Question

like earth, ecuadoria spins on its axis once every 24 hours. when the captain stands on the spring scale, the reading on the scale is 677.37 n . this spring-scale reading is less than the captain's true ecuadorian weight by what amount?

Answer 1

The captain's true Ecuadorian weight exceeds the reading on the spring scale by ΔW = (m)(4π²R/T²).

Step-by-step explanation:

In order to determine the amount by which the captain's true Ecuadorian weight exceeds the reading on the spring scale, we need to consider the effect of Earth's rotation on the apparent weight. Due to the centrifugal force caused by Earth's rotation, the apparent weight of an object at the equator is slightly less than its true weight. The difference between the true weight and the reading on the scale can be calculated using the equation:

ΔW = mω²R

Where:

1. ΔW is the difference between the true weight and the reading on the scale
2. m is the mass of the object
3. ω is the angular velocity of Earth's rotation
4. R is the radius of Earth

By substituting the known values into the equation, we can find the difference:

ΔW = (m)(ω²)(R)

Given that ω = 2π/T, where T is the period of rotation (24 hours or 86400 seconds) and R is the radius of Earth, we can calculate ω²:

ω² = (2π/T)²

Substituting ω² and R into the equation for ΔW:

ΔW = (m)(2π/T)²(R)

Finally, we can calculate the difference:

ΔW = (m)(4π²R/T²)

Answer 2

The apparent weight on a scale is reduced by the centripetal force due to Ecuadoria's rotation. To find how much less the scale reads than the captain's true weight, we would subtract the apparent from the true weight, which requires knowing Ecuadoria's acceleration due to gravity.

Step-by-step explanation:

The question deals with the concept of apparent weight at a rotating body, in this context, a fictional planet called Ecuadoria with a similar rotation period as Earth. When an object is weighed on a spring scale on a rotating body, the apparent weight is less than the true weight due to the centripetal force required for rotation. If the apparent weight (the reading on the scale) is less than the captain's true weight, we would subtract the apparent weight from the true weight to find the difference.

To calculate the captain's true weight, we would ordinarily take the mass of the captain and multiply by the acceleration due to gravity on Ecuadoria. However, since the acceleration due to gravity is not provided, and Earth's gravity is not specified to be the same in Ecuadoria, a direct calculation is not possible without additional information. Ideally, knowing Ecuadoria's gravity, we would use Fs = mg to find the true weight (where Fs is the scale reading and g is the acceleration due to gravity). The difference between real weight and the scale reading is attributed to the centripetal acceleration reducing apparent weight.