Answer:
the kinetic energy of body B is twice the kinetic energy of body A
Explanation:
The kinetic energy of a body is given by
K = ½ m v²
If two objects leave the same point, suppose that at the same height when they reach the ground they have the same velocity.
Therefore if the mass of body b is twice the mass of body A
= ½ (2
) v²
K_{b} = 2 (½ m_{a} v²)
K_{b} = 2 K_{a}
therefore the kinetic energy of body B is twice the kinetic energy of body A
Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
Vector B , which has a magnitude of 5.5 m
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A
Comparing we will get
Substituting in
So we have
Magnitude of vector A = 0.904
The work done by a constant force in a rectilinear motion is given by:
where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.
In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:
Therefore, the total work done is 578.123 J and the answer is option E