How do I put a starter in my Chevy 350

Which of the following ranges depicts the 2% tolerance range to the full 9 digits provided?

Lyrx [107]

Answer:

the only one that meets the requirements is option C .

Explanation:

The tolerance of a quantity is the maximum limit of variation allowed for that quantity.

To find it we must have the value of the magnitude, its closest value is the average value, this value can be given or if it is not known it is calculated with the formula

x_average = ∑ $x_{i}$ / n

The tolerance or error is the current value over the mean value per 100

Δx₁ = x₁ / x_average

tolerance = | 100 -Δx₁ 100 |

bars indicate absolute value

let's look for these values for each case

a)

x_average = (2.1700000+ 2.258571429) / 2

x_average = 2.2142857145

fluctuation for x₁

Δx₁ = 2.17000 / 2.2142857145

Tolerance = 100 - 97.999999991

Tolerance = 2.000000001%

fluctuation x₂

Δx₂ = 2.258571429 / 2.2142857145

Δx2 = 1.02

tolerance = 100 - 102.000000009

tolerance 2.000000001%

b)

x_average = (2.2 + 2.29) / 2

x_average = 2,245

fluctuation x₁

Δx₁ = 2.2 / 2.245

Δx₁ = 0.9799554

tolerance = 100 - 97,999

Tolerance = 2.00446%

fluctuation x₂

Δx₂ = 2.29 / 2.245

Δx₂ = 1.0200445

Tolerance = 2.00445%

c)

x_average = (2.211445 +2.3) / 2

x_average = 2.2557225

Δx₁ = 2.211445 / 2.2557225 = 0.9803710

tolerance = 100 - 98.0371

tolerance = 1.96%

Δx₂ = 2.3 / 2.2557225 = 1.024624

tolerance = 100 -101.962896

tolerance = 1.96%

d)

x_average = (2.20144927 + 2.29130435) / 2

x_average = 2.24637681

Δx₁ = 2.20144927 / 2.24637681 = 0.98000043

tolerance = 100 - 98.000043

tolerance = 2.000002%

Δx₂ = 2.29130435 / 2.24637681 = 1.0200000017

tolerance = 2.0000002%

e)

x_average = (2 +2,3) / 2

x_average = 2.15

Δx₁ = 2 / 2.15 = 0.93023

tolerance = 100 -93.023

tolerance = 6.98%

Δx₂ = 2.3 / 2.15 = 1.0698

tolerance = 6.97%

Let's analyze these results, the result E is clearly not in the requested tolerance range, the other values may be within the desired tolerance range depending on the required precision, for the high precision of this exercise the only one that meets the requirements is option C .

4 0

3 years ago

150 lb of force is applied at the end of a shaft that is 2 1/2 ft long. It is applied

Alona [7]

Answer:

Torque: 500

Power: 1700

Explanation:

Since the force is 150 lbs and if the shaft is rotating at 350 rpms, then you will need 1700 for the power in order to keep it running.

5 0

3 years ago

In python/Spyder

lianna [129]

Answer:

Answer to both the parts A and B are provided in the word document attached.

Explanation:

The explanation of the programs is provided in the attached file along with the screenshot of ATM application output.

Download docx

8 0

3 years ago

The assembly consists of a brass shell (1) fully bonded to a ceramic core (2). The brass shell [E = 93 GPa, α= 15.1 × 10−6/°C] h

marshall27 [118]

Answer:

ΔT = 62.11°C

Explanation:

Given:

- Brass Shell:

Inner Diameter d_i = 32 mm

Outer Diameter d_o = 39 mm

E_b = 93 GPa

α_b = 15.1*10^-6 / °C

- Ceramic Core:

Outer Diameter d_o = 32 mm

E_c = 310 GPa

α_c = 3.2*10^-6 / °C

- Unstressed @ T = 8°C

- Total Length of the cylinder L = 160 mm

Find:

Determine the largest temperature increase Δ⁢t that is acceptable for the assembly if the normal stress in the longitudinal direction of the brass shell must not exceed 60 MPa.

Solution:

- Since, α_b > α_c the brass shell is in compression and ceramic core is in tension. The stress in shell is given as б_a:

б_b = - 60 MPa

- The force equilibrium can be written as:

б_b*A_b + б_c*A_c = 0

Where, б_b is the stress in core

A_b is the cross sectional area of the shell

A_c is the cross sectional area of the core

б_b*pi*( d_o^2 - d_i^2) / 4 + б_c*pi*( d_i^2) / 4 = 0

б_b*( d_o^2 - d_i^2) + б_c*( d_i^2) = 0

б_c = - б_b*( d_o^2 - d_i^2) / ( d_i^2)

Plug in the values:

б_c = 60*( 0.039^2 - 0.032^2) / ( 0.032^2)

б_c = 29.121 MPa , б_b = - 60 MPa

- The total strains in both brass shell and ceramic core is given by:

ξ_b = α_b*ΔT + б_b / E_b

ξ_c = α_c*ΔT + б_c / E_c

- The compatibility relation is:

ξ_b = ξ_c

α_b*ΔT + б_b / E_b = α_c*ΔT + б_c / E_c

ΔT*(α_b - α_c ) = б_c / E_c - б_b / E_b

ΔT = [ б_c / E_c - б_b / E_b ] / (α_b - α_c )

Plug in values and solve:

ΔT = [ 0.029121 / 310 + 0.06 / 93 ]*10^6 / (15.1 - 3.2 )

ΔT = 62.11°C

8 0

4 years ago

When my animal is to old to physically carry her baby what process can I use to still produce her young from her

anygoal [31]

Taking care of" is a very human term in the sense of care giving, but so is "parents", because animals do not retain family structures in the form humans do. The vast majority of animals do not recognize family members. Among predators, an aging parent, such as the male lion in a pride, would continue to be taken care of by the females just as they always were, until a stronger outside male ousted it. An aging silverback in a group of gorillas would eventually be challenged and overthrown by one of his sons. When the elder became too infirm to care for itself, it would go off by itself to die. The only animal species that modify their behavior due to familial bonds to accommodate the needs of an aging member of the family are Cetaceans (whales) and Pachyderms (elephants). Whales have been seen to bouy up an ailing member of the pod (an extended family) to keep it from sinking. Elephants have the closest family ties of all animals, the eldest surviving female, the Matriarch of a herd, might be the great, great, great grandmother to the youngest and the others in varying relationships, are mothers, grandmothers, children, sisters, aunts and cousins, the older always taking care of and sheltering the younger. They also shelter and accommodate the older. If one member is sick or too old to keep up, the rest of the herd will slow down to allow the lagging one to keep up. This "caring" behavior has its limitations. The herd will not endanger itself to stay with an ailing member who will never be able to keep up. However, when the Matriarch herself is dying of old age, and the herd is in an area where food and water are abundant, the rest of the herd will stay with her, comfort her with their trunks, caressing her, talking to her, vocalizing over her, never leaving her side. While some are foraging, there will always be a group around her comforting her and expressing their love. (Scientists used to dismiss that term as inapplicable to animals, but the evidence is overwhelming that higher animal species have the same emotions humans do.) The herd will remain with her body, mourning, for several days, and years later will still return to fondle her bones with their trunks. Eventually one of the Matriarch's daughters will be chosen as the one to follow, or even two in which case the herd would divide into two new herds, the daughters and their offspring of each new Matriarch following their own family line.
9.7K viewsView upvotes

31

4

Related Questions (More Answers Below)

7 0

3 years ago