Banking angle is the angle between the normal horizontal path (N) and the curved banked path. The angle of banking of a curved path is defined as the angle through which the outer edge of the road is raised over the inner edge.
It is calculated as:
Here,
v = Linear velocity of object
r = Radius of the curved path
= Acceleration due to gravity
Rewriting the equation to solve for the radius we have that
Therefore the radius of a bobsled turn banked at 75° taken at 30m/s is 24.6m
Answer with Explanation:
We are given that
Mass of spring,m=3 kg
Distance moved by object,d=0.6 m
Spring constant,k=210N/m
Height,h=1.5 m
a.Work done to compress the spring initially=
b.
By conservation law of energy
Initial energy of spring=Kinetic energy of object
v=5.02 m/s
c.Work done by friction on the incline,
Answer:
0.56
Explanation:
Let the coefficient of friction is μ.
m = 4.3 kg, θ = 30 degree, initial velocity, u = 0, s = 2.7 m, t = 5.8 s
By the free body diagram,
Normal reaction, N = mg Cosθ = 4.3 x 9.8 x Cos 30 = 36.49 N
Friction force, f = μ N = 36.49 μ
Net force acting on the block,
Fnet = mg Sinθ - f = 4.3 x 9.8 x Sin 30 - 36.49 μ
Fnet = 21.07 - 36.49μ
Net acceleartion, a = Fnet / m
a = (21.07 - 36.49μ) / 4.3
use second equation of motion
s = ut + 1/2 a t^2
2.7 = 0 + 1/2 x (21.07 - 36.49μ) x 5.8 x 5.8 / 4.3
By solving we get
μ = 0.56
Answer: The current flowing through the circuit is 0.01A (or 10 mA)
Explanation:
Use Ohm's Law:
Given the values of U=30V and R=3000Ohm: