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6.7 Work of 5 Joules is done in stretching a spring from its natural length to 18cm beyond its natural length. What is the force

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6.7

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Work of 5 Joules is done in stretching a spring from its natural length to 18cm beyond its natural length. What is the force (in Newtons, to one decimal place) that holds the spring stretched at the same distance ( 18cm)? Watch your units! N A trough is 9 feet long and 1 foot high. The vertical cross-section of the trough parallel to an end is shaped like the graph of y = I" from x = -1 to x = 1 . The trough is full of water. Find the amount of work in foot-pounds required to empty the trough by pumping the water over the top. Note: In this problem, use 62 pounds per cubic foot as the weight of water. foot-pounds Question 9 0/1 pt 9 20 @ Details A circular swimming pool has the shape of a cylinder, with a diameter of 14 m, the sides are 4 m high, and the depth of the water is 3 m. How much work (in Joules, to the nearest integer) is required to pump all of the water over the side? The acceleration due to gravity is 9.8 -, and the density of water is 1000 kg m 3 . Joules A rectangular tank 3 feet long, 5 feet wide, and 5 feet deep is full of oil with weight density 40 - 1b Calculate the work required to pump all of the oil out over the top of the tank. ft 3 . ft-lb 80 m 10 m 30 m A river is blocked by a dam with the shape of an trapezoid. The dam is 10 m tall, 30 m at its base and 80 m on top. What is the total force acting on the dam if the water reaches the full height of the dam? Enter your answer in scientific notation to at least 2 significant figures. Use p = 1000 9 m3 and g = 9.8 $2 . If y = (5z) sin(52), then F = N dy dx y = (22+2) " Use logarithmic differentiation to find dy for y = (1 + 2 ) : . dy Use logarithmic differentiation to find dy dx dy dx y'= tan(5x), y(67) = 9 y = Question Help: Written Example Message instructor Submit Question X Question 11 Score on last try: 0 of 1 pts. See Details for more. You can retry this question below Solve the initial value problem. y" = 2 + sin(2x), y(0) - -3, y'(0) = -4, y"(0) = 3 y =

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