20. If 60 percent of the potential energy of water flowing over a dam can be converted to electricity, find the maximum power output, in kW (kilowatts), for a dam 50 m high where the water crosses at a rate of 300,000 kg/min(1 watt = 1 joule/sec, 1 kW = 1000 watts) ### Key Physics Formulas1. **Average Speed**: \[ v = \frac{d}{t} \] - **Explanation**: The average speed is calculated as the distance traveled (d) divided by the time taken (t).2. **Distance Under Acceleration**: \[ d = v_0 t + \frac{1}{2} a t^2 \] - **Explanation**: This formula calculates the distance (d) covered under uniform acceleration. \(v_0\) is the initial velocity, \(a\) is the acceleration, and \(t\) is the time.3. **Weight**: \[ W = mg \] - **Explanation**: Weight (W) is the force exerted by gravity on a mass (m), where \(g\) is the acceleration due to gravity.4. **Acceleration**: \[ a = \frac{F_{\text{net}}}{m} \] - **Explanation**: Acceleration (a) is the net force (\(F_{\text{net}}\)) acting on an object divided by its mass (m).5. **Gravity Law**: \[ F = \frac{G m_1 m_2}{r^2} \] - **Explanation**: This formula describes the gravitational force (F) between two masses (\(m_1\) and \(m_2\)) separated by a distance (r). \(G\) is the gravitational constant.6. **Momentum**: \[ p = mv \] - **Explanation**: Momentum (p) is the product of mass (m) and velocity (v).7. **Kinetic Energy**: \[ KE = \frac{1}{2} mv^2 \] - **Explanation**: The kinetic energy (KE) of an object is half the product of its mass (m) and the square of its velocity (v).8. **Potential Energy**: \[ PE = mgh \] - **Explanation**: Potential energy (PE) is the energy held by an object due to its position relative to other objects, calculated using mass (m), gravity (g), and height (h

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If 60 percent of the potential energy of water flowing over a dam can be converted to electricity, find the maximum power output, in kW (kilowatts), for a dam 50 m high where the water crosses at a rate of 300,000 kg/min (1 watt = 1 joule/sec, 1 kW = 1000 watts)

If 60 percent of the potential energy of water flowing over a dam can be converted to electricity, find the maximum power output, in kW (kilowatts), for a dam 50 m high where the water crosses at a rate of 300,000 kg/min (1 watt = 1 joule/sec, 1 kW = 1000 watts)

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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