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Alinara [238K]
3 years ago
13

a camera with a 100mm lens can be used to focus objects from 6pm to infinity onto screen. how much must the lens be moved to foc

us on the extremities of this range
Physics
1 answer:
Romashka [77]3 years ago
3 0

Answer:

i = f = 0.1 m until the lens moves towards the screen 0.1 m

Explanation:

For this exercise let's use the constructor equation

         1 / f = 1 / o + 1 / i

     

Where f is the focal distance, or the distance to the object and "i" the distance to the image

It indicates that the focal distance is 100 mm (f = 100 mm), when an object is at infinity the image is formed at its focal length

     1 / f = 1 / inf + 1 / i = 1 / i

      i = f = 100 mm

At this point the screen is placed

For the shortest distance the lens has to move a little so the distance to the image is

        f = 100 mm = 0.1 m

        o = 6 pm = 6 10⁻¹² m

        i = 100 + x

        1 / f = 1 / o + 1 / 100+ x

        1 /( 0.10 + x) = 1 / f - 1 / o

        1 / 0.100+ x = 1/0.100 - 1/6 10-12

        1 / 0.100 + x = 10 - 10¹¹ = -10¹¹

        0.100 + x = -10⁻¹¹

        x = -10⁻¹¹ -10⁻¹

        x = -10⁻¹

        x = - 0.1 m

This negative distance indicates that the lens moves towards the screen 0.1 m

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A proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.
lawyer [7]

The change in potential energy of the proton is  5.6 x 10^{-17} Joule

<h3>What is a Uniform Electric Field ?</h3>

A uniform electric field is where the electric field strength is the same at all points in the field. In the uniform field, the force experienced by a charge is the same no matter where it is placed in the field.

Given that a proton moves a distance 10 cm in a uniform electric field of 3.5 kN C, in the direction of the field.

  • The distance d = 10 cm = 0.1 m
  • Electric field E = 3.5 KN/C
  • Proton charge q = 1.6 x 10^{-19} C

The Work done = Fd

but F = Eq

Recall that Electric field E = F/q = V/d

Where V = potential difference.

Let us first calculate the V

E = V/d

V = Ed

Substitute all the parameters into the formula above

V = 3.5 × 10³ × 0.1

V = 350 v

from F/q = V/d

make F the subject of formula and substitute it in work formula

F = Vq/d

W.D = Vq/d x d

W.D = Vq

Substitute all the parameters into the formula above

W.D = 350 x 1.6 x 10^{-19}

W.D = 5.6 x 10^{-17} J

Work done = Energy = Potential Energy

Therefore, the change in potential energy of the proton is 5.6 x 10^{-17}<em> Joule</em>

<em />

Learn more about Electric Field here: brainly.com/question/14372859

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7 0
2 years ago
(d) If η = 40% and TH = 427°C, what is TC, in °C?
Brrunno [24]

Answer:

T_C=256.2^{\circ}C

Explanation:

Given that,

Efficiency of heat engine, \eta=40\%=0.4

Temperature of hot source, T_H=427^{\circ}C

We need to find the temperature of cold sink i.e. T_C. The efficiency of heat engine is given by :

\eta=1-\dfrac{T_C}{T_H}

T_C=(1-\eta)T_H

T_C=(1-0.4)\times 427

T_C=256.2^{\circ}C

So, the temperature of the cold sink is 256.2°C. Hence, this is the required solution.

3 0
3 years ago
When was the copernican treatise published
LiRa [457]
They were published in 1542.

5 0
4 years ago
A string with a length of 4.00 m is held under a constant tension. The string has a linear mass density of \mu=0.000600~\text{kg
yulyashka [42]

Answer:

T=245.76N

Explanation:

We know that the frequency of the nth harmonic is given by f_n=nf, where f is the fundamental harmonic. Since we have the values of two consecutive frequencies, we can do:

f_{n+1}-f_n=(n+1)f-nf=nf+f-nf=f

Which for our values means (we do not need the value of <em>n</em>, that is, which harmonics are the frequencies given):

f=f_{n+1}-f_n=480Hz-400Hz=80Hz

Now we turn to the formula for the vibration frequency of a string (for the fundamental harmonic):

f=\frac{1}{2L} \sqrt{\frac{T}{\mu}}

So the tension is:

T=\mu(2Lf)^2

Which for our values is:

T=(0.0006kg/m)(2(4m)(80Hz))^2=245.76N

6 0
3 years ago
Shrinking Loop. A circular loop of flexible iron wire has an initial circumference of 168 cm , but its circumference is decreasi
DedPeter [7]

Answer:

103.1 V

Explanation:

We are given that

Initial circumference=C=168 cm

\frac{dC}{dt}=-15cm/s

Magnetic field,B=0.9 T

We have to find the magnitude of the  emf induced in the loop after exactly time 8 s has passed since the circumference of the loop started to decrease.

Magnetic flux=\phi=BA=B(\pi r^2)

Circumference,C=2\pi r

r=\frac{C}{2\pi}

r=\frac{168}{2\pi} cm

\frac{dr}{dt}=\frac{1}{2\pi}\frac{dC}{dt}=\frac{1}{2\pi}(-15)=-\frac{15}{2\pi} cm/s

\int dr=-\int \frac{15}{2\pi}dt

r=-\frac{15}{2\pi}t+C

When t=0

r=\frac{168}{2\pi}

\frac{168}{2\pi}=C

r=-\frac{15}{2\pi}t+\frac{168}{2\pi}

E=-\frac{d\phi}{dt}=-\frac{d(B\pi r^2)}{dt}=-2\pi rB\frac{dr}{dt}

E=-2\pi(-\frac{5}{2\pi}t+\frac{168}{2\pi})B\times -\frac{15}{2\pi}

t=8 s

B=0.9

E=2\pi\times \frac{15}{2\pi}\times 0.9(-\frac{15}{2\pi}(8)+\frac{168}{2\pi})

E=103.1 V

6 0
3 years ago
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