<span>Target HR Zone 50-85% would be </span>90-153 beats per minuet. The average maximum heart rate 100% is 180 beats per minuet.
Answer:
the impulse experienced by the egg is 0.053 kgm/s.
Explanation:
Given;
mass of the egg, m = 60 g = 0.06 kg
initial velocity of the egg, u = 6 m/s
height moved by the egg, h = 50 cm = 0.5 m
Determine the final velocity of the egg as it moves upward;
v² = u² + 2(-g)h
v² = u² - 2gh
where;
v is the final velocity
-g is negative acceleration due gravity as it moves upward
v² = 6² - 2(9.8 x 0.5)
v² = 26.2
v = √26.2
v = 5.12 m/s
The impulse applied to the egg is the change in linear momentum;
J = ΔP
ΔP = mu - mv
ΔP = m(u - v)
ΔP = 0.06(6 - 5.12)
ΔP = 0.053 kgm/s
Therefore, the impulse experienced by the egg is 0.053 kgm/s.
Answer:
The bullet will rise 320 meters above the point of projection.
Explanation:
Assuming that air friction is negligent we can use the kinematic equation:
*acceleration is negative (-a) as it is acting in the opposite direction of the motion of the bullet.*
