Answer:
a) the elastic force of the pole directed upwards and the force of gravity with dissects downwards
Explanation:
The forces on the athlete are
a) at this moment the athlete presses the garrolla against the floor, therefore it acquires a lot of elastic energy, which is absorbed by the athlete to rise and gain potential energy,
therefore the forces are the elastic force of the pole directed upwards and the force of gravity with dissects downwards
b) when it falls, in this case the only force to act is batrachium by the planet, this is a projectile movement for very high angles
c) When it reaches the floor, it receives an impulse that opposes the movement created by the mat. The attractive force is the attraction of gravity.
Answer:
θ= 5 radian
Explanation:
Given data:
Radius r = 0.70 m
Initial angular speed ω_i = 2rev/s
Time t = 5 s
Final angular speed ω_f =0
so we have angular displacement
putting values
= 5 rad
Answer:
Slope = 2 m / 10 m = 1/5
For every 5 m of effort the object will be raised 1 m
W = work done on object = M g h increase in PE of object
E S = W where E is effort and S the distance thru which the effort acts
E S = M g H
E = 100 kg * 9.8 m/s^2 * 2 m / 10 m = 196 kg m / s^2 = 196 N
Check: total work = 2 * 9.8 * 100 = 1960 J
Force Needed = 1960 J / 2 m = 980 Newtons
Mechanical advantage = 980 / 196 = 5 as one would expect since the object is raised 1 m for every 5 m of force input
1). c ... 2). d ... 3). a ... 4). d ... 5). c ... 6). a
7). b-mass ... c-m/s ... d-Newton's 1st ... e-Newton's 2nd
