if we are walking on a perfectly smooth ground which has no friction our force would simply cancel out the force reverted by the ground and we would fall.
We need it to help push out feet off the ground
Hope those helps :)
Answer
A thin atmosphere does not supply much oxygen, and the heat from the sun would evaporate it, because mercury is close to the sun.
Answer:
the distance from charge A to C is r₁₃= 1.216 m
Explanation:
following Coulomb's law , the force exerted by 2 point charges between themselves is:
F= k*q₁*q₂/r₁₂² , where q is charge , r is distance and 1 and 2 represents the charge A and charge B respectively , k=constant
since C ( denoted as 3) is at equilibrium
F₁₃=F₂₃
k*q₁*q₃/r₁₃²=k*q₂*q₃/r₂₃²
q₁/r₁₃²=q₂/r₂₃²
r₁₃²/q₁=r₂₃²/q₂
r₂₃=r₁₃*√(q₂/q₁)
since C is at rest and is co linear with A and B ( otherwise it would receive a net force in either vertical or horizontal direction) , we have
r₁₃+r₂₃=d=r₁₂
r₁₃+r₁₃*√(q₂/q₁)=d
r₁₃*(1+√(q₂/q₁))=d
r₁₃=d/(1+√(q₂/q₁))
replacing values
r₁₃=d/(1+√(q₂/q₁)) = 3.00 m/(1+√(3.10 C/1.44 C)) = 1.216 m
thus the distance from charge A to C is r₁₃= 1.216 m
Explanation:
It is given that,
Velocity in East, 
Velocity in North, 
(a) The resultant velocity is given by :
(b) The width of the river is, d = 80 m
Let t is the time taken by the boat to travel shore to shore. So,
t = 16 seconds
(c) Let x is the distance covered by the boat to reach the opposite shore. So,
x = 48 meters
Hence, this is the required solution.
That depends on the mass of the object, and the unit of the '46.4' .
If the '46.4' is ' meters per second² ' , then the force required is
(mass of the object in kilograms) x (46.4) newtons .
