(b) maximum when breadth and width from the base. Pressure is maximum when area of base is minimum. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(b) maximum when breadth and width from the base. Pressure is maximum when area of base is minimum.
(c) Value of acceleration due to gravity on the surface of moon is 1/6 th of its value on the surface of earth. Thus gmoon = 1/6 ,g eath = 9.8/6 m s⁻². https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(c) Value of acceleration due to gravity on the surface of moon is 1/6 th of its value on the surface of earth. Thus gmoon = 1/6 ,g eath = 9.8/6 m s⁻².
(c) On the surface of earth value of 'g' is maximum at the poles and minimum at the equator. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(c) On the surface of earth value of ‘g’ is maximum at the poles and minimum at the equator.
(b) In a tug of war the work done by the winning team is positive because force applied by the winning team and the displacement of the team members is along same direction. However, work done by the losing team is negative because now displacement is in a direction opposite to that of force appliedRead more
(b) In a tug of war the work done by the winning team is positive because force applied by the winning team and the displacement of the team members is along same direction. However, work done by the losing team is negative because now displacement is in a direction opposite to that of force applied by the losing team.
(b) Radius of circular track R = 100 m and time taken by car to complete one lap of circular track t = 62.8 s. Total displacement of car on covering a lap = 0 ∴ Average velocity of car = total displacement/total time = 0/62.8 s = 0 Total distance covered by the car in one lap s = 2πR = 2 × 3.14 × 10Read more
(b) Radius of circular track R = 100 m and time taken by car to complete one lap of circular track t = 62.8 s.
Total displacement of car on covering a lap = 0
∴ Average velocity of car = total displacement/total time = 0/62.8 s = 0
Total distance covered by the car in one lap s = 2πR = 2 × 3.14 × 100 = 628 m
∴ Average speed of car = total distance/total time = s/t = 628 m/62.8 s = 10m s⁻¹ .
A girl stands on a box having 60 cm length, 40 cm breadth and 20 cm width in three ways. In which of the following cases, pressure exerted by the brick will be
(b) maximum when breadth and width from the base. Pressure is maximum when area of base is minimum. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(b) maximum when breadth and width from the base. Pressure is maximum when area of base is minimum.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
See lessIf g be the value of acceleration due to gravity on the surface of earth, then value of acceleration due to gravity on the surface of moon (gmoon) will be
(c) Value of acceleration due to gravity on the surface of moon is 1/6 th of its value on the surface of earth. Thus gmoon = 1/6 ,g eath = 9.8/6 m s⁻². https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(c) Value of acceleration due to gravity on the surface of moon is 1/6 th of its value on the surface of earth. Thus gmoon = 1/6 ,g eath = 9.8/6 m s⁻².
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
See lessThe value of acceleration due to gravity g
(c) On the surface of earth value of 'g' is maximum at the poles and minimum at the equator. https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
(c) On the surface of earth value of ‘g’ is maximum at the poles and minimum at the equator.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-9/
See lessIn a tug of war the work done by the winning team and the losing team are respectively
(b) In a tug of war the work done by the winning team is positive because force applied by the winning team and the displacement of the team members is along same direction. However, work done by the losing team is negative because now displacement is in a direction opposite to that of force appliedRead more
(b) In a tug of war the work done by the winning team is positive because force applied by the winning team and the displacement of the team members is along same direction. However, work done by the losing team is negative because now displacement is in a direction opposite to that of force applied by the losing team.
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-10/
See lessAcar runs at a uniform rate on a circular track of radius 100m taking 62.8 s on each lap. What is the average velocity and average speed of car for one complete lap?
(b) Radius of circular track R = 100 m and time taken by car to complete one lap of circular track t = 62.8 s. Total displacement of car on covering a lap = 0 ∴ Average velocity of car = total displacement/total time = 0/62.8 s = 0 Total distance covered by the car in one lap s = 2πR = 2 × 3.14 × 10Read more
(b) Radius of circular track R = 100 m and time taken by car to complete one lap of circular track t = 62.8 s.
Total displacement of car on covering a lap = 0
∴ Average velocity of car = total displacement/total time = 0/62.8 s = 0
Total distance covered by the car in one lap s = 2πR = 2 × 3.14 × 100 = 628 m
∴ Average speed of car = total distance/total time = s/t = 628 m/62.8 s = 10m s⁻¹ .
https://www.tiwariacademy.com/ncert-solutions/class-9/science/chapter-7/
See less