Answer:
correct answer is d
The pinhole has a size comparable with the laser wavelength, so the pinhole diffracts the passing laser beam.
Explanation:
The researcher does not have a refraction problem since the medium on both sides of the pinhole is the same with the same refractive index, the problem he is having is with the diffraction of the laser beam through the pinhole, let's analyze the diffraction process that is described by the expression
a sin θ = m λ
where a is the pinlole size, λ the wavelength of the laser and m an integer.
The laser extends from the maximum of diffraction to the first zero (m = 1) of diffraction
sin θ = λ/a
when analyzing this expression we have some interesting cases
* when λ « a. the sine approaches zero therefore we are in the case of optical geometry, in this case the laser passes through the hole without being diffracted
* when λ ≈ a. the sine function has values between 0 and 1, for which a diffraction of the laser beam occurs, which increases the diameter of the same
* when λ> a. The laser does not pass through the gap since the sine cannot have values greater than 1
After this analysis, we review the answers to the exercise to find that the correct answer is d
Answer:1) Via heat: ferromagnet materials will lose their magnetism if heated above a point known as the Curie temperature. ... With a strong enough magnetic field of opposite polarity, it is therefore possible to demagnetize the magnet [whether this comes from another permanent magnet, or a solenoid].
Explanation:Unlike permanent magnets, temporary magnets cannot remain magnetized on their own. Soft magnetic materials like iron and nickel will not attract paper clips after a strong external magnetic field has been removed.To restore a permanent magnet, you need to cool the metal (if heated) and expose it to a magnetic field. Coil your copper wire tightly around the piece of metal you would like to restore as a permanent magnet.
Answer:
Explanation:
Path difference for the destructive interference of a single slit:
For the first - order minimum, n = 1, and
.........(1)
For the second - order minimum, n = 2, and
.........(2)
Dividing equation (2) by equation (1):
Answer:
a) 21.6 KN/m
b)v = 18.56 m/s
Explanation:
For this problem, the spring should be strong enough to get the car just over the top so:
mgh = (1/2)kx^2
(400)*(9.8)*(10) = (0.5)*(k)*(2.0^2)
k = 39200/2.645
k = 19600 N/m
a) K increased by Increase by 10%, so:
k = 19600 * 1.10 = 21560 N/m = 21.6KN/m---->specified spring constant
b) Now, Maximum speed of a 350 kg car will be at the bottom of the roller coaster:
therefore, (1/2)mv^2 + mg(Y_1-Y_2) = (1/2)kx^2
(0.5)*(350)*(v^2) + (350)(9.8)(10-15) =(0.5)*(21560)*(2.0^2)
v^2 = 344.4
v = 18.56 m/s