Create 2 models, please.

An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 mpa m (36.4 ksi in.

GarryVolchara [31]

We will first determine using the given if an aircraft component will fracture with a given stress level (260 MPa), maximum internal crack length (6.0 mm) and fracture toughness (40 MPa m ), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. First, it is necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and 2 a = 4.0 mm). Therefore,

Y = K(Ic)/ sqrt(π a) = 40 MPa( m ) / (300 MPa) sqrt(( π ) ((4 × 10-3 m)/2)) = 1.68

We will now solve for the product Y σ π a for the other set of conditions, so as to ascertain whether or not this value is greater than the K(Ic) for the alloy. Thus,

Y sqrt(π a) = (1.68)(260 MPa) sqrt (( π )[(6 × 10^-3 m)/ 2])
= 42.4 MPa sqrt (m) (39 ksi in. )

Therefore, fracture will occur since this value ( 42.4 MPa sqrt(m)) is greater than the K(Ic) of the material, 40 MPa sqrt(m).

8 0

4 years ago

I have an object that I want to know when it was made. I calculate that the object contains 0.12g of Iodine-131 but started with

Sav [38]

Answer: 71.72 days

Explanation:

This problem can be solved using the <u>Radioactive Half Life Formula: </u>

$A=A_{o}.2^{\frac{-t}{h}}$ (1)

Where:

$A=0.12 g$ is the final amount of Iodine-131

$A_{o}=60 g$ is the initial amount of Iodine-131

$t$ is the time elapsed

$h=8 days$ is the half life of Iodine-131

Knowing this, let's substitute the values and find $t$ from (1):

$0.12 g=(60 g)2^{\frac{-t}{8 days}}$ (2)

$\frac{0.12 g}{60g}=2^{\frac{-t}{8 days}}$ (3)

Applying natural logarithm in both sides:

$ln(\frac{0.12 g}{60g})=ln(2^{\frac{-t}{8 days}})$ (4)

$-6.21=-\frac{t}{8 days}ln(2)$ (5)

Finding $t$ :

$t=71.72 days$

7 0

4 years ago

Motion with constant acceleration CER lab report <br><br> Can someone help!!

victus00 [196]

Answer:

Fan Speed For Low Position vs Time Graph : The slope is curved and it increases as you go up . The points start off close but they spread out as the time increases.

Fan Speed For Medium Position vs Time Graph : The speed increases quicker than the graph for low speed. The graph is less curved than the one for low speed. Also, the points spread out faster than they did for low speed as the time increases.

Fan Speed For High Position vs Time Graph: The Graph has a smaller curve then the low and medium speed. Also , the points are the furthest apart. The slope is not as spaced out as it was for the rest of the speed graphs .

Explanation: I just finished the lab:)

8 0

3 years ago

A sled and rider (combined mass of 69 kg) finish a downhill run with a speed of 30 m/s, then enter a flat (horizontal) area wher

NARA [144]

Answer:

Distance that sled move while slowing down= 195.65m

Explanation:

Distance traveled by sled after slowing down=S

acceleration at the start= -2.3 m/s2

initial speed u=30m/s

final speed=v

By using the equation

$v^{2} =u^{2} +2as\\$

Final speed is zero.

Therefore;

$0=30^{2} +2*(-2.3)*s\\2*2.3s=30^{2} \\s=30^{2} /(2*2.3)\\s=195.65 m$

6 0

4 years ago

5. Calculate how long in seconds it will take for a light pulse to travel from earth to the moon. The distance from earth to moo

solniwko [45]

Answer:

0.13 seconds

Explanation:

Since 1 Km = 0.621 miles

3.84 x 105 km = 3.84 x 105 × 0.621 = 23846.4 miles

Speed = distance/time

time= distance/speed

Time= 23846.4/186,000

Time= 0.13 seconds

3 0

3 years ago