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2 years ago

Plan to refuel when your fuel gauge reads __________.

Engineering

2 answers:

lana66690 [7]2 years ago

8 0

I would start planning to refuel when the fuel gauge reads under 1/4 of a tank remaining

bazaltina [42]2 years ago

3 0

Answer:

low on fuel or if it's red

Explanation:

common sense to be honest :/

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(40%) A bank wants to store the account number of its customers (an 8-digit number) in encrypted form on magnetic stripe ATM car

9966 [12]

Answer:

Explanation:

Attached is the solution to the question

3 0

3 years ago

If you want to obtain a Class E license before you turn 18, you must______. A. be in compliance with school attendance requireme

Makovka662 [10]

Answer:

The correct option is;

A. be in compliance with school attendance requirements

Explanation:

The requirements to acquire a special restricted driver's license for driver's license for under under 18 that have had a beginner's for up to 180 days include a minimum of 40 hours driving practice 10 of which should be in the dark. The application for the conditional drivers license is to be signed by the parent or guardian and the application is to be accompanied with proof of acceptable school attendance

At 17, after holding the special restricted drivers licence for a year without issues you can obtain the full drivers license.

7 0

3 years ago

The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm,

babunello [35]

Answer:

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

Explanation:

given data

diameter = 85 mm

length = 2 m

depth = 9mm

N = 60 rev/min

pressure p = 11 × $10^6$ Pa

viscosity n = 100 Pas

angle = 18°

so Qd will be

Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1

put here value and we get

Qd = 0.5 × π² × ( 85 $\times 10^{-3}$ )²× 9 $\times 10^{-3}$ × sin18 × cos18

Qd = 94.305 × $10^{-6}$ m³/s

and

Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2

Qb = 11 × $10^{6}$ × π × 85 $\times 10^{-3}$ × ( 9 $\times 10^{-3}$ )³ × sin²18 ÷ 12 × 100 × 2

Qb = 85.2 × $10^{-6}$ m³/s

so here

volume flow rate Qx = Qd - Qb ..............3

Qx = 94.305 × $10^{-6}$ - 85.2 × $10^{-6}$

Qx = 9.10 $9.10^5 \times 10^{-6}$ m³/s

8 0

3 years ago

An inductor has a 50.0-Ω reactance when connected to a 60.0-Hz source. The inductor is removed and then connected to a 45.0-Hz s

nignag [31]

Given:

$X_{L} = 50.0 \ohm$

frequency, f = 60.0 Hz

frequency, f' = 45.0 Hz

$V_rms} = 85.0 V$

Solution:

To calculate max current in inductor, $I_{L(max)$ :

At f = 60.0 Hz

$X_{L} = 2\pi fL$

$50.0 = 2\pi\times 60.0\times L$

L = 0.1326 H

Now, reactance $X_{L}$ at f' = 45.0 Hz:

$X'_{L} = 2\pi f'L$

$X'_{L} = 2\pi\times 45.0\times 0.13263 = 37.5\ohm$

Now, $I_{L(max)$ is given by:

$I_{L(max) = \sqrt {\frac{2V_{rms}}{X'_{L}}}$

$I_{L(max) = \sqrt {\frac{2\times 85.0}{37.5}} = 2.13 A$

Therefore, max current in the inductor, $I_{L(max)$ = 2.13 A

7 0

3 years ago

Consider two Carnot heat engines operating in series. The first engine receives heat from the reservoir at 1400 K and rejects th

Aleksandr-060686 [28]

Answer:

The temperature T= 648.07k

Explanation:

T1=input temperature of the first heat engine =1400k

T=output temperature of the first heat engine and input temperature of the second heat engine= unknown

T3=output temperature of the second heat engine=300k

but carnot efficiency of heat engine = $1 - \frac{Tl}{Th} \\$

where Th =temperature at which the heat enters the engine

Tl is the temperature of the environment

since both engines have the same thermal capacities <em> $n_{th}$ </em> therefore $n_{th} =n_{th1} =n_{th2}\\n_{th }=1-\frac{T1}{T}=1-\frac{T}{T3}\\ \\= 1-\frac{1400}{T}=1-\frac{T}{300}\\$

We have now that

$\frac{-1400}{T}+\frac{T}{300}=0\\$

multiplying through by T

$-1400 + \frac{T^{2} }{300}=0\\$

multiplying through by 300

- $420000+ T^{2} =0\\T^2 =420000\\\sqrt{T2}=\sqrt{420000} \\T=648.07k$

The temperature T= 648.07k

5 0

3 years ago