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09 Grade 11 New carriculum Textbooks☝️☝️☝️
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12 Happy Eid🥰🥰
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13 #physics
🚁" ሂሌኮፕተር አየር ላይ የሚንሳፈፍበት ሚስጥር ምንድነው?"
🫧እጅግ ቀላል እና በሳይዛቸው በጣም ትልቅ ነገሮች ውሃ ላይ በቀላሉ ይንሳፈፍሉ። የሰው ልጅም በቂ ውሃ በዋና ገንዳ(#Swimming pool) ውስጥ በእጅ እና በእግሩ ማንቀሳቀስ ከቻለ በቀላሉ የዋና ገዳው ውሃ ላይ ይንሳፈፋል።
🫧ፊኛ(#Balloon) እጅግ ትንሽ ክብደት ያለው ሲሆን በተቃራኒው ትልቅ ሳይዝ አለው። ይሄም ፊኛ #ከክብደቱ ጋር ተመጣጣኝ የሆነ አየር ወይም ጋዝ በቀላሉ #ማንቀሳቀስ ሰለሚችል አየር ላይ በቀላሉ ይንሳፈፋል።
🫧ውሃ ከፍተኛ ዴንሲቲ(#Density) ያለው ነገር በመሆኑ ሌሎች ነገሮች ውሃ ላይ በቀላሉ ይንሳፈፋሉ።
🫧ከመሬት ውጭ አየር ላይ ግን በቀላሉ መንሳፈፍ አይቻልም ይሄም የሆነው አየር ዝቅተኛ የሆነ ዴንሲቲ ስላለው ነው።
🫧ይሄንንም ተከትሎ አየር ላይ ተንሳፎ የሚንቀሳቀስ ነገርን መስራት ለሰው ልጆች እጅግ አስቸጋሪ ነገር ሆኖም መቆየቱን ታሪክ ያወሳል።
🫧ከ18ኛ ክፍለ ዘመን በዋላ ግን የሰው ልጆች እጅግ ሀይል ያላቸውን #ሞተሮች በመስራታቸው ምክንያት ይሄንን ሞተር በመጠቀም አየር ላይ የሚንሳፍፉ ነገሮችን ለመሰራት #ሳይንቲስቶች እንደተንቀሳቀሱ ይነገራል።
🫧የሂሊኮፕተሮች ክንፍ (#Wing) የሚያንቀሳቅሰው የአየር ወይም ጋዝ ክብደት ከሂሌኮፕተሩ ክብደት ጋር ተመጣጣኝ ካልሆነ ወይም ካልበለጠ ሄሊኮፕተሩ አየር ላይ መንሳፈፍ አይችልም።
🫧ሰለዚህ የሄሊኮፕተሮች ሞተር አጠቃላይ የሂሊኮፕተሩ #ክብደት ጋር ተመጣጣኝ የሆነ አየር ወይም ጋዝ ማንቀሳቀስ አለበት ለዚህም ነው የሄሊኮፕተሮች #ፈጠራ እጅግ አስቸጋሪ የሆነው ።
🫧የሄሊኮፕተር ክንፍ እጅግ በጥንቃቄ እና በረቀቅ መንገድ ዲዛይን የተደረገ ሲሆን ይሄንን ክንፍ የሚያንቀሳቅሰው ሞተርም እጅግ ከፍተኛ የሆነ ጉልበት እንዳለው ይነገራል።
🫧ይሄንንም ተከትሎ ሂሊኮፕተሮች የአለማችን #ውዱ ተንቀሳቃሽ ነገሮች ናቸው።
🫧ሂሌኮፕተሮች አየር ላይ ለመንሳፈፍ ክንፋቸውን ይጠቀማሉ። ሚዛናቸውን ለመጠበቅ እና ወደ ተለያዩ አቅጣጫዎች ለመንቀሳቀስ ምን #ይጠቀማሉ?
🫧አሁን ነገሮች ውስብስብ ይሆናሉ!! ሂልኮፕቶሮች ኖርማሊ ከላይ ባወራነው መንገድ ቢንሳፈፉ እንኳን በአንድ በተወሰነ ቦታ ላይ ተረጋግተው መንሳፈፍ አይችሉም። ወይም ዝንብለው 360 ድግሪ ይሽከረከራሉ።
🫧ለዚህም ሳይንቲስቶች አዲስ ሀሳብን አመጡ ሌላ ከዋላ አነስተኛ የሚሽከረከር ሞተር(#Tail motor) መግጠም የሚለው ሆነ ጉዳዩ እንዲ ነው።
🫧የሄሊኮፕተር ሞተር ሲሽከረከር ቶርክ (#Torque) የተሰኘ የእሽከርክሪት ሐይል መፍጠሩ ነው። ሂሌክፕተሩ ከተንሳፈፈ በዋላ #360 ድግሪ እንዲሽከረከር ያደረገው ለዚህም ተፃራሪ ቶርክ (#Opposite torque) ከዋላ በመፍጠር ሂልኮፕተሮች #ሚዛናቸውን ጠብቀው እንዲሳፈፉ ተደርገዋል።
🫧በመቀጠለም ይሄኑን የቶርክ ፊዚክስ በመጠቀም ሄሌኮፕተሮችን ወደ ምንፈለግው አቅጣጫ ማንቀሳቀስ አንድንችል ተደርገዋል።
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14 Deebii (Answers)
1. Using the equation of motion: v = u + at
where v is the final velocity, u is the initial velocity (0 m/s in this case), a is the acceleration (2 m/s^2), and t is the time (5 seconds).
v = 0 + 2 * 5 = 10 m/s
Therefore, the car is moving at 10 m/s after 5 seconds.
2. Using the kinematic equation for vertical motion: h = (u^2)/(2g)
where h is the maximum height, u is the initial velocity (20 m/s), and g is the acceleration due to gravity (9.81 m/s^2).
h = (20^2) / (2 * 9.81) ≈ 20.41 meters
Therefore, the ball reaches a height of approximately 20.41 meters.
3. Using Newton's second law: F = ma
where F is the force applied (10 N), m is the mass of the block (2 kg), and a is the acceleration.
a = F / m = 10 / 2 = 5 m/s^2
Therefore, the acceleration of the block is 5 m/s^2.
4. Using Coulomb's Law: F = k * |q1 * q2| / r^2
where F is the electric force, k is Coulomb's constant, q1 and q2 are the charges (+3 μC and -5 μC), and r is the distance between them (2 meters).
F = (9 * 10^9) * |3 * 5| / 2^2 = 337.5 N
The force is attractive since one charge is positive and the other is negative.
5. Using Snell's Law: n1 * sin(theta1) = n2 * sin(theta2)
where n1 and n2 are the refractive indices of the two media, theta1 is the angle of incidence (30 degrees), and theta2 is the angle of refraction.
1 * sin(30) = 1.5 * sin(theta2)
sin(theta2) = sin(30) / 1.5 ≈ 0.3333
theta2 = sin^(-1)(0.3333) ≈ 19.47 degrees
Therefore, the angle of refraction inside the glass block is approximately 19.47 degrees.
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15 Physics questions
1. A car accelerates from rest at a rate of 2 m/s^2 for 5 seconds. How fast is the car moving after 5 seconds?
2. A ball is thrown vertically upward with an initial velocity of 20 m/s. How high does the ball go before it starts falling back down?
3. A block of mass 2 kg is pushed along a frictionless surface with a force of 10 N. What is the acceleration of the block?
4. Two charges of +3 μC and -5 μC are placed 2 meters apart. What is the magnitude and direction of the electric force between them?
5. A light ray enters a glass block at an angle of 30 degrees to the normal. If the refractive index of the glass is 1.5, what is the angle of refraction inside the glass block?
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Grade 11 New carriculum Textbooks☝️☝️☝️
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👉History
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