Question

the samllest charge can be exist indenatly charge on​

Answer 1

Answer:

1.602×10−19

Explanation:

the charge is quantized; it comes in integer multiples of individual small units called the elementary charge, e, about 1.602×10−19 coulombs, which is the smallest charge which can exist freely (particles called quarks have smaller charges, multiples of 13e, but they are only found in combination,

Answer 2

Answer:

Let AB represents a pillar AB = 9m9m9m

Let AB represents a pillar AB = 9m9m9m Snake is at C

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = AD

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snake

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x)

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2 =9

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2 =9 2

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2 =9 2 +x

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2 =9 2 +x 2

Let AB represents a pillar AB = 9m9m9m Snake is at CBC = 2/m2/m2/m Peacock cought the snake at D and BD = xmxmxm DC = 27−x27 - x27−x = ADspeed of peacock = speed of snakeIn right △ABD→ (27−x)2=92+x2(27 - x) {}^{2} = 9 {}^{2} + x {}^{2}(27−x) 2 =9 2 +x 2