Explanation:
Assuming the wall is frictionless, there are four forces acting on the ladder.
Weight pulling down at the center of the ladder (mg).
Reaction force pushing to the left at the wall (Rw).
Reaction force pushing up at the foot of the ladder (Rf).
Friction force pushing to the right at the foot of the ladder (Ff).
(a) Calculate the reaction force at the wall.
Take the sum of the moments about the foot of the ladder.
∑τ = Iα
Rw (3.0 sin 60°) − mg (1.5 cos 60°) = 0
Rw (3.0 sin 60°) = mg (1.5 cos 60°)
Rw = mg / (2 tan 60°)
Rw = (10 kg) (9.8 m/s²) / (2√3)
Rw = 28 N
(b) State the friction at the foot of the ladder.
Take the sum of the forces in the x direction.
∑F = ma
Ff − Rw = 0
Ff = Rw
Ff = 28 N
(c) State the reaction at the foot of the ladder.
Take the sum of the forces in the y direction.
∑F = ma
Rf − mg = 0
Rf = mg
Rf = 98 N
Speed= distance/time
Speed= 150000m/7200s=20.83m/s(cor.to.2d.p.)
Answer:
Waves interact with objects and other waves.
Explanation:
Answer:
In coin card experiment smooth card is used so that the card can slide easily from glass
Answer:
v₁ = -0.8087 m / s
Explanation:
To solve this problem we can use the conservation of momentum, for this we define a system formed by the man, the skateboard and the brick, therefore the force during the separation is internal and the momentum is conserved
Initial instant. When they are united
p₀ = 0
Final moment. After throwing the brick
= (m_man + m_skate) v1 + m_brick v2
the moment is preserved
p₀ = p_{f}
0 = (m_man + m_skate) v₁ + m_brick v₂
v₁ = - 
the negative sign indicates that the two speeds are in the opposite direction
let's calculate
v₁ = - 
v₁ = -0.8087 m / s
