The velocity with which the jumper leaves the floor is 5.1 m/s.
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What is the initial velocity of the jumper?</h3>
The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.
Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height
¹/₂mv² = mgh
v² = 2gh
v = √2gh
where;
- v is the initial velocity of the jumper on the floor
- h is the maximum height reached by the jumper
- g is acceleration due to gravity
v = √(2 x 9.8 x 1.3)
v = 5.1 m/s
Learn more about initial velocity here: brainly.com/question/19365526
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Answer:
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Explanation:
Answer:
T=4.24 N.m
Explanation:
Torque is equal to force for distance for sinus of the angle between the direction of the force and the distance, the distance between the mass and the pivot is 1 m, and to obtain the force that is the mass for the gravity in this case, we need to know the component that produces a torque in the pivot
F=0.5 kg* 9.8 m/
= 4.9 N
and we decompose the force in parallel direction to the rod and perpendicular direction to the rod, the magnitude that produces torque is the perpendicular component, because the torque is in function of the sinus
so, we obtain -> Fy= 4.9 N*sin(60)= 4.24 N
and, T= (4.24 N)*(1 m)*(Sin(90))= 4.24 N.m
anothe way to do it is,
T= (4.9 N)*(1 m)*(Sin(60))= 4.24 N.m, and we obtain the same result
