<span> When headed uphill at a </span>curb<span>, turn the front </span>wheels<span> away from the </span>curb<span> and let </span>your vehicle<span> roll backwards slowly until the rear part of the front </span>wheel<span> rests against the </span>curb<span> using it as a block.</span>
Please elaborate more on your question so I can help you
Answer:
Kinetic energy, E = 133.38 Joules
Explanation:
It is given that,
Mass of the model airplane, m = 3 kg
Velocity component, v₁ = 5 m/s (due east)
Velocity component, v₂ = 8 m/s (due north)
Let v is the resultant of velocity. It is given by :


Let E is the kinetic energy of the plane. It is given by :


E = 133.38 Joules
So, the kinetic energy of the plane is 133.38 Joules. Hence, this is the required solution.
Based on Newton's principle, whenever objects A and B interact with each other, they exert forces upon each other.
When a horse pulls on a cart, t<span>he horse exerts a force only to the cart. But that force applies only to the cart, not to the horse.
The cart in turn exerts a force on the horse. But that force applies only to the horse, not the cart also.
</span>
There are two forces resulting from this interaction - a force on the horse and a force on the cart. T<span>he net force on the cart remains as it was --- a positive force in the direction of the horse's movement. Therefore, the cart begins to accelerate and move.</span><span>
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