Remember Coulomb's law: the magnitude of the electric force F between two stationary charges q₁ and q₂ over a distance r is
![F = \dfrac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cdfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
where k ≈ 8,98 × 10⁹ kg•m³/(s²•C²) is Coulomb's constant.
8.1. The diagram is simple, since only two forces are involved. The particle at Q₂ feels a force to the left due to the particle at Q₁ and a downward force due to the particle at Q₃.
8.2. First convert everything to base SI units:
0,02 µC = 0,02 × 10⁻⁶ C = 2 × 10⁻⁸ C
0,03 µC = 3 × 10⁻⁸ C
0,04 µC = 4 × 10⁻⁸ C
300 mm = 300 × 10⁻³ m = 0,3 m
600 mm = 0,6 m
Force due to Q₁ :
![F_{Q_2/Q_1} = \dfrac{k (6 \times 10^{-16} \,\mathrm C)}{(0,3 \, \mathrm m)^2} \approx \boxed{6,0 \times 10^{-5} \,\mathrm N} = 0,06 \,\mathrm{mN}](https://tex.z-dn.net/?f=F_%7BQ_2%2FQ_1%7D%20%3D%20%5Cdfrac%7Bk%20%286%20%5Ctimes%2010%5E%7B-16%7D%20%5C%2C%5Cmathrm%20C%29%7D%7B%280%2C3%20%5C%2C%20%5Cmathrm%20m%29%5E2%7D%20%5Capprox%20%5Cboxed%7B6%2C0%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%5Cmathrm%20N%7D%20%3D%200%2C06%20%5C%2C%5Cmathrm%7BmN%7D)
Force due to Q₃ :
![F_{Q_2/Q_3} = \dfrac{k (12 \times 10^{-16} \,\mathrm C)}{(0,6 \, \mathrm m)^2} \approx \boxed{3,0 \times 10^{-5} \,\mathrm N} = 0,03 \,\mathrm{mN}](https://tex.z-dn.net/?f=F_%7BQ_2%2FQ_3%7D%20%3D%20%5Cdfrac%7Bk%20%2812%20%5Ctimes%2010%5E%7B-16%7D%20%5C%2C%5Cmathrm%20C%29%7D%7B%280%2C6%20%5C%2C%20%5Cmathrm%20m%29%5E2%7D%20%5Capprox%20%5Cboxed%7B3%2C0%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%5Cmathrm%20N%7D%20%3D%200%2C03%20%5C%2C%5Cmathrm%7BmN%7D)
8.3. The net force on the particle at Q₂ is the vector
![\vec F = F_{Q_2/Q_1} \, \vec\imath + F_{Q_2/Q_3} \,\vec\jmath = \left(-0,06\,\vec\imath - 0,03\,\vec\jmath\right) \,\mathrm{mN}](https://tex.z-dn.net/?f=%5Cvec%20F%20%3D%20F_%7BQ_2%2FQ_1%7D%20%5C%2C%20%5Cvec%5Cimath%20%2B%20F_%7BQ_2%2FQ_3%7D%20%5C%2C%5Cvec%5Cjmath%20%3D%20%5Cleft%28-0%2C06%5C%2C%5Cvec%5Cimath%20-%200%2C03%5C%2C%5Cvec%5Cjmath%5Cright%29%20%5C%2C%5Cmathrm%7BmN%7D)
Its magnitude is
![\|\vec F\| = \sqrt{\left(-0,06\,\mathrm{mN}\right)^2 + \left(-0,03\,\mathrm{mN}\right)^2} \approx 0,07 \,\mathrm{mN} = \boxed{7,0 \times 10^{-5} \,\mathrm N}](https://tex.z-dn.net/?f=%5C%7C%5Cvec%20F%5C%7C%20%3D%20%5Csqrt%7B%5Cleft%28-0%2C06%5C%2C%5Cmathrm%7BmN%7D%5Cright%29%5E2%20%2B%20%5Cleft%28-0%2C03%5C%2C%5Cmathrm%7BmN%7D%5Cright%29%5E2%7D%20%5Capprox%200%2C07%20%5C%2C%5Cmathrm%7BmN%7D%20%3D%20%5Cboxed%7B7%2C0%20%5Ctimes%2010%5E%7B-5%7D%20%5C%2C%5Cmathrm%20N%7D)
and makes an angle θ with the positive horizontal axis (pointing to the right) such that
![\tan(\theta) = \dfrac{-0,03}{-0,06} \implies \theta = \tan^{-1}\left(\dfrac12\right) - 180^\circ \approx \boxed{-153^\circ}](https://tex.z-dn.net/?f=%5Ctan%28%5Ctheta%29%20%3D%20%5Cdfrac%7B-0%2C03%7D%7B-0%2C06%7D%20%5Cimplies%20%5Ctheta%20%3D%20%5Ctan%5E%7B-1%7D%5Cleft%28%5Cdfrac12%5Cright%29%20-%20180%5E%5Ccirc%20%5Capprox%20%5Cboxed%7B-153%5E%5Ccirc%7D)
where we subtract 180° because
terminates in the third quadrant, but the inverse tangent function can only return angles between -90° and 90°. We use the fact that tan(x) has a period of 180° to get the angle that ends in the right quadrant.
Answer:
Tarzan, who weighs 688N, swings from a cliff at the end of a convenient vine that is 18m long. From the top of the cliff to the bottom of the swing he descends by 3.2m.
Explanation:
Answer:
work output is always less than work input - the ratio is less than 1.
Explanation:
This principle comes from the fact that a machine or system cannot produce more work than is supplied to it, because this would violate the energy conservation law (work is a type of mechanical energy).
In theoretical machines called "ideal machines" the input work is the same as the output work, but these machines are only theoretical because in real applications there is always some type of energy loss, either in heat produced by a machine or processes for its operation, for this reason the output work is always less than the input work.
Regarding the ratio work output to work input:
![\frac{WO}{WI} < 1](https://tex.z-dn.net/?f=%5Cfrac%7BWO%7D%7BWI%7D%20%3C%201)
because work input WI is always greater than work output WO.
Answer:
![\omega_f=571.42\ rpm](https://tex.z-dn.net/?f=%5Comega_f%3D571.42%5C%20rpm)
Explanation:
It is given that,
Diameter of cylinder, d = 6.6 cm
Radius of cylinder, r = 3.3 cm = 0.033 m
Acceleration of the string, ![a=1.5\ m/s^2](https://tex.z-dn.net/?f=a%3D1.5%5C%20m%2Fs%5E2)
Displacement, d = 1.3 m
The angular acceleration is given by :
![\alpha =\dfrac{a}{r}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7Ba%7D%7Br%7D)
![\alpha =\dfrac{1.5}{0.033}](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B1.5%7D%7B0.033%7D)
![\alpha =45.46\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D45.46%5C%20rad%2Fs%5E2)
The angular displacement is given by :
![\theta=\dfrac{d}{r}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%7Bd%7D%7Br%7D)
![\theta=\dfrac{1.3}{0.033}](https://tex.z-dn.net/?f=%5Ctheta%3D%5Cdfrac%7B1.3%7D%7B0.033%7D)
![\theta=39.39\ rad](https://tex.z-dn.net/?f=%5Ctheta%3D39.39%5C%20rad)
Using the third equation of rotational kinematics as :
![\omega_f^2-\omega_i^2=2\alpha \theta](https://tex.z-dn.net/?f=%5Comega_f%5E2-%5Comega_i%5E2%3D2%5Calpha%20%5Ctheta)
Here, ![\omega_i=0](https://tex.z-dn.net/?f=%5Comega_i%3D0)
![\omega_f=\sqrt{2\alpha \theta}](https://tex.z-dn.net/?f=%5Comega_f%3D%5Csqrt%7B2%5Calpha%20%5Ctheta%7D)
![\omega_f=\sqrt{2\times 45.46\times 39.39}](https://tex.z-dn.net/?f=%5Comega_f%3D%5Csqrt%7B2%5Ctimes%2045.46%5Ctimes%2039.39%7D)
![\omega_f=59.84\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D59.84%5C%20rad%2Fs)
Since, 1 rad/s = 9.54 rpm
So,
![\omega_f=571.42\ rpm](https://tex.z-dn.net/?f=%5Comega_f%3D571.42%5C%20rpm)
So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.
The three ways a person can manipulate light
would be the following:,
filter, and the time the photograph is taken
<span>1.
</span>Angle
- <span>The </span>camera angle<span> <span>marks
the specific location at which the movie </span></span>camera<span> <span>or
video </span></span>camera<span> is
placed to take a shot.</span>
<span>2.
</span>Filter - Camera<span> <span>lens </span></span>filters<span> <span>still have many uses in digital photography,
and should be an important part of any photographer's </span></span>camera<span> bag.</span>
<span>3.
</span>Time
the photograph is taken - The golden hour, sometimes called the "magic
hour", is roughly the first hour of light after sunrise, and the last hour
of light before sunset, although the exact duration varies between seasons.
During these times the sun is low in the sky, producing a soft, diffused light
which is much more flattering than the harsh midday sun that so many of us are
used to shooting in.
I am hoping that these answers
have satisfied your queries and it will be able to help you in your endeavors, and
if you would like, feel free to ask another question.