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3 years ago

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what would my engine exhaust sound like if i made a custom muffler that was just smooth on the inside, would it have an echoing

sound? the headers and the pipes would be smooth too (after market) on a nissan 350z would that make it rumble instead of the trumpet/trumbone sound?

1 answer:

Elena L [17]3 years ago

8 0

Answer:

you know what sounds rlly nice? first have you seent he new corvettes? next u should here a nisan gtr. (like tanner foxes)

Explanation:

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For which of 'water' flow velocities at 200C can we assume that the flow is incompressible ? a.1000 km per hour b. 500 km per ho

ad-work [718]

Answer:d

Explanation:

Given

Temperature $=200^{\circ}\approc 473 K$

Also $\gamma for air=1.4$

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no. $=\frac{V}{\sqrt{\gamma RT}}$

(a) $1000 km/h\approx 277.78 m/s$

Mach no. $=\frac{277.78}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.63

(b) $500 km/h\approx 138.89 m/s$

Mach no. $=\frac{138.89}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.31

(c) $2000 km/h\approx 555.55 m/s$

Mach no. $=\frac{555.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=1.27

(d) $200 km/h\approx 55.55 m/s$

Mach no. $=\frac{55.55}{\sqrt{1.4\times 287\times 473}}$

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

5 0

3 years ago

Calculate the current in the 8-W resistor of Figure below by using Thevenin’s theorem. What will be its value of connections of

kvv77 [185]

solution from c hegg. hope it helps. see photos explanation

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3 years ago

A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for

Aleks [24]

Answer:

The distance between the station A and B will be:

$x_{A-B}=55.620\: km$

Explanation:

Let's find the distance that the train traveled during 60 seconds.

$x_{1}=x_{0}+v_{0}t+0.5at^{2}$

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

$x_{1}=\frac{1}{2}(0.6)(60)^{2}$

$x_{1}=1080\: m$

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

$v_{1}=v_{0}+at$

$v_{1}=(0.6)(60)=36\: m/s$

Then the second distance will be:

$x_{2}=v_{1}*1500$

$x_{2}=(36)(1500)=54000\: m$

The final distance is calculated whit the decelerate value:

$v_{f}^{2}=v_{1}^{2}-2ax_{3}$

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

$0=36^{2}-2(1.2)x_{3}$

$x_{3}=\frac{36^{2}}{2(1.2)}$

$x_{3}=540\: m$

Therefore, the distance between the station A and B will be:

$x_{A-B}=x_{1}+x_{2}+x_{3}$

$x_{A-B}=1080+54000+540=55.620\: km$

I hope it helps you!

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3 years ago

Which website suffixes are usually the least credible? Check all that apply.

zmey [24]

.com hope this helps

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3 years ago

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rt this question on a NEW page Give ONE word/tem for each of the following descriptions by choosing a word/term from the list be

Shalnov [3]

Answer:

oh I am sorry I can't understand your question.

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3 years ago