At a time when mining asteroids has become feasible, astronauts have connected a line between their 3460-kg space tug and a 6430-kg asteroid. They pull on the asteroid with a force of 366 N. Initially the tug and the asteroid are at rest, 493 m apart. How much time does it take for the ship and the asteroid to meet?

77.8s

Explanation:

Let d distance between the asteroid and space tug

So;d=Xtug+Xspace

Xtug=VtT+0.5atT^2

Xspace=VsT+0.5asT^2

Since Vt=Vs=0 initial velocity

Then

d=0.5(atT^2+asT^2)

T^2( at+as)=2d

T=√(2d/at+as)

But force F = mass M*acceleration a

Hence at=Ft/mt ,as=Fs/ms

But note Ft=F=Fs since the Same force acts on it

Hence T=√( 2d/F(1/mt+1/Ms))

T=√(2*493/366(1/3460+1/6430)

T=√(986/0.1627)=√(6060.195)=77.8s

r=rho l/a where a is the area, l is the length and rho is resistivity so the greater the area the lesser the resistance thats why the earth pin is thicker bcos current follow the shortest path, so less resistive .

c because the action was the roller coaster reaches bottom of hill (action) reaction would be blood tends to drain away from head

hope this !