To find the horizontal distance that the water bottle rocket travels, we first need to determine the horizontal component of the force applied by the air pump.
Given:
- Work done (W) = 5600 J
- Force (F) = 150 N
- Angle (θ) = 45°
The horizontal component (F_horizontal) of the force can be calculated using trigonometry: \[ F_{\text{horizontal}} = F \cdot \cos(\theta) \] \[ F_{\text{horizontal}} = 150 , \text{N} \cdot \cos(45^\circ) \] \[ F_{\text{horizontal}} = 150 , \text{N} \cdot \frac{\sqrt{2}}{2} \] \[ F_{\text{horizontal}} \approx 150 \cdot 0.7071 \approx 106.07 , \text{N} \]
Next, we can determine the distance (d) the rocket travels using the work-energy principle. The work done on the rocket is equal to the force times the distance in the direction of the force. We can express this as: \[ W = F_{\text{horizontal}} \cdot d \]
We can rearrange this to find the distance: \[ d = \frac{W}{F_{\text{horizontal}}} \] \[ d = \frac{5600 , \text{J}}{106.07 , \text{N}} \] \[ d \approx 52.8 , \text{m} \]
Rounding to two significant figures: \[ d \approx 53 , \text{m} \]
Among the given options, this rounds to \( 5.3 \times 10^1 , \text{m} \).
Thus, the horizontal distance the water bottle rocket travels is: 5.3 × 10¹ m.
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