Answer:
the last one, the third one, and the first one.
Answer:
D = 2.38 m
Explanation:
This exercise is a diffraction problem where we must be able to separate the license plate numbers, so we must use a criterion to know when two light sources are separated, let's use the Rayleigh criterion, according to this criterion two light sources are separated if The maximum diffraction of a point coincides with the first minimum of the second point, so we can use the diffraction equation for a slit
a sin θ = m λ
Where the first minimum occurs for m = 1, as in these experiments the angle is very small, we can approximate the sine to the angle
θ = λ / a
Also when we use a circular aperture instead of slits, we must use polar coordinates, which introduce a numerical constant
θ = 1.22 λ / D
Where D is the circular tightness
Let's apply this equation to our case
D = 1.22 λ / θ
To calculate the angles let's use trigonometry
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ (4.30 10⁻² / 140 10³)
θ = tan⁻¹ (3.07 10⁻⁷)
θ = 3.07 10⁻⁷ rad
Let's calculate
D = 1.22 600 10⁻⁹ / 3.07 10⁻⁷
D = 2.38 m
The answer is: (2) : <span>↘
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Answer:
i think its going to be 150 because its half of 300
Explanation:
Answer:
The magnitude of Force is 8.58×10⁵N and direction is upwards
Explanation:
The work beam does on the pile driver is given by
W=(FCos180°)Δx= -F(0.088m)
From work energy theorem
![W_{nc}=(KE_{f}-KE_{i})+(PE_{f}-PE_{i})\\W_{nc}=1/2m(vf^{2}-vi^{2})+mg(yf-yi)](https://tex.z-dn.net/?f=W_%7Bnc%7D%3D%28KE_%7Bf%7D-KE_%7Bi%7D%29%2B%28PE_%7Bf%7D-PE_%7Bi%7D%29%5C%5CW_%7Bnc%7D%3D1%2F2m%28vf%5E%7B2%7D-vi%5E%7B2%7D%29%2Bmg%28yf-yi%29)
Choosing y=0 at the the level where the driver first contacts the beam and vi=0 at yi=+3.40m and comes to rest again vf=0 at yf= -0.088m
So
![-F(0.088m)=1/2m(0-0)+2,200kg(9.81m/s^{2} )(-0.088m-3.40m)\\-F(0.088)=-75508.224\\F=75508.224/0.088\\F=8.58*10^{5} N](https://tex.z-dn.net/?f=-F%280.088m%29%3D1%2F2m%280-0%29%2B2%2C200kg%289.81m%2Fs%5E%7B2%7D%20%29%28-0.088m-3.40m%29%5C%5C-F%280.088%29%3D-75508.224%5C%5CF%3D75508.224%2F0.088%5C%5CF%3D8.58%2A10%5E%7B5%7D%20N)
The magnitude of Force is 8.58×10⁵N and direction is upwards