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Andrews [41]
3 years ago
10

Why does my car accelerate slowly when i press the gas

Engineering
1 answer:
olya-2409 [2.1K]3 years ago
6 0
If the pump stops working, malfunctions, or becomes clogged, it will not supply enough fuel to the injectors.
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What is the physical significance of the Reynolds number?. How is defined for external flow over a plate of length L.
yanalaym [24]

Answer:

Re=\dfrac{\rho\ v\ l}{\mu }

Explanation:

Reynolds number:

  Reynolds number describe the type of flow.If Reynolds number is too high then flow is called turbulent flow and Reynolds is  low then flow is called laminar flow .

Reynolds number is a dimensionless number.Reynolds number given is the ratio of inertia force to the viscous force.

Re=\dfrac{F_i}{F_v}

For plate can be given as

Re=\dfrac{\rho\ v\ l}{\mu }

Where  ρ is the density of fluid , v is the average velocity of fluid and μ is the dynamic viscosity of fluid.

Flow on plate is a external flow .The values of Reynolds number for different flow given as

Reynolds\ number\is \ >\ 5 \times 10 ^5\ then\ flow\ will\ be\ turbulent.

Reynolds\ number\is \

7 0
3 years ago
ANSWER QUICK<br>Why did Winston Churchill take over for Neville Chamberlain shortly after ww2? ​
kenny6666 [7]

Answer:neville chamberlain died

Explanation:

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4 years ago
PELASEE HELPPP WITH MARK BRAINLIST!!!! You are stopped at a red light, and a long string of cars is crossing in front of you. Wh
choli [55]

Answer:

1st one.

Explanation:

I think that because they the one who is going to get a ticket and also you will not gonna get in a car accident.

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3 years ago
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Answer:

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8 0
3 years ago
Suppose that the material to be used in a fuse melts when the current density rises to 459 A/cm2. What diameter of cylindrical w
I am Lyosha [343]

Answer:

The required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

Explanation:

We are given current density of 459 A/cm²  and we want to limit the current to 0.53 A in a fuse wire. We are asked to find the corresponding diameter of the fuse wire.

Recall that current density is given by

j = I/A

where I is the current flowing through the wire and A is the area of the wire

A = πr²

but r = d/2  so

A = π(d/2)²

A = πd²/4

so the equation of current density becomes

j = I/πd²/4

j = 4I/πd²

Re-arrange the equation for d

d² = 4I/jπ  

d = √4I/jπ

d = √(4*0.53)/(459π)

d = 0.0383 cm

Therefore, the required diameter of the fuse wire should be 0.0383 cm  to limit the current to 0.53 A with current density of 459 A/cm² .

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3 years ago
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