Answer: T2 = 7.07s
Explanation: The period of a loaded spring of spring constant k and mass m is given by
T= 2π √m/k
With 2π constant and k, it can be seen with little algebra that
T² is proportional to mass m
Hence (T1)²/m1 = (T2) ²/m2
Where T1 = 5, T2 =?, let m1 = m hence m2 = 2m.
By substituting, we have that
5²/m = (T2) ²/2m
25 / m = (T2) ²/2m
25 × 2m = (T2) ² × m
25 × 2 = (T2) ²
50 = (T2) ²
T2 = √50
T2 = 7.07s
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 lithosphere because it includes the outer region of the earth including the crust and outer mantle
