The time taken for him to move the bin 6.5 m is 2.30 s.
The given parameters;
- <em>weight of the load, w = 557 N</em>
- <em>force applied , F = 410 N</em>
- <em>angle of force, = 15°</em>
- <em>coefficient of kinetic friction = 0.46</em>
- <em>distance moved, d = 6.5 m</em>
The net horizontal force on the recycling bin is calculated as follows;

where;
- <em>m is the mass of the recycling bin</em>
- <em />
<em> is the frictional force </em>
W = mg

The net horizontal force on the recycling bin is calculated as;

The time taken for him to move the bin 6.5 m is calculated as follows;

Thus, the time taken for him to move the bin 6.5 m is 2.30 s.
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Se necesita 105 j de trabajo para mover 2,7 c de carga de la placa negativa a la placa positiva de un condensador de placa paralela. Qué diferencia de tensión existe entre las placas
Answer:
Solution:
we have given the equation of motion is x(t)=8sint [where t in seconds and x in centimeter]
Position, velocity and acceleration are all based on the equation of motion.
The equation represents the position. The first derivative gives the velocity and the 2nd derivative gives the acceleration.
x(t)=8sint
x'(t)=8cost
x"(t)=-8sint
now at time t=2pi/3,
position, x(t)=8sin(2pi/3)=4*squart(3)cm.
velocity, x'(t)=8cos(2pi/3)==4cm/s
acceleration, x"(t)==8sin(2pi/3)=-4cm/s^2
so at present the direction is in y-axis.
Answer:
H = start height (v = 0)
h = present height
v = present speed
assuming no friction
total energy = PE + KE
mgH = mgh + .5mv^2
if PE = KE then
mgH = mgh + mgh
h = H/2
potential energy = kinetic energy when object is at half its start height.
Explanation: