Answer:
Radius=15.773 m
Explanation:
Given data
v=29.5 km/h=8.2 m/s
μs=0.435
To find
Radius R
Solution
The acceleration is a centripetal acceleration which is experienced by the bicycle given by
This acceleration is only due to static force which given as
The maximum value of the static force is given as
where
FN is normal force equal to mass*gravity
Therefore when the car is on the verge of sliding
Therefore the minimum radius should be found by the bicycle move without sliding
So
Answer:
a) k = Mg / d
, b) v = √2gh
, c) v_{f} = , d) x² + 6d x -
dh = 0
e)the spring must compress a greater distance.
Explanation:
a) when the block of mass M is placed on the spring, we have an equilibrium condition,
∑ F = 0
- W = 0
k d = Mg
k = Mg / d
b) let's use the concepts of energy to find the velocity of the block just before the collision
starting point. Position when released
Em₀ = U = m g h
lowest point. Right at the point of shock
= K = ½ m v²2
as there is no friction, energy is conserved
Em₀ = Em_{f}
mg h = ½ m v²
v = √2gh
c) The velocity of the two blocks after the collision, we define a system formed by the two blocks, in such a way that the forces during the collision are internal and the moment is conserved
initial instant. Just before the crash
p₀ = 2M v + M 0
final instant. Just after the shock, before the spring compression begins
p_{f} = (2M + M) v_{f}
the moment is preserved
p₀ = p_{f}
2M v = 3M v_{f}
v_{f} = ⅔ v
v_{f} =
d) now we work with the joined system after the collision, let's use the concepts of energy
starting point. After shock, before beginning spring compression
Em₀ = K = ½ (3M)
Em₀ = 3/2 M (\frac{2}{3} \ \sqrt{2gh})²
Em₀ = 4/3 M gh
final point. With the spring fully compressed
Em_f = K_e + U = ½ k x² + (3M) g x
in this case we have taken the zero of gravitational potential energy at the point where the blocks collide, as there is no friction, the energy is conserved
Em₀ = Em_f
4/3 M g h = ½ k x² + 3M g x
½ k x² + 3Mg x - 4/3 Mgh = 0
we substitute the expression for k
(
) x² + 3Mg x -
Mgh = 0
+ 3 x -
h = 0
to find the value of the spring compression, the second degree equation must be solved
x² + 6d x - dh = 0
x = [-6d ± ] / 2
x = [-6d ± 6d ]/2
x = 3d ( -1± )
e) If the collision elastic force would not lose any part of the kinetic energy during the collision, therefore the speed of the block of mass M would be much higher and therefore the spring must compress a greater distance.
Answer:
the correct one is 2. the equipotential lines must be closer together where the field has more intensity
Explanation:
The equipotential line concept is a line or surface where a test charge can move without doing work, therefore the potential in this line is constant and they are perpendicular to the electric field lines.
In this exercise we have a charge and a series of equipotential lines, if this is a point charge the lines are circles around the charge, where the potential is given by
V = k q / r
also the electric field and the electuary potential are related
E =
therefore the equipotential lines must be closer together where the field has more intensity
When checking the answers, the correct one is 2