General Physics Reviewer.pdf

General Physics Reviewer STEM 12 Pointers to Review: 1. Conversion of Temperature 2. Conversion of Units of Measurements 3. Average Deviation 4. Vectors (Magnitude and Angle of Direction) 1. Conversio...

General Physics Reviewer STEM 12 Pointers to Review: 1. Conversion of Temperature 2. Conversion of Units of Measurements 3. Average Deviation 4. Vectors (Magnitude and Angle of Direction) 1. Conversion of Temperature Sample Problem: Dr. Emily is a climate scientist studying the effects of temperature on plant growth. Today, she is working in her laboratory, where the temperature is measured at 25°C. As part of her research, she needs to report the temperature in different units for her international collaborators. a. Convert the temperature from Celsius (°C) to Fahrenheit (°F). b. Convert the temperature from Celsius to Kelvin (K). Step-by-Step Solution: Step 1. Read and understand the problem. Then, state what is asked and what are the given. Asked: a. °C → °F b. °C → K Given: °C = 25°C Step 2. Write the formula for the following required. Remember that for Fahrenheit and Kelvin, we have the following formula: °F = °C x 1.8 + 32 𝐾 = °C + 273.15 Step 3. Input the given value into our formula. °F = °C x 1.8 + 32 °F = 25 x 1.8 + 32 °F = 77 °F 𝐾 = °C + 273.15 𝐾 = 25 + 273.15 𝐾 = 298.15 𝐾 Therefore, 25 °C is equal to 77 °F and 298.15 K. 2. Conversion of Units of Measurements Sample Problem: Maria is planning a hiking trip across the scenic trails of a national park. She has mapped out a route that is 15 kilometers long. To ensure she can accurately share her plans with her friends in the United States, she needs to convert the distance from kilometers to miles. Convert the distance from kilometers (km) to miles (mi). Note that 1km = 0.6214 mi Step-by-Step Solution: Step 1. Read and understand the problem. Then, state what is asked and what are the given. Asked: km → mi Given: 15 km Step 2: Determine the equivalence factor of km to mi. 1 km = 0.6214 mi STEP 3: Convert the given unit (km) to miles (mi) through dimensional analysis. 0.6214 𝑚𝑖 15𝑘𝑚 ( ) 1𝑘𝑚 In this equation, the values inside the parentheses are the equivalence factors. To understand why we place 1 km in the denominator and 0.621371 mi in the numerator, let’s examine the given value of 15 km. Since 15 km is in the numerator, we want to cancel the unit “km” in our calculation. To do this, we place the equivalence factor such that the same unit, “km,” appears in the denominator. This allows us to eliminate the kilometers and convert the value to miles. Therefore, we have, 0.6214 𝑚𝑖 0.6214 𝑚𝑖 15𝑘𝑚 ( ) → cancel km → 15𝑘𝑚 ( ) = 9.321→ since “mi” is the remaining unit, therefore, 1𝑘𝑚 1𝑘𝑚 15 𝑘𝑚 = 9.321 𝑚𝑖 3. Average Deviation Average deviation measures the dispersion of your measurements from the mean value. A lower average deviation indicates that the measurements are closely clustered around the mean, suggesting higher precision. Conversely, a higher average deviation signifies that the measurements are more spread out, indicating lower precision. In short, the LOWER average deviation, the MORE PRECISE the measurements are. And, the higher the average deviation is the less precise the measurements are. Example: Look at this table. Observe that the average deviation of vernier caliper (0.18) is lower compared to the average deviation of ruler (0.67). This means that the measurements using the vernier caliper is more precise compared to that of a ruler. 4. Vectors (Magnitude and Angle of Direction) First, we define and differentiate what are vectors and what are scalar quantities. Scalar quantities are quantities that has only magnitude but no direction. For example, when we read the time on a clock, we only say 10:00 AM or PM, and not 10:00 AM towards north. Therefore, time is a scalar quantity. Other examples are temperature, speed, velocity, distance, mass, and etc. Remember that scalars are graphed using a straight line. On the other hand, vector quantities are quantities that have both magnitude and direction. For example, when talking about displacement of a person, you may say that the person walked 5m north. Magnitude is 5m and direction is north. Other examples of vector quantities are velocity, force, weight, and etc. Remember that vectors are graphed through arrows. Note that the sum of two vectors is called the resultant vector (R). In solving problems related to vectors, we must first review the parts of a right triangle and the trigonometric functions: SOH, CAH, TOA. Now let’s try to solve this sample problem. Sample Problem: Sheila is captain of the Varsity cross country team. During the after-school practice on Tuesday, she led the team on the following run from school to a nearby park where they met the coach for a meeting. They ran 0.68 miles, north then 1.09 miles east. Determine the magnitude and angle of direction of the team's resultant displacement. Step-by-Step Solution: Step 1. Read and understand the problem. Then, state what is asked and what are the given. Asked: Magnitude and Angle of Direction (𝜃) of the team’s resultant displacement. Note that in this problem, the magnitude is referring to the resultant vector. Therefore, the problem is looking for: R=? 𝜃=? Given: V1 = 0.68 miles, north V2 = 1.09 miles east Step 2. Draw the graphical analysis of the two vectors. Remember, vectors are graphed using arrows. N V2 = 1.09 mi (opposite side) Observe. A right triangle is formed. Remember that the longest side of a right triangle is called the hypotenuse. V1 = 0.68 mi (adjacent side) 𝜃 =? This hypotenuse will now refer to the resultant vector (R). Look at the given angle. The side that is opposite to the W E given angle is the opposite side (O) and the side next to the angle is called the adjacent side (A). S Step 3. Identify the formula to use to calculate for the resultant vector and angle of direction. First, for the formula of resultant vector, use: 𝑅 2 = 𝑎2 + 𝑏 2 = (0.68𝑚𝑖)2 + (1.09𝑚𝑖)2 𝑅2 𝑅 = 0.4624𝑚𝑖 2 + 1.1881𝑚𝑖 2 2 𝑅 2 = 1.6505𝑚𝑖 2 Now, to remove the exponent (2), we must look for the square root of both sides. √𝑅 2 = √1.6505𝑚𝑖 2 𝑅 = 1.2847 𝑚𝑖 Therefore, the magnitude of the resultant vector is 1.2847 mi. Second, in solving for the angle of displacement, we can use either of the three trigonometric functions SOH, CAH, TOA. For this, let’s try to use the sin function (SOH). Therefore, we use: 𝑂 𝑠𝑖𝑛𝜃 = ( ) 𝐻 Input the given into our equation 1.09 𝑚𝑖 𝑠𝑖𝑛𝜃 = ( ) 1.2847 𝑚𝑖 Solve using your scientific calculator and cancel the like terms. 1.09 𝑚𝑖 𝑠𝑖𝑛𝜃 = ( ) 1.2847 𝑚𝑖 𝑠𝑖𝑛𝜃 = 0.8484 Remember that we are only looking for (𝜃). To remove “sin”, we use the inverse sin function in our calculators. (To do this, press “Shift” in your calculator then “sin”) We will have: 𝜃 = 𝑠𝑖𝑛−1 (0.8484) 𝜃 = 58.0380° Therefore, the angle of direction of the resultant vector is 58.0380°. Step 4. Complete the graph.