Physical forms are: gas,liquid,and solid
Answer:
fcosθ + Fbcosθ =Wtanθ
Explanation:
Consider the diagram shown in attachment
fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)
Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)
Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)
sum of x-direction forces = 0
fx+ Fbx=Wx
fcosθ + Fbcosθ =Wtanθ
Answer:
The airliner travels 1.65 km along the runway before coming to a halt.
Explanation:
Given
Resistive forces = (2.90 × 10⁵) N = 290000 N
Mass of the airliner = (1.70 × 10⁵) kg = 170000 kg
Velocity of airliner = 75 m/s
Let the distance over moved by the airliner be equal to d
According to the work-energy theorem, the work done by the resistive forces in stopping the airliner is equal to the travelling kinetic energy of the airliner.
Work done by the resistive forces = (290000) × d = (290,000d) J
Kinetic energy of the airliner = (1/2)(170000)(75²) = 478,125,000 J
290000d = 478,125,000
d = (478,125,000/290,000)
d = 1648.7 m = 1.65 km
Hope this helps!!!
Answer:
vₓ = 20 m/s, v_{y} = -15 m / s
Explanation:
This is a conservation of moment problem, since it is a vector quantity we can work each axis independently
The system is formed by the two drones, so the forces during the crash are internal and the moment is conserved
X axis
Initial moment. Before the crash
p₀ = m₁ v₀ₓ + m₂ v₀ₓ
Final moment. After the crash
p_{fx} = (m₁ + m₂) vₓ
p₀ₓ =
m₁ v₀ₓ + m₂ v₀ₓ = (m₁ + m₂) vₓ
vₓ = (m₁ + m₂) v₀ₓ / (m₁ + m₂)
vₓ = v₀ₓ = 20 m/s
Y Axis
Initial
p_{oy} = m₁ v_{oy}
Final
p_{fy} = (m₁ + m₂) v_{y}
p_{oy} = p_{fy}
the drom rises and when it falls it has the same speed because there is no friction v_{oy} = -60 m/s
m₁
= (m₁ + m₂) v_{y}
v_{y} = m₁ / (m₁ + m₂) v_{oy}
v_{y} = 1/4 60
v_{y} = -15 m / s
Vertical speed is down