Answer: The minimum acceleration for the air plane is 2.269m/s2.
Explanation: To solve such problem the equation of motion are applicable.
The initial velocity is 0 since the airplane was initially standing. We are going to use this equation
V^2=U^2+2as
33^2=0+2a (240)
a= 2.269m/s2
Answer:
N = 23.4 N
Explanation:
After reading that long sentence, let's solve the question
The contact force is the so-called normal in this case we can find it by writing the translational equilibrium equation for the y axis
N - w₁ -w₂ =
N = m₁ g + m₂ g
N = g (m₁ + m₂)
let's calculate
N = 9.8 (0.760 + 1.630)
N = 23.4 N
This is the force of the support of the two blocks on the surface.
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Conservation of momentum: total momentum before = total momentum after
Momentum = mass x velocity
So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
32+0=32
Therefore after the collision
4kg x 4.8m/s = 19.2
1kg x βm/s = β
19.2 + β = 32
Therefore β = 12.8 m/s
