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fgiga [73]
3 years ago
13

Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature

rise due to friction?
Engineering
1 answer:
Ronch [10]3 years ago
7 0

Answer:

The surface temperature increases when two bodies are rubbed against each other due to friction.

Explanation:

No object has a perfectly even surface. So, when two bodies with uneven surfaces are rubbed against each other, they experience friction.

Friction is a resistance experienced by the two bodies when they are moved against each other.

The friction between the two surfaces, converts the kinetic energy of the movement to the thermal energy.

Thus, resulting in rise in the surface temperature of the two bodies.

Therefore, when two bodies are rubbed against each other, the surface temperature increases due to friction.

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Nate needs to replace the cable to his lamp. He is stripping it to connect it to the termils. What should he remember to do with
pshichka [43]

Answer: i got you its d

Explanation:had the smae question as you

5 0
3 years ago
Air enters a constant-area combustion chamber at a pressure of 101 kPa and a temperature of 70°C with a velocity of 130 m/s. By
Norma-Jean [14]

Answer:

451 kj/kg

Explanation:

Velocity = 139m/s

Temperature = 70⁰C

T = 343K

M1 = v/√prt

= 130/√1.4x287x343

= 130/√137817.4

= 130/371.2

= 0.350

T1/To1 = 0.9760

From here we cross multiply and then make To1 the subject of the formula

To1 = T1/0.9760

To1 = 343/0.9760

To1 = 351.43

Then we go to the rayleigh table

At m = 0.35

To1/To* = 0.4389

To* = 351.43/0.4389

= 800k

M2 = 1

Maximum amount of heat

1.005(800-351.43)

= 450.8kj/kg

= 452kj/kg

8 0
3 years ago
Air expands through a turbine from 10 bar, 900 K to 1 bar, 500 K. The inlet velocity is small compared to the exit velocity of 1
Elis [28]

Answer:

- the mass flow rate of air is 7.53 kg/s

- the exit area is 0.108 m²

Explanation:

Given the data in the question;

lets take a look at the steady state energy equation;

m" = W"_{cv / [ (h₁ - h₂ ) -\frac{V_2^2}{2} ]

Now at;

T₁ = 900K, h₁ = 932.93 k³/kg

T₂ = 500 K, h₂ = 503.02 k³/kg

so we substitute, in our given values

m" = [ 3200 kW × \frac{1\frac{k^3}{s} }{1kW} ] / [ (932.93 - 503.02  )k³/kg  -\frac{100^2\frac{m^2}{s^2} }{2}|\frac{ln}{kg\frac{m}{s^2} }||\frac{1kJ}{10^3N-m}| ]

m" = 7.53 kg/s

Therefore, the mass flow rate of air is 7.53 kg/s

now, Exit area A₂ = v₂m" / V₂

we know that; pv = RT

so

A₂ = RT₂m" / P₂V₂

so we substitute

A₂ = {[ (\frac{8.314}{28.97}\frac{k^3}{kg.K})×500 K×(7.54 kg/s) ] / [(1 bar)(100 m/s )]} |\frac{1 bar}{10N/m^2}||10^3N.m/1k^3

A₂ = 0.108 m²

Therefore, the exit area is 0.108 m²

8 0
3 years ago
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sashaice [31]

Answer:

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8 0
3 years ago
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Mazyrski [523]

Answer: look left to right twice b/c somebody may try to run the light

Explanation:

3 0
3 years ago
Read 2 more answers
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