A disc is thrown through the air for 1.5 min with a power output of 12.5 W. How much work is done when throwing the disc?

Physics

1 answer:

klasskru [66]3 years ago

7 0

Answer:

work = 1125 [J]

Explanation:

To solve this problem we must remember the definition of power, which is defined as the relationship between work and time. The power can be calculated using the following equation:

Power = work/time

Power = 12.5 [w]

work = joules [J]

time = 1.5 [min] = 90 [s]

work = 12.5*90

work = 1125 [J]

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A street light is at the top of a 19 foot tall pole. A woman 5.25 feet tall walks towards the pole with a speed of 7ft/sec along

bezimeni [28]

Answer:

The tip of her shadow is moving at the speed of 9.66 ft/sec

Explanation:

The height of the street light = 19 feet

The height of the woman = 5.25 feet

Distance between the woman and the base of the pole, x = 35 ft

The speed of the woman towards the pole, dx/dt = 7ft/sec

The distance from the base of the streetlight to the tip of the woman's shadow = y

The distance from the woman to the tip of her shadow = y - x

The diagram illustrating this description is shown below

Using similar triangle:

$\begin{gathered} \frac{19}{5.25}=\text{ }\frac{y}{y-x} \\ 19(y-x)\text{ = 5.25y} \\ 19y-19x\text{ = 5.25y} \\ 19y-5.25y\text{ = 19x} \\ 13.75y\text{ = 19x} \\ y\text{ = }\frac{19}{13.75}x \\ y\text{ = }1.38x \end{gathered}$

Find the derivative of both sides with respect to time, t

$\begin{gathered} \frac{dy}{dt}=\text{ 1.38}\frac{dx}{dt} \\ \frac{dy}{dt}=\text{ 1.38(7)} \\ \frac{dy}{dt}\text{ = }9.66\text{ ft/sec} \end{gathered}$

The tip of her shadow is moving at the speed of 9.66 ft/sec