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Question 1 A spring that has a natural/resting length of 12 cm is stretched to a length of 22 cm by a 4 N force. How much work was required to do that?
Step 1: The spring is stretched \(22-12=10\) cm \(=0.1\) m. Step 2: Apply Hooke's Law \(F=kx\). Step 3: Compute \(k=\frac{4}{0.1}=40\,\text{N/m}\). Step 4: Use the work formula \(W=\frac12kx^2\). Step 5: \(W=\frac12(40)(0.1)^2=0.2\,\text{J}\). Answer: D. 0.2 J
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Question 2 The average value of \(f(x)=\frac{x+5}{x}\) on \([-4,-1]\) is approximately \(-1.31\).
Step 1: Use \(f_{avg}=\frac1{b-a}\int_a^bf(x)dx\). Step 2: Evaluate \(\frac13\int_{-4}^{-1}\frac{x+5}{x}dx\). Step 3: Integrate to obtain \(\frac13(3-5\ln4)\approx-1.31\). Step 4: The statement is true. Answer: B. True
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Question 3 If a spring is oscillating according to the velocity equation \(v(t)=\cos t\) (ft/s), then its displacement over its first \(2\pi\) seconds is:
Step 1: Displacement equals \(\int_0^{2\pi}v(t)dt\). Step 2: Compute \(\int_0^{2\pi}\cos t\,dt=[\sin t]_0^{2\pi}=0\). Step 3: Therefore the displacement is 0 ft. Answer: C. 0 ft
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Question 4 The distance a spring travels in its first \(2\pi\) seconds if \(v(t)=\sin t\) (m/s) is:
Step 1: Distance is \(\int_0^{2\pi}|\sin t|dt\). Step 2: Split at \(\pi\): \(\int_0^{\pi}\sin tdt+\int_{\pi}^{2\pi}-\sin tdt\). Step 3: Each integral equals 2. Step 4: Total distance is 4 m. Answer: B. 4 m
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