Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is
farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with


- m is the order number.
is the wavelength of the monochromatic light.- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light
is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :


The dot in the center of fringe E is
farther from the left slit than from the right slit.
data which is expressed in form of following way

here in above expression
= true value
= uncertainty in the value
now the relative uncertainty is given as

now by above formula we can say
a) 2.70 ± 0.05cm
here
True value = 2.70
uncertainty = 0.05
Relative uncertainty =
= 0.0185
b) 12.02 ± 0.08cm
here
True value = 12.02
uncertainty = 0.08
Relative uncertainty =
= 0.00665
Answer:
y = 17 m
Explanation:
For this projectile launch exercise, let's write the equation of position
x = v₀ₓ t
y =
t - ½ g t²
let's substitute
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
the maximum height the ball can reach where the vertical velocity is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
0 = v₀ sin θ - 9.8 t
Let's write our system of equations
45 = v₀ cos θ t
10 = v₀ sin θ t - ½ 9.8 t²
0 = v₀ sin θ - 9.8 t
We have a system of three equations with three unknowns for which it can be solved.
Let's use the last two
v₀ sin θ = 9.8 t
we substitute
10 = (9.8 t) t - ½ 9.8 t2
10 = ½ 9.8 t2
10 = 4.9 t2
t = √ (10 / 4.9)
t = 1,429 s
Now let's use the first equation and the last one
45 = v₀ cos θ t
0 = v₀ sin θ - 9.8 t
9.8 t = v₀ sin θ
45 / t = v₀ cos θ
we divide
9.8t / (45 / t) = tan θ
tan θ = 9.8 t² / 45
θ = tan⁻¹ ( 9.8 t² / 45
)
θ = tan⁻¹ (0.4447)
θ = 24º
Now we can calculate the maximum height
v_y² =
- 2 g y
vy = 0
y = v_{oy}^2 / 2g
y = (20 sin 24)²/2 9.8
y = 3,376 m
the other angle that gives the same result is
θ‘= 90 - θ
θ' = 90 -24
θ'= 66'
for this angle the maximum height is
y = v_{oy}^2 / 2g
y = (20 sin 66)²/2 9.8
y = 17 m
thisis the correct
Answer:
<h2>1.17 m/s²</h2>
Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula

f is the force
m is the mass
From the question we have

We have the final answer as
<h3>1.17 m/s²</h3>
Hope this helps you
Hello!

Use the equation for momentum:

Plug in the given mass and velocity into the equation:

