If I am in outer space, and I am using a rope to twirl an object that weighs 39.2 N on earth in a circle of radius 4 meters at a speed of 10 m/s, then the tension in the rope is nearest to...

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BlackTom AI · Accepted Answer

First, let me restate the problem to ensure we’re on the same page: in outer space, a rope is used to twirl an object that weighs 39.2 N on Earth, with a circular path of radius 4 m and a tangential speed of 10 m/s. The question asks for the rope tension, given the space setting. Now, let’s examine each answer choice in turn.

Option a: 980 N. This value resembles the weight of a mass on Earth with m ≈ 100 kg (since weight = m g ≈ 980 N). However, in outer space, gravity is effectively negligible, so the weight of the object does not contribute to the tension. The centripetal force needed is not determined by weight here, but by mass, speed, and radius, so 980 N is not correct for this configuration.

Option b: 25 N. This would be a plausible magnitude if we had a much smaller centripetal requirement, or if the mass were smaller. But using only intuition isn’t safe; we should compute using the actual mass, speed, and radius. Since the object’s weight on Earth is 39.2 N, its mass is m = W/g ≈ 39.2/9.8 ≈ 4.0 kg. With v = 10 m/s and r = 4 m, the required centripetal force is F = m v^2 / r = 4.0 × 100 / 4 = 100 N. So 25 N underestimates the needed centripetal force by a factor of four.

Option c: 24.5 N. This is very close to 25 N and would imply roughly a quarter of the computed centripetal requirement. As shown by the calculation above, the actual force needed is around 100 N, so this option underestimates by about 75 N and does not satisfy the centripetal requirement for the given mass, speed, and radius.

Option d: 100 N. This value matches the computation F = m v^2 / r with m ≈ 4.0 kg, v = 10 m/s, and r = 4 m: F = 4 × 100 / 4 = 100 N. In the space context, tension in the rope provides exactly the centripetal force that keeps the object moving in a circle, so this aligns with the physics of circular motion in zero gravity.

Putting it together, the key is recognizing that the relevant quantity is the centripetal force, which depends on mass, velocity, and radius, not on weight in a gravity-free environment. The given weight on Earth helps us find mass, but in outer space the weight does not contribute to tension; the calculation hinges on m v^2 / r, which yields 100 N. Hence, among the options, the one that matches the necessary centripetal force is the 100 N value, while the others misestimate due to either misapplying Earth-weight reasoning or underestimating the required force.

If I am in outer space, and I am using a rope to twirl an object that weighs 39.2 N on earth in a circle of radius 4 meters at a speed of 10 m/s, then the tension in the rope is nearest to...单项选择题

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