Answer:
C.) 1.5 kg
Explanation:
Start with the equation:

Plug in what you know, and solve:

Find matching soluation:
C.) 1.5 kg
Answer:
the acceleration is reduced by gravity
a = (15 / .35) - [9.8 * sin(65º)]
Explanation:
break the launch vector into two components, vertical and horizontal
Force Net Vertical=-9.8*.350+15cos65 N
force net horizonal=15sin65
initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i
using i,j vectors..
Answer:
241.8 N.
Explanation:
The force on branch provides a reaction to the ape's weight force plus the centripetal force needed to keep the gibbon in a circular motion of radius 0.60 m.
Centripetal force = mv^2/r
F = mg + mv²/r
F = m(g + v²/r)
where,
m = mass
= 9 kg
g = acceleration due to gravity
= 9.8 m/s²
v = 3.2 m/s
r = 0.60 m
F = 9 * (9.8 + 3.2²/0.60)
= 241.8 N.
Answer:
Minimum work = 5060 J
Explanation:
Given:
Mass of the bucket (m) = 20.0 kg
Initial speed of the bucket (u) = 0 m/s
Final speed of the bucket (v) = 4.0 m/s
Displacement of the bucket (h) = 25.0 m
Let 'W' be the work done by the worker in lifting the bucket.
So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.
Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

Therefore, the work done by the worker in lifting the bucket is given as:

Now, plug in the values given and solve for 'W'. This gives,

Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.
The slope of the line on a velocityversus time graph is equal to the acceleration of the object. If the object is moving with an acceleration of +4 m/s/s (i.e., changing its velocity by 4 m/s per second), then the slope of the line will be +4 m/s/s.