Answer:
Explanation:
The question relates to motion on a circular path .
Let the radius of the circular path be R .
The centripetal force for circular motion is provided by frictional force
frictional force is equal to μmg , where μ is coefficient of friction and mg is weight
Equating cenrtipetal force and frictionl force in the case of car A
mv² / R = μmg
R = v² /μg
= 26.8 x 26.8 / .335 x 9.8
= 218.77 m
In case of moton of car B
mv² / R = μmg
v² = μRg
= .683 x 218.77x 9.8
= 1464.35
v = 38.26 m /s .
The weight is not coming from the center of the mass because the force that act on it is not is equal is side.(2) section B donot have weight because the ruler bend down and section be raise up so no weight.
Since the light travels in the vacuum then its velocity will be C=3*10^8 m/sec.....then you will convert it to secs=28000*365*24*60*60=8.83008*10^11
then v=s/t then s=v*t=(3*10^8)(8.83008*10^11)=2.649024*10^20 meters
Particles in the liquid state of matter are close together, yet free to move around one another
Answer: Protons give the nucleus a positive charge.
Explanation:
Protons - Gives of positive charge
Neutrons - Gives of neutral charge (No charge)
Electrons - Gives off negitive charge