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3 years ago

Help Please!!!!!!! Whatever3443 Please help!

Engineering

2 answers:

KIM [24]3 years ago

8 0

Answer:

Thanks for the pointssssssssssssss

Explanation:

Olin [163]3 years ago

7 0

Answer:

I need some help understanding this question what is it you need to know about doc. whatever 3443?

744 with 3040 1230 0 215 141 184 247 353 whatever 3443 1230 0 215 140 246 337 400 492 581 673 737 for 4232 1230 0 215 50 154 218 which 4499 1230

Explanation:

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g If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0

Anettt [7]

A) The amount of space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C is; 0.6048 cm

B) The stress in the rails on a summer day when their temperature is 31.0 °C is; 86.4 × 10⁶ Pa

<h3>Linear Thermal Expansion</h3>

We are given;

Length; L = 14 m

Initial Temperature; T_i = −5 °C

Final Temperature; T_f = 31 °C

The formula for Linear Thermal Expansion is;

ΔL = L_i * α * ΔT

where;

L_i is initial length

α is thermal expansion

ΔL is change in length

ΔT is change in temperature

Now, the thermal expansion of steel from online tables is α = 1.2 × 10⁻⁵ C⁻¹

Thus;

ΔL = 14 * 1.2 × 10⁻⁵ * (31 - (-5))

ΔL = 6.048 × 10⁻³ m = 0.6048 cm

The formula to get the stress is;

σ = Y * α * ΔT

where;

Y is young's modulus of steel = 20 × 10¹⁰ Pa

α is thermal expansion

ΔT is change in temperature

Thus;

σ = 20 × 10¹⁰ × 1.2 × 10⁻⁵ × (31 - (-5))

σ = 86.4 × 10⁶ Pa

The complete question is;

Steel train rails are laid in 14.0-m long segments placed end to end. The rails are laid on a winter day when their temperature is −5 °C.

(a) How much space must be left between adjacent rails if they are just to touch on a summer day when their temperature is 31.0 °C?

(b) If the rails are originally laid in contact, what is the stress in them on a summer day when their temperature is 31.0 °C?

Read more about Linear Thermal Expansion at; brainly.com/question/6985348

4 0

2 years ago

What type of engineer would be most likely to develop a design for cars? chemical civil materials mechanical

Andreyy89

I don’t know but good luck

4 0

3 years ago

A cylinder fitted with a frictionless piston contains 2 kg of R-134a at 3.5 bar and 100 C. The cylinder is now cooled so that th

inna [77]

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = $\int\limits^a_b P \, dV$ = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

5 0

3 years ago

The pressure of a gas in a rigid container is 125kpa at 300k, what we be the new pressure if the temperature increases to 900k

kipiarov [429]

Answer:

375 KPa

Explanation:

From the question given above, the following data were obtained:

Initial pressure (P₁) = 125 KPa

Initial temperature (T₁) = 300 K

Final temperature (T₂) = 900 K

Final pressure (P₂) =?

The new (i.e final) pressure of the gas can be obtained as follow:

P₁/T₁ = P₂/T₂

125 / 300 = P₂ / 900

Cross multiply

300 × P₂ = 125 × 900

300 × P₂ = 112500

Divide both side by 300

P₂ = 112500 / 300

P₂ = 375 KPa

Thus, the new pressure of the gas is 375 KPa

7 0

3 years ago

0/5 pts

Brilliant_brown [7]

Explanation:

150 divide by 150 and that how you do the is you what to divide together 15/ 150 you welcome have a good day is you need something else

4 0

3 years ago