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

Given vectors D (3.00 m, 315 degrees wrt x-axis) and E (4.50 m, 53.0 degrees wrt x-axis), find the resultant R= D + E. (a) Write

fomenos

Answer:

(a) $\vec{R}= 4.83\ m\ \hat{i}+1.47\ m\ \hat{j}$

(b) (5.05 m, 16.93 degrees wrt x-axis)

Explanation:

Given:

$\vec{D}$ = (3.00 m, 315 degrees wrt x-axis)
$\vec{E}$ = (4.50 m, 53.0 degrees wrt x-axis)

Let us first fond out vector D and E in their rectangular form.

$\vec{D} = (3\cos 315^\circ\ \hat{i}+3\sin 315^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.12\ \hat{i}-2.12\ \hat{j})\ m\\$

Similarly,

$\vec{E} = (4.5\cos 53^\circ\ \hat{i}+4.5\sin 53^\circ\ \hat{j})\ m\\\Rightarrow \vec{D} = (2.71\ \hat{i}+3.59\ \hat{j})\ m\\\because \vec{R}=\vec{D}+\vec{E}\\\therefore \vec{R} = (2.12\ \hat{i}-2.12\ \hat{j})\ m+(2.71\ \hat{i}+3.59\ \hat{j})\ m\\\Rightarrow \vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (a):

We can write the resultant vector R as below:

$\vec{R} = (4.83\ \hat{i}+1.47\ \hat{j})\ m$

Part (b):

$Magnitude\ of\ resultant = \sqrt{4.83^2+1.47^2}\ m = 5.05\ m\\\textrm{Direction in angle with the x-axis} = \theta = \tan^{-1}(\dfrac{1.47}{4.83})= 16.93^\circ$

Since both the components of the resultant lie on the positive x and y axes. So, the resultant makes an acute angle with the positive x-axis.

So, R = (5.05 m, 16.93 degrees wrt x-axis)

3 0

3 years ago

You kick a ball with a speed of 4 m/s at an angle of 10ﾟ. How far away does the ball land?

kolbaska11 [484]

Answer: 0.55m

1. First find Vf
• 4 Sin(10)=0.7
•Vf= 0.7

2. Find time
•9.8=0.7/t
•9.8t=0.7
•t=0.07

3. Find distance
• 4 Cos(10)=d/0.14 < ————- double time
• d=0.55m

8 0

3 years ago

Read 2 more answers

What is specific latent heat mean in simple word?

Nuetrik [128]

Answer:

Amount of energy required to change the 1kg of a substance without changing its temperature.

7 0

2 years ago

Which of the following frequencies are higher harmonics of a string with fundamental frequency of 150 Hz? (Select all that apply

Tom [10]

400 5000 and 200 because the input was a good we

3 0

3 years ago

A uniform rectangular plate is hanging vertically downward from a hinge that passes along its left edge. By blowing air at 11.0

Serjik [45]

Answer:

The airspeed must be 7.78 m/s for the rectangular plate kept at 30°.

Explanation:

By looking at the images below wee see that the airspeed on one side of the rectangular plate decreases the statical pressure over this side. Since over the downside, the pressure still bein the atmospheric pressure. This difference in pressure produces a lift force in the plate. The list force is the net force obtained between the difference of the forces that produce the pressure over the upside and the downside:

$F_{lift}=F_{up} - F_{dw}=0.5*p*V^2$

Where up and down relate to what movement the forces produce. And p and V are the respective air density and velocity.

When the plate is kept horizontal the lift force balance the moment due to the weight of the plate and considering that both forces act at the same point:

$F_{lift}=0.5*p*V^2=W$

By replacing the known values it is possible to find the plate's weight:

$F_{lift}=0.5*1.2 \frac{kg}{m^{3}}*(11 m/s)^2=W$

$W=72.6 N$

When the plate kept to 30° from the vertical the moment equation balance is written as:

$F_{lift}=0.5*p*V^2=W*sen(30\°)$

The sine of 30° is due to the weight is 30° oriented, therefore the new value for the airspeed is:

$V=\sqrt(W*sen(30\°)/0.5p)$

$V=\sqrt(\frac{72.6 N * 0.5}{0.5*1.2 kg/m^3})$

$V=\sqrt(60.5 \frac{N}{kg/m^3})$

$V=\sqrt(60.5 \frac{kg.m/s^2}{kg/m^3})$

$V=\sqrt(60.5 \frac{m^2}{s^2})$

$V= 7.78 m/s$

7 0

3 years ago