Answer:
in order to find the average velocity use change in Displacement / change in time?
Explanation:
1. The problem statement, all variables and given/known data A person jumps from the roof of a house 3.4 meters high. When he strikes the ground below, he bends his knees so that his torso decelerates over an approximate distance of 0.70 meters. If the mass of his torso (excluding legs) is 41 kg. A. Find his velocity just before his feet strike the ground. B. Find the average force exerted on his torso by his legs during deceleration. 2. Relevant equations I can't even seem to figure that part out. Help please? 3. The attempt at a solution I don't know how to start this at all
Answer:
m = 15.15 kg
Explanation:
Newton's Second Law of motion states that when an unbalanced force is applied on a body, an acceleration is produced in it in the direction of force. The component of force along the horizontal direction here, will be given by the Newton's Second Law as:
Fx = ma
F Cosθ = ma
where,
F = Magnitude of Force = 85 N
θ = Angle with horizontal = 27°
m = mass of object = ?
a = acceleration of object = 5 m/s²
Therefore,
85 N Cos 27° = m(5 m/s²)
m = 75.73 N/5 m/s²
<u>m = 15.15 kg</u>
Yes its bigger than 5 millimeters
