Angle of incidence is 36° and so is the reflection. Both angles are equal.
Answer:
7/150
Explanation:
The following data were obtained from the question:
Object distance (u) = 75cm
Image distance (v) = 3.5cm
Magnification (M) =..?
Magnification is simply defined as:
Magnification (M) = Image distance (v)/ object distance (u)
M = v /u
With the above formula, we can obtain the magnification of the image as follow:
M = v/u
M = 3.5/75
M = 7/150
Therefore, the magnification of the image is 7/150.
Answer:
m,lkj,mkn,njn
Explanation:because she is telling you to do a project
Ok so here is the thing. It is necessary to introduce the atomic number Z into the following equation and the reason for that is that we are not working here with hydrogen (H). It will go like this:
<span>E=(2.18×10^-18 J)(Z^2 )|1/(ni^2 )-1/(nf^2 )| </span>
<span>E=(2.18×10^-18 J)(2^2 )|1/(6 ^2 )-1/(4 ^2 )|=3.02798×10^-19 J </span>
<span>After that we need to plug the E value calculated into the equation. Remember that the wavelength is always positive:</span>
<span>E=hc/λ 3.02798×10^-19 J=hc/λ λ=6.56×10^-7 m </span>
so 6.56×10^-7 m or better written 656 nm is in the visible spectrum
935,500 joules because when we use the KE formula KE=1/2mv^2;
KE=1/2(750)(50)^2
KE=375(2500)
KE=935,500 Joules
Hope it helps