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2 years ago

if you help then I will thank u by sooo much I will give tons of points but the answer has to be right.

Engineering

2 answers:

romanna [79]2 years ago

8 0

Answer:

The first one A, " These are lines of equal air pressure".

Explanation:

emmainna [20.7K]2 years ago

8 0

Answer: A
Isobars are lines on a weather map joining together places of equal atmospheric pressure.

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A pressure gage at the inlet to a gas compressor indicates that the gage pressure is 40.0 kPa. Atmospheric pressure is 1.01 bar.

bonufazy [111]

Answer:

Given

inlet Pga =40kpa = 40000pa

Patm=1.01bar = 1.01 x 100000pa =101000pa

exit Pab= 6.5 (inlet Pab)

But generally, Pab = Patm + Pga

1. the absolute pressure of the gas at the inlet, inlet Pab?

inlet Pab = Patm + inlet Pga

= 101000pa + 40000pa = 141kpa

the absolute pressure of the gas at the inlet, inlet Pab = 141kpa

2. the gage pressure of the gas at the exit? exit Pga?

exit Pab = Patm + exit Pga

exit Pga = exit Pab - Patm

= (6.5 x 141kpa) - 101kpa

= 815.5kpa

the gage pressure of the gas at the exit exit Pga=815.5kpa

5 0

3 years ago

A six-lane divided highway (three lanes in each direction) is on rolling terrain with two access points per mile and has 10- ft

tatuchka [14]

Answer

given,

6 lanes divided highway 3 lanes in each direction

rolling terrain

lane width = 10'

shoulder on right = 5'

PHF = 0.9

shoulder on the left direction = 3'

peak hour volume = 3500 veh/hr

large truck = 7 %

tractor trailer = 3 %

speed = 55 mi/h

LOS is determined based on V p

10' lane weight ; f_{Lw}=6.6 mi/h

5' on right ; f_{Lc} = 0.4 mi/hr

3' on left ; no adjustment

3 lanes in each direction f n = 3 mi/h

$v_p =\dfrac{V}{f_{HV}\times N\times f_p\times PHF}$

$f_{HV}=\dfrac{1}{1+P_T(E_T-1)+P_R(E_R-1)}$

$f_{HV}=\dfrac{1}{1+0.08(2.5-1)+0.02(2-1)}$

= 0.877

$v_p =\dfrac{3500}{0.877\times 3\times 0.95\times0.9}$

= 1,555 veh/hr/lane

$FFS = BFFS - F_{Lw}-F_{Lc}-F_{N}-F_{ID}$

= (55 + 5) - 6.6 - 0.4 -3 -0

= 50 mi/h

$D = \dfrac{V_P}{s}$

$D = \dfrac{1555}{55} =28.27$

level of service is D using speed flow curves and LOS for basic free moving of vehicle

5 0

3 years ago

Coal fire burning at 1100 k delivers heat energy to a reservoir at 500 k. Find maximum efficiency.

Marizza181 [45]

Answer:

55%

Explanation:

hot reservoir = 1100 K

cold reservoir = 500 K

This is a Carnot system

For a Carnot system, maximum efficicency of the system is given as

Eff = 1 - $\frac{Tc}{Th}$

where Tc = temperature of cold reservoir = 500K

Th = temperature of hot reservoir = 1100 K

Eff = 1 - $\frac{500}{1100}$

Eff = 1 - 0.45 = 0.55 or 55%

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2 years ago

Line(s) indicates passing is allowed if there are no oncoming cars.

anygoal [31]

Broken yellow b/c you can’t pass on a double solid yellow

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3 years ago

A three-point flexure test is conducted on a cylindrical specimen of aluminum oxide. The specimen radius is 5.0 mm and the dista

kondaur [170]

Answer:

Detailed solution is given in the attached diagram

7 0

3 years ago