Answer:
a. True
Explanation:
Equipotential lines are the imaginary lines in the space where actually the electric potential is same at each and every point.
Work is not required to move along such points of the equipotential line because the movement is always perpendicular to the electric field lines because these lines are always perpendicular to the electric field lines.
The electric potential for a point charge is given mathematically as:
where:
magnitude of the point charge
radial distance form the charge
permittivity of free space
Explanation:
It is given that,
Mass of Millersburg Ferry, m = 13000 kg
Velocity, v = 11 m/s
Applied force, F = 10⁶ N
Time period, t = 20 seconds
(a) Impulse is given by the product of force and time taken i.e.
(b) Impulse is also given by the change in momentum i.e.
(c) For new velocity,
Hence, this is the required solution.
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:
The moment of inertia decreased by a factor of 4
Explanation:
Given;
initial angular velocity of the ice skater, ω₁ = 2.5 rev/s
final angular velocity of the ice skater, ω₂ = 10.0 rev/s
During this process we assume that angular momentum is conserved;
I₁ω₁ = I₂ω₂
Where;
I₁ is the initial moment of inertia
I₂ is the final moment of inertia
Therefore, the moment of inertia decreased by a factor of 4