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3 years ago

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a pulley system with a diameter of 0.5 meters lifts an object 736 meters in 1.05 minutes. how many radians did the pulley move ?

how many revolutions did the pulley make ? what was the angular velocity? what was the linear velocity ?

1 answer:

muminat3 years ago

4 0

its b welcome

Explanation:

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Differentiate the following functions with respect to x <br>xsin x

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Would you expect to observe every supernova in our own Galaxy? Why or why not?

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A paper pinwheel is spinning in the wind. Which statement is correct about the forces responsible for the rotation?

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3 years ago

A top fuel dragster with a mass of 500.0 kg starts from rest and completes a quarter mile (402 m) race in a time of 5.0 s. The d

blsea [12.9K]

The average power needed to produce this final speed is 1069.1 hp.

Mass of the dragster, m = 500.0 kg,

Displacement travelled by the dragster, s = 402 m,

Time taken in this travel, t = 5.0 s,

Final velocity of the dragster, v = 130 m/s.

Let the initial velocity of the dragster be u and acceleration be a.

Using kinematical equation, s = ut + (1/2)at^2.

402 = u*5 + (1/2)*a*5^2

10*u + 25*a = 804. ...........(1)

Using kinematical equation, v = u +at.

130 = u + 5*a

5*u + 25*a = 650. .............(2)

Solving (1)and (2), we get,

u = 30.8 m/s.

According to work-energy theorem,

Work done = change in kinetic energy

W = (1/2)*m*(v^2 - u^2)

W = (1/2)*500*(130^2 - 30.8^2)

W = 3987840. J

Therefore power rating of the dragster is given by,

P ⇒ W/t. = 3987840/5 = 797568 watt.

P ⇒ 797568/746 = 1069.1 hp.

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You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a

zepelin [54]

Answer:

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 170.0 J/s from a furnace at 350.0 ∘C to a container of boiling water under 1 atmosphere.

Explanation:

Given Values:

L = 50 cm = 0.5 m

H = 170 j/s

To find the diameter of the rod, we have to find the area of the rod using the following formula.

Here Tc = 100.0° C

k = 50.2

H = k × A × $\frac{[T_{H -}T_{C} ] }{L}$

Solving for A

A = $\frac{H * L }{k * [ T_{H}- T_{C} ] }$

A = $\frac{170 * 0.5}{50.2 * [ 350 - 100 ]}$

A = $\frac{85}{12550}$ = 6.77 × $10^{-3}$ m²

Now Area of cylinder is :

A = $\frac{\pi }{4}$ d²

solving for d:

d = $\sqrt{\frac{4 * 0.00677 }{\pi } }$

d = 9.28 cm

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