energy is the correct answer to fill the blank bb :)
Answer:
Explanation:
v = Velocity of the elevator = 3.1 m/s
= Angle of the slope =
Vertical component is given by
The vertical component of the velocity is .
Horizontal component is given by
The horizontal component of the velocity is .
The potential energy of the ball before it falls is (mass) (gravity) (height) =
(0.5 kg) (9.8 m/s²) (4 m) = 19.6 joules
The kinetic energy of the ball when it hits the ground is (1/2) (mass) (speed)² =
(1/2) (0.5 kg) (5 m/s)² = (0.25 kg) (25 m²/s²) = 6.25 joules
a). The <em>energy lost</em> to air resistance during the fall is (19.6J - 6.25J) = <em>13.35 Joules. </em><em>Energy is never destroyed, so these missing joules had to go somewhere. This is the </em><em>work done on the ball by air resistance</em><em> during the fall of the ball.</em>
b). Air resistance worked on the ball all during the fall of 4 meters.
Work = (force) x (distance)
13.35 Joules = (force) x (4 meters)
Divide each side by (4 meters) :
Average force = (13.35 Joules / 4 meters)
<em>Average force = 3.3375 Newtons</em>
Let the rod be on the x-axes with endpoints -L/2 and L/2 and uniform charge density lambda (2.6nC/0.4m = 7.25 nC/m).
The point then lies on the y-axes at d = 0.03 m.
from symmetry, the field at that point will be ascending along the y-axes.
A charge element at position x on the rod has distance sqrt(x^2 + d^2) to the point.
Also, from the geometry, the component in the y-direction is d/sqrt(x^2+d^2) times the field strength.
All in all, the infinitesimal field strength from the charge between x and x+dx is:
dE = k lambda dx * 1/(x^2+d^2) * d/sqrt(x^2+d^2)
Therefore, upon integration,
E = k lambda d INTEGRAL{dx / (x^2 + d^2)^(3/2) } where x goes from -L/2 to L/2.
This gives:
E = k lambda L / (d sqrt((L/2)^2 + d^2) )
But lambda L = Q, the total charge on the rod, so
E = k Q / ( d * sqrt((L/2)^2 + d^2) )