Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( ) or Calories (Cal), where . On the other hand, power is the rate at which energy is used and is measured in watts ( ). Other useful units of power are kilowatts ( kW ) and megawatts ( ). If energy is used at a rate of 1 kW for 1 hr , the total amount of energy used is 1 kilowatt-hour ( kWh ), which is . Suppose the power function of a large city over a 24 -hr period is given by , where is measured in and corresponds to . (see accompanying figure). Complete parts (a) through ( d ) below.

Click the icon to view the graph of .
(Simplify your answer.)
c. Fission of 1 gram of uranium-235 (U-235) produces about of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 yr ?

The amount of uranium needed in a day is g.
(Simplify your answer. Round to the nearest integer as needed.)

Ask by Olson Mathis. in the United States

Jan 23,2025

Question

Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( ) or Calories (Cal), where . On the other hand, power is the rate at which energy is used and is measured in watts ( ). Other useful units of power are kilowatts ( kW ) and megawatts ( ). If energy is used at a rate of 1 kW for 1 hr , the total amount of energy used is 1 kilowatt-hour ( kWh ), which is . Suppose the power function of a large city over a 24 -hr period is given by , where is measured in and corresponds to . (see accompanying figure). Complete parts (a) through ( d ) below.

Click the icon to view the graph of .
(Simplify your answer.)
c. Fission of 1 gram of uranium-235 (U-235) produces about of energy. How many grams of uranium are needed to meet the energy needs of the city for 1 day? For 1 yr ?

The amount of uranium needed in a day is g.
(Simplify your answer. Round to the nearest integer as needed.)

Ask by Olson Mathis. in the United States

Jan 23,2025

UpStudy AI · Accepted Answer

To determine the amount of uranium-235 (U-235) needed to meet the energy demands of the city for one day, follow these steps:

1. **Calculate the Total Energy Consumed in One Day:**

The power function is given by:
   $$
   P(t) = 345 - 230 \sin\left(\frac{\pi t}{12}\right) \quad \text{(in MW)}
   $$
   
   To find the total energy consumed over a 24-hour period, integrate the power function with respect to time:
   $$
   E = \int_{0}^{24} P(t) \, dt = \int_{0}^{24} \left(345 - 230 \sin\left(\frac{\pi t}{12}\right)\right) dt
   $$
   
   Evaluate the integral:
   $$
   E = 345 \times 24 - 230 \times \left[\frac{12}{\pi} \left(-\cos\left(\frac{\pi t}{12}\right)\right)\right]_{0}^{24}
   $$
   
   Since $$\cos(2\pi) = \cos(0) = 1$$, the sine integral term cancels out:
   $$
   E = 345 \times 24 = 8280 \, \text{MWh}
   $$
   
   Convert MWh to kWh:
   $$
   8280 \, \text{MWh} = 8,280,000 \, \text{kWh}
   $$

2. **Determine the Amount of Uranium-235 Needed:**

Given that fission of 1 gram of U-235 produces 16,000 kWh:
   $$
   \text{Grams of U-235 needed} = \frac{8,280,000 \, \text{kWh}}{16,000 \, \text{kWh/g}} = 517.5 \, \text{g}
   $$
   
   Rounding to the nearest integer:
   $$
   \text{Grams of U-235 needed} = 518 \, \text{g}
   $$

**Answer:**

The amount of uranium needed in a day is 518 g.
The city needs 518 grams of uranium-235 each day to meet its energy requirements.

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