All categories

3 years ago

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.

You might be interested in

A time-dependent but otherwise uniform magnetic field of magnitude B0(t) is confined in a cylindrical region of radius 6.5 cm. I

vodka [1.7K]

Answer:

a = 603.59 m/s^2

Explanation:

from the data given . the rate of change in magnetic field is as follow

$\frac{dB}{dt} = 280 G/s = 280 \times 10^{-4} T/s$

from the faraday's law of induction , the expression for the induced emf in region of radius r as follow

$\epsilon = \frac{d \phi}{dt}$

$\int E.dl = \frac{d(BA)}{dt}$

$E(2\pi r)= \pi r^2 \frac{dB}{dt}$

$E = \frac{r}{2} \frac{dB}{dt}$

electric field at point P_1 as follow

$E = \frac{r}{2} \frac{dB}{dt}$

$E = \frac{1.5\times 10^{-2}}{2} 280 \times 10^{-4}$

$E = 6.3\times 10^{-6} V/m$

from newton 2nd law of motion, the acceleration of proton is

F = ma

qE = ma

$a = \frac{qE}{m}$

$a = \frac{1.6 \times 10^{-19} (6.3\times 10^{-6})}{1.67\times 10^{-27}}$

a = 603.59 m/s^2

5 0

3 years ago

The head of a grass string trimmer has 100 g of cord wound in a light, cylindrical spool with inside diameter 3.00 cm and outsid

faust18 [17]

Answer:

a).11.546J

b).2.957kW

Explanation:

Using Inertia and tangential velocity

a).

$w=2250*2\pi *\frac{1}{60}\\ w=235.61$

$I=\frac{1}{2}*m*((\frac{d_{i} }{2})^{2} +(\frac{d_{e} }{2})^{2})\\m=100g *\frac{ikg}{1000g}=0.1kg\\ d_{i}=3cm*\frac{1m}{100cm}=0.03m \\ d_{e}=18cm*\frac{1m}{100cm}=0.18m\\I=\frac{1}{2}*0.1kg*((\frac{0.03m}{2})^{2} +(\frac{0.18m}{2})^{2})\\I=0.41625x10^{-3}kg*m^{2}$