Ask question

Ask question

All categories

3 years ago

8

Consider a makeup mirror that produces a magnification of 1.5 when a person's face is 11.5 cm away what is the focal length of t

he makeup mirror in meters?

1 answer:

LiRa [457]3 years ago

4 0

<u>Answer:</u> The focal length of makeup mirror is 0.3448 m.

<u>Explanation:</u>

We are given:

m = 1.5

u = -11.5 cm

To calculate the image distance, we the equation for mirror:

$m=-\frac{v}{u}$

Putting values in above equation, we get:

$1.5=-(\frac{v}{-11.5})\\\\v=17.25cm$

To calculate the focal length of mirror, we use the mirror formula, which is:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}$

Putting values in above equation, we get:

$\frac{1}{f}=\frac{1}{17.25}+\frac{1}{-11.5}\\\\f=-34.48cm$

Negative sign represents the position of focal length.

Converting the focal length into meters, we use the conversion factor:

1 m = 100 cm

So, 34.48 cm = 0.3448 m

Hence, the focal length of makeup mirror is 0.3448 m.

You might be interested in

Lasers are now used in eye surgery. Given the wavelength of a certain laser is 514 nm and the power of the laser is 1.1 W, how m

Leno4ka [110]

Answer: $1.593*10^{17} photons$ released if the laser is used 0.056 s during the surgery

Explanation:

First, you have to calculate the energy of each photon according to Einstein's theoty, given by:

$E =\frac{hc}{\lambda}$

Where $\lambda$ is the wavelength, $h$ is the Planck's constant and $h$ is the speed of light

$h = 6.626*10^{-34} \frac{m^{2} kg }{s}$ -> Planck's constant

$c = 3*10^{8} \frac{m}{s}$ -> Speed of light

So, replacing in the equation:

$E =\frac{ 6.626*10^{-34} \frac{m^{2} kg }{s}*3*10^{8} \frac{m}{s}}{514*10^-9 m}$

Then, the energy of each released photon by the laser is:

$E = 3.867*10^{-19} \frac{J}{photons}$

After, you do the inverse of the energy per phothon and as a result, you will have the number of photons in a Joule of energy:

$\frac{1}{3.867*10^{-19}} = 2.586*10^{18} \frac{photons}{J}$

The power of the laser is 1.1 W, or 1.1 J/s, that means that you can calculate how many photons the laser realease every second:

$2.586*10^{18}\frac{photons}{J} * 1.1 \frac{J}{s} = 2.844*10^{18} \frac{photons}{s}$

And by doing a simple rule of three, if $2.844*10^{18} photons$ are released every second, then in 0.056 s:

$0.056 s*2.844*10^{18} \frac{photons}{s} = 1.593*10^{17} photons$ are released during the surgery

5 0

3 years ago

Why does a rolling sphere slows down?

VLD [36.1K]

The sphere slow down due to friction force between the surface of the sphere and the surface on that the sphere is rolling . The friction force acting against the motion of the sphere. Thats why it is slowed down. In fact not only a sphere, anything can not slow down untill a force act against it's motion.

3 0

3 years ago

Read 2 more answers

Part C Let’s start the analysis by looking at your “extreme usage” cases. Compare the two cases in detail—low usage period versu

Serga [27]

Answer:

Day 7 DataUsage notes (since last reading)day & datetimekWh readingkWh usedhours elapsedavg. kW usedb.Usage Extremes: Data CollectionFor this experiment, you’ll measure electrical usage during a time period when you expect to havevery light electrical usage (for instance, while you’re asleep at night or during the day when no oneis at home). Likewise you’ll measure electrical usage during a time period when you expect to have heavier than average electrical usage. This time period might be in the evening, when lights and other appliances are on. Both of these time periods should be at least 4 hours long, to increase the accuracy of your results. Record your results in the tables below for each situation. For each time period, you’ll need to takean initial and a final reading.Type your response here:Low Usage - Initial Readingday & datetimekWh readingLow Usage - Final ReadingEnergy Usage Notesday & datetimekWh readingkWh usedhours elapsedavg. kW usedHigh Usage - Initial Readingday & datetimekWh reading4

6 0

3 years ago

An object of mass 0.400.40 kg, hanging from a spring with a spring constant of 8.08.0 N/m, is set into an up-and-down simple har

sineoko [7]

Answer:

Acceleration will be equal to $2m/sec^2$

Explanation:

We have given mass of the object m = 0.4 kg

Spring constant k = 8 N/m

Maximum displacement of the spring is given x = 0.1 m

From newton's law force is equal to $F=ma$ .....eqn 1

By hook's law spring force is equal to $F=kx$ .....eqn 2

From equation 1 and equation 2

$ma=kx$

$0.4\times a=8\times 0.1$

$a=2m/sec^2$

So acceleration will be equal to $2m/sec^2$

8 0

3 years ago

What do you mean by specific heat capacity?

Dennis_Churaev [7]

Please mark me as Brainliest ......

6 0

3 years ago