(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s 2 , what is the reading on the spring scale?

Question

Deepak Patra · Answer

Therefore, the acceleration of the point at equator of Earth is 0.033 m/s 2 Round off to two significant figures, T = 246 N Therefore, the reading in spring scale is 246 N.

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(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s 2 , what is the reading on the spring scale?

(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s², what is the reading on the spring scale?

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(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s 2 , what is the reading on the spring scale?

(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s2, what is the reading on the spring scale?

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(a) What is the centripetal acceleration of an object on the Earth's equator due to the rotation of the Earth? (b) A 25.0-kg object hangs from a spring scale at the equator. If the free-fall acceleration due only to the Earth's gravity is 9.80 m/s², what is the reading on the spring scale?