Class 11th

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V
Vishal Baghel

Contributor-Level 10

t = 2 h g = 2 * 1 9 . 6 9 . 8 = 2 s e c

x = u t = 9 * 1 0 3 3 6 0 0 * 2 = 5 m

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3 months ago

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Syed Aquib Ur Rahman

Contributor-Level 10

Yes. Escape speed from a planet or celestial body is the same for all objects. It is irrespective of their mass. This is because the gravitational force and potential energy depend on the mass of the celestial body and distance. The object's own mass cancels out when deriving escape speed. 

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3 months ago

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Syed Aquib Ur Rahman

Contributor-Level 10

The escape speed from the Earth's surface is approximately 11.2 km/s. This means that the object must travel at least at this speed in the upwards direction to entirely escape Earth's gravitational pull without falling back.  

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3 months ago

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Syed Aquib Ur Rahman

Contributor-Level 10

Escape velocity or escape speed is the minimum velocity required for an object to move away from the gravitational pull of a celestial body or planet. This velocity for escape does not require any additional propulsion. 

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3 months ago

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alok kumar singh

Contributor-Level 10

| P ? + Q ? | = | P ? |

P 2 + Q 2 + 2 P Q c o s ? θ = P 2

Q + 2 P c o s ? θ = 0

c o s ? θ = - Q 2 P

t a n ? α = 2 P s i n ? θ 2 P c o s ? θ + Q =

? [ 2 P c o s ? θ + Q = 0 ]

α = 90 ?

 

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3 months ago

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Vishal Baghel

Contributor-Level 10

K E = P E 1 - P E 2 = m g h 1 - m g h 2

10

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3 months ago

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Vishal Baghel

Contributor-Level 10

V = 2 α 2 + α 1

= 10 * 10 - 6 + 5 * 10 - 5

= 60 * 10 - 6 / ? C

60

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3 months ago

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alok kumar singh

Contributor-Level 10

M ice   L f + m ice   ( 40 - 0 ) C w = m steam   L v + m steam   ( 100 - 40 ) C w

200 [ 80 + 40 ( 1 ) ] = M [ 540 + 60 ( 1 ) ]

200 ( 120 ) = M ( 600 )

M = 40 g m .

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3 months ago

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Vishal Baghel

Contributor-Level 10

n = W Q h = 1 - 300 900 = 2 3

Q h = 3 2 W = 1800 J

Q L = O h - W = 600 J

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