Answer:
Explanation:
xx=33 vvalues
Which of the following positions would be responsible for attaching I-beams to a crane?
1 answer:
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Answer:
Δx = 25 ft.
Explanation:
Assuming that the person on the ground starts running at the same time as the egg is dropped, we have two simultaneous trajectories:
1 ) Egg falling:
If the egg is dropped, and we neglect the air resistance, we can use the kinematic equation that relates the distance and fall time, as follows:
yf-y₀ = 1/2* g* t²
If we take the up direction as positive, we can solve for t as follows:
0-100 ft = 1/2* (-32.15 ft/s²)* t²
⇒ t = = 2.5 sec.
2) Person on the ground running away:
In order to be able to run away, and then return to catch the egg, running at constant speed, he must run during exactly the half of the time that the egg is falling, i.e., 1.25 sec.
We can get the distance at which he can reach, applying the definition of velocity:
v = (xf-x₀) / (tfi-t₀)
If we choose t₀=0 and x₀ = 0 , we can solve for xf, as follows:
xf = v*t = 20 ft/sec*1.25 sec = 25 ft.
Answer:
the ratio of the etched to the original crack tip radius is 30.24
Explanation:
Given the data in the question;
we determine the initial fracture stress using the following expression;
(σf)₁ = 2(σ₀)₁ α₁/(
)₁
----- let this be equation 1
where; (σ₀)₁ is the initial fracture strength
()₁ is the original crack tip radius
α₁ is the original crack length.
first, we determine the final crack length;
α₂ = α₁ - 16% of α₁
α₂ = α₁ - ( 0.16 × α₁)
α₂ = α₁ - 0.16α₁
α₂ = 0.84α₁
next, we calculate the final fracture stress;
the fracture strength is increased by a factor of 6;
(σ₀)₂ = 6( σ₀ )₁
Now, expression for the final fracture stress
(σf)₂ = 2(σ₀)₂ α₂/(
)₂
------- let this be equation 2
where ()₂ is the etched crack tip radius
value of fracture stress of glass is constant
Now, we substitute 2(σ₀)₁ α₁/(
)₁
from equation for (σf)₂ in equation 2.
0.84α₁ for α₂.
6( σ₀ )₁ for (σ₀)₂.
∴
2(σ₀)₁ α₁/(
)₁
= 2(6( σ₀ )₁)
0.84α₁/(
)₂
divide both sides by 2(σ₀)₁
α₁/(
)₁
= 6
0.84α₁/(
)₂
1/(
)₁
= 6
0.84/(
)₂
1/(
)₁
= 36
0.84/(
)₂
1 / ()₁ = 30.24 / (
)₂
()₂ = 30.24(
)₁
()₂/(
)₁ = 30.24
Therefore, the ratio of the etched to the original crack tip radius is 30.24