Conservation of energy explains that energy can only be transferred between different forms of energy
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
![n_1sin(\theta_1)=n_2sin(\theta_2)](https://tex.z-dn.net/?f=n_1sin%28%5Ctheta_1%29%3Dn_2sin%28%5Ctheta_2%29)
Where
![\theta_2](https://tex.z-dn.net/?f=%5Ctheta_2)
differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question,
![\Delta x](https://tex.z-dn.net/?f=%5CDelta%20x%20)
is the distance between both rays.
![\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705](https://tex.z-dn.net/?f=%5Ctheta_%7B2red%7D%3DAsin%28%5Cfrac%7Bsin%2830%29%7D%7B1.57%7D%29%5Capprox%2018.5705)
![\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21](https://tex.z-dn.net/?f=%5Ctheta_%7B2violet%7D%3DAsin%28%5Cfrac%7Bsin%2830%29%7D%7B1.60%7D%29%5Capprox%2018.21)
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
![d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m](https://tex.z-dn.net/?f=d_%7Bviolet%7D%3Dh.tan%28%5Ctheta_%7B2violet%7D%29%5Capprox%200.0132m)
For red we have:
![d_{red}=h.tan(\theta_{2red})\approx 0.0134m](https://tex.z-dn.net/?f=d_%7Bred%7D%3Dh.tan%28%5Ctheta_%7B2red%7D%29%5Capprox%200.0134m)
We finally have:
Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:
![F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}](https://tex.z-dn.net/?f=F_%7BAnet%7D%20%3D%20m_A%5Ccdot%20a_A%20%3D%20F%20-%20F_%7Bfr%7D%5C%5CF_%7BBnet%7D%20%3D%20m_B%5Ccdot%20a_B%20%3D%20F_%7Bfr%7D)
where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:
![a_B = a_A-0.5 \frac{m}{s^2}](https://tex.z-dn.net/?f=a_B%20%3D%20a_A-0.5%20%5Cfrac%7Bm%7D%7Bs%5E2%7D)
Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:
![30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N](https://tex.z-dn.net/?f=30kg%5Ccdot%20a_A%20%3D%2075N%20-%20F_%7Bfr%7D%5C%5C24kg%5Ccdot%20a_B%20%3D%20F_%7Bfr%7D%5C%5Ca_B%3D%20a_A-0.5%20%5Cfrac%7Bm%7D%7Bs%5E2%7D%5C%5C%5Cimplies%20%5C%5Ca_A%3D%5Cfrac%7B87%7D%7B54%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%2C%5C%2C%5C%2C%5C%2Ca_B%3D%5Cfrac%7B10%7D%7B9%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%5C%5CF_%7Bfr%7D%20%3D%2024kg%20%5Ccdot%20%5Cfrac%7B10%7D%7B9%7D%5Cfrac%7Bm%7D%7Bs%5E2%7D%3D%5Cfrac%7B80%7D%7B3%7Dkg%5Cfrac%7Bm%7D%7Bs%5E2%7D%5Capprox%2026.7N)
The force of friction acting on block B is approximately 26.7N.
This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.
Explanation:
1 inch = 25.4 mm
1 foot = 12 inches
1 mile = 5260 feet
1 cm = 0.01 m or 10 mm
Now Sammy's height is 5 feet and 5.3 inches.
(a) We need to find Sammy's height in inches.
Since, 1 foot = 12 inches
5 feet = 5 × 12 inches = 60 inches
Now, 5 feet and 5.3 inches = 60 inches + 5.3 inches = 65.3 inches
Sammy's height is 65.3 inches.
(b) We need to find Sammy's height in feet.
Since, 1 foot = 12 inches
![1\ \text{inch}=\dfrac{1}{12}\ \text{feet}](https://tex.z-dn.net/?f=1%5C%20%5Ctext%7Binch%7D%3D%5Cdfrac%7B1%7D%7B12%7D%5C%20%5Ctext%7Bfeet%7D)
So,
![5.3\ \text{inch}=\dfrac{5.3}{12}\ \text{feet}=0.4416\ \text{feet}](https://tex.z-dn.net/?f=5.3%5C%20%5Ctext%7Binch%7D%3D%5Cdfrac%7B5.3%7D%7B12%7D%5C%20%5Ctext%7Bfeet%7D%3D0.4416%5C%20%5Ctext%7Bfeet%7D)
5 feet and 5.3 inches = 5 feet + 0.4416 feet = 5.44 feet
Sammy's height is 5.44 feet.
If you search that click the first link ;)