Same speed, because mass is neglected. The things that affect the speed are the distance and speed of the rock.
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))
Answer:
2.572 m/s²
Explanation:
Convert the given initial velocity and final velocity rates to m/s:
- 65 km/h → 18.0556 m/s
- 35 km/h → 9.72222 m/s
The motorboat's displacement is 45 m during this time.
We are trying to find the acceleration of the boat.
We have the variables v₀, v, a, and Δx. Find the constant acceleration equation that contains all four of these variables.
Substitute the known values into the equation.
- (9.72222)² = (18.0556)² + 2a(45)
- 94.52156173 = 326.0046914 + 90a
- -231.4831296 = 90a
- a = -2.572
The magnitude of the boat's acceleration is |-2.572| = 2.572 m/s².
The answer is the last choice.
Its electrical potential energy stays the same because it has the same electric potential. The reason why is that moving the charge towards X does not change the distance of the negative charge between the plates. The Electrical potential energy of a particle is the result energy by virtue of its position from the electrical fields produce by the plates both positive and negative. Since the charge is still equidistant to each other (assuming based from the diagram) no change in terms of electrical energy consumption or work was done.
