Question

Two point masses are separated by a distance 10cm. the potential at a point 2cm from m1 is(m1=2g ; m2=8g)​

Answer 1

Answer:

Given that the distance reduces by two-thirds, means that the new distance is

d1=d−(32×d)=3d

Therefore new force is F1=Gd12m1m2=9Gd2m1m2=9F

Answer 2

Answer:

the potential at a point 2cm from m1 is -1.067× 10^-14.

Step-by-step explanation:

Electric capability distinction among factors in an electric powered circuit wearing a few current is the work accomplished to transport a unit rate from one factor to another.

Thus, capability distinction is the work carried out in keeping with unit rate and is given as V =W/Q.

Given :

m1= 2g = 0.002kg

m2= 8g = 0.008 kg

r= 10cm = 10× 10^-2 m = 0.1m

$u = - g(m1 \times m2) \div r$

$= ( - 6.67 \times {10}^{ - 11} \times 0.002 \times 0.008) \div 10 \times {10}^{ - 2}$

= - 1.067 × 10^-4 × 10^-11 / 0.1

$- 1.067 \times {10}^{ - 14}$

(#SPJ2)