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Question 1 A spring that has a natural/resting length of 5 cm is stretched to a length of 25 cm by a 5 N force. How much work was required to do that?
Step 1: The spring is stretched from \(5\) cm to \(25\) cm, so the stretch distance is \(25-5=20\) cm \(=0.2\) m. Step 2: Use Hooke's Law \(F=kx\). Step 3: Substitute \(F=5\) N and \(x=0.2\) m: \(k=\frac{F}{x}=\frac{5}{0.2}=25\). Step 4: Use the work formula for a spring: \(W=\frac{1}{2}kx^2\). Step 5: Substitute the values: \(W=\frac{1}{2}(25)(0.2)^2=0.5\) J. Answer: A. 0.5 J
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Question 2 The average value of \(f(x)=\ln(3x)\) on \([2,5]\) is approximately \(2.32\).
Step 1: Use the average value formula: \(f_{avg}=\frac{1}{b-a}\int_a^b f(x)\,dx\). Step 2: Substitute the interval \([2,5]\): \(f_{avg}=\frac{1}{5-2}\int_2^5 \ln(3x)\,dx\). Step 3: Evaluate the integral using 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: A. True
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Question 3 The distance a particle travels in its first 4 seconds of travel, if it moves according to the velocity equation \(v(t)=-t^2+4\) (in feet/sec), is:
Step 1: Distance is the integral of speed, so use \(\int_0^4 |v(t)|\,dt\). Step 2: The velocity is \(v(t)=-t^2+4=-(t-2)(t+2)\), so it changes sign at \(t=2\). Step 3: On \([0,2]\), \(v(t)\ge 0\). On \([2,4]\), \(v(t)\le 0\). Step 4: Set up the distance: \(\int_0^2(-t^2+4)\,dt-\int_2^4(-t^2+4)\,dt\). Step 5: Evaluate the first part: \(\int_0^2(-t^2+4)\,dt=\left[-\frac{t^3}{3}+4t\right]_0^2=\frac{16}{3}\). Step 6: Evaluate the second part: \(\int_2^4(-t^2+4)\,dt=\left[-\frac{t^3}{3}+4t\right]_2^4=-\frac{32}{3}\). Step 7: Distance is \(\frac{16}{3}-\left(-\frac{32}{3}\right)=16\). Answer: C. \(16\) ft
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Question 4 If a particle is traveling according to the velocity equation \(v(t)=2t^3+5\) (in m/s), then its displacement over its first 2 seconds of travel is equal to \(18\) m.
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: \(\int_0^2 2t^3\,dt=\left[\frac{t^4}{2}\right]_0^2=8\), and \(\int_0^2 5\,dt=10\). Step 5: Add the values: \(8+10=18\). Step 6: The statement is true. Answer: B. True
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