I believe the answer is x
Answer:
a) that laser 1 has the first interference closer to the central maximum
c) Δy = 0.64 m
Explanation:
The interference phenomenon is described by the expression
d sin θ = m λ
Where d is the separation of the slits, λ the wavelength and m an integer that indicates the order of interference
For the separation of the lines we use trigonometry
tan θ = sin θ / cos θ = y / x
In interference experiments the angle is very small
tan θ = sin θ = y / x
d y / x = m λ
a) and b) We apply the equation to the first laser
λ = d / 20
d y / x = m d / 20
y = m x / 20
y = 1 4.80 / 20
y = 0.24 m
The second laser
λ = d / 15
d y / x = m d / 15
y = m x / 15
y = 0.32 m
We can see that laser 1 has the first interference closer to the central maximum
c) laser 1
They ask us for the second maximum m = 2
y₂ = 2 4.8 / 20
y₂ = 0.48 m
For laser 2 they ask us for the third minimum m = 3
In this case to have a minimum we must add half wavelength
y₃ = (m + ½) x / 15
m = 3
y₃ = (3 + ½) 4.8 / 15
y₃ = 1.12 m
Δy = 1.12 - 0.48
Δy = 0.64 m
Answer:
a = - 9.8 j ^ m/s²
Explanation:
This is a projectile launch problem, they give us the initial velocity in the two components
v₀ₓ = 17.1 m / s
= 14.7 m / s
They indicate that the only acceleration that exists is the acceleration of gravity, which acts in the direction towards the center of the Earth, in general in a coordinate system it coincides with the direction of the y axis.
a = - g j ^
a = - 9.8 j ^ m /s²
Answer:
Work done by a tug boat, W = 1.735 x 10⁸ J
Explanation:
Given,
The of each tugboat, F = 1.5 x 10⁶ N
The angle of each tugboat forms with the resultant force, θ = 19°
The displacement of the supertanker, s = 710 m
The individual tugboat will be responsible for the displacement, d = 710/2
= 355 m
The displacement component in each tugboat direction = 355 · sin θ meter
Therefore, the work done by each tugboat is
W = F x S joules
Substituting the values in the above equation
W = 1.5 x 10⁶ x 355 · sin θ
= 1.735 x 10⁸ J
Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J
Firstly, we must find the equation the speed, it can be obtained by the derivative of x =<span>x = 4.0t2 - 32t + 36, it means v= 8.0t-32. the particule stops means v= 0, and 0= 8.0t-32, which implies 8t=32, so t=4s, and then x(t=4) = 4.0(4)^2 - 32(4) + 36= - 28m
so x = - 28 m</span>
