Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that
L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH
Two children push on heavy crate that rests on a basement floor. one pushes horizontally with a force of 150 N and the other pus
1 answer:
The force of friction is 330 N
Explanation:
We can solve this problem by applying Newton's second law of motion, which states that the net force acting on an object is equal to the product between the mass of the object and its acceleration:
where
is the net force
m is the mass
a is the acceleration
For the crate in this problem, we have 3 forces acting on it:
- The force applied by the 1st child, forward
- The force applied by the 2nd child,
- The force of friction, , acting backward
So the net force is
We also know that the crate remains stationary, which means that its acceleration is zero:
a = 0
Therefore, we can rewrite Newton's second law as
From which we find
Learn more about friction and Newton's second law:
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Speed of car A is given as
now we need to convert it into SI units
1 miles = 1609 m
1 hour = 3600 s
now we have
now its distance from Bambi is given as
time taken by it to hit the Bambi
Now other car is moving at speed 50 mph
so its speed in SI unit will be
now its distance from Bambi is given as
as we know that 1 feet = 0.3048 m
now the time to hit the other car is
So Car B will hit the Bambi first
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