Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( J ) or Calories (Cal), where $$ 1 \mathrm{Cal}=4184 \mathrm{~J} $$. On the other hand, power is the rate at which energy is used and is measured in watts (W; $$ 1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) $$. Other useful units of power are kilowatts ( kW ) and megawatts $$ \left(1 \mathrm{MW}=10^{6} \mathrm{~W} ight) $$. 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 $$ 3.6 imes 10^{6} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ P(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi / / 12) $$, where P is measured in MW and $$ \mathrm{t}=0 $$ corresponds to $$ 6: 00 $$ p.m. (see accompanying figure). Complete parts (a) through (d) below. Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$. (Simpiry your answer.) c. Fission of 1 gram of uranium- $$ 235(\mathrm{U}-235) $$ produces about $$ 16,000 \mathrm{kWh} $$ 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 518 g . (Simplify your answer. Round to the nearest integer as needed.) The amount of uranium needed in a year is $$ \square \mathrm{g} $$. (Simplify your answer.)

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Power and energy are often used interchangeably, but they are quite different. Energy is measured in units of joules ( J ) or Calories (Cal), where $$ 1 \mathrm{Cal}=4184 \mathrm{~J} $$. On the other hand, power is the rate at which energy is used and is measured in watts (W; $$ 1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}) $$. Other useful units of power are kilowatts ( kW ) and megawatts $$ \left(1 \mathrm{MW}=10^{6} \mathrm{~W}ight) $$. 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 $$ 3.6 	imes 10^{6} \mathrm{~J} $$. Suppose the power function of a large city over a 24 -hr period is given by $$ P(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi / / 12) $$, where P is measured in MW and $$ \mathrm{t}=0 $$ corresponds to $$ 6: 00 $$ p.m. (see accompanying figure). Complete parts (a) through (d) below. Click the icon to view the graph of $$ \mathrm{P}(\mathrm{t})=\mathrm{E}^{\prime}(\mathrm{t})=345-230 \sin (\pi \mathrm{t} / 12) $$. (Simpiry your answer.) c. Fission of 1 gram of uranium- $$ 235(\mathrm{U}-235) $$ produces about $$ 16,000 \mathrm{kWh} $$ 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 518 g . (Simplify your answer. Round to the nearest integer as needed.) The amount of uranium needed in a year is $$ \square \mathrm{g} $$. (Simplify your answer.)

UpStudy AI · Accepted Answer

To determine the amount of uranium-235 ($$\mathrm{U}-235$$) needed to meet the energy needs of the city for one year, we'll follow these steps:

1. **Given Information:**
   - **Energy Produced by 1 gram of $$\mathrm{U}-235$$:** $$16,000 \ \mathrm{kWh}$$
   - **Uranium Needed for One Day:** $$518 \ \mathrm{g}$$

2. **Calculate the Uranium Needed for One Year:**
   - **Number of Days in a Year:** $$365$$
   - **Total Uranium Needed:**
     $$
     \text{Uranium per Year} = 518 \ \mathrm{g/day} \times 365 \ \mathrm{days} = 189,070 \ \mathrm{g}
     $$

3. **Final Answer:**
   $$
   \text{The amount of uranium needed in a year is } 189070 \ \mathrm{g}.
   $$

**Answer:**  
The amount of uranium needed in a year is 189070 g.
The amount of uranium needed in a year is 189,070 grams.

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