<u>Answer:</u>
Option A is the correct answer.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
First walking 1.2 km north, displacement = 1.2 j km
Secondly 1.6 km east, displacement = 1.6 i km
Total displacement = (1.6 i + 1.2 j) km
Magnitude = 
Angle of resultant with positive X - axis =
= 36.87⁰ east of north.
These energy exchanges are not changes in kinetic energy. They are changes in bonding energy between the molecules. If heat is coming into a substance during a phase change, then this energy is used to break the bonds between the molecules of the substance. The example we will use here is ice melting into water.
The additional force needed to bring the car into equilibrium is frictional force.
Answer:
Explanation:
Given
mass of Flywheel 
mass of bus 
radius of Flywheel 
final speed of bus 
Conserving Energy i.e.
0.9(Rotational Energy of Flywheel)= change in Kinetic Energy of bus
Let
be the angular velocity of Flywheel





<span>The core finally cools into a white dwarf, then a black dwarf. This is what happens when a normal-sized star dies. If a really huge star dies, it has so much mass that after the helium is used up, it still has enough carbon to fuse it into heavy elements like iron. When the core turns to iron, it no longer burns.
please give me </span>Brainliest answer?