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question:Two cars, C and D, are racing along a straight road. Car C has a head start of distance DC and travels at a constant speed vC. Car D starts from the same starting line as Car C but accelerates uniformly with acceleration aD. If Car D catches up with Car C at time t*, at what distance from the starting line will this occur?
answer:At time t*, the distance traveled by Car C is: {eq}s_C = v_C t^* {/eq} The distance traveled by Car D is: {eq}s_D = frac{1}{2} a_D (t^*)^2 {/eq} Since Car D catches up with Car C at time t*, we have: {eq}s_C + D_C = s_D {/eq} Substituting the expressions for s_C and s_D, we get: {eq}v_C t^* + D_C = frac{1}{2} a_D (t^*)^2 {/eq} Solving for t*, we get: {eq}t^* = frac{v_C pm sqrt{v_C^2 + 2a_D D_C}}{a_D} {/eq} Since time cannot be negative, we take the positive root. Substituting t* into the expression for s_C, we get the distance from the starting line where Car D catches up with Car C: {eq}s = v_C cdot frac{v_C + sqrt{v_C^2 + 2a_D D_C}}{a_D} + D_C {/eq}
question:A bond has a par value of 1,500, a coupon rate of 7.5% paid annually, and 10 years remaining to maturity. The bond is currently trading at 1,350. If the bond is callable in 3 years at a call price of 1,400, what is the bond's yield to call? Round your answer to two decimal places.
answer:To calculate the yield to call, we use the following formula: {eq}Bond,Price = PMT times dfrac{1-(1+r)^{-t}}{r} + dfrac{F}{(1+r)^{t}} {/eq} Where: * PMT = coupon payment = (7.5% * 1,500) = 112.50 * t = time until call date = 3 years * F = call price = 1,400 * Bond price = 1,350 Substituting these values into the formula, we get: {eq}1,350 = 112.50 times dfrac{1-(1+r)^{-3}}{r} + dfrac{1,400}{(1+r)^{3}} {/eq} Solving for r using trial and error or a financial calculator, we get r = 4.42%. Therefore, the bond's yield to call is 4.42%.
question:Calculate the equilibrium partial pressure (atm) of H<sub>2</sub>S for the equilibrium: NH<sub>4</sub>HS (s) ⇌ NH<sub>3</sub> (g) + H<sub>2</sub>S (g) given that K<sub>p</sub> = 0.075 at 23 °C and the equilibrium partial pressure of ammonia is 0.274 atm.
answer:Using the K<sub>p</sub> expression: K<sub>p</sub> = (NH<sub>3</sub>)(H<sub>2</sub>S) = 0.075 We can solve for the partial pressure of H<sub>2</sub>S: (H<sub>2</sub>S) = K<sub>p</sub> / (NH<sub>3</sub>) = 0.075 / 0.274 = 0.274 atm Therefore, the equilibrium partial pressure of H<sub>2</sub>S is 0.274 atm.
question:DBD Ltd's current dividend per share is 1.80, and its share price is 10.00. The dividend is expected to grow at a constant annual rate in the foreseeable future. Market analysts estimate the stock's required return to be 15% annually. What is the implied constant annual growth rate of the dividend? a) 3.0% b) 2.5% c) 2.0% d) 1.5%
answer:The correct answer is Option b) 2.50%. Explanation: We can use the dividend discount model (Gordon Growth Model) to find the growth rate (g): Current stock price = Current dividend * (1 + g) / (Required return - g) 10.00 = 1.80 * (1 + g) / (0.15 - g) Solving for g, we get: 1.15g = 0.18 g = 0.18 / 1.15 g ≈ 0.1574 or 15.74% Since the growth rate cannot be greater than the required return, we have an error. The correct interpretation of the equation is that the growth rate is negative, as the dividend is expected to decrease over time: 0.15 - g = 1.80 * (1 + g) / 10.00 1.15g = -0.03 g = -0.03 / 1.15 g ≈ -0.0261 or -2.61% Rounded to one decimal place, the implied annual growth rate is -2.5%.