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denis23 [38]
3 years ago
14

Thermoplastics burn upon heating. a)-True b)- false?

Engineering
1 answer:
miv72 [106K]3 years ago
4 0

Yes, It is indeed true that Thermoplastics and thermosettingplastics burn upon heating.

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WARNING:<br><br> when people put links in the answer it is a virus DO NOT DOWNLOAD IT
kobusy [5.1K]

Answer:

Thank you for this!

Explanation:

I was about to click it on a question I saw.

6 0
2 years ago
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Explain why surface temperature increases when two bodies are rubbed against each other. What is the significance of temperature
Ronch [10]

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.

7 0
3 years ago
An interrupted line of sight means changes in ......and .... are necessary for re-establishing a ......... to the driver’s path
Mrrafil [7]

Answer:

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Explanation:

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5 0
2 years ago
1. True/False The Pressure Relief valve maintains the minimum pressure in the hydraulic circuit​
elena55 [62]
Yeah it’s true. Good luck!!
3 0
3 years ago
You are working as an electrical technician. One day, out in the field, you need an inductor but cannot find one. Looking in you
telo118 [61]

Answer:

a) the inductance of the coil is 6 mH

b) the emf generated in the coil is 18 mV  

Explanation:

Given the data in the question;

N = 570 turns

diameter of tube d = 8.10 cm = 0.081 m

length of the wire-wrapped portion l =  35.0 cm = 0.35 m

a) the inductance of the coil (in mH)

inductance of solenoid

L = N²μA / l

A = πd²/4  

so

L = N²μ(πd²/4) / l

L = N²μ(πd²) / 4l

we know that μ = 4π × 10⁻⁷ TmA⁻¹

we substitute

L = [(570)² × 4π × 10⁻⁷× ( π × (0.081)² )] / 4(0.35)

L =  0.00841549 / 1.4

L = 6 × 10⁻³ H    

L = 6 × 10⁻³ × 1000 mH

L = 6 mH

Therefore, the inductance of the coil is 6 mH

b)

Emf ( ∈ ) = L di/dt

given that; di/dt = 3.00 A/sec

{∴ di = 3 - 0 = 3 and dt = 1 sec}

Emf ( ∈ ) = L di/dt

we substitute

⇒ 6 × 10⁻³ ( 3/1 )

= 18 × 10⁻³ V

= 18 × 10⁻³ × 1000

= 18 mV  

Therefore, the emf generated in the coil is 18 mV  

7 0
2 years ago
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