Answer:
The force pulling the roller along the ground is 128.55 N
Explanation:
A force of 200 N acting at an angle of 50° with the ground level
This force is pulled a garden roller
We need to find the force pulling the roller along the ground
The force that pulling the roller along the ground is the horizontal
component of the force acting
→ The force acting is 200 N at direction 50° with ground (horizontal)
→ The horizontal component = F cosФ
→ F = 200 N , Ф = 50
→ The horizontal component = 200 cos(50) = 128.55 N
128.55 N is the horizontal component of the force that pulling the
roller along the ground
<em>The force pulling the roller along the ground is 128.55 N</em>
5-ohm
Extra
Variable
120-ohm
Variable
Pg. 614
V^2=u^2 +2aS
U is found first by considering that first 8 secs and using v=u+at. {different v and u though}
V=-u+gt.
Magnitude of u = magnitude of v if there is no resistance ( because the conservation of energy says the k. E. must be the same when it passes you as when it left your hand).... up is negative here, down is positive.
V+v=gt
2v= g x 8
V=4xg.= the initial velocity for the next calculation
V^2=(4g)^2+(2xgx21)
So v can be calculated.
Answer:
The normal force will be lower than the gravitational force acting on the car. Therefore the answer is N < mg, which is <em>option B</em>.
Explanation:
Over a round hill, the centripetal force acting toward the the radius of the hill supports the gravitational force (mg) of the car. This notion can be expressed mathematically as follows:
At the top of a round hill
At the foot of a round hill
