Two sections, A and B, are 0.5 km apart along a 0.05 m diameter rough concrete pipe. A is 115 m higher than B, the water tempera
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The modulus of elasticity is 28.6 X 10³ ksi
<u>Explanation:</u>
Given -
Length, l = 5in
Force, P = 8000lb
Area, A = 0.7in²
δ = 0.002in
Modulus of elasticity, E = ?
We know,
Modulus of elasticity, E = σ / ε
Where,
σ is normal stress
ε is normal strain
Normal stress can be calculated as:
σ = P/A
Where,
P is the force applied
A is the area of cross-section
By plugging in the values, we get
σ =
σ = 11.43ksi
To calculate the normal strain we use the formula,
ε = δ / L
By plugging in the values we get,
ε =
ε = 0.0004 in/in
Therefore, modulus of elasticity would be:
Thus, modulus of elasticity is 28.6 X 10³ ksi
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:
attached below
Explanation:
