i think the answer is B-House Rules Comittee
Answer:
The film thickness is 4.32 * 10^-6 m
Explanation:
Here in this question, we are interested in calculating the thickness of the film.
Mathematically;
The number of fringes shifted when we insert a film of refractive index n and thickness L in the Michelson Interferometer is given as;
ΔN = (2L/λ) (n-1)
where λ is the wavelength of the light used
Let’s make L the subject of the formula
(λ * ΔN)/2(n-1) = L
From the question ΔN = 8 , λ = 540 nm, n = 1.5
Plugging these values, we have
L = ((540 * 10^-9 * 8)/2(1.5-1) = (4320 * 10^-9)/1 = 4.32 * 10^-6 m
Answer:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
Explanation:
The graph shown in the figure is a velocity-time graph, which means that:
- On the x-axis, the time is plotted
- On the y-axis, the velocity is plotted
Therefore, this means that the object is not moving when the line is horizontal (because at that moment, the velocity is constant, so the object is not moving). This occurs in the following intervals:
Between 2.0 s and 4.0 s (B and C)
Between 5.0 s and 8.0 s (D and E)
Between 10.0 s and 11.0 s (F and G)
From the graph, it would be possible to infer additional information. In particular:
- The area under the graph represents the total distance covered by the object
- The slope of the graph represents the acceleration of the object
Answer:b
Explanation:
im in fifth grade so i guessd. gl tho heh.
<span>stuntman's vertical velocity is
vy= 6 sin 15=1.55m/s.
Height that he goes up is,
vertical kinetic energy = mgh
1/2 v² = gh, h=v²/(2g)=0.12 m.
time wanted to go up= vy/9.8=0.16s, Time to fall through a height
0.12+2.9=3.02m is
t=sqrt(3.02*2/9.8)=0.78 s
Total time needed to go up and down is 0.78+0.16=0.94 s.
to calculate the horizontal range,
Horizontal velocity = 6 cos 15=5.8 m/s.
Distance which he can cover is
5.8*0.94=5.44 m.
If the distance between the two building is less than 5.44 m then he will be safe and he can jump that distance.</span>
