Use v = u + at
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Answer:
, where the minus indicates the direction is opposite to that of the throw.
Explanation:
a)
Since MKS stands for meter-kilogram-second and we know that:



We can write that:



These are conversion factors, equal to 1, so multiplying our results by them won't change their value, only their units.
So we have that:



b)
Newton's 2nd Law tells us that F=ma, and the definition of acceleration is
, so we have:

Taking the throw direction as the positive one, for our values we have:

Answer:
The straight line that is obtained, intercept it on the y-axis and the value of displacement will obtained.
Explanation:
Answer:

Explanation:
Given:
- mass of vehicle,

- radius of curvature,

- coefficient of friction,

<u>During the turn to prevent the skidding of the vehicle its centripetal force must be equal to the opposite balancing frictional force:</u>

where:
coefficient of friction
normal reaction force due to weight of the car
velocity of the car

is the maximum velocity at which the vehicle can turn without skidding.
The statement about "<span>efficiency compared the output work to the output force" is false. Efficiency can be compared from the input work to the output work.</span>