EBUS CAL12 Chapter 6 Quiz 2-3 (Lesson 4,5)

Step 1: The spring is stretched from \(12\) cm to \(22\) cm, so the stretch distance is \(22-12=10\) cm \(=0.1\) m. Step 2: Use Hooke's Law \(F=kx\). Step 3: Substitute \(F=4\) N and \(x=0.1\) m: \(k=\frac{F}{x}=\frac{4}{0.1}=40\). Step 4: Use the spring work formula: \(W=\frac{1}{2}kx^2\). Step 5: Substitute the values: \(W=\frac{1}{2}(40)(0.1)^2=0.2\) J. Answer: C. 0.2 J

Step 1: Use the average value formula: \(f_{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx\). Step 2: Substitute \([2,5]\): \(f_{avg}=\frac{1}{5-2}\int_2^5 \ln(3x)\,dx\). Step 3: Use integration by parts: \(\int \ln(3x)\,dx=x\ln(3x)-x+C\). Step 4: Evaluate from \(2\) to \(5\): \([x\ln(3x)-x]_2^5=(5\ln 15-5)-(2\ln 6-2)\). Step 5: Divide by \(3\): \(f_{avg}=\frac{1}{3}(5\ln 15-2\ln 6-3)\approx 2.32\). Step 6: The statement is true. Answer: B. True

Step 1: Displacement is the integral of velocity. Step 2: Set up the integral: \(\int_0^2(2t^3+5)\,dt\). Step 3: Split the integral: \(\int_0^2 2t^3\,dt+\int_0^2 5\,dt\). Step 4: Evaluate: \(\int_0^2 2t^3\,dt=\left[\frac{t^4}{2}\right]_0^2=8\). Step 5: Evaluate the constant part: \(\int_0^2 5\,dt=10\). Step 6: Add the results: \(8+10=18\). Step 7: The statement is true. Answer: B. True

Step 1: Distance is the integral of speed, so use \(\int_0^{2\pi}|v(t)|\,dt\). Step 2: Since \(v(t)=\sin t\), the distance is \(\int_0^{2\pi}|\sin t|\,dt\). Step 3: Split the interval where \(\sin t\) changes sign: \(\int_0^{\pi}\sin t\,dt+\int_{\pi}^{2\pi}(-\sin t)\,dt\). Step 4: Evaluate: \(\int_0^{\pi}\sin t\,dt=2\), and \(\int_{\pi}^{2\pi}(-\sin t)\,dt=2\). Step 5: Total distance is \(2+2=4\) m. Answer: C. 4 m