A. Condensation
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Answer:
The outline of the energy transfer are;
a) Kinetic energy → Clockwork spring → Potential energy
b) Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Please find attached the drawings of the energy transfer created with MS Visio
Explanation:
The energy transfer diagrams are diagrams that can be used to indicate the part of a system where energy is stored and the form and location to which the energy is transferred
a) The energy transfer diagram for the winding up a clockwork car is given as follows;
Mechanical kinetic energy is used to wind up (turn) the clockwork car such that the kinetic energy is transformed into potential energy and stored in the wound up clockwork as follows;
Kinetic energy → Clockwork spring → Potential energy
b) Letting a wound up clockwork car run results in the conversion of mechanical potential energy into kinetic (energy due tom motion) energy as follows;
Potential energy in clockwork car → Clockwork spring coil unwound → Clockwork car run
c) The energy stored in the battery of a battery powered car is chemical potential energy. When the battery powered car runs, the chemical potential energy produces an electromotive force which is converted into kinetic energy as electric current flows from the batteries
Therefore, we have;
Chemical potential energy → Batteries in the car → Electric motors → Kinetic energy
Answer:
a
The number of fringe is z = 3 fringes
b
The ratio is 
Explanation:
a
From the question we are told that
The wavelength is 
The distance between the slit is 
The width of the slit is 
let z be the number of fringes that appear between the first diffraction-envelope minima to either side of the central maximum in a double-slit pattern is and this mathematically represented as

Substituting values
z = 3 fringes
b
From the question we are told that the order of the bright fringe is n = 3
Generally the intensity of a pattern is mathematically represented as
![I = I_o cos^2 [\frac{\pi d sin \theta}{\lambda} ][\frac{sin (\pi a sin \frac{\theta}{\lambda } )}{\pi a sin \frac{\theta}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%5B%5Cfrac%7B%5Cpi%20d%20sin%20%5Ctheta%7D%7B%5Clambda%7D%20%5D%5B%5Cfrac%7Bsin%20%28%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%20%7D%20%29%7D%7B%5Cpi%20a%20sin%20%5Cfrac%7B%5Ctheta%7D%7B%5Clambda%7D%20%7D%20%5D)
Where
is the intensity of the central fringe
And Generally 
![I = I_o co^2 [ \frac{\pi (\frac{n \lambda}{d} )}{\lambda} ] [\frac{\frac{sin (\pi a (\frac{n \lambda}{d} ))}{\lambda} }{\frac{\pi a (\frac{n \lambda}{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20co%5E2%20%5B%20%5Cfrac%7B%5Cpi%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%5D%20%5B%5Cfrac%7B%5Cfrac%7Bsin%20%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%7D%7B%5Cfrac%7B%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (n \pi)[\frac{\frac{sin(\pi a (\frac{n \lambda}{d} ))}{\lambda} )}{ \frac{ \pi a (\frac{n \lambda }{d} )}{\lambda} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%28n%20%5Cpi%29%5B%5Cfrac%7B%5Cfrac%7Bsin%28%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%7D%7Bd%7D%20%29%29%7D%7B%5Clambda%7D%20%29%7D%7B%20%5Cfrac%7B%20%5Cpi%20a%20%28%5Cfrac%7Bn%20%5Clambda%20%7D%7Bd%7D%20%29%7D%7B%5Clambda%7D%20%7D%20%5D)
![I = I_o cos^2 (3 \pi) [\frac{sin (\frac{3 \pi }{6} )}{\frac{3 \pi}{6} } ]](https://tex.z-dn.net/?f=I%20%3D%20I_o%20cos%5E2%20%283%20%5Cpi%29%20%5B%5Cfrac%7Bsin%20%28%5Cfrac%7B3%20%5Cpi%20%7D%7B6%7D%20%29%7D%7B%5Cfrac%7B3%20%5Cpi%7D%7B6%7D%20%7D%20%5D)


Answer:
2,87 * 
Explanation:
When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:
Px= 5,30 *
* 301 = 1,5953 kg*m/s
Py= 5,30 *
* 301 = 1,5953 kg*m/s
The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).
For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.
Pz = 1,5953 = m * 554
m = 2,87 *
kg
B. Impedes the flow of electrons