Answer:
The typical weight of a human heart is approximately 0.93 lbs.
Explanation:
Based on this,
the heart's weight constitutes about 0.5% of total body mass.
Total human weight = 185 lbs
Let the entire body weight be represented as w and the heart's weight as 
.
We aim to determine the heart's weight for a human
Using the provided information
Where, h = heart weight
w = human weight
The final weight of a human heart is 0.93 lbs.
Let T be the force exerted on the rope by her. This force induces tension in the rope, which exerts an upward force on the crates, while the weight of the crate pulls downward. Thus, the net force acting on the crate can be expressed as mg - T, acting in the downward direction. According to Newton's law, we can set up the equation: mg - T = ma. Given that a = 0 (the speed remains constant), this simplifies our equation to mg - T = 0, which leads to T = mg. Therefore, T = 25 x 9.8 = 245 N, indicating that the force she needs to apply is 245 N.
Response:
The ball remained airborne for 3.896 seconds
Explanation:
Given that
g = 9.8 m/s², representing gravitational acceleration,
If the angle of launch is 45°, the horizontal range will be maximized.
Both horizontal and vertical launch velocities are equal, each equating to
v_h = v cos θ
v_h = 27 × cos 45°
= 19.09 m/s.
The duration to reach maximum height is half of the flight time.
v = u + at ∵ v = 0 (at maximum height)
19.09 - 9.8 t₁ = 0
t₁ = 1.948 s
The total time in the air equals twice the time to reach maximum height
2 t₁ = 3.896 s
The horizontal distance covered is
D = v × t
D = 3.896×19.09
= 74.375 m
The ball was in the air for 3.896 seconds
Objects in vertical motion are an illustration of non-uniform motion. At the peak of the circle, centripetal force is balanced by the object's weight. Therefore, the minimum speed required at this top point is given by v =
=
= 5.23 m/s. As the sphere descends from the top to the bottom of the circle, according to the law of conservation of energy, potential energy can be expressed as
, where h signifies the diameter of the circle (2r). Hence, the expression will be written as
where u is the velocity at the lowest point. Consequently, the modified equation is
= 
= 
= 11.71 m/s. The collision of the dart with the bullet is an inelastic one. According to the conservation of momentum: v = 
= 
= 
= 58.55 m/s. Thus, the dart's minimum initial speed for the combined system to complete a circular loop post-collision is 58.55 m/s.
