For the work-energy theorem, the work needed to stop the bus is equal to its variation of kinetic energy:

where
W is the work
Kf is the final kinetic energy of the bus
Ki is the initial kinetic energy of the bus
Since the bus comes at rest, its final kinetic energy is zero:

, so the work done by the brakes to stop the bus is

And the work done is negative, because the force applied by the brake is in the opposite direction to that of the bus motion.
Given that the block have two applied masses 250 g at East and 100 g at South. In order to make a situation in which block moves towards point A, we have to apply minimum number of masses to the blocks. In order to prevent block moving toward East, we have to apply a mass at West, equal to the magnitude of mass at East but opposite in direction. Therefore, mass of 250 g at West is the required additional mass that has to be added. There is already 100 g of mass acting at South, that will attract block towards South or point A. No need to add further mass in North-South direction.
Answer:
A. We have that radius r = 4.00m intensity I = 8.00 W/m^
total power = power/ Area ( 4πr2)= 8.00 w/m^2( 4π ( 4.00 m)2=1607.68 W
b) I = total power/ 4πr2= 8.00 W/m2 ( 4.00 m/ 9.5 m)2= 1.418 W/m2
c) E = total power x time= 1607 . 68 W x 1s= 1607.68 J
The periodic sign of magnesium is ------> mg
That would be <span>the national chairperson
-I hope this helped.</span>