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3 years ago

You must yield the right-of-way to all of the following EXCEPT:

Engineering

1 answer:

Elena-2011 [213]3 years ago

7 0

Answer:

You must yield the right-of-way to a police car, fire engine, ambulance, or other emergency vehicle that uses a siren and flashing lights. Pull as close to the right of the road as possible and stop until the emergency vehicle(s) has passed. However, don't stop in an intersection because that would cause a wreck.

Explanation:

please have a good day. and also mark as brainllest bye

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A particle moving on a straight line has acceleration a = 5-3t, and its velocity is 7 at time t = 2. If s(t) is the distance fro

Vikki [24]

Given acceleration a = 5-3t, and its velocity is 7 at time t = 2, the value of s2 - s1 = 7

<h3>How to solve for the value of s2 - s1</h3>

We have

= $\frac{dv}{dt} =v't = 5-3t\\\\\int\limits^a_b {v'(t)} \, dt$

$= \int\limits^a_b {(5-3t)} \, dt$

$5t - \frac{3t^2}{2} +c$

v2 = 5x2 - 3x2 + c

= 10-6+c

= 4+c

$s(t) = \frac{5t^2}{2} -\frac{t^3}{2} +3t + c$

S2 - S1

$=(5*\frac{4}{2} -\frac{8}{2} +3*2*c)-(\frac{5}{2} *1^2-\frac{1^2}{2} +3*1*c)$

= 6 + 6+c - 2+3+c

12+c-5+c = 0

7 = c

Read more on acceleration here: brainly.com/question/605631

5 0

2 years ago

A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th

SCORPION-xisa [38]

Answer:

$F_{thrust}$ ≅ 111 KN

Explanation:

Given that;

A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8

mass = 85,000 kg

drag co-efficient (C) = 0.37

(velocity (v)= 230 m/s

density (ρ) = 1.0 kg/m³

To calculate the thrust; we need to determine the relation of the drag force; which is given as:

$F_{drag}$ = $\frac{1}{2}$ × CρAv²

where;

ρ = density of air wind.

C = drag co-efficient

A = Area of the jet

v = velocity of the jet

From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0

SO, $F_{drag}-F_{thrust} = 0$

We can as well say:

$F_{drag}= F_{thrust}$

We can now replace $F_{thrust} with F_{drag}$ in the above equation.

Therefore, $F_{thrust}$ = $\frac{1}{2}$ × CρAv²

The A which stands as the area of the jet is given by the formula:

$A=\frac{\pi d^2}{4}$

We can now have a new equation after substituting our A into the previous equation as:

$F_{thrust}$ = $\frac{1}{2}$ × Cρ $(\frac{\pi d^2}{4})v^2$

Substituting our data from above; we have:

$F_{thrust}$ = $\frac{1}{2}$ × $(0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2$

$F_{thrust}$ = $\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2$

$F_{thrust}$ = 110,990N

$F_{thrust}$ in N (newton) to KN (kilo-newton) will be:

$F_{thrust}$ = $(110,990N)*\frac{1KN}{1,000N}$

$F_{thrust}$ = 110.990 KN

$F_{thrust}$ ≅ 111 KN

In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.

5 0

3 years ago

How is the difference between science and engineering Best stated?

stiv31 [10]

Answer:Science is the body of knowledge that explores the physical and natural world. Engineering is the application of knowledge in order to design, build and maintain a product or a process

Explanation:

8 0

2 years ago

Suppose we have a database for an investment firm, consisting of the following attributes: B (broker), O (office of a broker), I

Snowcat [4.5K]

Answer:

Given, FDs are:

S -> D

I -> B

IS -> Q

B -> O

"I" and "S" must be there in any candidate key because they do not appear on the right side of any functional dependency.

The only candidate key is: IS

IS -> ISBDQO

Decomposition of R into 3NF: (I, B), (S, D), (B, O), (I, S, Q)

Decomposition of R into BCNF:

Decompose R by I → B into R1 = (I, B) and R2 = (I, O, S, Q, D).

R1 is in BCNF

Decompose R2 by S → D into R21 = (S, D) and R22 = (O, I, S, Q).

R21is in BCNF

Decompose R22 by I → O into R221 = (I, O) and R222 = (I, S, Q).

R221 is in BCNF.

R222 is in BCNF.

The decomposition is: (I, B), (S, D), (I, O), (I, S, Q)

We can also write it as: (I, B), (S, D), (B, O), (I, S, Q)

Explanation:

The answer above is rendered in a very explanatory way.

8 0

3 years ago

A pump with a power of 5 kW (pump power, and not useful pump power) and an efficiency of 72 percent is used to pump water from a

almond37 [142]

Answer:

a) The mass flow rate of water is 14.683 kilograms per second.

b) The pressure difference across the pump is 245.175 kilopascals.

Explanation:

a) Let suppose that pump works at steady state. The mass flow rate of the water ( $\dot m$ ), in kilograms per second, is determined by following formula: