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Hitman42 [59]
2 years ago
12

Provide one example of a bad collision, and suggest an engineering solution to avoid the collision.

Engineering
1 answer:
JulsSmile [24]2 years ago
7 0

Answer:

1). Keep your distance. Drive far enough behind the car in front of you so you can stop safely. ...

Drive strategically. Avoid situations that could force you to suddenly use your brakes. ...

Don't get distracted. ...

Don't drive when drowsy or under the influence.

2). By far the deadliest accident type is the head-on collision. Head-on collisions consider both vehicle's speed at the time of the crash, which means even an accident at lower speeds can be catastrophic

Explanation:

first is how to avoid the collision and second is bad collision

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A wing generates a lift L when moving through sea-level air with a velocity U. How fast must the wing move through the air at an
vredina [299]

Answer:

V1 = 1.721 * V2

Explanation:

To start with, we assume that both lift forces are equal, such that

L2 = L1

1 is that of the level at 10000 m, and 2 is that of the level at sea level.

Next, we try and substitute the general formula for both forces such that

C(l).ρ1/2.V1².A = C(l).ρ2/2.V2².A

On further simplification, we have

ρ1.V1² = ρ2.V2², making V1 subject of formula, we have

V1 = √(ρ2/ρ1). V2²

Using the values of density for air at 10000 m and at sea level(source is US standard atmosphere), we have

V1 = √(1.225/0.4135) * V2

V1 = √2.9625 * V2

V1 = 1.721 * V2

4 0
2 years ago
Which emission is created by the heat in the combustion process?
Musya8 [376]
Water vapor and carbon dioxide!
5 0
3 years ago
Read 2 more answers
An add tape of 101 ft is incorrectly recorded as 100 ft for a 200-ft distance. What is
baherus [9]

Answer:

the correct distance is 202 ft

Explanation:

The computation of the correct distance is shown below:

But before that correction to be applied should be determined

= (101 ft - 100 ft) ÷ (100 ft) × 200 ft

= 2 ft

Now the correct distance is

= 200 ft +  2 ft

= 202 ft

Hence, the correct distance is 202 ft

The same would be relevant and considered too

4 0
3 years ago
Select the correct answer.
juin [17]
Orthographic projection, common method of representing three-dimensional objects, usually by three two-dimensional drawings in each of which the object is viewed along parallel lines that are perpendicular to the plane of the drawling.
4 0
2 years ago
Two balanced Y-connected loads in parallel, one drawing 15kW at 0.6 power factor lagging and the other drawing 10kVA at 0.8 powe
NemiM [27]

Answer:

(a) attached below

(b) pf_{C}=0.85 lagging

(c) I_{C} =32.37 A

(d) X_{C} =49.37 Ω

(e) I_{cap} =9.72 A and I_{line} =27.66 A

Explanation:

Given data:

P_{1}=15 kW

S_{2} =10 kVA

pf_{1} =0.6 lagging

pf_{2}=0.8 leading

V=480 Volts

(a) Draw the power triangle for each load and for the combined load.

\alpha_{1}=cos^{-1} (0.6)=53.13°

\alpha_{2}=cos^{-1} (0.8)=36.86°

S_{1}=P_{1} /pf_{1} =15/0.6=25 kVA

Q_{1}=P_{1} tan(\alpha_{1} )=15*tan(53.13)=19.99 ≅ 20kVAR

P_{2} =S_{2}*pf_{2} =10*0.8=8 kW

Q_{2} =P_{2} tan(\alpha_{2} )=8*tan(-36.86)=-5.99 ≅ -6 kVAR

The negative sign means that the load 2 is providing reactive power rather than consuming  

Then the combined load will be

P_{c} =P_{1} +P_{2} =15+8=23 kW

Q_{c} =Q_{1} +Q_{2} =20-6=14 kVAR

(b) Determine the power factor of the combined load and state whether lagging or leading.

S_{c} =P_{c} +jQ_{c} =23+14j

or in the polar form

S_{c} =26.92°

pf_{C}=cos(31.32) =0.85 lagging

The relationship between Apparent power S and Current I is

S=VI^{*}

Since there is conjugate of current I therefore, the angle will become negative and hence power factor will be lagging.

(c) Determine the magnitude of the line current from the source.

Current of the combined load can be found by

I_{C} =S_{C}/\sqrt{3}*V

I_{C} =26.92*10^3/\sqrt{3}*480=32.37 A

(d) Δ-connected capacitors are now installed in parallel with the combined load. What value of capacitive reactance is needed in each leg of the A to make the source power factor unity?Give your answer in Ω

Q_{C} =3*V^2/X_{C}

X_{C} =3*V^2/Q_{C}

X_{C} =3*(480)^2/14*10^3 Ω

(e) Compute the magnitude of the current in each capacitor and the line current from the source.

Current flowing in the capacitor is  

I_{cap} =V/X_{C} =480/49.37=9.72 A

Line current flowing from the source is

I_{line} =P_{C} /3*V=23*10^3/3*480=27.66 A

8 0
3 years ago
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