Question

Can u please ans (b) the lever one and (c)the graph one

Answer 1

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

Answer 2

Explanation:

1)2N

2)40

3)20m

are the correct answers