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kow [346]
3 years ago
10

10 properties of metals? ​

Engineering
2 answers:
Mrrafil [7]3 years ago
6 0

Answer:

1. they are malleable

2. they are ductile

3. they are good conductor of electricity and heat

4. they are solid at room temperature except mercury

5. they are sonorous

6. they are hard

7. they can be hammered into small sheets

8. they are lustrous

9. they are strong

10. they have high density

Misha Larkins [42]3 years ago
4 0

Answer:

Physical Properties of Metals:

Metals can be hammered into thin sheets. ...

Metals are ductile. ...

Metals are a good conductor of heat and electricity.

Metals are lustrous which means they have a shiny appearance.

Metals have high tensile strength. ...

Metals are sonorous. ...

Metals are hard.

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Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while t
Andrej [43]

Answer:

q' = 5826 W/m

Explanation:

Given:-

- The length of the rectangular fin, L = 0.15 m

- The surface temperature of fin, Ts = 250°C

- The free stream velocity of air, U = 80 km/h

- The temperature of air, Ta = 27°C

- Parallel flow over both surface of the fin, assuming turbulent conditions through out.

Find:-

What is the rate of heat removal per unit width of the fin?

Solution:-

- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.

- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):

    Dynamic viscosity , v = 27.85 * 10^-6 m^2/s  

    Thermal conductivity, k = 0.0346 W / m.K

    Prandlt number Pr = 0.69

- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:

                          Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}

Where,    Re_L: The average Reynolds number for the entire length of fin:

                          Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909

Therefore,

                         Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378

- The convection coefficient (h) can now be determined from:

                          h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}

- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:

                          q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}

- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m

- The heat loss per unit width (q') due to radiation:

                  q' = 2*a*T_s^4*L

Where, a: Stefan boltzman constant = 5.67*10^-8

                  q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}

- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.  

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3 years ago
(I really need help ASAP please!! this is for science her is the problem)
grandymaker [24]

Answer:

Explanation:

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2 years ago
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A car is traveling at 36 km/h on an acceleration lane to a freeway. What acceleration is required to obtain a speed of 72 km/h i
dmitriy555 [2]

First, write down the information given and the change units if necessary (we must have similar units to operate on).

Initial speed, u = 36 km/h = 10 m/s

Final speed, v = 72 km/h = 20 m/s

Distance, s = 100 m

We know that

{v}^{2}  -  {u}^{2}  = 2as \\  {20}^{2}  -  {10}^{2}  = 2 \times a \times 100 \\ 400 - 100 = 200 \times a \\ a =  \frac{300}{200 }  =  \frac{3}{2}  \: m {s}^{ - 2}

Now, we substitute v, u, and a in the formula

v = u + at \\ 20 = 10 +  \frac{3}{2} t \\  \frac{3}{2} t = 10 \\ 3t = 20 \\ t =  \frac{20}{3}  = 6.67 \: seconds

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