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3 years ago

Where are the ar manufacturers not fitting the engine in the high end sport cars

Engineering

1 answer:

fomenos3 years ago

8 0

Answer:

it depends on the but i would recommend check in the front next to the turbo intake.

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In highways the far left lane is usually the _____

Ivan

Fastest

(Known as the fast lane)

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3 years ago

Read 2 more answers

In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water

Orlov [11]

Answer:

The amount of energy transferred to the water is 4.214 J

Explanation:

The given parameters are;

The mass of the object that drops = 5 kg

The height from which it drops = 86 mm (0.086 m)

The potential energy P.E. is given by the following formula

P.E = m·g·h

Where;

m = The mass of the object = 5 kg

g = The acceleration de to gravity = 9.8 m/s²

h = The height from which the object is dropped = 0.086 m

Therefore;

P.E. = 5 kg × 9.8 m/s² × 0.086 m = 4.214 J

Given that the potential energy is converted into heat energy, that raises the 1 g of water by 1°C, we have;

The amount of energy transferred to the water = The potential energy, P.E. = 4.214 J.

6 0

3 years ago

Write a class named FBoard for playing a game, where player x is trying to get her piece to row 7 and player o is trying to make

OverLord2011 [107]

Answer:

12345=V

Explanation:

7 0

4 years ago

Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tin

Marina CMI [18]

Solution:

Given that :

Volume flow is, $Q_1 = 1000 \ mm^3/s$

So, $Q_2= \frac{1000}{100}=10 \ mm^3/s$

Therefore, the equation of a single straight vessel is given by

$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$ ......................(i)

So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is

$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4}$

or $d_1=10 \ d_2$

Now for parallel pipes

$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$ ...........(ii)

Solving the equations (i) and (ii),

$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$

$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$

$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$

$=\frac{10^6}{10^7}$

Therefore,

$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$

or $H_{f_2}=10 \ H_{f_1}$

Thus the answer is option A). 10

7 0

3 years ago

The position of a particle is given by s = 0.27t

Natali [406]

Sorry bro people do this22.2 pls

8 0

2 years ago