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Posted by Surbhi Suman 4 years, 4 months ago
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Yogita Ingle 4 years, 4 months ago
Given,
distance covered in 2nd s = 20 m
distance covered in 4th s = 30 m
We know,
Snth = u + a(2n - 1)/2
=> S2nd = u + a(2*2 - 1)
=> 20 = u+ 3a/2
=>20 = (2u+3a)/2
=>40 = 2u + 3a ------------- (1)
Again,
S4th = u + a(2*4 -1)/2
=>30 = u + 7a/2
=>30 = (2u + 7a)/2
=>60 = 2u + 7a --------------(2)
(1) - (2)
=>2u + 3a -2u -7a = 40 - 60
=> -4a = -20
=>a = 5
Putting the value of a in eq (1), we get,
=>2u + 3(5) = 40
=>2u = 40 -15
=>u = 25/2
Now,
S6th = u+ a(2n -1)
=>S6th = 25/2 + 5(2*6 -1)/2
=>S6th = 25/2 + 5*11/2
=>S6th = 25/2 + 55/2
=>S6th = 80/2
=>S6th = 40
Hence , distance covered in the 6th s = 40 m
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Meghna Thapar 4 years, 4 months ago
The SI unit of work is the joule (J), which is defined as the work expended by a force of one newton through a displacement of one metre. Absolute units: (i) SI System - Joule is the unit of work in SI system. ... (ii) CGS system – CGS unit of work is erg. Work done is said to be 1 erg if a force of one dyne displaces the body by 1 cm in the direction of force applied.
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Meghna Thapar 4 years, 4 months ago
The astronomical unit (symbol: AU) is a unit of length, roughly the distance from Earth to the Sun and equal to about 150 million kilometres (93 million miles). However, that distance varies as Earth orbits the Sun, from a maximum (aphelion) to a minimum (perihelion) and back again once a year. The most common is to measure the apparent angular diameter of the planet – how big it looks against the sky – very precisely using a telescope. Combining this with a measure of its distance (deduced from its orbit around the Sun) reveals the planet's actual size.
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Sl. No. |
Differentiating Property |
Velocity |
Speed |
1 |
Definition |
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2 |
Type of quantity |
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3 |
Magnitude |
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4 |
Change of direction |
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5 |
Interrelation |
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6 |
Unit (SI) |
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Pooja ** 4 years, 3 months ago
1Thank You