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CaHeK987 [17]
3 years ago
9

An airplane travels 2100 km at 1000km/hE. It encounters a wind and slows to 800 km/h E for the next 1300 km. What is the average

velocity of the airplane for this trip?
Chemistry
1 answer:
Deffense [45]3 years ago
7 0

Answer:

The average velocity of the airplane for this trip is 1684.21 km/h

Explanation:

Average velocity is the rate of change of displacement with time. That is,

Average velocity = \frac{Displacement }{Change in time} = Δx / Δt = \frac{x2 - x1}{t2 - t1}

Now we will calculate the time taken by the airplane for the first motion before it encounters a wind.

From,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{2100km }{1000km/h}

Time = 2.1h

This is the time taken before the airplane encounters a wind.

Hence, t1 = 2.1h

Now, For the time taken by the airplane when it encounters a wind

Also from,

Velocity = \frac{Distance traveled}{Time taken}

Time = \frac{Distance traveled}{Velocity}

Therefore, Time = \frac{1300km }{800km/h}

Time = 1.625h

Hence, t2 = 1.625h

Now, to calculate the average velocity

Average velocity = \frac{x2 - x1}{t2 - t1}

x1= 2100, x2= 1300, t1= 2.1h and t2= 1.625h

Hence, Average velocity = \frac{1300 - 2100}{1.625 - 2.1}

Average velocity = 1684.21 km/h

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Answer:

The principal elements are Iron, Nickel, Chromium

It has 0.37% to 0.43% Carbon

It is hardenable.

It can be hardened by cold working, annealing, quenching.

Explanation:

A)

The chemical composition of AISI 4340 Steel are as follows:

Iron  ---  95% to 96%

Nickel  ---  1.6% to 2.0%

Chromium  ---  0.7% to 0.9%

Manganese  ---  0.6% to 0.8%

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Molybdenum  ---  0.2% to 0.3%

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Phosphorous  --- 0.0350%

So, the principal alloy elements from this composition are <u>Iron, Nickel and Chromium</u>

B)

The carbon content in this alloy is <u>0.37% to 0.43%</u>

C)

<u>Yes, it can be hardened</u>.

D)

<u>It can be Hardened by Cold Working, Annealing or Quenching</u>.

3 0
3 years ago
F a sample of butene (C4H8) that has a mass of 136.6 g is combusted in excess oxygen, what is the mass of CO2 that is produced?
alukav5142 [94]

The balanced chemical reaction will be:

C4H8 + 6 O2 --> 4 CO2 + 4 H2O

We are given the amount of butene being combusted. This will be our starting point.

136.6 g C4H8 (1 mol  C4H8/ 56.11 g C4H8) (4 mol CO2/1 mol <span>C4H8</span>) ( 44.01 g CO2/ 1 mol CO2) = 428.6 g CO2
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Answer:

2

0.4167\ \text{M}^{-1}\text{min}^{-1}

Explanation:

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{t_{1/2}}B=4\ \text{min}

Concentration

{[A]_0}_A=1.2\ \text{M}

{[A]_0}_B=0.6\ \text{M}

We have the relation

t_{1/2}\propto \dfrac{1}{[A]_0^{n-1}}

So

\dfrac{{t_{1/2}}_A}{{t_{1/2}}_B}=\left(\dfrac{{[A]_0}_B}{{[A]_0}_A}\right)^{n-1}\\\Rightarrow \dfrac{2}{4}=\left(\dfrac{0.6}{1.2}\right)^{n-1}\\\Rightarrow \dfrac{1}{2}=\left(\dfrac{1}{2}\right)^{n-1}

Comparing the exponents we get

1=n-1\\\Rightarrow n=2

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8 0
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Q = 2·10⁻⁹; <span> the ion product.</span>

8 0
3 years ago
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