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3 years ago

Three consequences of unbalanced wheels on a motor vehicle

1 answer:

8 0

Unstable car and loss of grip on the road. A hazard to other vehicles and pedestrians as well as a very shaky car and risk of your wheels rolling off and riding on rims.

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A nozzle receives an ideal gas flow with a velocity of 25 m/s, and the exit at 100 kPa, 300 K velocity is 250 m/s. Determine the

Margaret [11]

Given Information:

Inlet velocity = Vin = 25 m/s

Exit velocity = Vout = 250 m/s

Exit Temperature = Tout = 300K

Exit Pressure = Pout = 100 kPa

Required Information:

Inlet Temperature of argon = ?

Inlet Temperature of helium = ?

Inlet Temperature of nitrogen = ?

Answer:

Inlet Temperature of argon = 360K

Inlet Temperature of helium = 306K

Inlet Temperature of nitrogen = 330K

Explanation:

Recall that the energy equation is given by

$C_p(T_{in} - T_{out}) = \frac{1}{2} \times (V_{out}^2 - V_{in}^2)$

Where Cp is the specific heat constant of the gas.

Re-arranging the equation for inlet temperature

$T_{in} = \frac{1}{2} \times \frac{(V_{out}^2 - V_{in}^2)}{C_p} + T_{out}$

For Argon Gas:

The specific heat constant of argon is given by (from ideal gas properties table)

$C_p = 520 \:\: J/kg.K$

So, the inlet temperature of argon is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{520} + 300$

$T_{in} = \frac{1}{2} \times 119 + 300$

$T_{in} = 360K$

For Helium Gas:

The specific heat constant of helium is given by (from ideal gas properties table)

$C_p = 5193 \:\: J/kg.K$

So, the inlet temperature of helium is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{5193} + 300$

$T_{in} = \frac{1}{2} \times 12 + 300$

$T_{in} = 306K$

For Nitrogen Gas:

The specific heat constant of nitrogen is given by (from ideal gas properties table)

$C_p = 1039 \:\: J/kg.K$

So, the inlet temperature of nitrogen is

$T_{in} = \frac{1}{2} \times \frac{(250^2 - 25^2)}{1039} + 300$

$T_{in} = \frac{1}{2} \times 60 + 300$

$T_{in} = 330K$

Note: Answers are rounded to the nearest whole numbers.

5 0

3 years ago

What effect does air have on the acceleration of aircraft during flight?

scoundrel [369]

The effect would be the altitude of the air, the higher you go up the closer you are to space we’re there’s no oxygen and everything moves slow so when your trying to fly across the world it could feel like your moving slower

5 0

3 years ago

Experimental Design Application Production engineers wish to find the optimal process for etching circuit boards quickly. They c

Veseljchak [2.6K]

Answer:

Hello your question is incomplete attached below is the missing part and answer

options :

Effect A

Effect B

Effect C

Effect D

Effect AB

Effect AC

Effect AD

Effect BC

Effect BD

Effect CD

Answer :

A = significant

B = significant

C = Non-significant

D = Non-significant

AB = Non-significant

AC = significant

AD = Non-significant

BC = Non-significant

BD = Non-significant

CD = Non-significant

Explanation:

The dependent variable here is Time

Effect of A = significant

Effect of B = significant

Effect of C = Non-significant

Effect of D = Non-significant

Effect of AB = Non-significant

Effect of AC = significant

Effect of AD = Non-significant

Effect of BC = Non-significant

Effect of BD = Non-significant

Effect of CD = Non-significant

8 0

3 years ago

State the four advantages of levers

dezoksy [38]

Answer:

Here are 2 sense i cant find 4

Explanation:

Levers are used to multiply force, In other words, using a lever gives you greater force or power than the effort you put in.

In a lever, if the distance from the effort to the fulcrum is longer than the distance from the load to the fulcrum, this gives a greater mechanical advantage.

3 0

3 years ago

Calculate the radius of gyration for a bar of rectangular cross section with thickness t and width w for bending in the directio

SVEN [57.7K]

Answer:

a)R= sqrt( wt³/12wt)

b)R=sqrt(tw³/12wt)

c)R= sqrt ( wt³/12xcos45xwt)

Explanation:

Thickness = t

Width = w

Length od diagonal =sqrt (t² +w²)

Area of raectangle = A= tW

Radius of gyration= r= sqrt( I/A)

Moment of inertia in the direction of thickness I = w t³/12

R= sqrt( wt³/12wt)

Moment of inertia in the direction of width I = t w³/12

R=sqrt(tw³/12wt)

Moment of inertia in the direction of diagonal I= (w t³/12)cos 45=( wt³/12)x 1/sqrt (2)

R= sqrt ( wt³/12xcos45xwt)

4 0

3 years ago

Three consequences of unbalanced wheels on a motor vehicle​

Three consequences of unbalanced wheels on a motor vehicle