A battery powers a circuit for a small noisy fan. The fan's motor gets warm as it turns. What energy transformations are

Physics

2 answers:

Alex73 [517]3 years ago

6 0

Answer:

Electric potential energy in the battery transforms to electric energy. Electric energy transforms to motion energy (spinning fan blades), sound energy (noise), and thermal energy (a warm motor).

Explanation:

Edmentum sample answer. :))

Pepsi [2]3 years ago

3 0

Answer:

Chemical energy to electrical energy to mechanical and thermal energy

Explanation:

In the rotating fan, the chemical energy stored in the battery is used to generate electrical energy.

The electrolytes within the battery undergoes chemical reactions that produces electric current.

The current is being used to drive the motor of the fan to produce a mechanical energy as it rotates.
In like manner, heat energy is also produced.
So, energy is simply being transformed from one form to the other in this system.

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A rescue pilot drops a survival kit while her plane is flying at an ultitude of 2500m with a forward velocity of 95m. If the air

never [62]

Answer:

Approximately $2.1\; \rm km$ , assuming that $g = -9.8\; \rm m \cdot s^{-2}$ .

Explanation:

Let $t$ denote the time required for the package to reach the ground. Let $h(\text{initial})$ and $h(\text{final})$ denote the initial and final height of this package.

$\displaystyle h(\text{final}) = \frac{1}{2}\, g\, t^2 + h(\text{initial})$ .

For this package:

Initial height: $h(\text{initial}) = 2500\; \rm m$ .
Final height: $h(\text{final}) = 0\; \rm m$ (the package would be on the ground.)

Solve for $t$ , the time required for the package to reach the ground after being released.

$\displaystyle t^{2} = \frac{2\, (h(\text{final}) - h(\text{initial}))}{g}$ .

$\begin{aligned} t &= \sqrt{\frac{2\, (h(\text{final}) - h(\text{initial}))}{g} \\ &\approx \sqrt{\frac{2\times (0\; \rm m - 2500\; \rm m)}{(-9.8\; \rm m \cdot s^{-2})}} \approx 22.588\; \rm s\end{aligned}$ .

Assume that the air resistance on this package is negligible. The horizontal ("forward") velocity of this package would be constant (supposedly at $95\; \rm m \cdot s^{-1}$ .) From calculations above, the package would travel forward at that speed for about $22.588\; \rm s$ . That corresponds to approximately: $95\; \rm m \cdot s^{-1} \times 22.588\; \rm s \approx 2.1 \times 10^{3}\; \rm m = 2.1\; \rm km$ .