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tresset_1 [31]
2 years ago
15

Running ropes must be taken out of service if they have _____ broken wires in one strad in one lay

Engineering
2 answers:
astraxan [27]2 years ago
5 0

Answer:

  3

Explanation:

3 broken wires in one strand in one lay are cause for removal from service.

zheka24 [161]2 years ago
5 0

Answer:

<h2> 3</h2>

  • Running ropes must be taken out o service if they have <u>3</u> broken wires in one strand in one lay.
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A circuit has a source voltage of 15V and two resistors in series with a total resistance of 4000Ω .If RI has a potential drop o
anastassius [24]

Answer:

1500Ω

Explanation:

Given data

voltage = 15 V

total Resistance = 4000Ω

potential drop V = 9.375 V

To find out

R2

Solution

we know R1 +R2 = 4000Ω

So we use here Ohm's law to find out current I

current = voltage / total resistance

I = 15 / 4000 = 3.75 × 10^{-3} A

Now we apply Kirchhoffs Voltage Law for find out R2

R2 = ( 15 - V ) / current

R2 = ( 15 - 9.375 ) / 3.75 × 10^{-3}

R2 = 1500Ω

6 0
2 years ago
A 60-m-long steel wire is subjected to a 6-kN tensile load. Knowing that E = 200 GPa and that the length of the rod increases by
Talja [164]

Answer:

(a) 6.91 mm (b) 160 MPa

Explanation:

Solution

Given that:

E = 200 GPa

The rod length = 48 mm

P =P¹ = 6 kN

Recall that,

1 kN = 10^3 N

1 m =10^3 mm

I GPa = 10^9 N/m²

Thus

The rod deformation is stated as follows:

δ = PL/AE-------(1)

σ = P/A----------(2)

Now,

(a) We substitute the values in equation and obtain the following:

48 * 10 ^⁻3 m =  (6 * 10³ N) (60 m)/A[ 200 * 10^9 N/m^2]

Thus, we simplify

A = (6 * 10³) (60)/ ( 200 * 10^9) (48 * 10 ^⁻3)m²

A =0.0375 * 10 ^⁻3 m²

A =37.5 mm²

A = π/4 d²

Thus,

d² = 4A /π

After inserting the values we have,

d = √37.5 * 4/3.14 mm

= 6.9116 mm

or d = 6.91 mm

Therefore, the smallest that should be used is 6.91 mm

(B) To determine the corresponding normal stress that is caused by the tensile force, we input the values in equation (2)

Thus,

σ = P/A

σ= 6 * 10 ^ 3 N/ 37. 5 * 10 ^⁻6 m²

σ= 160 MPa

Note: I MPa = 10^6 N/m²

Hence the the corresponding normal stress is σ= 160 MPa

5 0
3 years ago
Determine the resistance of 3km of copper having a diameter of 0,65mm if the resistivity of copper is 1,7x10^8
LUCKY_DIMON [66]

Answer:

Resistance of copper = 1.54 * 10^18 Ohms

Explanation:

<u>Given the following data;</u>

Length of copper, L = 3 kilometers to meters = 3 * 1000 = 3000 m

Resistivity, P = 1.7 * 10^8 Ωm

Diameter = 0.65 millimeters to meters = 0.65/1000 = 0.00065 m

Radius, r = \frac {diameter}{2}

Radius = \frac {0.00065}{2}

Radius = 0.000325 m

To find the resistance;

Mathematically, resistance is given by the formula;

Resistance = P \frac {L}{A}

Where;

  • P is the resistivity of the material.
  • L is the length of the material.
  • A is the cross-sectional area of the material.

First of all, we would find the cross-sectional area of copper.

Area of circle = πr²

Substituting into the equation, we have;

Area  = 3.142 * (0.000325)²

Area = 3.142 * 1.05625 × 10^-7

Area = 3.32 × 10^-7 m²

Now, to find the resistance of copper;

Resistance = 1.7 * 10^{8} \frac {3000}{3.32 * 10^{-7}}

Resistance = 1.7 * 10^{8} * 903614.46

<em>Resistance = 1.54 * 10^18 Ohms </em>

3 0
2 years ago
What part connect to the tie rod that helps the brake chamber out that deals with slack adjuster
scoray [572]

Answer:

none

Explanation:

7 0
1 year ago
Which size of impurity atom, smaller impurity atom or larger impurity atom, when located near a dislocation, will nullify some o
svp [43]

Answer:

Smaller impurity atom will nullify some of the compressive strain of a dislocation in a crystal. Because, smaller impurity atoms located near a dislocation creates tensile strain on atoms around it thereby partially nullifying compressive strain at the dislocation.

4 0
3 years ago
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