Class 11 phy ch 6 solutions
WebAnswer: (a) Volume of cube, V = (1 cm) 3 = (10 -2 m) 3 = 10 -6 m 3 (b) Surface area = curved area + area on top /base = 2πrh + 2πr 2 = 2πr (h + r) r = 2 cm = 20 mm h = 10 cm = 100 mm Surface area = 2πr (h + r) = 2 x 3.14 x 20 (100 + 20) = 15072 mm 2 Hence, the answer is 15072 mm 2 (c) Speed of vehicle = 18 km/h 1 km = 1000 m 1 hr = 60 x 60 = … Web= bc n ^ a →. ( b → × c →) = a. ( b c n ^) = abc cosθ n ^ = abc cos 0 0 = abc The volume of the parallelepiped. Q6. Find the components along the x, y, and z axes of the angular momentum l of a particle whose position vector is r with components x, y, and z and whose momentum is p with components px, py and pz.
Class 11 phy ch 6 solutions
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WebClass 11 Physics NCERT Solutions for Chapter 14 Oscillations. These NCERT Solutions provide you with the answers to the question from the textbook, important questions from previous year question papers and … Web1. Estimate the fraction of molecular volume to the actual volume occupied by oxygen gas at STP. Take the diameter of an oxygen molecule to be 3 Å. Solution: Diameter of an oxygen molecule, d = 3 Å Radius, r = d / 2 r = 3 / 2 = 1.5 Å = 1.5 x 10 -8 cm The actual volume occupied by 1 mole of oxygen gas at STP = 22400 cm 3
WebNCERT Solutions for Class 11 Physics Chapter 6 Work Energy and Power. NCERT Solutions for Class 11 Physics Chapter 6 Work … WebSolution 6. a) Applied force must be greater than net force which is acting downwards to make to move up. Freq = µN + mgsin30°. N = mgcos30°. M = 2kg , g = 9.8m/s2 , μ = 0.2. …
WebClass 11 Physics Notes are free and will always remain free. We will keep adding updated notes, past papers, guess papers and other materials with time. We will also introduce a … WebApr 9, 2024 · class xi Physics Elasticity Chapter Problem Set 1 solutions
WebRelative position of second stone w.r.t. first is given by x 2 (t) – x 1 (t) = 15t. Since there is a linear relationship between x 2 (t) – x 1 (t) and t, therefore the graph is a straight line. For maximum separation, t = 8 s So maximum separation is 120 m. After 8 second, only the second stone would be in motion.
WebBalbharati 11th Physics (11th) Chapter 6: Mechanical Properties of Solids solutions Concepts covered in Mechanical Properties of Solids are Elastic Behavior of Solids, … hop to it towing vtWebCBSE Class 11 Physics Chapter 11 – Thermal Properties of Matter NCERT Solutions Matter refers to those object that occupies space and has a mass. Thermal properties of matter define the state of matter based on their temperature. Apart from this, there are other thermal properties of matter. hopton attackWebAnswer: (a) When the pebble is moving upward, the acceleration g is acting downward, so the force is acting downward is equal to F = mg = 0.05 kg x 10 ms -2 = 0.5 N. (b) In this case also F = mg = 0.05 x 10 = 0.5 N. (downwards). (c) The pebble is not at rest at highest point but has horizontal component of velocity. look north newcastle newsWebSolution 6. a) Applied force must be greater than net force which is acting downwards to make to move up. Freq = µN + mgsin30°. N = mgcos30°. M = 2kg , g = 9.8m/s2 , μ = 0.2. On substituting, Freq = 13N. b) Net force acting down the incline is given by. look north male reportersWebNCERT Solutions for Class 11 Physics Chapter 6 – Work, Energy, and Power. Physics is a crucial part of Physics which aims to explain the behavior of objects. Moreover, it … look north news hullWebSolution 11. Method 1: The gravitational field inside a sphere of radius R at a distance x from center is. So, force on particle of mass m is. Method 2: Acceleration due to gravity … look north lincolnshireWebCBSE Class 11 Physics Chapter 11 – Thermal Properties of Matter NCERT Solutions. Matter refers to those object that occupies space and has a mass. Thermal properties of … look north magazine