<em>Inertia</em> is the property of all matter by which it tends to remain in constant, uniform motion until it's acted on by an external force.
Answer:
The true course: 
north of east
The ground speed of the plane: 96.68 m/s
Explanation:
Given:
= velocity of wind = 
= velocity of plane in still air = 
Assume:
= resultant velocity of the plane
= direction of the plane with the east
Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.
Let us find the direction of this resultant velocity with respect to east direction:
This means the the true course of the plane is in the direction of north of east.
The ground speed will be the magnitude of the resultant velocity of the plane.
Hence, the ground speed of the plane is 96.68 km/h.
Answer: Instantaneous speed.
Explanation:
Maybe you can split up the questions. I will try to answer your first question.
1. In an elastic collision, momentum is conserved. The momentum before the collision is equal to the momentum after the collision. This is a consequence of Newton's 3rd law. (Action = Reaction)
2. Momentum: p = m₁v₁ + m₂v₂
m₁ mass of ball A
v₁ velocity of ball A
m₂ mass of ball B
v₂ velocity of ball B
Momentum before the collision:
p = 2*9 + 3*(-6) = 18 - 18 = 0
Momentum after the collision:
p = 2*(-9) + 3*6 = -18 + 18 = 0
3: mv + m(-v) = m(-v) + m(v)
the velocities would reverse.
4.This question is not factual since the energy of an elastic collision must also be conserved. The final velocities should be: v₁ = -1 m/s and v₂ = 5 m/s. That said assuming the given velocities were correct:
before collision
p = 10*3 + 5*(-3) = 30 - 15 = 15
after collision:
p = 10*(-2) + 5 * v₂ = 15
v₂ = 7
5.You figure out.
Answer:
a) 0.036 J b) 0.036J c) 0.036 d) 1.9m/s e) 0.18 m
Explanation:
Mass of the dart = 0.02kg, the spring was compressed to 6cm
Work needed to compress the spring = 1/2*k*x ^2 where k is the force constant of the spring in N/m, x is the distance it was compressed in m
Work needed to compress the spring = 0.5 * 20* 0.06^2 since 6cm = 6 / 100 = 0.06 m
Work needed to compress the spring = 0.036J
b) the total energy stored in the spring = the work done to compress the spring = 0.036J
c) kinetic energy of the dart as it leaves the the spring = elastic potential energy stored in the spring = the work done in compressing the = 0.036J using the law of conservation of energy; energy is neither created nor destroyed but transformed from one form to another.
d) 1/2mv^2 = 0.036
mv^2 = 0.036*2
v^2 = 0.036*2 / 0.02 = 3.6
v = √3.6 = 1.897 approx 1.9m/s
e) kinetic energy of the dart = work done against gravity to get the body to height h
Work done against gravity = potential energy conserved at height = -mgh g is negative because the motion is upward while gravity acts downward
0.036 = 0.02 * 9.81 * h
0.036 / ( 0.02*9.81) = h
h = 0.18 m
