A 50 kg athlete running at speed v grabs a light rope that hangs from a 10-meter-high platform and swings to a maximum of 1.8 m
1 answer:
The solution is: 1.8 meters.
Explanation:
An athlete swinging on a rope can be modeled as a pendulum.
The pendulum attains its highest point—the moment it reverses direction—where its velocity drops to zero. At this apex, due to conservation of mechanical energy, its kinetic energy is completely converted to potential energy. Conversely, at the bottom position where the athlete holds the rope, all energy is purely kinetic.
Therefore, we can write:
Solving for height (h):
It’s clear that the maximum height depends neither on the rope’s length nor the mass of the athlete, implying the 100kg athlete starting with the same initial speed will reach the same maximum height as the 50kg athlete.
Since the 50kg athlete swings to a height of 1.8 m, the 100kg athlete’s maximum height will also be 1.8 m.
Explanation:
An athlete swinging on a rope can be modeled as a pendulum.
The pendulum attains its highest point—the moment it reverses direction—where its velocity drops to zero. At this apex, due to conservation of mechanical energy, its kinetic energy is completely converted to potential energy. Conversely, at the bottom position where the athlete holds the rope, all energy is purely kinetic.
Therefore, we can write:
Solving for height (h):
It’s clear that the maximum height depends neither on the rope’s length nor the mass of the athlete, implying the 100kg athlete starting with the same initial speed will reach the same maximum height as the 50kg athlete.
Since the 50kg athlete swings to a height of 1.8 m, the 100kg athlete’s maximum height will also be 1.8 m.
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Answer:
Consequently, the temperature difference across the material will be
Explanation:
In this case, we apply the Fourier Law of heat conduction expressed by the following equation:
(1)
Where k = thermal conductivity = 0.2 W/ mK
A= 1m^2 denotes the cross-sectional area
Q= 3KW signifies the heat transfer rate
is the temperature difference we need to determine
represents the thickness of the material
Initially, we convert 3KW to W, resulting in:
With all variables accounted for, we can substitute and calculate:
Thus, the temperature difference across the material will be
Answer:
a = 18.28 ft/s²
Explanation:
the values provided are:
duration of force application, t= 10 s
Work done = 10 Btu
mass of the object = 15 lb
acceleration, a =? ft/s²
1 Btu = 778.15 ft.lbf
thus, 10 Btu = 7781.5 ft.lbf
m = 0.466 slug
So,
the work is equivalent to the change in kinetic energy
The acceleration of the object is therefore
a = 18.28 ft/s²
the constant acceleration of the object is calculated to be 18.28 ft/s²