Answer:
Explanation:
Given
Distance to grandmother's house=100 mi
it is given that during return trip Julie spend equal time driving with speed 30 mph and 70 mph
Let Julie travel x mi with 30 mph and 100-x with 70 mph

x=30 mi
Therefore
Julie's Average speed on the way to Grandmother's house
=42 mph
On return trip

Answer:
4. Electrons move from higher energy states to lower energy states.
Explanation:
When electrons fall from a higher (excited) energy state to a lower energy state, it loses/gives out energy.
This energy is given out by the emission of photons (quanta of light) by the electron.
In short, the key value added of CDR data over census or survey approaches is the potential to access current and comprehensive evidence on population size, density, and dynamics, information that is fundamentally necessary for managing any humanitarian emergency or disease-related disaster but which is often
Answer:
physical feature of a wave is related to the depth of the wave base is The circular orbital motion
B. The wave base is the depth, and the still water level is the horizontal level
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))