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To determine on which planet the space probe would have the highest speed after falling from a height of 25 meters, we will use the kinematic equation for the final velocity of a freely falling object under gravity:
[tex]\[ v = \sqrt{2gh} \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the height fallen.
Given:
- Mass of the probe is [tex]\( 250 \)[/tex] kg (mass does not affect the speed in this context).
- The height fallen, [tex]\( h \)[/tex], is [tex]\( 25 \)[/tex] meters.
- The values of [tex]\( g \)[/tex] for the planets in question are provided in the table.
Now, we will calculate the final velocities on each planet considering the given height of 25 meters:
### 1. Venus
[tex]\[ g_\text{Venus} = 8.9 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Venus} = \sqrt{2 \cdot 8.9 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 21.095 \, \text{m/s} \][/tex]
### 2. Earth
[tex]\[ g_\text{Earth} = 9.8 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Earth} = \sqrt{2 \cdot 9.8 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 22.136 \, \text{m/s} \][/tex]
### 3. Uranus
[tex]\[ g_\text{Uranus} = 8.7 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Uranus} = \sqrt{2 \cdot 8.7 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 20.857 \, \text{m/s} \][/tex]
### 4. Saturn
[tex]\[ g_\text{Saturn} = 9.0 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Saturn} = \sqrt{2 \cdot 9.0 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 21.213 \, \text{m/s} \][/tex]
Comparing the final velocities:
- Venus: [tex]\( 21.095 \, \text{m/s} \)[/tex]
- Earth: [tex]\( 22.136 \, \text{m/s} \)[/tex]
- Uranus: [tex]\( 20.857 \, \text{m/s} \)[/tex]
- Saturn: [tex]\( 21.213 \, \text{m/s} \)[/tex]
The highest speed is achieved on Earth, with a final velocity of [tex]\( 22.136 \, \text{m/s} \)[/tex].
Therefore, the planet on which the space probe would have the highest speed after falling 25 meters is:
C. Earth.
[tex]\[ v = \sqrt{2gh} \][/tex]
Where:
- [tex]\( v \)[/tex] is the final velocity,
- [tex]\( g \)[/tex] is the acceleration due to gravity,
- [tex]\( h \)[/tex] is the height fallen.
Given:
- Mass of the probe is [tex]\( 250 \)[/tex] kg (mass does not affect the speed in this context).
- The height fallen, [tex]\( h \)[/tex], is [tex]\( 25 \)[/tex] meters.
- The values of [tex]\( g \)[/tex] for the planets in question are provided in the table.
Now, we will calculate the final velocities on each planet considering the given height of 25 meters:
### 1. Venus
[tex]\[ g_\text{Venus} = 8.9 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Venus} = \sqrt{2 \cdot 8.9 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 21.095 \, \text{m/s} \][/tex]
### 2. Earth
[tex]\[ g_\text{Earth} = 9.8 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Earth} = \sqrt{2 \cdot 9.8 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 22.136 \, \text{m/s} \][/tex]
### 3. Uranus
[tex]\[ g_\text{Uranus} = 8.7 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Uranus} = \sqrt{2 \cdot 8.7 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 20.857 \, \text{m/s} \][/tex]
### 4. Saturn
[tex]\[ g_\text{Saturn} = 9.0 \, \text{m/s}^2 \][/tex]
[tex]\[ v_\text{Saturn} = \sqrt{2 \cdot 9.0 \, \text{m/s}^2 \cdot 25 \, \text{m}} = 21.213 \, \text{m/s} \][/tex]
Comparing the final velocities:
- Venus: [tex]\( 21.095 \, \text{m/s} \)[/tex]
- Earth: [tex]\( 22.136 \, \text{m/s} \)[/tex]
- Uranus: [tex]\( 20.857 \, \text{m/s} \)[/tex]
- Saturn: [tex]\( 21.213 \, \text{m/s} \)[/tex]
The highest speed is achieved on Earth, with a final velocity of [tex]\( 22.136 \, \text{m/s} \)[/tex].
Therefore, the planet on which the space probe would have the highest speed after falling 25 meters is:
C. Earth.
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