Answer:
6 m/s is the missing final velocity
Explanation:
From the data table we extract that there were two objects (X and Y) that underwent an inelastic collision, moving together after the collision as a new object with mass equal the addition of the two original masses, and a new velocity which is the unknown in the problem).
Object X had a mass of 300 kg, while object Y had a mass of 100 kg.
Object's X initial velocity was positive (let's imagine it on a horizontal axis pointing to the right) of 10 m/s. Object Y had a negative velocity (imagine it as pointing to the left on the horizontal axis) of -6 m/s.
We can solve for the unknown, using conservation of momentum in the collision: Initial total momentum = Final total momentum (where momentum is defined as the product of the mass of the object times its velocity.
In numbers, and calling
the initial momentum of object X and
the initial momentum of object Y, we can derive the total initial momentum of the system: ![P_{total}_i=P_{xi}+P_{yi}= 300*10 \frac{kg*m}{s} -100*6\frac{kg*m}{s} =\\=(3000-600 )\frac{kg*m}{s} =2400 \frac{kg*m}{s}](https://tex.z-dn.net/?f=P_%7Btotal%7D_i%3DP_%7Bxi%7D%2BP_%7Byi%7D%3D%20300%2A10%20%5Cfrac%7Bkg%2Am%7D%7Bs%7D%20-100%2A6%5Cfrac%7Bkg%2Am%7D%7Bs%7D%20%3D%5C%5C%3D%283000-600%20%29%5Cfrac%7Bkg%2Am%7D%7Bs%7D%20%3D2400%20%5Cfrac%7Bkg%2Am%7D%7Bs%7D)
Since in the collision there is conservation of the total momentum, this initial quantity should equal the quantity for the final mometum of the stack together system (that has a total mass of 400 kg):
Final momentum of the system: ![M * v_f=400kg * v_f](https://tex.z-dn.net/?f=M%20%2A%20v_f%3D400kg%20%2A%20v_f)
We then set the equality of the momenta (total initial equals final) and proceed to solve the equation for the unknown(final velocity of the system):
![2400 \frac{kg*m}{s} =400kg*v_f\\\frac{2400}{400} \frac{m}{s} =v_f\\v_f=6 \frac{m}{s}](https://tex.z-dn.net/?f=2400%20%5Cfrac%7Bkg%2Am%7D%7Bs%7D%20%3D400kg%2Av_f%5C%5C%5Cfrac%7B2400%7D%7B400%7D%20%5Cfrac%7Bm%7D%7Bs%7D%20%3Dv_f%5C%5Cv_f%3D6%20%5Cfrac%7Bm%7D%7Bs%7D)
<h2>Answer: about the same size of the gap or slit</h2>
Diffraction happens when a wave (mechanical or electromagnetic wave, in fact, any wave) meets an obstacle or a slit .When this occurs, the wave bends around the corners of the obstacle or passes through the opening of the slit that acts as an obstacle, forming multiple patterns with the shape of the aperture of the slit.
Note that the principal condition for the occurrence of this phenomena is that the obstacle must be comparable in size (similar size) to the size of the wavelength.
In other words, when the gap (or slit) size is larger than the wavelength, the wave passes through the gap and does not spread out much on the other side, but when the gap size is equal to the wavelength, maximum diffraction occurs.
Therefore:
<h2>Waves diffract the most when their wavelength is <u>about the same size of the gap
</u></h2>
<u />
When light travels from a medium with higher refractive index to a medium with lower refractive index, the critical angle is the angle of incidence above which light is reflected only (no refraction occurs), and the value of this critical angle is given by
![\theta_c = \arcsin ( \frac{n_2}{n_1} )](https://tex.z-dn.net/?f=%5Ctheta_c%20%3D%20%5Carcsin%20%28%20%5Cfrac%7Bn_2%7D%7Bn_1%7D%20%29)
where n2 is the refractive index of the second medium and n1 is the refractive index of the first medium.
In this problem, the first medium is the glass (
![n_1 = 1.50](https://tex.z-dn.net/?f=n_1%20%3D%201.50)
), while the second medium is oil (
![n_2 =1.46](https://tex.z-dn.net/?f=n_2%20%3D1.46)
), therefore the critical angle is given by
<span>Formula-
E = hc/ {wavelength}
Given in quation-wavelength=12.4 cm
We know the value of h and c.
h=6.626 * 10^-34 Js
c= 2.997 * 10^8 m/s
wavelength=12.4 cm converted to 0.124 m
E = [ 6.626 * 10^-34 Js * 2.997 * 10^8 m/s ] / 0.124 m
E = 1.60 * 10^-24 J
1.60 * 10^-24 J of energy of exactly one photon of this microwave radiation have.</span>