PE Integrated Concepts (a) Assuming 95.0% efficiency for

Chapter 15, Problem 64PE

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QUESTION:

PE Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00 × 10? -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00 × 2? 10? N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00×10? N force to overcome air resistance and friction? See Figure 20.44.

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QUESTION:

PE Integrated Concepts (a) Assuming 95.0% efficiency for the conversion of electrical power by the motor, what current must the 12.0-V batteries of a 750-kg electric car be able to supply: (a) To accelerate from rest to 25.0 m/s in 1.00 min? (b) To climb a 2.00 × 10? -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00 × 2? 10? N of force to overcome air resistance and friction? (c) To travel at a constant 25.0-m/s speed, exerting a 5.00×10? N force to overcome air resistance and friction? See Figure 20.44.

ANSWER:

Step-by-step solution Given: Efficiency=95.0% Voltage = 12.0-V Mass = 750-kg Velocity = 25.0 m/s TIme= 1.00 min, 2.00 min Height = 2.00 × 10 -m Force = 5.00×10 N 2 1) Calculate the current supplied 12.0-V batteries of a 750-kg electric car be able to supply= 2) Calculate the current supplied To climb a 2.00 × 10 -m - high hill in 2.00 min at a constant 25.0-m/s speed while exerting 5.00×10 N of force to overcome air resistance and friction = 2 3) Calculate the current supplied to travel at a constant 25.0-m/s speed, exerting a 5.00×10 N force to overcome air resistance and friction= Step 1 of The power is defined as the rate flow of energy, P= E/t………………………(1) Here, E is the energy flowing and t is time duration for which energy flows. Also in an electric circuit, the rate at which energy is produced through a current carrying element is, P= I V………………………..(2) Here, I is the current and V is the voltage supplied.

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