Egg cell , sperm cell , embryo
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
Given Information:
diameter = d = 15 mm
Length = L = 20 mm
Axial load = P = 300 N
Eₚ = 2.70x10⁹ Pa
vₚ = 0.4
Required Information:
Change in length = ?
Change in diameter = ?
Answer:
Change in length = 0.01257 mm
Change in diameter = -0.003772 mm
Explanation:
Stress is given by
σ = P/A
Where P is axial load and A is the area of the cross-section
A = 0.25πd²
A = 0.25π(0.015)²
A = 0.000176 m²
σ = 300/0.000176
σ = 1697792.8 Pa
The longitudinal stress is given by
εlong = σ/Eₚ
εlong = 1697792.8/2.70x10⁹
εlong = 0.0006288 mm/mm
The change in length can be found by using
δ = εlong*L
δ = 0.0006288*20
δ = 0.01257 mm
The lateral stress is given by
εlat = -vₚ*εlong
εlat = -0.4*0.0006288
εlat = -0.0002515 mm/mm
The change in diameter can be found by using
Δd = εlat*d
Δd = -0.0002515*15
Δd = -0.003772 mm
Therefore, the change in length is 0.01257 mm and the change in diameter is -0.003772 mm
Answer:
Hooke's law describes the elastic properties of materials only in the range in which the force and displacement are proportional. Hooke's law states that the applied force F equals a constant k times the displacement or change in length x, or F = kx. the maximum extent to which a solid may be stretched without permanent alteration of size or shape, is called elastic limit
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Answer:
factor that bug maximum KE change is 0.52284
Explanation:
given data
vertical distance = 6.5 cm
ripples decrease to = 4.7 cm
solution
We apply here formula for the KE of particle that executes the simple harmonic motion that is express as
KE = (0.5) × m × A² × ω² .................1
and kinetic energy is directly proportional to square of the amplitude.
so
.............2
= 0.52284
so factor that bug maximum KE change is 0.52284