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Ann [662]
3 years ago
9

I need solution fast plesss​

Engineering
1 answer:
alexandr1967 [171]3 years ago
6 0

Answer:

I hope this helps mark me as braniest

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A steel rod, which is free to move, has a length of 200 mm and a diameter of 20 mm at a temperature of 15oC. If the rod is heate
kherson [118]

Explanation:

thermal expansion ∝L = (δL/δT)÷L ----(1)

δL = L∝L + δT ----(2)

we have δL = 12.5x10⁻⁶

length l = 200mm

δT = 115°c - 15°c = 100°c

putting these values into equation 1, we have

δL = 200*12.5X10⁻⁶x100

= 0.25 MM

L₂ = L + δ L

= 200 + 0.25

L₂ = 200.25mm

12.5X10⁻⁶ *115-15 * 20

= 0.025

20 +0.025

D₂ = 20.025

as this rod undergoes free expansion at 115°c, the stress on this rod would be = 0

3 0
3 years ago
We have a tube with a diameter of 5 inches that is 1 foot long. The tube then reduces the diameter to 3 inches. According to the
Ksju [112]
Can I get a picture? Of the question so I can understand it more?
6 0
3 years ago
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One kilogram of water fills a 150-L rigid container at an initial pressure of 2 MPa. The container is then cooled to 40∘C. Deter
lukranit [14]

The pressure of water is 7.3851 kPa

<u>Explanation:</u>

Given data,

V = 150×10^{-3} m^{3}

m = 1 Kg

P_{1} = 2 MPa

T_{2}  = 40°C

The waters specific volume is calculated:

v_{1} = V/m

Here, the waters specific volume at initial condition is v_{1}, the containers volume is V, waters mass is m.

v_{1} = 150×10^{-3} m^{3}/1

v_{1} = 0.15 m^{3}/ Kg

The temperature from super heated water tables used in interpolation method between the lower and upper limit for the specific volume corresponds 0.15 m^{3}/ Kg and 0.13 m^{3}/ Kg.

T_{1}= 350+(400-350) \frac{0.15-0.13}{0.1522-0.1386}

T_{1} = 395.17°C

Hence, the initial temperature is 395.17°C.

The volume is constant in the rigid container.

v_{2} = v_{1}= 0.15 m^{3}/ Kg

In saturated water labels for T_{2}  = 40°C.

v_{f} = 0.001008 m^{3}/ Kg

v_{g} = 19.515 m^{3}/ Kg

The final state is two phase region v_{f} < v_{2} < v_{g}.

In saturated water labels for T_{2}  = 40°C.

P_{2} = P_{Sat} = 7.3851 kPa

P_{2} = 7.3851 kPa

7 0
3 years ago
Solid Isomorphous alloys strength
Sphinxa [80]

Answer:

Explanation:

ℎ

3 0
3 years ago
According to the amortization table, Demarco and Tanya will pay a total of in interest over the life of their loan.
Ymorist [56]

Answer:

(Interest rate/number of payments)*$170000= interest for the first month.

Interest amounts for all the months of repayment plus $170000=Total loan cost

Explanation:

Interest is the amount you pay for taking a loan from a bank on top of the original amount borrowed.

Factors affecting how much interest is paid are; the principal amount, the loan terms, repayment schedule, the repayment amount and the rate of interest.

The interest paid=(rate of interest/number of payments to make)*principal amount borrowed.

You divide the interest with number of payments done in a year where monthly are divided by 12.Multiplying it by loan balance in the first month which is your principal amount gives the interest rate to pay for that month.

You new loan balance will be= Principal -(repayment-interest)

Do this for the period the loan should take.

Add all the interest amount to original borrowed amount to get total cost of the loan after the period of time.

8 0
3 years ago
Read 2 more answers
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