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3 years ago

When you come to an intersection, follow the _________ before you proceed.

Engineering

2 answers:

Elodia [21]3 years ago

7 0

Answer:

When you come to an intersection, follow this important rule (left, right, left). Look to the left first because vehicles coming from the left are closer to you. Look to the right. Take one more look to the left before you pull out.

Explanation:

Minchanka [31]3 years ago

7 0

Can you help me by solving this question

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Object has a mass of 0.5 kg and is 1.9 meters above the ground. What is PEg?

Ugo [173]

Answer:

Polyethylene glycol (PEG) is a biocompatible, synthetic, hydrophilic polyether compound that has many applications, mostly in the medical industry, but also in the chemical and industrial sectors.

Explanation:

6 0

2 years ago

Give a big-Oh characterization, in terms of n, of the running time of the Ex1-EX5 functions as shown in following Code Fragment

mash [69]

Answer:

Hello there, Please follow the step by step explanations for answer.

Explanation:

Hello there, Please follow the step by step explanations for answer.

Ex1 ( A):

it depends on n

=> O(n)

2.)

O(n/2) which is equal to O(n)

3.)O(n^2)

4.)O(2n) ==> O(n)

5.)O(n^3)

Also, see file attachment on this question for more clarity. Thanks and all the best.

8 0

3 years ago

Air enters a 200 mm diameter adiabatic nozzle at 195 deg C, 500 kPa and 100 m/s. It exits at 85 kPa. If the exit diameter is 158

ahrayia [7]

Answer:

$v_2 = 160.23 m/s$

$T_2 = 475.797 k$

Explanation:

given data:

Diameter = $d_1 = 200mm$

$t_1 =195 degree$

$p_1 =500 kPa$

$v_1 = 100m/s$

$p_2 = 85kPa$

$d_2 = 158mm$

from continuity equation

$A_1v_1 = A_2v_2$

$v_2 = \frac{\frac{\pi}{4}d_1^2 v_1^2}{\frac{\pi}{4}d_2^2}$

$v_2 = \frac{d_2v_1}{d_2^2}$

$v_2 = [\frac{d_1}{d_2}]^2 v_1$

$= [\frac{0.200}{0.158}]^2 \times 100$

$v_2 = 160.23 m/s$

by energy flow equation

$h_1 + \frac{v_1^2}{2} +gz_1 +q =h_2 + \frac{v_2^2}{2} +gz_2 +w$

$z_1 =z_2$ and q =0, w =0 for nozzle

therefore we have

$h_1 -h_2 =\frac{v_1^2}{2} -\frac{v_2^2}{2}$

$dh = \frac{1}{2} (v_1^2 -v_2^2)$

but we know dh = Cp dt

hence our equation become

$Cp(T_2 -T_1) = \frac{1}{2} (v_1^2 -v_2^2)$

$Cp (T_2 -T_1) = 7836.94$

$(T_2 -T_1) = \frac{7836.94}{1.005*10^3}$

$(T_2 -T_1) = 7.797$

$T_2 = 7.797 +468 = 475.797 k$

8 0

3 years ago

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pogonyaev

Are you ok? You don’t seem ok...

6 0

4 years ago

Read 2 more answers

The A/C system refrigerant charge amount needs to be determined on a late-model vehicle. Technician A says that the emissions la

suter [353]

Technician A is correct when he says that the emissions label contains the refrigerant charge amount while Technician B is wrong.

<h3>What is emission label?</h3>

An emissions label is known to be a kind of sticker that is often used by EPA to have manufacturers show that their non-road engines meet the needed emission requirements.

In terms of AC system, they are found under the hood of the AC unit and as such, Technician A is correct when he says that the emissions label contains the refrigerant charge amount while Technician B is wrong.

Learn more about refrigerant charge from

brainly.com/question/12823745

#SPJ1

4 0

2 years ago