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3 years ago

2. The Federal Motor Carrier Safety Regulations (FMCSRs) establish basic safety rules

Engineering

1 answer:

Serga [27]3 years ago

6 0

A. Drivers and other employees of motor carrier

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Air (cp = 1.005 kJ/kg·°C) is to be preheated by hot exhaust gases in a cross-flow heat exchanger before it enters the furnace. A

uysha [10]

Answer:

Q=67.95 W

T=119.83°C

Explanation:

Given that

For air

Cp = 1.005 kJ/kg·°C

T= 20°C

V=0.6 m³/s

P= 95 KPa

We know that for air

P V = m' R T

95 x 0.6 = m x 0.287 x 293

m=0.677 kg/s

For gas

Cp = 1.10 kJ/kg·°C

m'=0.95 kg/s

Ti=160°C ,To= 95°C

Heat loose by gas = Heat gain by air

[m Cp ΔT] for air =[m Cp ΔT] for gas

by putting the values

0.677 x 1.005 ( T - 20)= 0.95 x 1.1 x ( 160 -95 )

T=119.83°C

T is the exit temperature of the air.

Heat transfer

Q=[m Cp ΔT] for gas

Q=0.95 x 1.1 x ( 160 -95 )

Q=67.95 W

7 0

3 years ago

Tech a says that the weight of the flywheel smoothest out the engines power pulses. Tech B says that the flexplate and torque co

lakkis [162]

Answer:

both statement is correct

Explanation:

Flywheel engine uses to reduce fluctuations.

And

FlexPlate is a metal disk that connects the output from the engine to the input of the torque converter. This will replace the flywheel

so that both statement is correct

4 0

3 years ago

A particular motor rotates at 3000 revolutions per minute. What is its speed in rad/sec, and how many seconds does it takes to m

Leno4ka [110]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

$\omega=\dfrac{2\pi N}{60}\ rad/s$

Now by putting the values in the above equation

$\omega=\dfrac{2\pi \times 3000}{60}\ rad/s$

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

$\omega=\dfrac{ 3000}{60}\ rad/s$

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

$\dfrac{1}{50}=0.02\ s$

4 0

3 years ago

The in-situ dry density of a sand is 1.72Mg/m3. The maximum and minimum drydensities, determined by standard laboratory tests, a

Stells [14]

Answer:

Relative density = 0.7 or 70%

Explanation:

The following information was provided by this question

Pd = 1.72mg/mg³

Pd max = 1.81 mg/mg³

Pd min = 1.54 mg/mg³

We substitute into the formula. This formula is contained in the attachment.

[(1/1.54)-(1/1.72)]/[1/1.54 - 1/1.81]

= 0.649350 - 0.581395 / 0.649350 - 0.552486

= 0.067955/0.096864

= 0.7015

= 0.7

The relative density is Therefore 0.7 or 70% when converted to percentage

8 0

3 years ago

SMAW and GMAW use constant current arc welding machines.<br><br> True<br><br> False

Oksi-84 [34.3K]

It is true. Good luck

4 0

3 years ago