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FrozenT [24]
3 years ago
8

Marta is standing 4 ft behind a fence 6 ft 6 in. tall. When she looks over the fence, she can just see the top edge of a buildin

g. She knows that the building is 32 ft 6 in. behind the fence. Her eyes are 5 ft from the ground. How tall is the building? Give your answer to the nearest half foot.

Physics
1 answer:
klasskru [66]3 years ago
6 0

Answer: 93ft

Explanation: Martha's eyesight make an angle with the fence, this is solved as a triangle problem.

We first convert all the dimensions to inches using the fact that 12 inches make 1 ft.

To solve for the angle

Tan § = 18/48 = 0.375

§ = tan^-1 of 0.375 = 20.55°

This same angle elevates to the edge of the building

Tan 20.55 = 399/x

x = 399/tan 20.55

x = 1040 inches

Height of building = x + height of fence

Height of building = 1040 + 78 = 1118 inches

Converting back to ft it becomes

1118/12 = 93 ft 2 inches

Approximately 93ft

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3 years ago
A flywheel with a diameter of 1.42 m is rotating at an angular speed of 207 rev/min. (a) What is the angular speed of the flywhe
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Answer:

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Explanation:

a. Its angular speed in radians per second  ω = angular speed in rev/min × 2π/60 = 207 rev/min × 2π/60 = 21.68 rad/s

b. The linear speed of a point on the flywheel is gotten from v = rω where r = radius of flywheel = 1.42 m

So, v = rω = 1.42 m × 21.68 rad/s = 30.78 m/s

c. Using α = (ω₁ - ω)/t where α = angular acceleration of flywheel, ω = initial angular speed of wheel in rev/min = 21.68 rad/s = 207 rev/min, ω₁ = final angular speed of wheel in rev/min = 1410 rev/min = 147.65 rad/s, t = time in minutes = 80.5/60 min = 1.342 min

α = (ω₁ - ω)/t

  = (1410 - 207)/(80.5/60)

  = 60(1410 - 207)/80.5

  = 60(1203)80.5

  = 896.65 rev/min² ≅ 897 rev/min²

d. Using θ = ωt + 1/2αt²

where θ = number of revolutions of flywheel. Substituting the values of the variables from above, ω = 207 rev/min, α = 896.65 rev/min² and  t = 80.5/60 min = 1.342 min

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