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3 years ago

A current vehicle registration expires at _____ of the first owner listed on the registration form

Engineering

1 answer:

wariber [46]3 years ago

3 0

Answer:

Midnight on the birthday. Hope this helps!

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6. Driving with parking lights only (in place of headlights) is against the law. A. True B. False

trasher [3.6K]

Answer:

B false it is illegal to only have got fog lights on though and bright headlights because it can distract other drivers going last and if the y are distracted then that will cause a collision

Hope this helps :)

Explanation:

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3 years ago

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Evan notices a small fire in his workplace. Since the fire is small and the atmosphere is not smoky he decides to fight the fire

Norma-Jean [14]

Answer:

not calling the firemean

Explanation:

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3 years ago

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$ ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

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3 years ago

For arwa and yeahahhahahhhhhhhhhhhhh

Setler79 [48]

Chicken nugget baby sauce

3 0

3 years ago

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A generator operating at 50 Hz delivers 1 pu power to an infinite bus through a transmission circuit in which resistance is igno

olga55 [171]

Answer:

critical clearing angle = 70.3°

Explanation:

Generator operating at = 50 Hz

power delivered = 1 pu

power transferable when there is a fault = 0.5 pu

power transferable before there is a fault = 2.0 pu

power transferable after fault clearance = 1.5 pu

using equal area criterion to determine the critical clearing angle

Attached is the power angle curve diagram and the remaining part of the solution.

The power angle curve is given as

= Pmax sinβ

therefore : 2sinβo = Pm

2sinβo = 1

sinβo = 0.5 pu

βo = $sin^{-1} (0.5) = 30$ ⁰

also ; 1.5sinβ1 = 1

sinβ1 = 1/1.5

β1 = $sin^{-1} (\frac{1}{1.5} )$ = 41.81⁰

∴ βmax = 180 - 41.81 = 138.19⁰

attached is the remaining solution

The critical clearing angle = $cos^{-1} 0.3372$ ≈ 70.3⁰

3 0

3 years ago