In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
where D=5.00 m is the distance of the screen from the slits, and
is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
And from the relationship between frequency and wavelength,
, we can find the frequency of the light:
<em>"A concave lens is thinner at the center than it is at the edges."</em>
If this isn't on the list of choices, that's tough. We can't help you choose the best one if we don't know what any of them is.
The approximate speed of the sound wave traveling through the solid material is 1012m/s.
<h3>
Wavelength, Frequency and Speed</h3>
Wavelength is simply the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.
From the wavelength, frequency and speed relation,
λ = v ÷ f
Where λ is wavelength, v is velocity/speed and f is frequency.
Given the data in the question;
- Frequency of sound wave f = 440Hz = 440s⁻¹
- Wavelength of the wave λ = 2.3m
To determine the approximate speed of the wave, we substitute our given values into the expression above.
λ = v ÷ f
2.3m = v ÷ 440s⁻¹
v = 2.3m × 440s⁻¹
v = 1012ms⁻¹
v = 1012m/s
Therefore, the approximate speed of the sound wave traveling through the solid material is 1012m/s.
Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701
Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, 
= 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
We will apply the conservation of linear momentum to answer this question.
Whenever there is an interaction between any number of objects, the total momentum before is the same as the total momentum after. For simplicity's sake we mostly use this equation to keep track of the momenta of two objects before and after a collision:
m₁v₁ + m₂v₂ = m₁v₁' + m₂v₂'
Note that v₁ and v₁' is the velocity of m₁ before and after the collision.
Let's choose m₁ and v₁ to represent the bullet's mass and velocity.
m₂ and v₂ represents the wood block's mass and velocity.
The bullet and wood will stick together after the collision, so their final velocities will be the same. v₁' = v₂'. We can simplify the equation by replacing these terms with a single term v'
m₁v₁ + m₂v₂ = m₁v' + m₂v'
m₁v₁ + m₂v₂ = (m₁+m₂)v'
Let's assume the wood block is initially at rest, so v₂ is 0. We can use this to further simplify the equation.
m₁v₁ = (m₁+m₂)v'
Here are the given values:
m₁ = 0.005kg
v₁ = 500m/s
m₂ = 5kg
Plug in the values and solve for v'
0.005×500 = (0.005+5)v'
v' = 0.4995m/s
v' ≅ 0.5m/s
