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evablogger [386]
3 years ago
10

At a museum you can see the four cable lines that are used to pull cable cars up and down a hill. Each cable travels at a speed

of 12.65 miles per​ hour, caused by a rotating wheel whose diameter is 6.5 feet. How fast is the wheel​ rotating? Express your answer in revolutions per minute.
Mathematics
1 answer:
borishaifa [10]3 years ago
6 0

Answer:

54.514 rev/min

Step-by-step explanation:

Data provided in the question:

Speed of the cable car = 12.65 miles per hour

Diameter of the rotating wheel, d = 6.5 feet

Now,

The speed of rotation of the wheel will be equal to the speed of the cable car

thus,

The speed of rotation of the wheel = 12.65 miles per hour

also,

1 miles = 5280 feets

1 hour = 60 minutes

Thus,

The speed of rotation of the wheel = \frac{12.65\times5280}{60}

= 1113.2 ft/ min

Circumference of the wheel = πd

= 6.5π feet

= 20.42 feet

Therefore,

The speed of rotation of the wheel in  revolutions per minute

=  1113.2 ÷ 20.42 rev/min

= 54.514 rev/min

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An object is launched from a launching pad 144 ft. above the ground at a velocity of 128ft/sec. what is the maximum height reach
ollegr [7]

Answer:

18) a. h(x) = -16x² + vx + h(0) ⇒ h(x) = -16x² + 128x + 144

b. The maximum height = 400 feet

c. Attached graph

19) The rocket will reach the maximum height after 4 seconds

20) The rocket hits the ground after 9 seconds

Step-by-step explanation:

* Lets study the rule of motion for an object with constant acceleration

# The distance S = ut ± 1/2 at², where u is the initial velocity, t is the time

  and a is the acceleration of gravity

# The vertical distances h in x second is h(x) - h(0), where h(0)

   is the initial height of the object above the ground

∵ h(x) = vx + 1/2 ax², where h is the vrtical distance, v is the initial

  velocity, a is the acceleration of gravity (32 feet/second²) and x

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18)a.

∵ The value of a = -32 ft/sec² ⇒ negative because the direction

   of the motion

  is upward

∴ h(x) - h(0) = vx - (1/2)(32)(x²) ⇒ (1/2)(32) = 16

∴ h(x) = vx - 16x² + h(0)

∴ h(x) = -16x² + vx + h(0) ⇒ proved

* Find the height of the object after x seconds from the ground

∵ h(0) = 144 and v = 128 ft/sec

∴ h(x) = -16x² + 128x + 144

b.

* At the maximum height h'(x) = 0

∵ h'(x) = -32x + 128

∴ -32x + 128 = 0 ⇒ subtract 128 from both sides

∴ -32x = -128 ⇒ ÷ -32

∴ x = 4 seconds

- The time for the maximum height = 4 seconds

- Substitute this value of x in the equation of h(x)

∴ The maximum height = -16(4)² + 128(4) + 144 = 400 feet

c. Attached graph

19)

- The object will reach the maximum height after 4 seconds

20)

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∵ h(x) = -16x² + 128x + 144

∴ 0 = -16x² + 128x + 144 ⇒ divide the two sides by -16

∴ x² - 8x - 9 = 0 ⇒ use the factorization to find the value of x

∵ x² - 8x - 9 = 0

∴ (x - 9)( x + 1) = 0

∴ x - 9 = 0 OR x + 1 = 0

∴ x = 9 OR x = -1

- We will rejected -1 because there is no -ve value for the time

* The time for the object to hit the ground is 9 seconds

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