A then c then b then d but it might be mising a step
Answer:
(A) 374.4 J
(B) -332.8 J
(C) 0 J
(D) 41.6 J
(E) 351.8 J
Explanation:
weight of carton (w) = 128 N
angle of inclination (θ) = 30 degrees
force (f) = 72 N
distance (s) = 5.2 m
(A) calculate the work done by the rope
- work done = force x distance x cos θ
- since the rope is parallel to the ramp the angle between the rope and
the ramp θ will be 0
work done = 72 x 5.2 x cos 0
work done by the rope = 374.4 J
(B) calculate the work done by gravity
- the work done by gravity = weight of carton x distance x cos θ
- The weight of the carton = force exerted by the mass of the carton = m x g
- the angle between the force exerted by the weight of the carton and the ramp is 120 degrees.
work done by gravity = 128 x 5.2 x cos 120
work done by gravity = -332.8 J
(C) find the work done by the normal force acting on the ramp
- work done by the normal force = force x distance x cos θ
- the angle between the normal force and the ramp is 90 degrees
work done by the normal force = Fn x distance x cos θ
work done by the normal force = Fn x 5.2 x cos 90
work done by the normal force = Fn x 5.2 x 0
work done by the normal force = 0 J
(D) what is the net work done ?
- The net work done is the addition of the work done by the rope, gravitational force and the normal force
net work done = 374.4 - 332.8 + 0 = 41.6 J
(E) what is the work done by the rope when it is inclined at 50 degrees to the horizontal
- work done by the rope= force x distance x cos θ
- the angle of inclination will be 50 - 30 = 20 degrees, this is because the ramp is inclined at 30 degrees to the horizontal and the rope is inclined at 50 degrees to the horizontal and it is the angle of inclination of the rope with respect to the ramp we require to get the work done by the rope in pulling the carton on the ramp
work done = 72 x 5.2 x cos 20
work done = 351.8 J
<span>Climatic changes can definitely influence changes in life forms.</span>
To determine the Force it is necessary that Samantha starts from the consideration of the dynamic balance between the forces acting on the car. For this, the centripetal force must be equal to the friction force (otherwise the car would lose contact with the ground), the expression of these two forces would be


Here,
= Coefficient of kinematic friction
m = mass
g = Acceleration due to gravity
v = Velocity
r = Radius
From this relationship then she should assume the following
1) The car travels at a constant speed
2) There is indeed that frictional force between the car and the road
3) There is no loss of mass during displacement
4) The centripetal force must be equal to the frictional force so that the car does not lose contact or slip
5) The radius must be constant
Answer:Although any object in motion through space (for example a thrown baseball, kicked football, fired bullet, thrown arrow, stone released from catapult) are projectiles, they are commonly found in warfare and sports.
Explanation: