what is the friction force for a 15kg aluminum table which is at rest on a steel floor, knowing that the coefficient of static friction=0.61

power out to power in less than 100%

the closer the planet is to the sun, the larger the gravity

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach

t = 1/400 h   ⇔ 0.0025 h ⇔ 9s ⇔ 0.15 m

any of the answers are correct.