What property is a characteristic of a pure substance that can be observed with out changing the substance into something else
1 answer:
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Answer:
1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) the two diameters have the same order of magnitude and are very close to each other
Explanation:
You have some problems in the writing of your exercise, we will try to answer.
1) The equation to be used in geometric optics is the constructor equation
where p and q are the distance to the object and the image, respectively, f is the focal length
* For the normal eye and with presbyopia
the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)
f'₀ = 1.5 cm
this is the focal length for this type of eye
* Eye with myopia
the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm
1 / f = 1/15 + 1 / 1.5
1 / f = 0.733
f = 1.36 cm
this is the focal length for the myopic eye.
In general, the two focal lengths are related
f’₀ / f = 1.5 / 1.36
f’₀ / f = 1.10
The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit
a sin θ= m λ
the first zero occurs for m = 1, as the angles are very small
tan θ = y / f = sin θ / cos θ
for some very small the cosine is 1
sin θ = y / f
where f is the distance of the lens (eye)
y / f = lam / a
in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor
y / f = 1.22 λ / D
y = 1.22 λ f / D
where D is the diameter of the eye
D = 2R₀
D = 2 0.1
D = 0.2 cm
the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation
* normal eye
the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation
\frac{1}{f} = \frac{1}{p} + \frac{1}{q}
sustitute
= 0.7066
f = 1.415 cm
therefore the diffraction is
y = 1.22 550 10⁻⁹ 1.415 / 0.2
y = 4.75 10⁻⁶ m
this is the radius, the diffraction diameter is
d = 2y
d_normal = 9.49 10⁻⁶ m
* myopic eye
In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm
= 0.733
f = 1.36 cm
diffraction is
y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2
y = 4.56 10-6 m
the diffraction diameter is
d_myope = 2y
d_myope = 9.16 10-6 m
= 9.49 /9.16
\frac{d_{normal}}{d_{myope}} = 1.04
we can see that the two diameters have the same order of magnitude and are very close to each other
1) The equivalent resistance is
2) The voltage in the circuit is 120 V
3) The total current in the circuit is 44.0 A
Explanation:
1)
Resistors are said to be in parallel when they have the same potential difference at their terminals.
The formula to calculate the equivalent resistance for three resistors in parallel is:
where are the resistances of the three resistors.
In this problem, the resistance of each resistor is:
Substituting,
So the equivalent resistance is
2)
In this problem we are told that the three resistors are connected in parallel. This means that their terminals are connected to the same point of the circuit: therefore, this means that they also have the same potential difference across them.
Therefore, the voltage in each branch containing each resistor is the same, and it is equal to the voltage of the battery, which is
So, the voltage in the circuit is 120 V.
3)
The total current in the circuit can be found by using Ohm's law:
where
V is the voltage
R is the equivalent resistance of the circuit
I is the current
In this circuit, we have:
V = 120 V
Re-arranging the equation for I, we find the current:
Learn more about current and potential difference:
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Answer:
0.572 Hz
Explanation:
given,
length of simple pendulum, l = 0.76 m
mass of the bob, m = 365 g = 0.365 Kg
angle made from the vertical, = 12°
frequency, f = ?
f = 0.572 Hz
The frequency at which pendulum vibrates is equal to 0.572 Hz