If you drop a <span>6.0x10^-2 kg ball from height of 1.0m above hard flat surface, and a</span>fter the ball had bounce off the flat surface, the kinetic energy of the ball would be mgh - 0.14 = 0.45.
Answer:
(C) zero (there is no net horizontal component of the E-field)
Explanation:
If we subdivide the bar into small pieces, each piece (dx) contains a charge (dq), the electric field of each piece is equivalent to the field of a punctual electric charge, and has a direction as shown in the attached figure. For each piece (dx) in the negative axis there is another symmetric piece (dx) in the positive axis, and as we see in the figure for symmetry the sum of their electric fields gives a resultant in the Y axis (because its components in X are cancelled by symmetry).
Then the resultant of the electric field will be only in Y.
(C) zero (there is no net horizontal component of the E-field)
Answer:
W = 506.75 N
Explanation:
tension = 2300 N
Rider is towed at a constant speed means there no net force acting on the rider.
hence taking all the horizontal force and vertical force in consideration.
net horizontal force:
F cos 30° - T cos 19° = 0
F cos 30° = 2300 × cos 19°
F = 2511.12 N
net vertical force:
F sin 30° - T sin 19°- W = 0
W = F sin 30° - T sin 19°
W = 2511.12 sin 30° - 2300 sin 19°
W = 506.75 N
The acceleration and distance is related to the following expression:
y=v0*t + a*t^2/2 ; v0=0
y=44.1*100/2 = 2205m
hence, the speed will be
v=0 + a*t = 441m/s
from that height it will just be subjected to the gravitational acceleration
0=v_acc^2 -2g*y_free
y_free = v_acc^2/2g = 9922.5m
<span>y_max = y_acc+y_free = 441+9922.5 =10363.5m</span>