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3 years ago

Always refill your gas tank well before

Engineering

1 answer:

Scorpion4ik [409]3 years ago

8 0

I believe it’s c because you don’t want your gas to run real low, so I think it’s best to do it when your fuel.

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Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th

jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

5 0

3 years ago

You were hoping to receive a scholarship for volleyball and are entering your personal information and statistics on a secure we

kozerog [31]

I think it is fair to say 5 and 3/4 inches.

6 0

3 years ago

Read 2 more answers

Find the following for an input of 120 VAC(RMS), 60 hertz, given a 10:1 stepdown transformer, and a full-wave bridge rectifier.

atroni [7]

Answer:

(i) 169.68 volt

(ii) 16.90 volt

(iii) 16.90 volt

(iv) 108.07 volt

(v) 2.161 A

Explanation:

Turn ratio is given as 10:1

We have given that input voltage $v_p=120volt$

(i) We know that peak voltage is give by $v_{peak}=\sqrt{2}v_p=\sqrt{2}\times 120=169.68volt$

(ii) We know that for transformer $\frac{v_p}{v_s}=\frac{n_p}{n_s}$

So $\frac{169.08}{v_s}=\frac{10}{1}$

$v_s=16.90volt$

So peak voltage in secondary will be 16.90 volt

(iii) Peak voltage of the rectifier will be equal to the peak voltage of the secondary

So peak voltage of the rectifier will be 16.90 volt

(iv) Dc voltage of the rectifier is given by $v_{dc}=\frac{2v_m}{\pi }=\frac{2\times 1.414\times 120}{3.14}=108.07volt$

(v) Now dc current is given by $i_{dc}=\frac{v_{dc}}{R}=\frac{108.07}{50}=2.1614A$

4 0

3 years ago

Two standard spur gears have a diametrical pitch of 10, a center distance 3.5 inches and a velocity ratio of 2.5. How many teeth

lubasha [3.4K]

Answer:50 , 20

Explanation:

Given

Diametrical Pitch $\left ( P_D\right )=\frac{T}{D}$

where T= no of teeths

D=diameter

module(m) of gears must be same

$m=\frac{D}{T}=\frac{1}{P_D}=0.1$

Let $T_1 & T_2$ be the gears on two gears

Therefore Center distance is given by

$m\frac{\left ( T_1+T_2\right )}{2}=3.5$

thus

$0.1\frac{\left ( T_1+T_2\right )}{2}=3.5$

$T_1+T_2=70----1$

and Velocity ratio is given by

$VR=\frac{No\ of\ teeths\ on\ Driver\ gear}{No.\ of\ teeths\ on\ Driven\ gear}$

$2.5=\frac{T_1}{T_2}----2$

From 1 & 2 we get

$T_1=50, T_2=20$

6 0

3 years ago

Summary of Possible Weather and Associated Aviation Impacts for Geographic/Topographic Categories Common in the Western United S

ruslelena [56]

Answer: answer provided in the explanation section.

Explanation:

Weather phenomenons that would impart Aviation Operations in Santa Barbara -

1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.

2. The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.

3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.

4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.

This and more are some factors to look into when considering wheather conditions that would affect aviation operations.

I hope this was a bit helpful. cheers

5 0

3 years ago