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a = -2 m/s²

formula of distance travelled in nth second by an object is

Sn = u + a(2n-1)/2

distance travelled in 7th second is

S = 13 - 2(2*7-1)/2

(2) let distance between them is r

formula of force is

F = kQ1Q2/r²

where k = 1/4π∈

force between 3Q and Q is

F = 3Q²×k/r² (repulsive) -----------------------(1)

let charge distribution on the sphere is as after connecting them by a thin wire and let there is no loss of charge

3Q-q................................Q+q

r

charge on the both sphere is same after connecting them

so

3Q-q = Q+q

q = Q

so charge on each sphere is 2Q

so force

F1 = 2Q×2Qk/r²

F1 = 4Q²×k/r² (repulsive) ---------------------(2)

(2) ÷ (1)

F1/F = 4/3

F1 = 4F/3

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U = 13m/s j

a = -2 m/s² j

velocity will be zero at t = 13/2 = 6.5s

We need to find distance travelled in 7th second.

It is equal to distance travelled in 7 seconds minus distance travelled in 6 seconds.

Since it is changing its direction at t=6.5s, displacement will be 0 if we find the displacement in the 7th second. To find the distance travelled, find the distance travelled from 6 to 6.5 second and 6.5s to 7 second. Both will be same as time is same. So find the displacement from 6 to 6.5s and multiply it with 2.

S = (ut+0.5at²)₆.₅ - (ut+0.5at²)₆

⇒ S = (13×6.5 + 0.5×(-2)×6.5²) - (13×6 + 0.5×(-2)×6²)

⇒ S = 13(6.5-6) - 0.5×2×(6.5²-6²)

⇒ S = 13×0.5 - 1×(42.25-36)

⇒S = 6.5 - 6.25

⇒ S = 0.25m

So distance travelled in 7th second = 2S = 2×0.25 =** 0.5m**

2.

Initial charges = Q and 3Q

When they are connected by a wire, charge distribution will happen according to the potential of the spheres. Since both are identical spheres, charge will be equally divided between the two.

The charges on both spheres will be (Q+3Q)/2 = 2Q

a = -2 m/s² j

velocity will be zero at t = 13/2 = 6.5s

We need to find distance travelled in 7th second.

It is equal to distance travelled in 7 seconds minus distance travelled in 6 seconds.

Since it is changing its direction at t=6.5s, displacement will be 0 if we find the displacement in the 7th second. To find the distance travelled, find the distance travelled from 6 to 6.5 second and 6.5s to 7 second. Both will be same as time is same. So find the displacement from 6 to 6.5s and multiply it with 2.

S = (ut+0.5at²)₆.₅ - (ut+0.5at²)₆

⇒ S = (13×6.5 + 0.5×(-2)×6.5²) - (13×6 + 0.5×(-2)×6²)

⇒ S = 13(6.5-6) - 0.5×2×(6.5²-6²)

⇒ S = 13×0.5 - 1×(42.25-36)

⇒S = 6.5 - 6.25

⇒ S = 0.25m

So distance travelled in 7th second = 2S = 2×0.25 =

2.

Initial charges = Q and 3Q

When they are connected by a wire, charge distribution will happen according to the potential of the spheres. Since both are identical spheres, charge will be equally divided between the two.

The charges on both spheres will be (Q+3Q)/2 = 2Q