Respuesta :
Answer:
To find the average force exerted on the window by the hailstones, we can use the impulse-momentum theorem, which states:
\[ \text{Average force} = \frac{\text{Change in momentum}}{\text{Time interval}} \]
First, let's calculate the total momentum change. The momentum change for each hailstone is given by \( \Delta p = 2mv \), where \( m \) is the mass of the hailstone and \( v \) is its speed. Since the collisions are elastic, the change in momentum for each hailstone is twice its initial momentum (before the collision), which is given by \( mv \). So, \( \Delta p = 2mv \).
Now, the total number of hailstones hitting the window is \( 446 \). Therefore, the total momentum change due to all the hailstones hitting the window is:
\[ \text{Total momentum change} = 446 \times 2 \times (m \times v) \]
Given that \( m = 8 \) g and \( v = 4.2 \) m/s, we need to convert \( m \) to kilograms and calculate the product \( m \times v \):
\[ m = 8 \text{ g} = 8 \times 10^{-3} \text{ kg} \]
\[ \text{Total momentum change} = 446 \times 2 \times (8 \times 10^{-3} \times 4.2) \]
Next, we'll calculate the time interval:
\[ \text{Time interval} = 69 \text{ s} \]
Now, we can find the average force using the impulse-momentum theorem:
\[ \text{Average force} = \frac{\text{Total momentum change}}{\text{Time interval}} \]
Substitute the values to find the average force in newtons (N).
First, let's calculate the total momentum change:
\[ \text{Total momentum change} = 446 \times 2 \times (8 \times 10^{-3} \times 4.2) \]
\[ \text{Total momentum change} = 446 \times 2 \times (0.008 \times 4.2) \]
\[ \text{Total momentum change} = 446 \times 2 \times 0.0336 \]
\[ \text{Total momentum change} = 29.9072 \]
Now, let's calculate the average force using the impulse-momentum theorem:
\[ \text{Average force} = \frac{\text{Total momentum change}}{\text{Time interval}} \]
\[ \text{Average force} = \frac{29.9072}{69} \]
\[ \text{Average force} \approx 0.433 \, \text{N} \]
So, the average force exerted on the window by the hailstones is approximately \( 0.433 \, \text{N} \).
