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3 years ago

Some extremely hazardous materials used in welding operations include.

1 answer:

3 0

Many of the substances in welding smoke, such as chromium, nickel, arsenic, asbestos, manganese, silica, beryllium, cadmium, nitrogen oxides, phosgene, acrolein, fluorine compounds, carbon monoxide, cobalt, copper, lead, ozone, selenium, and zinc, can be extremely toxic.

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A woodcutter wishes to cause the tree trunk to fall uphill, even though the trunk is leaning downhill. With the aid of the winch

Sedbober [7]

Answer:

The correct answer will be "400.4 N". The further explanation is given below.

Explanation:

The given values are:

Mass of truck,

m = 600 kg

g = 9.8 m/s²

On equating torques at the point O,

⇒ $T\times Cos(10+5)\times (1.3+4)=mg\times Sin(5)\times 4$

So that,

On putting the values, we get

⇒ $T\times Cos(15^{\circ})\times 5.3=600\times 9.8\times Sin(5^{\circ})\times 4$

⇒ $T=400.4 \ N$

8 0

3 years ago

Using the celsius_to_kelvin function as a guide, create a new function, changing the name to kelvin_to_celsius, and modifying th

aleksandr82 [10.1K]

Answer:

# kelvin_to_celsius function is defined

# it has value_kelvin as argument

def kelvin_to_celsius(value_kelvin):

# value_celsius is initialized to 0.0

value_celsius = 0.0

# value_celsius is calculated by

# subtracting 273.15 from value_kelvin

value_celsius = value_kelvin - 273.15

# value_celsius is returned

return value_celsius

# celsius_to_kelvin function is defined

# it has value_celsius as argument

def celsius_to_kelvin(value_celsius):

# value_kelvin is initialized to 0.0

value_kelvin = 0.0

# value_kelvin is calculated by

# adding 273.15 to value_celsius

value_kelvin = value_celsius + 273.15

# value_kelvin is returned

return value_kelvin

value_c = 0.0

value_k = 0.0

value_c = 10.0

# value_c = 10.0 is used to test the function celsius_to_kelvin

# the result is displayed

print(value_c, 'C is', celsius_to_kelvin(value_c), 'K')

value_k = 283.15

# value_k = 283.15 is used to test the function kelvin_to_celsius

# the result is displayed

print(value_k, 'is', kelvin_to_celsius(value_k), 'C')

Explanation:

Image of celsius_to_kelvin function used as guideline is attached

Image of program output is attached.

4 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

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

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

3 years ago

A quantity of water within a piston-cylinder assembly executes a Carnot power cycle. During isothermal expansion, the water is h

Harman [31]

Answer:

Answer with Explanation is in the following attachments.

Explanation:

7 0

3 years ago

A cylindrical specimen of this alloy 12 mm in diameter and 188 mm long is to be pulled in tension. Assume a value of 0.34 for Po

kicyunya [14]

This question is incomplete, the missing image in uploaded along this answer below.

Answer:

The required stress is 200 Mpa

Explanation:

Given the data in the question;

diameter D = 12 mm = 12 × 10⁻³ m

Length L = 188 mm = 188 × 10⁻³ m

Poisson's ratio v = 0.34

Reduction in diameter Δd = 0.0105 mm = 0.0105 × 10⁻³ m

The transverse strain will;

εˣ = Δd / D

εˣ = -0.0105 × 10⁻³ / 12 × 10⁻³ m

εˣ = -0.00088

The longitudinal strain will be;

$E^z$ = - ( εˣ / v )

$E^z$ = - ( -0.00088 / 0.34 )

$E^z$ = - ( - 0.002588 )

$E^z$ = 0.0026

Now, Using the values for strain, we get the value of stress from the graph provided in the question, ( first image uploaded below.

From the graph, in the Second image;

The stress is 200 Mpa

Therefore, The required stress is 200 Mpa

8 0

3 years ago