Answer:
3. When energy is used, it can transform to new types but can never disappear.
Explanation: it can transform into heat, light, ect and will never disapear unlessed turned of.
Answer:
A
B
![h = 1.3061 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%201.3061%20%5C%20%20m%20%20)
C
![v = 5.06 \ m/s](https://tex.z-dn.net/?f=%20%20v%20%3D%20%205.06%20%5C%20%20m%2Fs%20)
D
![d = 4.0273 \ m](https://tex.z-dn.net/?f=d%20%3D%204.0273%20%5C%20%20m%20%20)
Explanation:
Considering the first question
From the question we are told that
The spring constant is ![k = 32.50 N/m](https://tex.z-dn.net/?f=k%20%20%3D%20%2032.50%20N%2Fm)
The potential energy is ![PE = 0.640 \ J](https://tex.z-dn.net/?f=PE%20%20%3D%20%200.640%20%5C%20J)
Generally the potential energy stored in spring is mathematically represented as ![PE = \frac{1}{2} * k * x^2](https://tex.z-dn.net/?f=PE%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20k%20%20%2A%20%20x%5E2)
=>
=>
=>
Considering the second question
From the question we are told that
The mass of the dart is m = 0.050 kg
Generally from the law of energy conservation
![PE = mgh](https://tex.z-dn.net/?f=PE%20%3D%20%20mgh)
=> ![0.640 = 0.050 * 9.8 * h](https://tex.z-dn.net/?f=0.640%20%20%20%3D%20%200.050%20%2A%20%209.8%20%20%2A%20%20h)
=> ![h = 1.3061 \ m](https://tex.z-dn.net/?f=h%20%20%3D%20%201.3061%20%5C%20%20m%20%20)
Considering the third question
The height at which the dart was fired horizontally is ![H = 3.90\ m](https://tex.z-dn.net/?f=H%20%20%3D%20%20%203.90%5C%20%20m)
Generally from the law of energy conservation
![PE = KE](https://tex.z-dn.net/?f=PE%20%3D%20KE%20)
Here KE is kinetic energy of the dart which is mathematical represented as
![KE = \frac{1}{2} * mv^2](https://tex.z-dn.net/?f=KE%20%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%20%20mv%5E2)
=> ![0.640 = \frac{1}{2} * 0.050 * v^2](https://tex.z-dn.net/?f=0.640%20%3D%20%20%5Cfrac%7B1%7D%7B2%7D%20%20%2A%200.050%20%2A%20%20v%5E2%20)
=> ![v^2 = 25.6](https://tex.z-dn.net/?f=%20%20v%5E2%20%3D%2025.6%20)
=> ![v = 5.06 \ m/s](https://tex.z-dn.net/?f=%20%20v%20%3D%20%205.06%20%5C%20%20m%2Fs%20)
Considering the fourth question
Generally the total time of flight of the dart is mathematically represented as
![t = \frac{ 2 * H }{g}](https://tex.z-dn.net/?f=t%20%20%3D%20%20%5Cfrac%7B%202%20%2A%20%20H%20%7D%7Bg%7D)
=> ![t = \frac{ 2 * 3.90 }{9.8 }](https://tex.z-dn.net/?f=t%20%20%3D%20%20%5Cfrac%7B%202%20%2A%203.90%20%7D%7B9.8%20%7D)
=> ![t = 0.7959 \ s](https://tex.z-dn.net/?f=t%20%20%3D%20%200.7959%20%5C%20s%20)
Generally the horizontal distance from the equilibrium position to the ground is mathematically represented as
![d = v * t](https://tex.z-dn.net/?f=d%20%3D%20%20v%20%20%2A%20%20%20t)
=> ![d = 5.06 * 0.7959](https://tex.z-dn.net/?f=d%20%3D%205.06%20%20%2A%20%20%200.7959)
=> ![d = 4.0273 \ m](https://tex.z-dn.net/?f=d%20%3D%204.0273%20%5C%20%20m%20%20)
Answer:
-17.5 nC
Explanation:
charge A = -30 nC
charge B = -5 nC
After adding them it would be the average of the two charges because of the getting same voltage difference. so
c = (-30+(-5)) / 2 nC
c= -17.5 nC
answer is -17.5 nC
We make a graphic of this problem to define the angle.
The angle we can calculate through triangle relation, that is,
![sin\theta = \frac{c}{QP}\\sin\theta = \frac{c}{R}\\\theta=sin^{-1}\frac{c}{R}](https://tex.z-dn.net/?f=sin%5Ctheta%20%3D%20%5Cfrac%7Bc%7D%7BQP%7D%5C%5Csin%5Ctheta%20%3D%20%5Cfrac%7Bc%7D%7BR%7D%5C%5C%5Ctheta%3Dsin%5E%7B-1%7D%5Cfrac%7Bc%7D%7BR%7D)
With this function we should only calculate the derivate in function of c
![\frac{d\theta}{dc} = \frac{1}{\sqrt{1-\frac{c^2}{R^2}}}(\frac{c}{R})'\\\frac{d\theta}{dc} = \frac{1}{\sqrt{R^2-c^2}}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7B1-%5Cfrac%7Bc%5E2%7D%7BR%5E2%7D%7D%7D%28%5Cfrac%7Bc%7D%7BR%7D%29%27%5C%5C%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7BR%5E2-c%5E2%7D%7D)
That is the rate of change of
.
b) At this point we need only make a substitution of 0 for c in the equation previously found.
![\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{\sqrt{R^2-0}}\\\frac{d\theta}{dc}\big|_{c=0} = \frac{1}{R}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%5Cbig%7C_%7Bc%3D0%7D%20%3D%20%5Cfrac%7B1%7D%7B%5Csqrt%7BR%5E2-0%7D%7D%5C%5C%5Cfrac%7Bd%5Ctheta%7D%7Bdc%7D%5Cbig%7C_%7Bc%3D0%7D%20%3D%20%5Cfrac%7B1%7D%7BR%7D)
Hence we have finally the rate of change when c=0.
Answer:
Option d
The minimum angular separation between two objects that the Hubble Space Telescope can resolve is
.
Explanation:
The resulting image in a telescope that will be gotten from an object is a diffraction pattern instead of a perfect point (point spread function (PSF)).
That diffraction pattern is gotten because the light encounters different obstacles on its path inside the telescope (interacts with the walls and edges of the instrument).
The diffraction pattern is composed by a central disk, called Airy disk, and diffraction rings.
The angular resolution is defined as the minimal separation at which two sources can be resolved one for another, or in other words, when the distance between the two diffraction pattern maxima is greater than the radius of the Airy disk.
The angular resolution can be determined in analytical way by means of the Rayleigh criterion.
(1)
Where
is the wavelength and D is the diameter of the telescope.
Notice that it is necessary to express the wavelength in the same units than the diameter.
⇒
Finally, equation 1 can be used.
Hence, the minimum angular separation between two objects that the Hubble Space Telescope can resolve is
.