Newton's Second law of motion: It states that rate of change of momentum is directly proportional to unbalance external force. And change of momentum is in the direction of unbalance force. Mathematical Derivation: Initial momentum , P i = mu Final momentum , P f = mv Change in momentum ,&#x2206p = P f - P i        = mv - mu But we know that rate of change of momentum is directly proportional to the applied unbalanced force. ∴ F &#x221D &#x2206p/t or F &#x221D (mv - mu ) / t/ Or F &#x221D m. ( v -u )/t Or F &#x221D m. a , Where a= Acceleration Or F = k. m. a , Or F = 1. m. a , In SI system K=1 ∴ F = m. a

Newton's First Laws of Motion { Law of Inertia }- It states that a body continue in its state of rest or of uniform motion until it is comepelled by any external unbalance forces. This laws is also known as law of inertia. Inertia is the proerty of a body which cause it to maintain its state of motion or rest. Unbalanced Forces : When two unequal forces acts on an object and the object change its state then we say that the forces are unbalance. Unbalance forces can be understood from example of tug of war. Watch this video.

Learning Hindi is easy if we are going in right way. Many of us are struggling for Hindi learning. We already have created Youtube videos for learning hindi . However We need practice to write correctly. This is the list of Hindi words in which a student of class 5 have created mistakes are. This is a tested list of words. I think many students do mistakes while writing these Hindi words. I suggest you to practice these words and post your mistakes if possible. There are Videos at the end for learning Hindi उचित  पछताएगा न्याय गिर बकरी पेचिश पीलिया त्वचा योग्य मानव इकाई व्यवस्थित वर्णमाला उच्चारण व्यवस्थापना उच्चारित  इकाईयां तुम्हारे  विवाह  धृष्टता दूँगी पक्षी विशेष समाज ब्राह्मण पहरा कार्य प्राप्त खरीद शीतल जादू स्वच्छ वायु बीमारी पढ़ने अकाल समस्याए  प्रयोग अविष्कार देवदार रसायनों आत्मकथा परिशान बचाया चंगुल कूदना भावना सहभागिता पढ़ाई निगरानी चिंता मौखिक संस्कृत इसलिए देवनागरी लिपि उचित विमाला ओत

In triangle ABC, AB > AC. E is the mid-point of BC and AD is perpendicular to BC. Prove that AB 2 + AC 2 = 2.AE 2 + 2.BE 2 Solution: In &#x2206 ABD AB 2 = AD 2 + BD 2 Or AB 2 = AD 2 + (BE + ED ) 2 Or AB 2 = AD 2 + BE 2 + 2. BE . ED + ED 2 Or AB 2 = AE 2 + BE 2 + 2. BE . ED -----(i) In &#x2206 ADC AC 2 = AD 2 + CD 2 OR AC 2 = AD 2 + ( CE - ED) 2 Or AC 2 = AD 2 + CE 2 - 2. CE . ED + ED 2 Or AC 2 = AE 2 + CE 2 - 2. CE . ED ------(ii) Adding equation 1 and situation 2 we have: AB 2 + AC 2 = AE 2 + AE 2 + BE 2 + CE 2 + 2. BE . ED - + 2. CE . ED Or AB 2 + AC 2 = AE 2 + AE 2 + BE 2 + BE 2 + 2. BE . ED - 2. BE . ED (As BE = CE ) ∴ AB 2 + AC 2 = 2 AE 2 + 2BE 2 Proved