Can u please ans (b) the lever one and (c)the graph one
Answers
Let the pivot point is O.
For the lever to be balanced:
AO × mass of AO = BO × mass of BO
=> 50 × mass of AO = 70 × mass of BO
=> 5 × mass of AO = 7 × mass of BO
=> mass of AO / mass of BO = 7/5
If we take mass of AO and BO as 7x and 5x
total.mass = 420
=> 12x = 420
=> x = 420/12 = 35
mass of AO =.7x = 35×7 = 245g
mass of BO = 5x = 35×5 = 175g
Now if it is pivoted at centre, O' at 60cm
mass of AO' = mass of AO + mass of 10cm of BO = 245 + 175/70 × 10 = 245 + 25 = 270g
mass of BO' = mass of AB - mass of AO' = 420 - 270 = 150g
Let the minimum weight be m. It will be placed at B.
Now,
AO' × mass of AO' = BO' × mass of BO' + BO' × m
=> 60 × 270 = 60 × 150 + 60 × m
Cancel all 60
=> 270 = 150 + m
=> m = 270 - 150
=> m = 120g
Minimum mass required is 120g
______________________________
The graph one:
mass = 200g = 0.2 kg
acceleration due to gravity (g) = 10 m/s^2
Force = mg = 0.2 × 10 = 2N
(i) Force = a = 2N
So a = 2
(ii) area = 40
Work done = Fs = Area under force-displacement graph = 40 J
(iii) area of graph = ab = 40
=> 2 × b = 40
=> b = 40/2 = 20
Thus b = 20m
The lever one:
Let the pivot point is O.
For the lever to be balanced:
AO × mass of AO = BO × mass of BO
=> 50 × mass of AO = 70 × mass of BO
=> 5 × mass of AO = 7 × mass of BO
=> mass of AO / mass of BO = 7/5
If we take mass of AO and BO as 7x and 5x
total.mass = 420
=> 12x = 420
=> x = 420/12 = 35
mass of AO =.7x = 35×7 = 245g
mass of BO = 5x = 35×5 = 175g
Now if it is pivoted at centre, O' at 60cm
mass of AO' = mass of AO + mass of 10cm of BO = 245 + 175/70 × 10 = 245 + 25 = 270g
mass of BO' = mass of AB - mass of AO' = 420 - 270 = 150g
Let the minimum weight be m. It will be placed at B.
Now,
AO' × mass of AO' = BO' × mass of BO' + BO' × m
=> 60 × 270 = 60 × 150 + 60 × m
Cancel all 60
=> 270 = 150 + m
=> m = 270 - 150
=> m = 120g
Minimum mass required is 120g
______________________________
The graph one:
mass = 200g = 0.2 kg
acceleration due to gravity (g) = 10 m/s^2
Force = mg = 0.2 × 10 = 2N
(i) Force = a = 2N
So a = 2
(ii) area = 40
Work done = Fs = Area under force-displacement graph = 40 J
(iii) area of graph = ab = 40
=> 2 × b = 40
=> b = 40/2 = 20
Thus b = 20m.
HOPE SO IT WILL HELP.....