C) The longer the line, the greater the magnitude of the vector. As for the direction, just think of a compass.
The correct answer is D. Amount of time and area of physical contact between the substances.
Explanation:
Heat transfer refers to the flow of thermal energy or heat between two or more objects. This process involves multiple factors and implies heat from the hottest object goes to the coldest one until there is an equilibrium. To begin, heat transfer depends on the amount of thermal energy in the objects because objects must have a different amount of thermal energy for heat to flow.
Besides this, the amount of energy that flows depends on the time and the contact between the substances of objects. Indeed, objects need to be in contact or close to each other for heat to transfer, and the time needs to be enough for the process to occur. For example, if you place a pot over the fire just for a few seconds it is likely the heat transferred is minimal, which does not occur if you leave the pot more time. At the same time if the pot is in close contact with fire more heat will be transferred.-
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))