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2. When a shot putter exerts a net force of 140 N on a shot, the shot has an acceleration of \( 19 \mathrm{~m} / \mathrm{s}^{3} \). What is the mass of the shot? 3. A 7.3 kg bowling ball sliding down a smooth ramp has a net force of 20 N exerted on it. What is its acceleration?

Ask by Curry Hodges. in the United States
Feb 06,2025

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Answer

**Problem 2:** The mass of the shot is approximately **7.37 kg**. **Problem 3:** The bowling ball accelerates at approximately **2.74  m/s²**.

Solution

Let's solve each problem step by step. --- ### **Problem 2** **Given:** - **Net Force (\( F \))**: 140 N - **Acceleration (\( a \))**: \( 19 \, \text{m/s}^2 \) *(Assuming the acceleration unit is \( \text{m/s}^2 \) instead of \( \text{m/s}^3 \))* **To Find:** - **Mass (\( m \))** of the shot. **Solution:** Newton's Second Law states: \[ F = m \times a \] Solving for mass (\( m \)): \[ m = \frac{F}{a} \] Plugging in the values: \[ m = \frac{140 \, \text{N}}{19 \, \text{m/s}^2} \approx 7.37 \, \text{kg} \] **Answer:** The mass of the shot is approximately **7.37 kg**. --- ### **Problem 3** **Given:** - **Mass (\( m \))** of the bowling ball: 7.3 kg - **Net Force (\( F \))**: 20 N **To Find:** - **Acceleration (\( a \))** of the bowling ball. **Solution:** Again, using Newton's Second Law: \[ F = m \times a \] Solving for acceleration (\( a \)): \[ a = \frac{F}{m} \] Plugging in the values: \[ a = \frac{20 \, \text{N}}{7.3 \, \text{kg}} \approx 2.74 \, \text{m/s}^2 \] **Answer:** The bowling ball accelerates at approximately **2.74  m/s²**. ---

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To determine the mass of the shot put, you can use Newton's second law, which states that \( F = ma \). Rearranging this formula gives \( m = \frac{F}{a} \). Plugging in the values, \( m = \frac{140 \, \text{N}}{19 \, \text{m/s}^2} \) computes to approximately 7.37 kg! So, the shot weighs a little over 7.3 kg – like lifting a kitten, just a bit heavier! For the bowling ball, we can again lean on Newton's second law. You can rearrange the formula \( F = ma \) to find acceleration: \( a = \frac{F}{m} \). Substituting the values given, \( a = \frac{20 \, \text{N}}{7.3 \, \text{kg}} \) results in approximately 2.74 m/s². Imagine it rolling down the alley at that rate – it’s bound to knock some pins over!

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