**Solar Energy Intensity and Calculation of Energy on a Surface**The **intensity of electromagnetic waves from the sun** is approximately **1.4 kW/m²** just above the Earth's atmosphere. On a clear summer day at noon, about **eighty percent** of this intensity reaches the Earth's surface.To visualize a real-world application of this information, consider the following scenario:**Suppose you think of your back as a rectangle measuring 32.0 cm by 49.0 cm.**### Example Calculation**Problem:**How many joules of solar energy fall on your back as you work on your tan for **1.20 hours**?**Instructions:**- Express your answer with the appropriate units.**Calculation Steps:**1. **Convert the dimensions of the back from centimeters to meters:** - 32.0 cm = 0.32 m - 49.0 cm = 0.49 m2. **Calculate the area of the back in square meters:** \( \text{Area} = 0.32 \, \text{m} \times 0.49 \, \text{m} = 0.1568 \, \text{m}² \)3. **Calculate the effective intensity reaching the surface:** - \( \text{Effective Intensity} = 0.8 \times 1.4 \, \text{kW/m}² = 1.12 \, \text{kW/m}² \)4. **Convert the intensity from kW/m² to W/m²:** - 1.12 kW/m² = 1120 W/m²5. **Calculate the energy received over the time period:** - **Energy (Joules) \( = \text{Intensity} \times \text{Area} \times \text{Time} \)** - Time in hours: 1.20 hr = 1.20 × 3600 s = 4320 s - \( \text{Energy} = 1120 \, \text{W/m}² \times 0.1568 \, \text{m}² \times 4320 \, \text{s} \)6. **Determine the final energy value:** - Energy \( = 757,401.6 \, \text{

The intensity of electromagnetic waves from the sun is 1.4 kW/m² just above the earth's atmosphere. Eighty percent of this reaches the surface at noon on a clear summer day. Suppose you think of your back as a 32.0 cm x 49.0 cm rectangle. Part A How many joules of solar energy fall on your back as you work on your tan for 1.20 hr? Express your answer with the appropriate units. μᾶ Value Units ? Units input for part A

The intensity of electromagnetic waves from the sun is 1.4 kW/m² just above the earth's atmosphere. Eighty percent of this reaches the surface at noon on a clear summer day. Suppose you think of your back as a 32.0 cm x 49.0 cm rectangle. Part A How many joules of solar energy fall on your back as you work on your tan for 1.20 hr? Express your answer with the appropriate units. μᾶ Value Units ? Units input for part A

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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Concept explainers

Electromagnetic Waves

Electromagnetic waves are waves that consist of magnetic field and electric field components perpendicular to each other. These field components are also perpendicular to the direction of propagation of the wave.

Question

**Solar Energy Intensity and Calculation of Energy on a Surface**

The **intensity of electromagnetic waves from the sun** is approximately **1.4 kW/m²** just above the Earth's atmosphere. On a clear summer day at noon, about **eighty percent** of this intensity reaches the Earth's surface.

To visualize a real-world application of this information, consider the following scenario:

**Suppose you think of your back as a rectangle measuring 32.0 cm by 49.0 cm.**

### Example Calculation

**Problem:**
How many joules of solar energy fall on your back as you work on your tan for **1.20 hours**?

**Instructions:**
- Express your answer with the appropriate units.

**Calculation Steps:**
1. **Convert the dimensions of the back from centimeters to meters:**
- 32.0 cm = 0.32 m
- 49.0 cm = 0.49 m

2. **Calculate the area of the back in square meters:**
\( \text{Area} = 0.32 \, \text{m} \times 0.49 \, \text{m} = 0.1568 \, \text{m}² \)

3. **Calculate the effective intensity reaching the surface:**
- \( \text{Effective Intensity} = 0.8 \times 1.4 \, \text{kW/m}² = 1.12 \, \text{kW/m}² \)

4. **Convert the intensity from kW/m² to W/m²:**
- 1.12 kW/m² = 1120 W/m²

5. **Calculate the energy received over the time period:**
- **Energy (Joules) \( = \text{Intensity} \times \text{Area} \times \text{Time} \)**
- Time in hours: 1.20 hr = 1.20 × 3600 s = 4320 s
- \( \text{Energy} = 1120 \, \text{W/m}² \times 0.1568 \, \text{m}² \times 4320 \, \text{s} \)

6. **Determine the final energy value:**
- Energy \( = 757,401.6 \, \text{

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