Answer:
rod end A is strongly attracted towards the balls
rod end B is weakly repelled by the ball as it is at a greater distance
Explanation:
When the ball with a negative charge approaches the A end of the neutral bar, the charge of the same sign will repel and as they move they move to the left end, leaving the rod with a positive charge at the A end and a negative charge of equal value at end B.
Therefore rod end A is strongly attracted towards the balls and
rod end B is weakly repelled by the ball as it is at a greater distance
Answer:
V0=27.4 m/s; t=0.8 s
Explanation:
Final position y=37.0 m, time = 2.3 s; Initial position is set to be zero. We calculate the initial speed with the kinematics equation:
We solve for initial speed

Now, using the same expression we estimated time to first reach 18.5 m :
Second order equation with solutions
t1=0.8 s and t2=4.8 s
The first time corresponds to the first reach.
<span>None of the choices makes a correct statement. The third choice is close,
but misleading.
The pencil appears broken because light bends away from a straight line
when it crosses the boundary between air and water.</span>
Answer:
The magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²
Explanation:
The total linear acceleration is the vector sum of the tangential acceleration and radial acceleration.
The radial acceleration is given by;

where;
a is the angular acceleration and
r is the radius of the circular path

Determine time of the rotation;

Determine angular velocity
ω = at
ω = 1.6 x 0.707
ω = 1.131 rad/s
Now, determine the radial acceleration

The magnitude of total linear acceleration is given by;

Therefore, the magnitude of the total linear acceleration is 0.27 m/s²
b. 0.27 m/s²