<h2>Answer: about the same size of the gap or slit</h2>
Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.
Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.
In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.
Therefore:
<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap
</u></h2>
<u />
What is the atomic number of an element whose atoms each 47 protons, 60 neutrons, and 47 electrons?
The correct answer is number 47 or Ag (Silver).
Hope I Helped!
The answer is B because it literally says it right there, and I don't wanna make it impossibly confusing for a middle school student, lol.
Answer:
a) A = 0.07129 m²
b) A / A ’= 1.77
Explanation:
In this exercise we are asked to find the area in SI units, so let's start by reducing the dimensions to SI units.
width a = 8.5 inch (2.54 cm 1 inch) (1 m / 100 cm)
a = 0.2159 m
length l = 13 inch (2.54 cm / 1 inch) (1 m / 100 cm)
l = 0.3302 m
The area of a rectangle is
A = l a
A = 0.3302 0.2159
A = 0.07129 m²
b) we have a second sheet with reduced dimensions
a ’= 3/4 a
l ’= ¾ l
Let's find the area of this glossy sheet
A ’= l’ a ’
A ’= ¾ l ¾ a
A ’= 9/16 l a
To find the factor we divide the two quantities
A / A ’= l a 16 / (9 l a
A / A ’= 1.77
Answer:
a) —0.5 j m/s
b) 4.5 i + 2.25 j m
Explanation:
<u>Givens:</u>
v_0 =3.00 i m/s
a= (-3 i — 1.400 j ) m/s^2
The maximum x coordinate is reached when dx/dt = 0 or v_x = 0 ,thus :
<em>v_x = v_0 + at = 0 </em>
(3.00 i m/s) + (-3 i m/s^2)t=0
t = (3 m/s)/-3 i m/s^2
t = -1 s
Therefore the particle reaches the maximum x-coordinate at time t = 1 s.
Part a The velocity-of course- is all in the y-direction,therefore:
v_y =v_0+ at
We have that v_0 = 0 in the y-direction.
v_y = (-0.5 j m/s^2)(1 s)
= —0.5 j m/s
Part b: While the position of the particle at t = 1 s is given by:
r=r_0+v_0*t+1/2*a*t^2
Where r_0 = 0 since the particle started from the origin.
Its position at t = 1 s is then given by :
r =(3.00 i m/s)(1 s)+1/2(-3 i — 1.400 j )(1 s)^2
=4.5 i + 2.25 j m
