Answer:
s
Explanation:
From the question we are told that
The outer ring with a radius of 30 m
inner Gravity Approximately 9.80 m/s'
Outer Gravity Approximately 5.35 m/s.
Generally the equation for centripetal force is given mathematically as
Centripetal acceleration enables Rotation therefore?
Considering the outer ring,
Therefore solving for Period T
Generally the equation for solving Period T is mathematically given as
s
Answer:
v₁ = 1146.6 m / s
Explanation:
This problem must be solved in two parts, a first part of the crash or another with energy after the crash.
Let's start by analyzing the crash.
Definition the system as formed by the clay block and the wooden block, in this case the moment is conserved because the forces are of action and reaction type
Let's use subscript 1 for the clay block and subscript 2 for the wood block
Initial. Before the crash
p₀. = m₁ v₁
Final. Right after the crash
= (m₁ + m₂) v
p₀ = p_{f}
m₁ v₁ = (m₁ + m₂) v
v₁ = v (m₁ + m₂) / m₁
Now the body is formed by the joint of the plug + the wooden block
W = ΔK
-fr x = Kf - K₀
The final velocity is zero, so the final kinetic energy is zero
The equation for friction force is
fr = μ N
From Newton's second law, on the horizontal surface
N- W = 0
N = (m₁ + m₂) g
fr = μ (m₁ + m₂) g
We substitute
- μ (m₁ + m₂) g x = 0 - ½ (m₁ + m₂) v
v = 2 μ g x
v = 2 0.650 9.8 7.50
v = 95.55 m / s
We substitute in the moment equation
v₁ = v (m₁ + m₂) / m₁
v₁ = 95.55 (10 + 110) / 10
v₁ = 1146.6 m / s
Answer:
The bumper will be able to move by 0.01155m.
Explanation:
The magnitude of deceleration of the car in the front end collision.
This is the deceleration of the car that is generated to stop due to a front end collision.
4 km/h = 1.11 m/s
Now, the initial speed of the bumper in the relation of car, Vi = 0
Now, the initial speed of the bumper in the relation of car, Vf = 1.11 m/s
Use the below equation:
Thus, the bumper can move relative to the car is 0.01155 m .