All categories

3 years ago

A 3.50 cm tall object is held 24.8 cm from a lens of focal length 16.0 cm. What is the image height?

Physics

1 answer:

Mandarinka [93]3 years ago

6 0

Answer:

Explanation:

Using the formula to first get the image distance

1/f = 1/u+1/v

f = focal length of the lens

u = object distance

v = image distance

Given f = 16.0 cm, u = 24.8 cm

1/v = 1/16 - 1/24.8

1/v = 0.0625-0.04032

1/v = 0.02218

v = 1/0.02218

v = 45.09 cm

To get the image height, we will us the magnification formula.

Mag = v/u = Hi/H

Hi = image height = ?

H = object height = 3.50 cm

45.09/24.8 = Hi/3.50

Hi = (45.09*3.50)/24.8

Hi = 6.36 cm

The image height is 6.36 cm

You might be interested in

16 to 19 year old male and females are how much more likely to be involved in a crash?

Sedbober [7]

I say around 40% - 60%

https://www.dmv.ca.gov/portal/dmv/detail/teenweb/more_btn6/traffic/traffic
http://www.teendriversource.org/stats/support_teens/detail/57
http://www.rmiia.org/auto/teens/Teen_Driving_Statistics.asp

(I just corrected the question. Sorry if it is still incorrect.)

6 0

4 years ago

a car whose initial speed is 30 seconds m / s slows uniformly to a stop in 5.00 seconds what was the cars displacement

CaHeK987 [17]

Answer:

Explanation:

4 0

3 years ago

SVEN [57.7K]

Answer:

Frequency = 1,550Hz

Explanation:

To solve this we can use the equation: $f=\frac{v}{\lambda}$

(frequency = velocity/wavelength).

We are given the information that the wavelength is 22cm and the speed is 340m/s. The first step is to make sure everything is in the correct units (SI units), and to convert them if needed. The SI Units for velocity and wavelength are m/s and m respectively. This means we need to convert 22cm into meters, which we can do by dividing by 100, (as there are 100cm in a meter). 22/100 = 0.22m

Now we can substitute these values into the formula and calculate to solve:

$f=\frac{340}{0.22} \\\\f=1545.454...$

Simplify to 3 significant figures:

f = 1,550Hz

(Which I believe is just below a G6 if you were interested)

Hope this helped!

4 0

3 years ago

A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci

MA_775_DIABLO [31]

Answer:

$v_B=3.78\times 10^5\ m/s$

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is $6.63\times 10^{-27}\ kg$

Speed of nucleus at A is $v_A=6.2\times 10^5\ m/s$

Potential at point A, $V_A=1.5\times 10^3\ V$

Potential at point B, $V_B=4\times 10^3\ V$

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

$\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s$