Answer:
V = 10 km / 1 hr = 10 km/hr
V = -10 j km / hr if one were to use i, j, k as unit vectors with the usual orientation
To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
option C
Explanation:
given,
Force by the engine on plane in West direction = 350 N
Frictional force on the runway = 100 N in east
force exerted by the wind = 100 N in east
net force and direction = ?
consider west to be positive and east be negative.
when airplane will be moving there will be frictional as well as wind resistance will be acting in opposite direction of airplane
Net force = 350 N - 100 N - 100 N
= 150 N
as our answer comes out to be positive so the airplane will be moving in West
hence, the correct answer is option C
Oil, grease and dry lubricants
