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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

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>

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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.

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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