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 .
Answer:
after 6 second it will stop
he travel 36 m to stop
Explanation:
given data
speed = 12 m/s
distance = 100 m
decelerates rate = 2.00 m/s²
so acceleration a = - 2.00 m/s²
to find out
how long does it take to stop and how far does he travel
solution
we will apply here first equation of motion that is
v = u + at ......1
here u is speed 12 and v is 0 because we stop finally
put here all value in equation 1
0 = 12 + (-2) t
t = 6 s
so after 6 second it will stop
and
for distance we apply equation of motion
v²-u² = 2×a×s ..........2
here v is 0 u is 12 and a is -2 and find distance s
put all value in equation 2
0-12² = 2×(-2)×s
s = 36 m
so he travel 36 m to stop
INCREASE in temperature of the material practically increase the energy of the particles. which increases their motion due to increase in energy . thus when the temperature is decreased the energy level decreases which causes the particle's motion to slow down.. the motion of the particle is highly reduced when the temperature is lowered