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2 years ago

Before you attempt to change a tire yourself, you should _____.

Engineering

1 answer:

olchik [2.2K]2 years ago

7 0

Answer: read your vehicle owner's manual for any special directions or warnings.

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A 227 pound compressor is supported by four legs that contact the floor of a machine shop. At the bottom of each leg there is a

Ganezh [65]

Answer:

1.312 in

Explanation:

Data provided in the question:

Weight of the compressor, W = 227 pound

Number of legs = 4

Maximum pressure = 42 psi

Now,

Let F be the force taken by the legs

Therefore,

W = 4F

227 pound = 4F

F = 56.75 pounds

Also,

Force = Pressure × Area

56.75 pounds = 42 psi × πr² [ r is the diameter of one leg]

r² = 0.4301

r = 0.656

therefore,

diameter = 2r = 2 × 0.656

= 1.312 in

6 0

2 years ago

Your task is to fill in the missing parts of the C code to get a program equivalent to the generated assembly code. Recall that

Rudik [331]

Answer:

See Explaination

Explanation:

//Function

long loop (long x, long n)

{

//Declare a variable named result and initialize it to zero

long result = 0;

//Declare a variable named mask

long mask;

//For loop

for(mask = 1; mask != 0; mask = mask << (n & 0xFF))

{

//Calculate

result | = (x&mask);

}

//Return result

return result;

}

6 0

2 years ago

Hot exhaust gases of an internal combustion engine are to be used to produce saturated water vapor at 2 MPa pressure. The exhaus

Anastaziya [24]

Answer:

The flowrate of water is 0.03556kg/s

Explanation:

Exhaust gases inlet temperature T1=4000C

Water inlet temperature T3=150C Exit Pressure of water as saturated vapor P4=2MPa

Mass flow rate of exhaust gases Heat lost to the surroundings Qgases=32kg/min

Mass flow rate of exhaust gases is 15 times that of the water

Heat exchangers typically involve no work interactions (w = 0) and negligible...

7 0

2 years ago

Read 2 more answers

A 20 dBm power source is connected to the input of a directional coupler having a coupling factor of 20 dB, a directivity of 35

lukranit [14]

Answer:

$P_O = 0.989 watt = 19.9 dBm$

Explanation:

Given data:

P_1 power = 20 dBm = 0.1 watt

coupling factor is 20dB

Directivity = 35 dB

We know that

coupling factor $= 10 log \frac{P_1}{P_f}$

solving for final power

$20 = 10 log\frac{P_1}{P_f}$

$2 = log \frac{P_1}{P_f}$

$100 = \frac{0.1}{P_f}$

$P_f = 0.001 watt = 0 dBm$

Directivity $D = 10 \frac{P_f}{P_b}$

$35 = 10 \frac{0.001}{P_b}$

$P_b = 3.162 \times 10^{-7} wattt$

output Power $= P_1 -P_f - P_b$

$= 0.1 - 0.001 - 3.162 \times 10^{-7}$

$P_O = 0.989 watt = 19.9 dBm$

6 0

2 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

2 years ago