Average speed = (total distance covered) / (time to cover the distance)
Ian's total distance covered = (2km + 0.5km + 2.5km) = 5 km.
His time to cover the distance = 3 hours.
Average speed = (5 km) / (3 hrs)
Average speed = (5/3) (km/hr)
<em>Average speed = 1.67 km/hr</em>
There's no such thing as one balanced force or one unbalanced force.
If ALL of the forces in a GROUP of forces acting on the same object
all add up to zero, then we say that the GROUP of forces is balanced.
If they don't, then the GROUP of forces is unbalanced.
Two or more forces can be balanced or unbalanced.
One force can't.
Answer:
2 m/s
Explanation:
The first part of the question the car is going in reverse or negative along the x axis. Then the second part the car is moving forward along the x axis. So the car would only have velocity in the current direction of movement. So our equation for velocity is as follows.
v = d/t
v = 10 m/5 s
v = 2 m/s
Abundant energy: Fusing atoms together in a controlled way releases nearly four million times more energy than a chemical reaction such as the burning of coal, oil or gas and four times as much as nuclear fission reactions (at equal mass)
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))
