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3 years ago

How does a turbo charger work

Engineering

2 answers:

yKpoI14uk [10]3 years ago

8 0

Answer:

The turbocharger on a car applies a very similar principle to a piston engine. It uses the exhaust gas to drive a turbine. This spins an air compressor that pushes extra air (and oxygen) into the cylinders, allowing them to burn more fuel each second.

Explanation:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

Explanation:

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A cannon fires a ball vertically upward from the Earth’s surface. Which one of the following statements concerning the net force

trasher [3.6K]

Answer:

a) The net force on the ball is instantaneously equal to zero newtons at the top of the flight path.

Explanation:

At an instantenous time at the top of the flight path, the upward force due to the Canon explosion on the ball is just equal to the weight of the ball, this will equate the net force on the ball to zero. At this point the velocity of the ball is zero before it decends down to earth under its own weight.

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3 years ago

2. The unthreaded part of a bolt or screw is called the

aksik [14]

Answer:

The grip

Explanation:

the head of all headed bolt (except countersunk head bolt)

3 0

2 years ago

Read 2 more answers

Select the correct answer.

MAVERICK [17]

Answer:crane and engine I guess

Explanation:

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3 years ago

Hi, any kind of help on these questions will be appreciated.

Zielflug [23.3K]

Answer:

IDK

Explanation:

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3 years ago

A reservoir delivers water to a horizontal pipeline 39 long The first 15 m has a diameter of 50 mm, after which it suddenly beco

allsm [11]

Answer:

The difference of head in the level of reservoir is 0.23 m.

Explanation:

For pipe 1

$d_1=50 mm,f_1=0.0048$

For pipe 2

$d_2=75 mm,f_2=0.0058$

Q=2.8 l/s

$Q=2.8\times 10^{-3]$

We know that Q=AV

$Q=A_1V_1=A_2V_2$

$A_1=1.95\times 10^{-3}m^2$

$A_2=4.38\times 10^{-3} m^2$

$So V_2=0.63 m/s,V_1=1.43 m/s$

head loss (h)

$h=\dfrac{f_1L_1V_1^2}{2gd_1}+\dfrac{f_2L_2V_2^2}{2gd_2}+0.5\dfrac{V_1^2}{2g}$

Now putting the all values

$h=\dfrac{0.0048\times 15\times 1.43^2}{2\times 9.81\times 0.05}+\dfrac{0.0058\times 24\times 0.63^2}{2\times 9.81\times 0.075}+0.5\dfrac{1.43^2}{2\times 9.81}$

So h=0.23 m

So the difference of head in the level of reservoir is 0.23 m.

8 0

3 years ago