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astraxan [27]
2 years ago
5

A bicycle has wheels of 0.8 m diameter. The bicyclist accelerates from rest with constant acceleration to 22 km/h in 11.6 s. Wha

t is the angular acceleration of the wheels?
Physics
1 answer:
larisa [96]2 years ago
7 0

Answer:

α = 1.32 rad/s²

Explanation:

given,

diameter of the bicycle = 0.8 m

radius of the bicycle = 0.4 m

initial speed of the bicyclist,u = 0 m/s

final speed of the bicyclist,v = 22 Km/h = 22 x 0.278

                                           = 6.12 m/s

time,t = 11.6 s

acceleration =\dfrac{v-u}{t}

                     =\dfrac{6.12-0}{11.6}

                    =0.53 m/s²

we know,

    a = α r

\alpha=\dfrac{a}{r}

\alpha=\dfrac{0.53}{0.4}

  α = 1.32 rad/s²

the angular acceleration of the wheels is equal to α = 1.32 rad/s²

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T^2 = \frac{4\pi^2 d^3}{2}

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7 0
3 years ago
A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate
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Answer:

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The Area of the triangle is given by A=h\times b where h=\sqrt{l^2-b^2} (by using the Pythagoras' Theorem) and b is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

A=\sqrt{l^2-b^2}b

The rate of change of the area is given by its time derivative

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\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}

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3 0
2 years ago
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