Lurking alligators. An alligator waits for prey by floating with only the top of its head exposed, so that the prey cannot easily see it. One way it can adjust the extent of sinking is by controlling the size of its lungs. Another way may be by swallowing stones (gastrolithes) that then reside in the stomach. Figure 14-41 shows a highly simplified model (a “rhombohedron gater”) of mass 130 kg that roams with its head partially exposed. The top head surface has area 0.20 m 2 . If the alligator were to swallow stones with a total mass of 1.0% of its body mass (a typical amount), how far would it sink? Figure 14-41 Problem 40.
Lurking alligators. An alligator waits for prey by floating with only the top of its head exposed, so that the prey cannot easily see it. One way it can adjust the extent of sinking is by controlling the size of its lungs. Another way may be by swallowing stones (gastrolithes) that then reside in the stomach. Figure 14-41 shows a highly simplified model (a “rhombohedron gater”) of mass 130 kg that roams with its head partially exposed. The top head surface has area 0.20 m 2 . If the alligator were to swallow stones with a total mass of 1.0% of its body mass (a typical amount), how far would it sink? Figure 14-41 Problem 40.
Lurking alligators. An alligator waits for prey by floating with only the top of its head exposed, so that the prey cannot easily see it. One way it can adjust the extent of sinking is by controlling the size of its lungs. Another way may be by swallowing stones (gastrolithes) that then reside in the stomach. Figure 14-41 shows a highly simplified model (a “rhombohedron gater”) of mass 130 kg that roams with its head partially exposed. The top head surface has area 0.20 m2. If the alligator were to swallow stones with a total mass of 1.0% of its body mass (a typical amount), how far would it sink?
A small spherical bead of mass 3.00 g is released from rest at t =0 from a point under the surface of a viscous fluid. The terminal speed is observed to be vT = 2.00 cm/s.Find a) the value of the constant k that appears in F = -kv, b) the time t at which the bead reaches 0.632 vT and c) the value of the resistive force when the bead reaches terminal speed.
When pushing off horizontally from a pool wall with her legs, the resulting force of the wall on the swimmer can be considered to act parallel to the length of her femur (thighbone), compressing it 2.98 × 10–5 m.The bone is equivalent to a uniform cylinder that is 36.0 cm long and 1.82 cm in radius.Young's modulus for bone is 1.60 ✕ 1010 N/m2.
(a)
Calculate the force exerted on the bone.
N
(b)
If her mass is 75.0 kg and water resistance is negligible, what is her acceleration (in m/s2)?Assume her weight is precisely supported by the buoyant force of the water.
m/s2
include diagram
For safety in climbing, a mountaineer uses a nylon rope that is 50 m long and 1.0 cm in diameter. When supporting a 90 kg climber, the rope elongates 1.6 m. Find the Young’s modulus of the rope.
Conceptual Physics: The High School Physics Program
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