Answer:
The amount of work the factory worker must to stop the rolling ramp is 294 joules
Explanation:
The object rolling down the frictionless ramp has the following parameters;
The mass of the object = 10 kg
The height from which the object is rolled = 3 meters
The work done by the factory worker to stop the rolling ramp = The initial potential energy, P.E., of the ramp
Where;
The potential energy P.E. = m × g × h
m = The mass of the ramp = 10 kg
g = The acceleration due to gravity = 9.8 m/s²
h = The height from which the object rolls down = 3 m
Therefore, we have;
P.E. = 10 kg × 9.8 m/s² × 3 m = 294 Joules
The work done by the factory worker to stop the rolling ramp = P.E. = 294 joules
The monomer of glucose makes up all carbohydrates
Explanation:
It is given that,
Displacement of the delivery truck, 
(due east)
Then the truck moves, 
(due south)
Let d is the magnitude of the truck’s displacement from the warehouse. The net displacement is given by :
d = 4.03 km
Let 
is the direction of the truck’s displacement from the warehouse from south of east.
So, the magnitude and direction of the truck’s displacement from the warehouse is 4.03 km, 37.4° south of east.
