In a wheel and axle arrangement with an axle diameter of 25 cm supporting a load of 2 tonnes, if the mechanical advantage is 30, what is the diameter of the wheel?

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Multiple Choice

In a wheel and axle arrangement with an axle diameter of 25 cm supporting a load of 2 tonnes, if the mechanical advantage is 30, what is the diameter of the wheel?

Explanation:
In a wheel and axle system, the mechanical advantage (MA) is determined by the ratio of the radius of the wheel to the radius of the axle. The formula for mechanical advantage in this context can be expressed as: \[ MA = \frac{r_{wheel}}{r_{axle}} \] where \( r_{wheel} \) is the radius of the wheel and \( r_{axle} \) is the radius of the axle. The axle diameter is given as 25 cm, which translates to a radius of: \[ r_{axle} = \frac{25 \text{ cm}}{2} = 12.5 \text{ cm} \] To find the radius of the wheel, rearranging the mechanical advantage formula gives us: \[ r_{wheel} = MA \times r_{axle} \] Since the mechanical advantage is provided as 30, we can substitute into the equation: \[ r_{wheel} = 30 \times 12.5 \text{ cm} = 375 \text{ cm} \] To convert the wheel's radius back to diameter: \[ d_{wheel} = 2 \times r_{wheel} =

In a wheel and axle system, the mechanical advantage (MA) is determined by the ratio of the radius of the wheel to the radius of the axle. The formula for mechanical advantage in this context can be expressed as:

[

MA = \frac{r_{wheel}}{r_{axle}}

]

where ( r_{wheel} ) is the radius of the wheel and ( r_{axle} ) is the radius of the axle.

The axle diameter is given as 25 cm, which translates to a radius of:

[

r_{axle} = \frac{25 \text{ cm}}{2} = 12.5 \text{ cm}

]

To find the radius of the wheel, rearranging the mechanical advantage formula gives us:

[

r_{wheel} = MA \times r_{axle}

]

Since the mechanical advantage is provided as 30, we can substitute into the equation:

[

r_{wheel} = 30 \times 12.5 \text{ cm} = 375 \text{ cm}

]

To convert the wheel's radius back to diameter:

[

d_{wheel} = 2 \times r_{wheel} =

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