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topjm [15]
3 years ago
6

an object is placed 15.8 cm in front of a thin converging lens with an unknown focal length. if a real image forms behind the le

ns with an image distance of 4.2cm what is the focal length of the lens
Physics
1 answer:
storchak [24]3 years ago
6 0
<span>On what:

f (is the focal length of the lens) = ? 
p (is the distance from the object to the lens) =15.8 cm
p' (is the distance from the image to the spherical lens) = 4.2 cm

</span><span>Using the Gaussian equation, to know where the object is situated (distance from the point).
</span>
\frac{1}{f} = \frac{1}{p} + \frac{1}{p'}
\frac{1}{f} = \frac{1}{15.8} + \frac{1}{4.2}
\frac{1}{f} = \frac{2.1}{33.18} + \frac{7.9}{33.18}
\frac{1}{f} =  \frac{10}{33.18}
Product of extremes equals product of means:
10*f = 1*33.18
10f = 33.18
f =  \frac{33.18}{10}
\boxed{\boxed{f = 3.318\:cm}}\end{array}}\qquad\quad\checkmark

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Ishani and John now try a problem involving a charging capacitor. An uncharged capacitor with C = 6.81 μF and a resistor with R
DENIUS [597]

Answer:

Q=81.72\times10^{-6}C

I=2.1\times10^{-5}A

Explanation:

The maximum charge on the capacitor will be, at the end of the process, given by the formula (and for our values):

Q=CV=(6.81\times10^{-6}F)(12V)=81.72\times10^{-6}C

The maximum current on the resistor will be, at the beginning of the process, given by the formula (and for our values):

I=\frac{V}{R}=\frac{12V}{5.8\times10^{5}\Omega}=2.1\times10^{-5}A

5 0
2 years ago
A 360-nm thick oil film floats on the surface of a pool of water. the indices of refraction of the oil and the water are 1.5 and
jenyasd209 [6]
The correct answer is 432, and 720.
The thickness of a film is t= 360nm
the refractive index of oil n₀t = (m +1/2) λ
For m =0
λ = 4n₀t
= 4(1.50)(360)
= 2160nm
for m = 1
λ = 4n₀t
= 4(1.50)(360)/3
= 720nm
m = 2
λ = 4n₀t/5 = 4(1.50)(360)/5
= 432nm
The wavelength which are most strongly reflected are
432nm, 720nm.
7 0
3 years ago
Calculate the average times it took the car to travel 0. 25 and 0. 50 meters. Record the averages, to two decimal places, in Tab
Illusion [34]

The average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

<h3>How to calculate the Average speed?</h3>

The average speed can be calculated by adding the speed of each trial divided by the number of trials,

For 0.25 m the average speed will be:

S_{avg} = \dfrac{2.24 + 2.21 + 2.23}{ 3}\\\\S_{avg} = 2.22

For the 0.50 m, the average speed will:

S_{avg} = \dfrac {3.16 + 3.08 + 3.15} {3 }\\\\S_{avg}  = 3.13\rm \  s

Therefore, the average speed of the given car is 2.22 s and 3.13 s for 0.25 m and 0.50 m distance respectively.

Learn more about Average speed:

brainly.com/question/26386984

6 0
2 years ago
• List four uses of the concave mirror.<br> State five uses of lenses in everyday activities
Zina [86]
FOUR USES OF CONCAVE MIRROR:Satellite dishes,headlights of a car, telescopes used for astronomical studies, and shaving mirrors because of there curved and reflective surface.

FIVE USES OF LENSES: Camera lens ,microscopes ,magnifying glass,eyeglasses,projector
6 0
2 years ago
A lamp can work on 50V mains taking 2 amps. What value of resistance must be connected in series with it, so that it can be oper
wolverine [178]
The resistance of the lamp is apparently  50V/2A  =  25 ohms.

When the circuit is fed with more than 50V, we want to add
another resistor in series with the 25-ohm lamp so that the
current through the combination will be 2A.

In order for 200V to cause 2A of current, the total resistance
must be      200V/2A = 100 ohms.

The lamp provides 25 ohms, so we want to add another 75 ohms 
in series with the lamp.  Then the total resistance of the circuit is
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The power delivered by the 200V mains is (200V) x (2A) = 400 watts.

The lamp dissipates ( I² · R ) = (2² · 25 ohms) = 100 watts.

The extra resistor dissipates  ( I² · R) = (2² · 75 ohms) = 300 watts.

Together, they add up to the 400 watts delivered by the mains.

CAUTION:
300 watts is an awful lot of power for a resistor to dissipate !
Those little striped jobbies can't do it. 
It has to be a special 'power resistor'. 
300 watts is even an unusually big power resistor.
If this story actually happened, it would be cheaper, easier,
and safer to get three more of the same kind of lamp, and
connect THOSE in series for 100 ohms.  Then at least the
power would all be going to provide some light, and not just
wasted to heat the room with a big moose resistor that's too
hot to touch.
3 0
3 years ago
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