Anonymous
Jul 6
In experiment a ball bearing is rolled off table.At the time it leaves the table a 2nd ball bearing is dropped to floor.1st ball bearing fall to floor 64cm away from table why does both fall same tim?
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ChatGPT
Jul 6
Assuming both ball bearings are dropped at the same time, we can calculate the time it takes for the first ball bearing to hit the ground using the equation:
h = 0.5 * g * t^2
Where:
h = height of the table (64 cm = 0.64 m)
g = acceleration due to gravity (9.81 m/s^2)
t = time taken for the ball bearing to hit the ground
Rearranging the equation to solve for t:
t = sqrt(2h / g)
t = sqrt(2 * 0.64 / 9.81)
t ≈ 0.36 seconds
Since the second ball bearing is dropped at the same time, it will also take approximately 0.36 seconds to hit the ground. The distance it falls can be calculated using the equation:
d = 0.5 * g * t^2
d = 0.5 * 9.81 * 0.36^2
d ≈ 0.64 meters
Therefore, the second ball bearing will also fall approximately 64 cm away from the table.
h = 0.5 * g * t^2
Where:
h = height of the table (64 cm = 0.64 m)
g = acceleration due to gravity (9.81 m/s^2)
t = time taken for the ball bearing to hit the ground
Rearranging the equation to solve for t:
t = sqrt(2h / g)
t = sqrt(2 * 0.64 / 9.81)
t ≈ 0.36 seconds
Since the second ball bearing is dropped at the same time, it will also take approximately 0.36 seconds to hit the ground. The distance it falls can be calculated using the equation:
d = 0.5 * g * t^2
d = 0.5 * 9.81 * 0.36^2
d ≈ 0.64 meters
Therefore, the second ball bearing will also fall approximately 64 cm away from the table.
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Google Gemini AI
Jul 6
Given:
- Distance traveled by 1st ball bearing horizontally: 64 cm
- Acceleration due to gravity: g (9.8 m/s²)
Objective:
To find the time of flight of the 2nd ball bearing.
Solution:
The horizontal distance traveled by the 1st ball bearing is given by:
```
Distance = Velocity - Time
```
Since the ball bearing is rolling off the table, its initial horizontal velocity is non-zero. Let's assume it is v.
Therefore,
```
64 cm = v - t
```
Converting the units to meters:
```
0.64 m = v - t
```
Now, let's consider the 2nd ball bearing. Its initial velocity is zero, and it falls vertically under gravity. The distance it falls in time t is given by:
```
Distance = (1/2) - g - t²
```
We are not given the exact height from which the 2nd ball bearing is dropped, but we can still find its time of flight by setting the distance traveled by both ball bearings equal to each other:
```
0.64 m = (1/2) - g - t²
```
Solving for t:
```
t² = (0.64 m) / (0.5 - g)
t² = 0.13 s²
t = √0.13 s
t ≈ 0.36 s
```
Therefore, the time of flight of the 2nd ball bearing is approximately 0.36 seconds.
- Distance traveled by 1st ball bearing horizontally: 64 cm
- Acceleration due to gravity: g (9.8 m/s²)
Objective:
To find the time of flight of the 2nd ball bearing.
Solution:
The horizontal distance traveled by the 1st ball bearing is given by:
```
Distance = Velocity - Time
```
Since the ball bearing is rolling off the table, its initial horizontal velocity is non-zero. Let's assume it is v.
Therefore,
```
64 cm = v - t
```
Converting the units to meters:
```
0.64 m = v - t
```
Now, let's consider the 2nd ball bearing. Its initial velocity is zero, and it falls vertically under gravity. The distance it falls in time t is given by:
```
Distance = (1/2) - g - t²
```
We are not given the exact height from which the 2nd ball bearing is dropped, but we can still find its time of flight by setting the distance traveled by both ball bearings equal to each other:
```
0.64 m = (1/2) - g - t²
```
Solving for t:
```
t² = (0.64 m) / (0.5 - g)
t² = 0.13 s²
t = √0.13 s
t ≈ 0.36 s
```
Therefore, the time of flight of the 2nd ball bearing is approximately 0.36 seconds.
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