Aref hints that he is like a butterfly. Fill in the blank to explain how Aref is like a butterfly?
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Subject:
Engineering -
Author:
pixie32 -
Created:
1 year ago
Answers 2
Quick Answer:Aref is like a butterfly because he undergoes personal transformation and growth.
Full Explanation:Comparing Aref to a butterfly suggests that he undergoes a process of personal transformation and growth. Butterflies start their lives as caterpillars and go through a metamorphosis, transforming into beautiful winged creatures. Similarly, Aref may experience a journey of self-discovery and development, going through changes that shape his character and perspective.
Just as a butterfly emerges from a cocoon with newfound beauty and freedom, Aref may undergo a similar transformation, shedding old beliefs or limitations and embracing a new sense of identity and freedom. The comparison may imply that Aref is evolving, growing, and spreading his wings to explore new possibilities and experiences.
Furthermore, butterflies are often associated with grace, delicacy, and vibrant colors. This comparison could suggest that Aref possesses these qualities, whether in terms of his personality, his resilience, or his ability to navigate challenges with grace and beauty.
In summary, by likening Aref to a butterfly, it suggests that he undergoes personal transformation, growth, and a journey of self-discovery. This comparison may also imply that he possesses qualities of grace and resilience as he navigates through life's challenges.
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When the system is in equilibrium, the sum of the moment about a point is
zero.
Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar \ is \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}
Reasons:
Length of the beam = l
Mass of the beam = m_B
Location \ of \ the \ two \ pillars = \dfrac{l}{3} \ from \ either \ end
Mass of the duck = m_D
Required:
Force shown by the scale under the right pillar.
Solution:
The location of the duck = On the left end of the beam
When the system is in equilibrium, we have; ∑M = 0
Taking moment about the left pillar, we get;
Clockwise moment = m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3}
Anticlockwise moment = m_B \times g \times \dfrac{l}{6}
At equilibrium, clockwise moment = Anticlockwise moment
Therefore;
m_D \times g \times \dfrac{l}{3} + F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6}
F \times \dfrac{l}{3} = m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}
F = \dfrac{m_B \times g \times \dfrac{l}{6} - m_D \times g \times \dfrac{l}{3}}{\dfrac{l}{3} } = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}
Force \ shown \ by \ the \ scale \ under \ the \ right \ pillar, \ F = \dfrac{(m_B - 2 \cdot m_D) \cdot g}{2}
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