Question

Constants $\alpha = 5.67 \times 10^{-8} \frac{W}{m^2 \cdot K^4}$ $g = 9.8 \frac{m}{s^2}$ $G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2}$ $m_E = 6.0 \times 10^{24} kg$ $r_E = 6.4 \times 10^6 m$ $m_M = 7.3 \times 10^{22} kg$ $r_M = 1.7 \times 10^6 m$ $m_{proton} = 1.67 \times 10^{-27} kg$ $m_{electron} = 9.11 \times 10^{-31} kg$ $q_{electron} = -1.6 \times 10^{-19} C$ $q_{proton} = 1.6 \times 10^{-19} C$ $k = 9.0 \times 10^9 \frac{Nm^2}{C^2}$ $c = 3 \times 10^8 \frac{m}{s}$ $h = 6.6 \times 10^{-34} Js$ $h = 4.14 \times 10^{-15} eV \cdot s$ $f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$ $f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$ $f = \frac{1}{2\pi} \sqrt{\frac{1}{LC}}$ $H_0 = 4\pi \times 10^{-7} \frac{Tm}{A}$ $H_0 = 2 \times 10^{-7} \frac{Tm}{A}$ $\epsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{Nm^2}$ $C_{water} = 4186 \frac{J}{kg \cdot K}$ $C_{ice} = 2090 \frac{J}{kg \cdot K}$ $R = 8.31 \frac{J}{mol \cdot K}$ $R = 1.38 \times 10^{-23} \frac{J}{K}$ $R = 1.097 \times 10^7 m^{-1}$ $1 cm^3 = 1 \times 10^{-6} m^3$ $1 L = 1 \times 10^{-3} m^3$ rounding it. charges point away from each other, while charges. 2. Use the vector addition (tip-to-tail) method by placing the tip of the previous vec- tors at the tail of the next one to find the direction of the resultant force. 3. Set up a coordinate system (x-y plane). 4. If any forces are not acting along the coordinate system, break down the compo- nents of these forces along the axes of the coordinate system. 5. Based on the direction of the components of forces, add or subtract the components to find the resultant component along the x- and y-axis ($F_x$ and $F_y$). 6. Use Pythagorean theorem to find the magnitude of the resultant force $F = \sqrt{(F_x^2 + F_y^2)}$. 2. Mathematically calculate the magnitude of Coulomb's force acting on charge $Q$ for the following case using the superposition principle. The magnitude of each charge $q$ is $3 \mu C$, the magnitude of charge $Q$ is $2 \mu C$, and the sides of the square are 4 cm. Hint: Draw in force vectors that are acting between each $Q$ and $q$ set of charges. Recall that to calculate a resultant force, you must break them into x- and y-components. A coordi- nate system has been suggested for ease of calculation. Remember, look for symmetry. One of these systems has no force along the y-axis on charge $Q$. For the lab you will only attempt part a. Part b is only for self-practice and will not count towards your grade. $F = F_1 + F_2 + F_3$ a) (Image of a square with charges at corners and center, showing force vectors $F_1, F_2, F_3$ on the center charge $Q$) b) (Image of a square with charges at corners and center, showing x-axis)

Name: constants alpha 567 times 10 8 fracwm2 cdot k4 g 98 fracms2 g 667 times 10 11 fracnm2kg2 me 60 times 1024 kg re 64 times 106 m mm 73 times 1022 kg rm 17 times 106 m mproton 167 times 10 38909
Uploaded: 2023-04-07T00:44:53-08:00
Duration: 4 min 59 s
Channel: Madhur L
Description: constants alpha 567 times 10 8 fracwm2 cdot k4 g 98 fracms2 g 667 times 10 11 fracnm2kg2 me 60 times 1024 kg re 64 times 106 m mm 73 times 1022 kg rm 17 times 106 m mproton 167 times 10 38909

          Constants
$\alpha = 5.67 \times 10^{-8} \frac{W}{m^2 \cdot K^4}$
$g = 9.8 \frac{m}{s^2}$
$G = 6.67 \times 10^{-11} \frac{Nm^2}{kg^2}$
$m_E = 6.0 \times 10^{24} kg$
$r_E = 6.4 \times 10^6 m$
$m_M = 7.3 \times 10^{22} kg$
$r_M = 1.7 \times 10^6 m$
$m_{proton} = 1.67 \times 10^{-27} kg$
$m_{electron} = 9.11 \times 10^{-31} kg$
$q_{electron} = -1.6 \times 10^{-19} C$
$q_{proton} = 1.6 \times 10^{-19} C$
$k = 9.0 \times 10^9 \frac{Nm^2}{C^2}$
$c = 3 \times 10^8 \frac{m}{s}$
$h = 6.6 \times 10^{-34} Js$
$h = 4.14 \times 10^{-15} eV \cdot s$
$f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}$
$f = \frac{1}{2\pi} \sqrt{\frac{g}{L}}$
$f = \frac{1}{2\pi} \sqrt{\frac{1}{LC}}$
$H_0 = 4\pi \times 10^{-7} \frac{Tm}{A}$
$H_0 = 2 \times 10^{-7} \frac{Tm}{A}$
$\epsilon_0 = 8.85 \times 10^{-12} \frac{C^2}{Nm^2}$
$C_{water} = 4186 \frac{J}{kg \cdot K}$
$C_{ice} = 2090 \frac{J}{kg \cdot K}$
$R = 8.31 \frac{J}{mol \cdot K}$
$R = 1.38 \times 10^{-23} \frac{J}{K}$
$R = 1.097 \times 10^7 m^{-1}$
$1 cm^3 = 1 \times 10^{-6} m^3$
$1 L = 1 \times 10^{-3} m^3$

rounding it.
charges point away from each other, while
charges.

2. Use the vector addition (tip-to-tail) method by placing the tip of the previous vec-
tors at the tail of the next one to find the direction of the resultant force.

3. Set up a coordinate system (x-y plane).

4. If any forces are not acting along the coordinate system, break down the compo-
nents of these forces along the axes of the coordinate system.

5. Based on the direction of the components of forces, add or subtract the components
to find the resultant component along the x- and y-axis ($F_x$ and $F_y$).

6. Use Pythagorean theorem to find the magnitude of the resultant force
$F = \sqrt{(F_x^2 + F_y^2)}$.

2. Mathematically calculate the magnitude of Coulomb's force acting on charge $Q$ for the
following case using the superposition principle. The magnitude of each charge $q$ is
$3 \mu C$, the magnitude of charge $Q$ is $2 \mu C$, and the sides of the square are 4 cm. Hint:
Draw in force vectors that are acting between each $Q$ and $q$ set of charges. Recall that to
calculate a resultant force, you must break them into x- and y-components. A coordi-
nate system has been suggested for ease of calculation. Remember, look for symmetry.
One of these systems has no force along the y-axis on charge $Q$. For the lab you will
only attempt part a. Part b is only for self-practice and will not count towards
your grade.
$F = F_1 + F_2 + F_3$

a)
(Image of a square with charges at corners and center, showing force vectors $F_1, F_2, F_3$ on the center charge $Q$)

b)
(Image of a square with charges at corners and center, showing x-axis)

$constants alpha 567 times 10 8 fracwm2 cdot k4 g 98 fracms2 g 667 times 10 11 fracnm2kg2 me 60 times 1024 kg re 64 times 106 m mm 73 times 1022 kg rm 17 times 106 m mproton 167 times 10 38909$

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