Answer:
ANSWER
A = 5 cm = 0.05 m
T = 0.2 s
ω=2π/T=2π/0.2=10πrad/s
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Answer:
[tex]here \: amplitude = 5cm (convert \: to \: si \: unit) = .05m \\ t = .2sec \\ omega = \frac{2\pi}{t} \\ = \frac{2\pi}{.2 } = 10\pi \frac{rad}{s} \\ we \: want \: find \: a \: and \: v \\ we \: know \: that \: a = - {omega}^{2} x \\ v = omega \sqrt{ {r}^{2} - {x}^{2} } \\(1)x = .05m \\ a = - {10\pi}^{2} \times .05 = - 5 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.05}^{2} } = 0 \\ 2)x = 3cm = .03m \\ a = {(10\pi)}^{2} \times .03 = - 3 {\pi}^{2} \frac{m}{ {s}^{2} } \\ v = 10\pi \sqrt{ {.05}^{2} - {.03}^{2} } = 10\pi \times .04 = .4\pi \frac{m}{s} \\ 3)x = 0 \\ a = - {(10\pi)}^{2} \times 0 = 0 \\ v = 10\pi \times \sqrt{ {.05}^{2} - {0}^{2} } = - 10\pi \times .05 = .5\pi \frac{m}{s} \\ thank \: you[/tex]
An object is moving in a negative direction with a negative acceleration, for example an object dropped out a window. What does it mean that both velocity and acceleration are negative?
Answer:
yes bc they are falling :]
Explanation:
Scientists might use a diagram to model the water cycle. What are two
benefits of this model?
Answer:
B, and D
Explanation:
The two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.
What is water cycle?The water cycle is defined as a cycle of events that involves precipitation as rain and snow, drainage in streams and rivers, and return to the atmosphere by evaporation and transpiration. Water moves through this cycle between the earth's oceans, atmosphere, and land.
It can also be defined as the route all water takes as it travels around Earth in various conditions.
There are basically six stages of water cycle.
EvaporationSublimationCondensationPrecipitationInfiltrationRunoffThus, the two benefits of this model is it specify a process that is very complex. It can represent changes that occur very slowly. Thus option B and D is correct.
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How do you play Simon says?
what is the difference between each distance traveled and displacement travled
Displacement is a vector magnitude that depends on the position of the body which is individualistic for the trajectory.
While, Distance is a scalar magnitude that measures over the trajectory.
What is correct regarding an ideal isotropic antenna? a. An ideal isotropic antenna is a highly efficient antenna used extensively in today’s communication systems b. An ideal isotropic antenna is a specialized antenna used to direct EM signal energy towards a specific direction c. An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation d. None of the above are correct statements
Answer:
C An ideal isotropic antenna is a theoretical antenna that does not exist in practice, but is useful in explaining power density and unguided EM signal attenuation
Explanation:
Because practically speaking there is no ideal isotropic antenna it is only imaginary radiating in all directions and use as an arbitrary point for antenna gain
Which lists the elements in order from most conductive to least conductive?
potassium (K), selenium (Se), germanium (Ge)
germanium (Ge), potassium (K), selenium (Se)
selenium (Se), germanium (Ge), potassium (K)
potassium (K), germanium (Ge), selenium (Se)
Answer:
Answer: potassium (K) germanium (Ge) selenium (Se)
Explanation:
I just took the test, the farther to the right on the periodic table you go the less conductive it gets.
Answer:
Answer: potassium (K) germanium (Ge) selenium (Se)
Explanation:
The most conductive to least conductive is from left to right
a car moves 8 m in 4 s at a comstant velocity. what is the cars acceleration
Answer:
Zero.
Explanation:
Acceleration is defined as change in velocity per unit time. For constant, the acceleration is zero.
[tex]\hat{a} =\frac{ \delta V}{t}[/tex]as [tex]\delta[/tex][tex]v\\[/tex] is zero.
Force is the amount _____ or _____ on an object
Motion is the action of _____ from one place to another place.
Answer:
force is the amount of work or pressure given to an object
motion is the action of moving one place to another place
Convert the following to SI units:_______. a. 9.12μs9.12μs b. 3.42 km c. 44 cm/ms d. 80 km/h
Answer:
Explanation:
A. We know that 1microsecond= 10^-6s
So = 9.12*10^-6=9.12*10^-6s
B 1km= 1000m
So 3.42km= 3240m
C.1ms= 10^-3s
And 1cm= 10^-2m
So 44*10^-2m/10^-3s=440m/s
D. 80km/hr= 80*1000/3600= 22.2m/s
Because 1km= 1000m
1hr= 3600s
What is the latitude of the vertical (direct) rays of the sun?
Answer:
23.5 degree North
Explanation:
The latitude of the vertical (direct) rays of the sun is 23.5 degree North. This is however due to the sun's vertical rays being located directly above 23.5 degree South which is the position of the Tropic of Capricorn.
The sun ray is highest at this pap tion and it occurs during Summer solstice which happens between June 21 / 22 of every year.
Light travels at 3 × 108 m/s, and it takes about 8 min for light from the sun to travel to Earth. Based on this, the order of magnitude of the distance from the sun to Earth is g
Answer:
The order of magnitude of the distance from the sun to Earth is 10⁸ km.
Explanation:
The order of magnitude of the distance from the sun to Earth can be calculated as follows:
[tex] c = \frac{x}{t} [/tex]
Where:
c: is the speed of light = 3x10⁸ m/s
t: is the time = 8 min
Hence, the distance is:
[tex] x = c*t = 3 \cdot 10^{8} m/s*8 min*\frac{60 s}{1 min} = 1.44 \cdot 10^{11} m = 1.44 \cdot 10^{8} km [/tex]
Therefore, the order of magnitude of the distance from the sun to Earth is 10⁸ km.
I hope it helps you!
Give three examples of unbalanced forces in your everyday life. HELP FAST PLZ
Answer:
1. Kicking a soccer ball
2. Playing tug of war
3. Bouncing a Ball
Explanation:
Why si unit develop all over the world?
Answer:
SI unit is used in most places around the world, so our use of it allows scientists from disparate regions to use a single standard in communicating scientific data without vocabulary confusion.
Explanation:
hope it helps
Si unit is used almost all around the world, so our use of it allows
Scientist from disparate regions to use a single standered in communicating scientific data without vocabulary confusion
The position of a particle is given by the function x = \left(2t^3 - 6t^2 + 12\right) m, where t is in s. Question:At what time is the acceleration zero? (with working out)
Answer:
t = 1secExplanation:
Given the position of a particle expressed by the equation x = (2t^3 - 6t^2 + 12)m, where t is in seconds, the acceleration function can be gotten by taking the second derivative of the function with respect to t as shown;
a = d/dt(dx/dt)
First let us get dx/dt
dx/dt = 3(2)t³⁻¹-2(6)t²⁻¹+0
dx/dt = 6t²-12t
a = d/dt(dx/dt)
a = d/dx(6t²-12t)
a = 2(6)t²⁻¹-12t¹⁻¹
a = 12t - 12t⁰
a = 12t-12
If the acceleration is zero, then;
12t-12 = 0
add 12 to both sides
12t-12+12 = 0+12
12t = 12
t = 12/12
t = 1sec
Hence the time when acceleration is zero is 1sec
Part A. 13 in Express your answer to two significant figures and include the appropriate units. SubmitPrev Previous Answers Request Answer X Incorrect; Try Again; 5 attempts remaining Part B. 77 ft/s Express your answer to two significant figures and include the appropriate units ,, ? Value m/s
Answer:
Part A
[tex]x = 0.33 \ m[/tex]
Part B
[tex]y = 23.49 \ m/s[/tex]
Explanation:
From the question we are told to convert
13 inches to meters
Now
1 in = 0.0254 \ m
13 \ in = x m
=> [tex]x = \frac{13 * 0.0254 }{1}[/tex]
=> [tex]x = 0.33 \ m[/tex]
For part B we are told to covert 77ft/s to meters /srconds
So
1 ft/s = 0.305 \ m/s
77 ft/s = y
=> [tex]y = \frac{77 * 0.305 }{1}[/tex]
=> [tex]y = 23.49 \ m/s[/tex]
26.0 g of copper pellets are removed from a 300∘C oven and immediately dropped into 120 mL of water at 21.0 ∘C in an insulated cup.What will the new water temperature be?
Answer:
The new water temperature is 26.4 °C
Explanation:
Given;
mass of copper, [tex]M_{cu}[/tex] = 26 g = 0.026 kg
temperature of copper, t = 300 °C
volume of water, V = 120 mL = 0.12 L
temperature of water, t = 21 °C
density of water, ρ = 1 kg/L
mass of water = density x volume
mass of water = (1 kg/L) x 0.12 L = 0.12 kg
heat lost by copper = heat gained by water
Both copper and water reach final temperature, T
Heat gained by water, [tex]Q_w[/tex] = [tex]m_w[/tex]cΔθ = [tex]m_w C(T - t)[/tex]
[tex]Q_w = m_w C(T - t)\\\\Q_w = 0.12*4200(T-21)\\\\Q_w = 504(T-21)[/tex]
Heat lost by copper is given by;
[tex]Q_{cu} = m_{cu}C(300-T)\\\\Q_{cu} = 0.026*385(300-T)\\\\Q_{cu} = 10.01(300 - T)[/tex]
[tex]Q_{cu} = Q_w[/tex]
504(T- 21) = 10.01(300 - T)
504 T - 10584 = 3003 - 10.01 T
504 T + 10.01 T= 3003 + 10584
514.01 T = 13587
T = (13587) / 514.01
T = 26.4 °C
Therefore, the new water temperature is 26.4 °C
The new water temperature using the given parameters is;
26.49°C
We are given;
Mass of copper; m_cu = 26 g = 0.026 kg
Temperature; T_cu = 300 °C
Volume of water; V = 120 mL = 0.12 L
Temperature of water, T_w = 21 °C
Density of water; ρ = 1 kg/L
Let us find the mass of water from the formula;
m_w = ρ × V
m_w = 1 × 0.12
m_w = 0.12 kg
Now, from the principle of conservation of energy, we can say that the total heat lost by a hot body is equal to the total heat gained by a cold body.
The hot body here is copper while the cold body is water. Thus;
Heat lost by copper = Heat gained by water
Formula for heat lost by copper is;
Q_cu = m_cu * c_cu * (T_f - T_cu)
Formula for heat gained by water is;
Q_w = m_w * c_w * (T_f - T_w)
Where;
T_f is final temperature reached by copper and water
c_cu is specific heat capacity of copper = 385 J/kg.°C
c_w is specific heat capacity of water = 4200 J/kg.°C
Thus;
Q_cu = 0.026 × 385 × (300 - T_f)
Q_cu = 3030 - 10.01T_f
similarly;
Q_w = 0.12 × 4200 × (T_f - 21)
Q_w = 504T_f - 10584
Thus;
3030 - 10.01T_f = 504T_f - 10584
3030 + 10584 = 504T_f + 10.01T_f
13614 = 514.01T_f
T_f = 13614/514.01
T_f = 26.49°C
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How do scientists use the evidence they gather?
Answer:
When conducting research, scientists use the scientific method to collect measurable, empirical evidence in an experiment related to a hypothesis (often in the form of an if/then statement), the results aiming to support or contradict a theory.
A ball easily moves but not a bus when we push them. why?
Answer: The amount of matter is higher in a bus than a ball.
Explanation:
Place gamma rays, infrared, microwaves, radio waves, ultraviolet, visible light, and x-rays in order from largest wavelength to smallest wavelength.
Answer:
Going by EM SPECTRUM WE HAVE
radio waves, microwaves, infrared, VISIBLE LIGHT, ultraviolet, X-rays, GAMMA RAYS
Explanation:
BECAUSE
V= WAVELENGTH/ FREQUENCY
AS FREQUENCY INCREASES WAVELENGTH DECREASE AN VICE VERSA
The deck of a bridge is suspended 235 feet above a river. If a pebble falls off the side of the bridge, the height, in feet, of the pebble above the water surface after t seconds is given by y = 235 − 16t^2.
Required:
a. Find the average velocity (in ft/s) of the pebble for the time period beginning when t = 2 and lasting the following amount of time.
1. 0.1 sec:________
2. 0.05 sec:_______
3. 0.01 sec:_______
b. Estimate the instantaneous velocity (in ft/s) of the pebble after 3 seconds. ft/s.
Answer:
(a) 1, average velocity = -65.6 m/s
2, average velocity = -64.8 m/s
3, average velocity = -64.16 m/s
(b) The instantaneous velocity is -96 m/s
Explanation:
(a)
Average velocity is given by;
[tex]y(t_2,t_1) = \frac{y(t_2) - y(t_1)}{t_2-t_1}[/tex]
(1)
[tex]y(2.1,2) = \frac{(235-16*2.1^2) - (235-16*2^2)}{2.1-2}\\\\ y(2.1,2) = -65.6 \ m/s[/tex]
(2)
[tex]y(2.05,2) = \frac{(235-16*2.05^2) - (235-16*2^2)}{2.05-2}\\\\ y(2.05,2) = -64.8 \ m/s[/tex]
(3)
[tex]y(2.01,2) = \frac{(235-16*2.01^2) - (235-16*2^2)}{2.01-2}\\\\ y(2.01,2) = -64.16 \ m/s[/tex]
b. y = 235 - 16t²
The instantaneous velocity is given by;
v = dy /dt
dy / dt = -32t
when t = 3 s
v = -32(3)
v = -96 m/s
(a)The average velocity for the 0.1 sec,0.05 sec,0.01 sec will be -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.
What is the average velocity?The total displacement traveled by an object divided by the total time taken is the average velocity.
Average velocity is given by;
[tex]\rm y(t_2,t_1)=\frac{y(t_2)-y(t_1)}{t_2-t_1} \\\\[/tex]
The average velocity for case 1;
[tex]\rm y(2.1,2)=\frac{(235-16\times 2\times 1^2)-(235-16 \times 2^2)}{2.1-2} \\\\ \rm y(2.1,2)=\-65.6 \ m/sec[/tex]
The average velocity for case 2;
[tex]\rm y(2.015,2)=\frac{(235-16\times 2.05^2)-(235-16 \times 2^2)}{2.05-2} \\\\ \rm y(2.1,2)=\-64.8 \ m/sec[/tex]
The average velocity for case 3;
[tex]\rm y(2.01,2)=\frac{(235-16\times 2.01^2)-(235-16 \times 2^2)}{2.01-2} \\\\ \rm y(2.1,2)=\-64.16 \ m/sec[/tex]
Hence the average velocity for the 0.1 sec,0.05 sec,0.01 sec will be -65.6 m/s, -64.8 m/s and -64.16 m/s respectively.
(b) The instantaneous velocity will be -96 m/s.
The given equation in the problem is;
[tex]\rm y = 235 - 16t^2[/tex]
The instantaneous velocity is given as;
[tex]v = \frac{dy}{dt} \\\\ \frac{dy}{dt} = -32t\\\\ t = 3 s\\\\ v = -32\times 3\\\\ v = -96 m/s[/tex]
Hence the instantaneous velocity will be -96 m/s.
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Circle all true statements:
a) Water at 90°C is warmer than water at 202°F
b) 40K corresponds to -40°C
c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.
d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.
e) 0°F corresponds to -32°C.
Answer:
The true statements are:
c) Temperature differing by 25 on the Fahrenheit scale must differ by 45 on the Celsius scale.
d) Temperatures which differ by 10 on the Celsius scale must differ by 18 on the Fahrenheit scale.
Explanation:
option a is false, 90°C is equal to 194°F which is not warmer than 202°F
option b is false, 40K is equal to -233°C
option e is false 0°F corresponds to -17.8°C.
The acceleration of a piston is given by : rw^2 cos(wt-pi/4) find amplitude.
Given :
The acceleration of a piston is given by :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
To Find :
The amplitude of the acceleration of piston .
Solution :
Now , acceleration is :
[tex]a=r\omega^2cos\ (\omega t-\dfrac{\pi}{4})[/tex]
Now , amplitude is maximum of any function .
So , maximum value of [tex]cos\ \theta[/tex] is equal to 1 .
Therefore , amplitude of the acceleration of piston is :
[tex]A=r\omega^2[/tex]
Hence , this is the required solution .
Determine the slope of this graph from zero seconds to five seconds.
Explanation:
(m2-m1)/t
25-0/5
25/5
5m/s
Given that the lines are parallel, what would be the value of angle "a" in the diagram found below?
Explanation:
Hey there!!
a = 40° { corresponding angles on a parallel lines are equal}.
As A and B are parallel, a and 40° are corresponding angles.
Hope it helps...
The value of angle "a" will be 40°
What is corresponding angle ?The angles which are on the same side of one of two lines cut by a transversal and are on the same side of the transversal is called corresponding angles.
since , angle a and angle B = 40° (given) are on the same relative position at the intersection where a straight line crosses two others and since both the lines are parallel to each other hence ,both the angles are said to be corresponding angles .This implies both the angles must be equal .
angle a = 40°
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.A cart rolling down an incline for 5.0 seconds has an acceleration of 1.6 m/s2. If the cart has a beginning speed of 2.0 m/s, and its final velocity of 10 m/s, what was the car's displacement?
Use the formula,
[tex]\Delta x=v_it+\dfrac12at^2[/tex]
where [tex]\Delta x[/tex] is the cart's displacement (from the origin), [tex]v_i[/tex] is its initial speed, [tex]a[/tex] is its acceleration, and [tex]t[/tex] is time.
[tex]\Delta x=\left(2.0\dfrac{\rm m}{\rm s}\right)(5.0\,\mathrm s)+\dfrac12\left(1.6\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)^2[/tex]
[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]
Alternatively, since acceleration is constant, we have
[tex]\dfrac{v_f+v_i}2=\dfrac{\Delta x}t[/tex]
That is, we have these two equivalent expressions for average velocity, where [tex]v_f[/tex] is the cart's final velocity. Solve for [tex]\Delta x[/tex]:
[tex]\dfrac{10\frac{\rm m}{\rm s}+2.0\frac{\rm m}{\rm s}}2=\dfrac{\Delta x}{5.0\,\mathrm s}[/tex]
[tex]\implies\boxed{\Delta x=30\,\mathrm m}[/tex]
An isolated system contains two objects with charges q1q1 and q2q2. If the charge on object 1 is doubled, what is the charge on object 2? g
Answer:
The new charge on the second object is q₂ - q₁
Explanation:
Given;
charge on the first object, q₁
charge on the second object, q₂
Since there is no external force on isolated system, the total charge in the system is given by;
Q = q₁ + q₂
q₂ = Q - q₁
When the charge on object 1 is doubled, q₁' = 2q₁
let q₂' be the new charge on object 2
q₂' = Q - q₁'
q₂' = (q₁ + q₂) - 2q₁
q₂' = q₁ + q₂ - 2q₁
q₂' = q₂ - q₁
Therefore, the new charge on the second object is q₂ - q₁
The charge on object 2 will be "q₂ - q₁".
Charge on object:The charge's quantity on such an item represents the degree of imbalance between its electrons as well as protons.
Just to calculate the overall charge of a positively (+) charged item, divide the total no. of valence electrons by the overall number of protons.
According to the question,
Object 1's charge = q₁
Object 2's charge = q₂
Let,
New charge of object 2 be "q₂'".
The total charge in the system will be:
Q = q₁ + q₂
or,
q₂ = Q - q₁
hence,
The charge on object 2 be:
→ q₂' = Q - q₁'
By substituting the values,
= (q₁ + q₂) - 2q₁
= q₁ + q₂ - 2q₁
= q₂ - q₁
Thus the answer above is appropriate.
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μN/(kg⋅ns) in the correct SI of:________
Answer:
μN/ (kg ns) = 10³ N / (kg s)
Explanation:
In this exercise they ask us if the notation is correct. Let's write the different terms in the SI systems
force is N
time is in seconds
the unit given for the force is 1 N = 10⁶ μN
the unit of time is 1 s = 10⁹ ns
the correct way to give the answer should be: N / (kg s)
so the notation should be changed
μN /kg ns = μN / (kg ns) (1N / 10⁶ μN) (10⁹ ns / 1 s) =
μN/ (kg ns) = 10³ N / (kg s)
Can air make shadows
Answer:no
Explanation:the way to see air is by steaming something sometimes you might not see the air but you can see the shadow
If a cheetah could maintain its top speed of 120 km/h for 20 minutes, how far would it run?
Two positive charges q1 = q2 = 2.0 μC are located at x = 0, y = 0.30 m and x = 0, y = -0.30 m, respectively. Third point charge Q = 4.0 μC is located at x = 0.40 m, y = 0.What is the net force ((a)magnitude and (b)direction) on charge q1 exerted by the other two charges?
Answer:
F = 0.111015 N
Explanation:
For this exercise the force is given by Coulomb's law
F = k q₁q₂ / r₂₁²
we calculate the electric force of the other two particles on the charge q1
Charges q₁ and q₂
the distance between them is
r₁₂ = y₁ -y₂
r₁₂ = 0.30 + 0.30
r₁₂ = 0.60 m
let's calculate
F₁₂ = 9 10⁹ 2 10⁻⁶ 2 10⁻⁶ / 0.60 2
F₁₂ = 1 10⁻¹ N
directed towards the positive side of the y-axis
Charges 1 and 3
Let's find the distance using the Pythagorean Theorem
r₁₃ = RA [(0.40-0) 2 + (0-0.30) 2]
r₁₃ = 0.50 m
F₁₃ = 9 10⁹ 2 10⁻⁶ 4 10⁻⁶ / 0.50²
F₁₃ = 1.697 10⁻² N
The direction of this force is on the line that joins the two charges (1 and 3), let's use trigonometry to find the components of this force
tan θ = y / x
θ = tan⁻¹ y / x
θ = tan⁻¹ 0.3 / 0.4
tea = 36.87º
The angle from the positive side of the x-axis is
θ ’= 180 - θ
θ ’= 180 - 36.87
θ ’= 143.13º
sin143.13 = F_13y / F₁₃
F_13y = F₁₃ sin 143.13
F{13y} = 1.697 10⁻² sin 143.13
F_13y = 1.0183 10⁻² N
cos 143.13 = F_13x / F₁₃
F₁₃ₓ = F₁₃ cos 143.13
F₁₃ₓ = 1.697 10⁻² cos 143.13
F₁₃ₓ = -1.357 10-2 N
Now we can find the components of the resultant force
Fx = F13x + F12x
Fx = -1,357 10-2 +0
Fx = -1.357 10-2 N
Fy = F13y + F12y
Fy = 1.0183 10-2 + 1 10-1
Fy = 0.110183 N
We use the Pythagorean theorem to find the modulus
F = Ra (Fx2 + Fy2)
F = RA [(1.357 10-2) 2 + 0.110183 2]
F = 0.111015 N
Let's use trigonometry for the angles
tan tea = Fy / Fx
tea = tan-1 (0.110183 / -0.01357)
tea = 1,448 rad
to find the angle about the positive side of the + x axis
tea '= pi - 1,448
Tea = 1.6936 rad
The magnitude of the net force on q1 exerted by the other two charges is 0.357 N.
The direction of the net force on q1 is 50⁰.
The given parameters;
q1 = 2.0 μC located at (0, 0.3) mq2 = 2.0 μC located at (0, -0.3) mq3 = 4.0 μC located at (0.4, 0) mThe force on q1 due to q2 occurs only in y-direction and can be calculated using Coulomb's law as shown below;
[tex]F_1_2 = \frac{kq^2}{r^2}j = \frac{(9\times 10^9) \times (2\times 10^{-6})^2 }{(0.3 +0.3)^2} j \\\\\F_{12} = (0.1)j[/tex]
The force on q1 due to q3 occurs both in x-direction and y-direction, and it is calculated as follows;
[tex]distance \ between \ q1 \ and \ q3\ , r_{13} = \sqrt{0.3^2 \ + \ 0.4^2} = 0.5 \ m\\\\F_{13} = \frac{kq^2}{r_{13}^2} (\frac{0.4i}{0.5} \ + \ \frac{0.3j}{0.5} )\\\\F_{13} = \frac{kq^2}{r_{13}^2} (0.8i + 0.6j)\\\\F_{13} = \frac{9\times 10^9 \times (2\times 10^{-6}) \times (4\times 10^{-6})}{0.5^2} (0.8i + 0.6j)\\\\F_{13} = 0.288(0.8i + 0.6j)\\\\F_{13} = 0.23i + 0.173j[/tex]
The net force is calculated as follows;
[tex]F_{net} = F_{12} \ + \ F_{13}\\\\F_{net} = (0.1j) \ + \ (0.23i + 0.173j)\\\\F_{net} = (0.23i + 0.273j)[/tex]
The magnitude of the net force on q1 is calculated as follows;
[tex]|F| = \sqrt{(0.23^2) \ + \ (0.273^2)} \\\\|F| = 0.357 \ N[/tex]
The direction of the net force on q1 is calculated as follows;
[tex]tan(\theta )= \frac{F_y}{F_x} \\\\\theta = tan^{-1} (\frac{0.273}{0.23} )\\\\\theta = 50^0[/tex]
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