Ask question

Ask question

All categories

3 years ago

15

Find the image position for a picture placed 3.0 cm outside the focal point of a converging lens with a 4.0 cm focal length. a.

6.2 cm c. 2.8 cm b. 9.3 cm d. 11.0 cm

1 answer:

horrorfan [7]3 years ago

5 0

<span>Answer: Using 1/f = 1/d' + 1/d ...(where d' object distance and d is image distance) 1/4 = 1/7 + 1/d 1/4 - 1/7 = 1/d 3/28 = 1/d d = 28/3 d = 9.33 cm</span>

You might be interested in

A rugby player sits on a scrum machine that weighs 200 Newtons. Given that the coefficient of static friction is 0.67, the coeff

Trava [24]

a. 850 N is the minimum force needed to get the machine/player system moving, which means this is the maximum magnitude of static friction between the system and the surface they stand on.

By Newton's second law, at the moment right before the system starts to move,

• net horizontal force

∑ F[h] = F[push] - F[s. friction] = 0

• net vertical force

∑ F[v] = F[normal] - F[weight] = 0

and we have

F[s. friction] = µ[s] F[normal]

It follows that

F[weight] = F[normal] = (850 N) / (0.67) = 1268.66 N

where F[weight] is the combined weight of the player and machine. We're given the machine's weight is 200 N, so the player weighs 1068.66 N and hence has a mass of

(1068.66 N) / g ≈ 110 kg

b. To keep the system moving at a constant speed, the second-law equations from part (a) change only slightly to

∑ F[h] = F[push] - F[k. friction] = 0

∑ F[v] = F[normal] - F[weight] = 0

so that

F[k. friction] = µ[k] F[normal] = 0.56 (1268.66 N) = 710.45 N

and so the minimum force needed to keep the system moving is

F[push] = 710.45 N ≈ 710 N

4 0

2 years ago

Hooke’s law states that the distance that a spring is stretched by hanging object varies directly as the mass of the object. If

trasher [3.6K]

Answer:

$d_{2}$ = 33.33 cm

Explanation:

Given:

When mass, $m_{1}$ =21 kg

distance travelled is $d_{1}$ = 140 cm

When mass, $m_{2}$ =5 kg

distance travelled is $d_{2}$ = ?

Hooke's law state that within elastic limit, when an external force is applied to a body, the body gets deformed and when the force is released the gets back to its original form.

Therefore according to the question,

$\frac{d_{1}}{m_{1}}=\frac{d_{2}}{m_{2}}$

$\frac{140}{21}=\frac{d_{2}}{5}$

$d_{2}$ = 33.33 cm

Distance travelled is 33.33 cm when mass is 5 kg.

8 0

2 years ago

A light source of wavelength λ illuminates a metal with a work function (a.k.a., binding energy) of BE=2.00 eV and ejects electr

slega [8]

<h2>Answer: 1.011 eV</h2>

Explanation:

The described situation is the photoelectric effect, which consists of the emission of electrons (electric current) that occurs when light falls on a metal surface under certain conditions.

If we consider the light as a stream of photons and each of them has energy, this energy is able to pull an electron out of the crystalline lattice of the metal and communicate, in addition, a <u>kinetic energy. </u>

This is what Einstein proposed:

Light behaves like a stream of particles called photons with an energy $E$ :

$E=h.f$ (1)

So, the energy $E$ of the incident photon must be equal to the sum of the Work function $\Phi$ of the metal and the kinetic energy $K$ of the photoelectron:

$E=\Phi+K$ (2)

Where $\Phi$ is the <u>minimum amount of energy required to induce the photoemission of electrons from the surface of a metal, and </u><u>its value depends on the metal. </u>

In this case $\Phi=2eV$ and $K_{1}=4eV$

So, for the first light source of wavelength $\lambda_{1}$ , and applying equation (2) we have:

$E_{1}=2eV+4eV$ (3)

$E_{1}=6eV$ (4)

Now, substituting (1) in (4):

$h.f=6eV$ (5)

Where:

$h=4.136(10)^{-15}eV.s$ is the Planck constant

$f$ is the frequency

Now, the <u>frequency has an inverse relation with the wavelength </u>

$\lambda_{1}$ :

$f=\frac{c}{\lambda_{1}}$ (6)

Where $c=3(10)^{8}m/s$ is the speed of light in vacuum

Substituting (6) in (5):

$\frac{hc}{\lambda_{1}}=6eV$ (7)

Then finding $\lambda_{1}$ :

$\lambda_{1}=\frac{hc}{6eV }$ (8)

$\lambda_{1}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{6eV}$

We obtain the wavelength of the first light suorce $\lambda_{1}$ :

$\lambda_{1}=2.06(10)^{-7}m$ (9)

Now, we are told the second light source $\lambda_{2}$ has the double the wavelength of the first:

$\lambda_{2}=2\lambda_{1}=(2)(2.06(10)^{-7}m)$ (10)

Then: $\lambda_{2}=4.12(10)^{-7}m$ (11)

Knowing this value we can find $E_{2}$ :

$E_{2}=\frac{hc}{\lambda_{2}}$ (12)

$E_{2}=\frac{(4.136(10)^{-15} eV.s)(3(10)^{8}m/s)}{4.12(10)^{-7}m}$ (12)

$E_{2}=3.011eV$ (13)

Knowing the value of $E_{2}$ and $\lambda_{2}$ , and knowing we are working with the same work function, we can finally find the maximum kinetic energy $K_{2}$ for this wavelength:

$E_{2}=\Phi+K_{2}$ (14)

$K_{2}=E_{2}-\Phi$ (15)

$K_{2}=3.011eV-2eV$

$K_{2}=1.011 eV$ This is the maximum kinetic energy for the second light source

7 0

3 years ago

Under what atmospheric condition is fog most often formed in the san joaquin valley?

GalinKa [24]

It seems that you have missed the necessary options for us to answer this question so I had to look for it. Anyway,here is the answer. The atmospheric condition in which <span>fog is most often formed in the san Joaquin valley is stable stability. Hope this helps.</span>

4 0

3 years ago

Graham and hunter are circus performers. a cable lifts graham into the air at a constant speed of 1.5 ft/s. when graham’s arms a

siniylev [52]

Answer:

The equations are :

h = 18 + 1.5t

h = 5 + 24t - 16t^2

Explanation:

Speed = distance/time

1.5ft/s = 18ft/t

1.5t=18

The equation to determine height h= 18 + 1.5t

Graham moved another 5ft further with a velocity of 24ft/s

The equation to determine his height will be:

h= 5 + 24t -16t^2

3 0

3 years ago

Read 2 more answers