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trapecia [35]
3 years ago
15

1. A ski-plane with a total mass of 1200 kg lands towards the west on a frozen lake at 30.0

Physics
1 answer:
igor_vitrenko [27]3 years ago
6 0

Answer:

d = 229.5 m

Explanation:

It is given that,

Total mass of a ski-plane is 1200 kg

It lands towards the west on a frozen lake at 30.0  m/s.

The coefficient of kinetic friction between the skis and the ice is 0.200.

We need to find the distance covered by the plane before coming to rest. In this case,

\mu mg=ma\\\\a=\mu g\\\\a=0.2\times 9.8\\\\a=1.96\ m/s^2

It is decelerating, a = -1.96 m/s²

Now using the third equation of motion to find the distance covered by the plane such that :

v^2-u^2=2ad\\\\d=\dfrac{-u^2}{2a}\\\\d=\dfrac{-(30)^2}{2\times -1.96}\\\\d=229.59\ m

So, the plane slide a distance of 229.5 m.  

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George has a mass of 30 kg. He sits 1.4 m from the axis of rotation of a see-saw. If Susie has a mass of 27 kg, how far from the
aivan3 [116]

Answer:

d ₂ = 1.56 m : horizontal distance from where Susie sits to the axis of rotation.

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M = F * d   Formula (1)

F is the magnitude of said force (N)

d: distance to the axis of rotation

The direction of the moment is determined by the direction of rotation and can be assumed positive in a counterclockwise direction and negative in the clockwise direction.

Forces on the see-saw

W: Weight: in the vertical direction and down

W₁= m₁*g =  30 kg*9.8 m/s²= 294 N

W₂= m₂*g =  27 kg*9.8 m/s²= 264.6 N    

Distances of the forces to the axis of rotation

d₁ =  1.4 m :distance from where W₁ is located to the axis of rotation

d₂ :  distance from where W is located to the axis of rotation

Problem development

For static equilibrium the sum of moments around the axis of rotation is zero:

∑M =0

W₁*d₁ - W₂*d₂ = 0

W₁*d₁ = W₂*d₂

294*1.4  = 264.6*d₂

411.6= 264.6*d₂

d₂ =( 411.6) / (264.6)

d ₂= 1.56 m

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VladimirAG [237]

Answer:3800\ W

Explanation:

Given

Lawn mover running for t=20\ min

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