is 0.625

Explanation:

1) The parameters given are as follows;

Speed, u = 90 km/h = 25 m/s

Distance, s it takes to come to rest = 50 m

Mass, m = 1.8 tonnes = 1,800 kg

From the equation of motion, we have;

v² - u² = 2·a·s

Where:

v = Final velocity = 0 m/s

a = acceleration

∴ 0² - 25² = 2 × a × 50

a = -6.25 m/s²

Force, F = mass, m × a = 1,800 × (-6.25) = -11,250 N

The coefficient of friction, μ, is given as follows;

$\mu =\dfrac{u^2}{2 \times g \times s} = \dfrac{25^2}{2 \times 10 \times 50} = 0.625$

Weight transfer is given as follows;

$W_{t}=\dfrac{0.625 \times 0.9}{3}\times \dfrac{6.25}{10}\times 18000 = 2109.375 \, N$

Therefore, we have for the car at rest;

Taking moment about the Center of Gravity CG;

$F_R$ × 1.3 = 1.7 × $F_F$

$F_R$ + $F_F$ = 18000

$F_R + \dfrac{1.3 }{1.7} \times F_R = 18000$

$F_R$ = 18000*17/30 = 10200 N

$F_F$ = 18000 N - 10200 N = 7800 N

Hence with the weight transfer, we have;

The normal reactions at the rear wheel $F_R$ = 10200 N - 2109.375 N = 8090.625 N

The normal reactions at the front wheel $F_F$ = 7800 N + 2109.375 N = 9909.375 N

2) The least coefficient of friction, μ, is given as follows;

$\mu = \dfrac{F}{R} = \dfrac{11250}{18000} = 0.625$

The least coefficient of friction, μ = 0.625.

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4 years ago

An airplane is cruising at a velocity of 800 km/h in air whose density is 0.526 . The airplane has a wing planform area of 90 .

Aliun [14]

Complete question:

An airplane is cruising at a velocity of 800 km/h in air whose density is 0.526 kg/m³ . The airplane has a wing planform area of 90 m² . The lift and drag coefficients on cruising conditions are estimated to be 2.0 and 0.06, respectively. The power that needs to be supplied to provide enough trust to overcome wing drag is

Answer:

The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.

Explanation:

Given;

velocity of the airplane, v = 800 km/h = 222.22 m/s

density of air, ρ = 0.526 kg/m³

wing planform area, A = 90 m²

lift coefficients, CL = 2.0

drag coefficients, CD = 0.06

$F_D = C_D*A* \rho*\frac{v^2}{2} = 0.06*90* 0.526*\frac{222.22^2}{2} =70,131.93 \ N$

Power supplied = FD* V

$Power = 70131.93*222.22 = 15,584,717.63 \ W = 15,584.72 \ kW$

This is approximately 15,600 kW.

Therefore, The power that needs to be supplied to provide enough trust to overcome wing drag is 15,600 kW.

The correct option is C

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3 years ago