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Stella [2.4K]
3 years ago
5

2. Fracture mechanics. A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plan

e- strain fracture toughness of 98.9 MPa √ m and a yield strength of 860 MPa . The flaw size resolution limit of the flaw de tection appa ratus is 3.0 mm . If the design stress is one- half the yield strength and the value of Y is 1.0, determine whether a critical flaw for this plate is subject to detection.
Physics
1 answer:
ikadub [295]3 years ago
5 0

Answer:

the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

Explanation:

This problem asks that we determine whether or not a critical flaw in a wide plate is subject to detection  given the limit of the flaw detection apparatus (3.0 mm), the value of KIc (98.9 MPa m), the design stress (sy/2 in which s y = 860 MPa), and Y = 1.0.

ac=1/\pi (\frac{Klc}{Ys} )^{2}\\ ac=1/\pi(\frac{98.9}{(1)(860/2)} )^{2}\\  ac=0.0168m\\ac=16.8mm

Therefore, the critical flaw is subject to detection since this value of ac (16.8 mm) is greater than the 3.0 mm resolution limit.  

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Explain Rutherford's experiment?
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Answer:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

Explanation:

Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.

When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.

The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.

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3 years ago
Which branch of science involves the study of matter and energy
Olin [163]
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6 0
3 years ago
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If you have 100 W expended over 20 s how much energy did it take?
netineya [11]

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100 W  =  (energy) / (20 sec)

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8 0
3 years ago
A roller coaster has a "hump" and a "loop" for riders to enjoy (see picture). The top of the hump has a radius of curvature of 1
aivan3 [116]

Answer:

Part a)

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Part b)

v = 12.1 m/s

So this speed is independent of the mass of the rider

Explanation:

Part a)

By force equation on the rider at the position of the hump we can say

mg - F_n = ma_c

now we will have

mg - F_n = \frac{mv^2}{R}

F_n = mg - \frac{mv^2}{R}

now we have

F_n = 100(9.81) - \frac{100(9^2)}{12}

F_n = 981 - 675

F_n = 306 N

Part b)

At the top of the loop if the minimum speed is required so that it remains in contact so we will have

F_n + mg = ma_c

F_n = 0 at minimum speed

mg = \frac{mv^2}{R}

v = \sqrt{Rg}

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5 0
3 years ago
Our eyes are typically 6 cm apart. Suppose you are somewhat unique, and yours are 9.50 cm apart. You see an object jump from sid
Serhud [2]

Answer: 12.67 cm, 8 cm

Explanation:

Given

Normal distance of separation of eyes, d(n) = 6 cm

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Angle created during the jump, θ = 0.75°

To solve this, we use the formula,

θ = d/r, where

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r = 9.5 / 0.75

r = 12.67 cm

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r = 6 / 0.75

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Thus, the object is 12.67 cm far away in your own "unique" eyes, and just 8 cm further away to the normal person eye

8 0
3 years ago
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