Answer:
Spring constant, k = 24.1 N/m
Explanation:
Given that,
Weight of the object, W = 2.45 N
Time period of oscillation of simple harmonic motion, T = 0.64 s
To find,
Spring constant of the spring.
Solution,
In case of simple harmonic motion, the time period of oscillation is given by :
m is the mass of object
m = 0.25 kg
k = 24.09 N/m
or
k = 24.11 N/m
So, the spring constant of the spring is 24.1 N/m.
Answer:
The minimum possible coefficient of static friction between the tires and the ground is 0.64.
Explanation:
if the μ is the coefficient of static friction and R is radius of the curve and v is the speed of the car then, one thing we know is that along the curve, the frictional force, f will be equal to the centripedal force, Fc and this relation is :
Fc = f
m×(v^2)/(R) = μ×m×g
(v^2)/(R) = g×μ
μ = (v^2)/(R×g)
= ((25)^2)/((100)×(9.8))
= 0.64
Therefore, the minimum possible coefficient of static friction between the tires and the ground is 0.64.
Because atoms is something that pops or has bubbles in it
Answer:
(a) 0.063 m/s
(b) 1.01 m/s
Explanation:
rate of volume flow, V = 4 x 10^-6 m^3/s
(a) radius, r = 4.5 x 10^-3 m
Let the speed of blood is v.
So, V = A x v
where A be the area of crossection of artery
4 x 10^-6 = 3.14 x 4.5 x 10^-3 x 4.5 x 10^-3 x v
v = 0.063 m/s
Thus, the speed of flow of blood is 0.063 m/s .
(b) Now r' = r / 4 = 4.5 /4 x 10^-3 m = 1.125 x 10^-3 m
Let the speed is v'.
So, V = A' x v'
4 x 10^-6 = 3.14 x 1.125 x 10^-3 x 1.125 x 10^-3 x v'
v' = 1.01 m/s
Thus, the speed of flow of blood is 1.01 m/s .
