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An object with a mass 5 kg is dropped from the top of a building if the object falls from 95th to 65th floor between second and fourth seconds of its falling what is the height of each floor assum g is equal to 10 m/s²?
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An object with a mass 5 kg is dropped from the top of a building if th...
**Solution:**

To find the height of each floor, we need to calculate the total height fallen by the object and divide it by the number of floors.

Given:
- Mass of the object (m) = 5 kg
- Acceleration due to gravity (g) = 10 m/s²
- Time interval (t) = 4 s - 2 s = 2 s
- Distance fallen (s) = The difference in height between the 95th and 65th floor

**Calculating the distance fallen:**

To calculate the distance fallen by the object, we can use the equation of motion:

s = ut + (1/2)at²

Here, the initial velocity (u) is zero as the object is dropped from rest.

s = 0 + (1/2)(10)(2)²
s = 20 m

Therefore, the object falls a distance of 20 meters between the second and fourth seconds.

**Calculating the height of each floor:**

To find the height of each floor, we need to divide the total distance fallen by the number of floors.

Number of floors = 95 - 65 + 1 = 31 (since we include both the 95th and 65th floor)

Height of each floor = Distance fallen / Number of floors
= 20 m / 31
≈ 0.645 m

Therefore, the height of each floor is approximately 0.645 meters.

In conclusion, when an object with a mass of 5 kg is dropped from the top of a building and falls from the 95th to the 65th floor between the second and fourth seconds, the height of each floor is approximately 0.645 meters.
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