A concrete bridge is built of 315-cm-long concrete slabs with an expansion joint between them. the slabs just touch on a 115∘f day, the hottest day for which the bridge is designed.
The expansion joints in a concrete bridge allow for thermal expansion, preventing buckling by accommodating changes in size due to temperature variation. This allows the bridge to withstand frequent temperature changes without significant structural damage.
Explanation:The subject of this question involves a phenomenon known as thermal expansion, which is a concept in physics. When the temperature of a material changes, the material can expand or contract. Concrete, like many other materials, expands when heated and contracts when cooled.
On a sunny day, especially a very hot one like 115 degrees Fahrenheit, the heat from the sun causes the concrete slabs in the bridge to expand. Thanks to the expansion joints between the slabs, the bridge can change length without buckling under this thermal stress. The expansion joints are designed to accommodate this precise thermal expansion, allowing the slabs to expand and contract freely with changes in temperature.
In the case of the Auckland Harbour Bridge in New Zealand, the bridge is equipped with thermal expansion joints that prevent it from buckling due to thermal expansion. Similarly, the change in length of the bridge due to temperature changes is usually small and distributed across multiple expansion joints, making the process almost unnoticeable in daily use.
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18. In a solid conductor such as a metal, what is the cause of an electric current? A. A flow of positive and negative charges. B. A flow of positive charges only. C. +A flow of negative charges only. D. A flow of protons. E. A flow of neutrons.
a chandelier weighing 100 newtons is suspended by two wires that make an angle of 60° with the ceiling. Find the tension in both wires.
58 newtons
87 newtons
116 newtons
231 newtons
A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. know that car a has a mass of 30 mg and car b has a mass of 20 mg, and the braking force is 40 kn on each car.
The Physics related question involves concepts of Newton's Third Law and momentum. When brakes are applied, a force is exerted against the car's motion. The impact differs based on the car's mass, and its initial momentum which is the product of its mass and velocity.
Explanation:The subject of the question is related to Physics, specifically focusing on the topic of Newton's laws of motion and momentum. To understand the scenario, we should first comprehend that when the brakes are applied to the cars, a force is exerted which slows them down. This relates to Newton's Third Law - for every action, there's an equal and opposite reaction.
Considering the given scenario, when the brakes apply a force of 40 kN on each car, this force works against their forward momentum, slowing them down. The impact of the force will differ based on the mass of each car, with the lighter car likely slowing down faster than the heavier one. This can be calculated if we know the exact measure of force and the braking distance.
Additionally, the momentum of the cars before the brakes are applied could also be calculated using the formula momentum = mass x velocity, which in this case would have different results for each car due its mass difference, despite identical velocities. Overall, the problem illustrates how both the Newton's third law and principles of momentum come into play in everyday physics scenarios like that of braking cars in motion.
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which of the following statements applies to current?
a. it is the flow of electric charge
b. it is measured in ohms
c. it is an excess accumulation of charge
d. it is made of neutrons
Answer:A
Explanation:
Two meteoroids are heading for earth. their speeds as they cross the moon's orbit are 1.0 km/s . part a the first meteoroid is heading straight for earth. what is its speed of impact?
Answer : [tex]v_{2} = 11.2\ km/s[/tex]
Explanation : The total energy of the meteoroids
Total Energy = Kinetic Energy + Potential Energy
[tex]E = \dfrac{1}{2}mv^{2} - \dfrac{GMm}{R}[/tex]
Total energy remain constant.
Now, for both meteoroids
[tex]\dfrac{1}{2}m(v_{2}^{2} - v_{1}^{2}) = GMm\left(\dfrac{1}{R}-\dfrac{1}{r}\right)[/tex]
Where, R = radius of the earth
r = distance of the moon
M = mass of the earth
m = mass of the meteoroid
[tex]v_{2}^{2} = 2GM\left(\dfrac{1}{R}-\dfrac{1}{r}\right)+ v_{1}^{2}.....(I)[/tex]
Now, put the values in equation (I)
[tex]v_{2}^{2} = 1 + 2\times6.67\times10^{-11}\times6\times10^{24}\left(\dfrac{1}{6.37} - \dfrac{1}{384}\right)\times10^{-6}[/tex]
[tex]v_{2}^{2} = 12361\times 10^{4}\ m/s[/tex]
[tex]v_{2} = 11118.003\ m/s[/tex]
[tex]v_{2} = 11.2\ km/s[/tex]
Hence, this is the required solution.
The speed of the meteorite at its impact on Earth will be [tex]\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] or [tex]\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
Further Explanation:
The meteorite in the outer space at a particular distance from the Earth has its energy stored in the form of the kinetic energy as well as the potential energy.
The kinetic energy of the meteorite is given by:
[tex]{E_k} = \frac{1}{2}m{v^2}[/tex]
The potential energy of the meteorite at a particular height from Earth is given by:
[tex]{E_p} = - \frac{{GMm}}{r}[/tex]
Here, [tex]M[/tex] is the mass of the Earth, [tex]m[/tex] is the mass of satellite and [tex]r[/tex] is the distance of the meteorite from the centre of the Earth.
The total energy of the meteorite is given as:
[tex]\begin{aligned}{E_T}&={E_k} + {E_p}\\&=\frac{1}{2}m{v^2}-\frac{{GMm}}{r}\\\end{aligned}[/tex]
As the meteorite moves from the moon’s orbit towards the Earth, the potential energy of the meteorite gets converted into the kinetic energy.
Using conservation of energy for the meteorite from the moon’s orbit to the Earth’s surface:
[tex]\begin{aligned}\frac{1}{2}mv_1^2-\frac{{GMm}}{r}&=\frac{1}{2}mv_2^2-\frac{{GMm}}{R} \hfill\\v_1^2-v_2^2&=2GM\left({\frac{1}{r}-\frac{1}{R}}\right) \hfill\\{v_2}&=\sqrt {v_1^2-2GM\left({\frac{1}{r}-\frac{1}{R}}\right)}\hfill\\\end{aligned}[/tex]
Here, [tex]{v_1}[/tex] is the speed of the meteorite in moon’s orbit, [tex]r[/tex] is the distance of the moon’s orbit from centre of Earth and [tex]R[/tex] is the radius of Earth.
The meteorite passes the moon’s orbit at the speed of [tex]1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] .
We know that the distance of the Moon’s orbit from the centre of Earth is [tex]3.84 \times{10^5}\,{\text{km}}[/tex] and the radius of the Earth is [tex]6371\,{\text{km}}[/tex] .
Substitute the values in the equation of Energy.
[tex]\begin{aligned}v_{f}&=\sqrt{(1)^{2}-2(6.67\times10^{-11})(5.97\times10^{24})\left(\dfrac{1}{384}-\dfrac{1}{6.37}\right)10^{-6}}\\&=\sqrt{1.229\times10^{8}}\text{ m/s}\\&=11088.24\text{ m/s}\\&\approx11.1\text{ km/s}\end{aligned}[/tex]
Thus, the speed of the meteorite striking the Earth is [tex]\boxed{11.1\,{{{\text{km}}}\mathord{\left/{\vphantom{{{\text{km}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] or [tex]\boxed{11088.24\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}}[/tex] .
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Answer Details:
Grade: College
Subject: Physics
Chapter: Gravitation
Keywords:
Meteorite, moon’s orbit, 1km/s, two meteorite, heading for, speed of impact, 11.1 km/s, 384000 km, total energy, heading straight.
Phosphorus bursts into flame when exposed to oxygen. What element might be used when storing pure phosphorus to keep it from reacting?
Argon
Explanation;Phosphorus is an element found in group 15 of the periodic table, it is one of the element that belongs to nitrogen family.The element phosphorus exhibits allotrophy, namely; white phosphorus, red phosphorus and black phosphorus. Phosphorus, particularly the white phosphorus combines with oxygen so easily that it catches fire automatically. Thus, as a safety precaution, white phosphorus is stored under water in laboratories. It may be also stored in argon, since argon is an inert element which belongs to the noble gases group.Answer:
Noble gases. (Argon)
Explanation:
Noble gases are found in the group eight element on the periodic table and are known for their inert nature.
Meaning that they don't react easily with other elements.
Phosphorus is a reactive element that when exposed to air (oxygen) catches fire.
If phosphorus is stored under Noble gases escpecially Argon it won't react with the surrounding air in the laboratory.
give the reason why the horizontal component of the velocity of a projectile remains the same at every point of its flight
Describe how gravity aids in the formation of stars
Final answer:
Gravity aids in star formation by leading to cloud collapse and nucleus heating.
Explanation:
Gravity aids in the formation of stars by causing the collapse of clouds of gas and dust, leading to increased density in the core regions where nuclear fusion begins.
As the gas collapses, the gravitational force overcomes the thermal pressure, allowing the core to contract and heat up, initiating the process of star formation.
The gravitational attraction of the newly forming stars also plays a crucial role in increasing the density of the surrounding cloud material, facilitating the formation of organic molecules that can be incorporated into newly formed planetary systems.
The highest element in the hierarchical breakdown of the wbs is
The highest element in the hierarchical breakdown of the Work Breakdown Structure (WBS) is the project itself. This project is then divided into definable and measurable tasks or subtasks forming a hierarchical structure.
Explanation:The highest element in the hierarchical breakdown of the Work Breakdown Structure (WBS) is the project itself. WBS is a tool used in project management to divide a project into manageable chunks or portions. It begins with the final deliverable or project at the top level. This is subsequently broken down into definable and measurable tasks or subtasks forming a multi-level structure. A good example could be a software development project. The entire project would be the highest element and might be divided into phases like planning, design, coding, testing, and implementation. Each of these could be further broken down into more specific tasks.
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Which structures allow for the removal of wastes from the developing fetus?
The fetus relies upon its mother as it develops. These are some of the things it needs:
protection against knock and bumps, and temperature changes oxygen for respiration nutrients (food and water)The developing fetus also needs its waste substances removing.
The fetus is protected by the uterus and the amniotic fluid, a liquid contained in a bag called the amnion.
Volume or intensity of a sound wave is measured by the size of the _________ and reported in decibels.
Answer:
Amplitude
Explanation:
The intensity (or volume) of a sound wave is proportional to the square of the sound wave:
[tex]I\propto A^2[/tex]
where A is the amplitude of the wave.
The amplitude of a sound wave is measured as the maximum displacement of the wave with respect to its equilibrium position.
The intensity is measured in Decibels (dB), which is related to the logarithmic ratio between the power of two different sounds, i.e. the difference in decibels between two sounds is given by
[tex]10 Log (\frac{P_2}{P_1})[/tex]
where P1 and P2 are the power of the two sounds.
Kara built a model of how molecules behave when they are in solid form. She filled a clear plastic box with marbles to the very top and put the lid on the box. When she shakes the box, the marbles can barely move. Which of the following is a limitation of Kara's model?
Answer:
The marbles must be held in a box, but solids hold together without a container.
Explanation:
study island!!!
Answer:
The marbles must be held in a box, but solids hold together without a container.
Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.
The magnitude of the fourth force will be 90.1956 N and the angle will be 75.535° made from the horizontal in the first quadrant.
Given to us
Force A, [tex]F_A = 100\ N[/tex]Angle made by Force A from the horizontal, [tex]\theta_A = 30^o[/tex]Force B, [tex]F_B = 80 \ N[/tex]Angle made by Force B from the Vertical, [tex]\theta_B = 30^o[/tex]Force C, [tex]F_C = 40 \ N[/tex]Angle made by Force C from the Horizontal, [tex]\theta_C = 53^o[/tex]For Angle made by Force B,As the angle given to us is from the vertical axis, we will make it from the horizontal,
[tex]\theta_B = (90^o -30^o) = 60^o[/tex]
Now, the magnitude of a fourth force on the stone should make the vector sum of the four forces zero. therefore,
[tex]\sum F_{horizontal} = 0[/tex]
[tex]F_A Cos(\theta_A) -F_B Cos(\theta_B) -F_C Cos(\theta_C) = F_{Dh}[/tex]
[tex]100 Cos(30^o) -80Cos(60^o) -40Cos(53^o) = F_{Dh}\\\\F_{Dh} = 22.5299\ N[/tex]
[tex]\sum F_{Vertical} = 0[/tex]
[tex]F_A Sin(\theta_A) +F_B Sin(\theta_B) -F_C Sin(\theta_C) = F_{Dv}[/tex]
[tex]100 Sin(30^o) +80Sin(60^o) -40Sin(53^o) = F_{Dv}\\\\F_{Dv} =87.3365\ N[/tex]
Resultant Force[tex]R = \sqrt{(F_{Dh})^2+(F_{Dv})^2}\\\\R = \sqrt{(87.3365)^2+(22.5299)^2}\\\\R = \sqrt{7627.6642+507.5963}\\\\R = \sqrt{813526}\\\\R = 90.1956\ N[/tex]
The angle of the resultant,[tex]Tan (\theta_D) = \dfrac{F_{Dv}}{F_{Dh}}\\\\Tan(\theta_D) = \dfrac{87.3365}{22.5299}\\\\Tan(\theta_D) = 3.87647\\\theta_D = Tan^{-1} 3.87647\\\theta_D =75.535^o[/tex]
Hence, the magnitude of the fourth force will be 90.1956 N and the angle will be 75.535° made from the horizontal in the first quadrant.
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A car slows down from 27.7 m/s to 10.9 m/s in 2.37s. What is the acceleration?
A car slows down from 27.7 m/s to 10.9 m/s in 2.37s.Then the acceleration is -7.09 m/sec².
What is acceleration ?Acceleration is the rate at which an object's velocity with respect to time changes. They are vector quantities, accelerations. The direction of the net force acting on an object determines the direction of its acceleration.
An object has positive acceleration if it is accelerating and travelling in the right direction. Positive acceleration was demonstrated in the first example by the speeding car. The acceleration is occurring in the same direction as the car's motion, which is forward and speeding up.
The equation a = v/t denotes acceleration (a), which is the change in velocity (v) over the change in time (t).
The final velocity 'v' =10.9 m/s
The time 't' = 2.37 secs.
Then acceleration = -7.09 m/sec² or, decceleration is 7.09 m/sec².
Thus, the acceleration is -7.09 m/sec².
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A merry-go-round is spinning at a rate of 4.0 revolutions per minute. cora is sitting 1.0 m from the center of the merry-go-round and cameron is sitting right on the edge, 2.0 m from the center. what is the relationship between the rotational speeds of the two children?
Answer:
Same.
Explanation:
The rotational speed of an object is given by :
[tex]\omega=\dfrac{\theta}{t}[/tex]
[tex]\theta[/tex] is the angular displacement
t is the time taken
The angular speed of a merry- go- round is 4 revolutions per minute. There are two persons Cora and Cameron. Cora is sitting 1.0 m from the center of the merry-go-round and Cameron is sitting right on the edge, 2.0 m from the center.
The rotational speeds of both of the children remains the same because it is independent of the distance from the center.
El Sol está en promedio a 93 millones de millas de la Tierra. ¿A cuántos metros equivale esto? Expréselo usando potencias de diez.
La distancia promedio entre la Tierra y el Sol es de aproximadamente 93 millones de millas. En metros, esta distancia es de aproximadamente 1.5*10^11. Se utiliza la notación científica 1.5 × 10¹¹ metros para representar esta distancia de forma más concisa y manejable.
Explanation:El Sol está, en promedio, a 93 millones de millas de la Tierra. Sabemos que 1 milla equivale a 1.60934 kilómetros y que un kilómetro es igual a 10^3 metros. Por lo tanto, podemos decir que una milla es igual a 1.60934*10^3 metros.
Para convertir la distancia desde la Tierra hasta el Sol a metros, multiplicamos 93,000,000 millas por 1.60934*10^3 metros/milla. Esto da como resultado aproximadamente 1.5*10^11 metros. Así que, en notación científica, la distancia desde la Tierra hasta el Sol es de aproximadamente 1.5 × 10¹¹ metros.
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6. You stopped swimming laps to take a 10 second quick count and counted 25 beats. Explain how to find your beats per minute.
If you stop swimming laps to take a count to 10 seconds then, and you counted for 25 beats, then that is 150 beats per minute.
What is unit conversion?Conversion of units is defined as the process of multiplying or dividing different units and measurements of the same quantity. Conversion is the process of transforming the value of one form to another, such as inches to millimeters or liters to gallons in mathematics. Units are used to calculate length, weight, capacity, temperature, and speed.
Now, according to the question, the best way to solve the question is,
If you paused for ten seconds, simply multiply 25 by six because each minute equals 60 seconds.
When you multiply the 25 beats to 60 seconds, the answer you get is 150 beats per second.
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Why is the thickness greater for the bass strings of a guitar than for the treble strings?
If a 70-kg swimmer pushes off a pool wall with a force of 250 N what is her acceleration
A bicycle wheel is rotating at 49 rpm when the cyclist begins to pedal harder, giving the wheel a constant angular acceleration of 0.45 rad/s2 . what is the wheel's angular velocity, in rpm, 9.0 s later?
Answer:
Angular velocity, [tex]\omega_f=9.180\ rad/s[/tex]
Explanation:
It is given that
Initial angular velocity of the wheel, [tex]\omega_o=49\ rpm[/tex]
Since, 1 revolutions per minute = 0.10471 radians per second
[tex]\omega_o=49\ rpm=5.1307\ rad/s[/tex]
Angular acceleration of the wheel, [tex]\alpha =0.45\ rad/s^2[/tex]
Time taken, t = 9 s
Let [tex]\omega_f[/tex] is the final velocity of the wheel 9 seconds later. Using the equation of kinematics to find it :
[tex]\omega_f=\omega_o+\alpha t[/tex]
[tex]\omega_f=5.1307+0.45\times 9[/tex]
[tex]\omega_f=9.180\ rad/s[/tex]
So, the final angular velocity of the wheel is 9.180 rad/s. Hence, this is the required solution.
The bicycle wheel's angular velocity, in rpm is 87.68 rpm.
Given the following data:
Time = 9.0 secondsAngular acceleration = 0.45 [tex]rad/s^2[/tex]Initial angular velocity of the bicycle wheel = 49 rpm.Conversion:
1 rpm = 0.1047 rad/s
49 rpm = X rad/s
Cross-multiplying, we have:
[tex]X = 49[/tex] × [tex]0.1047[/tex]
X = 5.1303 rad/s
To find the bicycle wheel's angular velocity, in rpm, we would use the first equation of kinematics:
[tex]w_f = w_o + at[/tex]
Where:
[tex]w_f[/tex] is the final angular velocity.[tex]w_o[/tex] is the initial angular velocity.a is the angular acceleration.t is the time.Substituting the given parameters into the formula, we have;
[tex]w_f = 5.1303 + 0.45(9)\\\\w_f = 5.1303 + 4.05[/tex]
Final angular velocity = 9.1803 rad/s
Next, we would convert the value in rad/s to rpm:
Conversion:
1 rpm = 0.1047 rad/s
X rpm = 9.1803 rad/s
Cross-multiplying, we have:
[tex]9.1803 = 0.1047X\\\\X = \frac{9.1803}{0.1047}[/tex]
X = 87.68 rpm.
Final angular velocity = 87.68 rpm.
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You probably have noticed that, as you increase ω, there will be a value, ωcrit, for which r(ω) goes to infinity. find ωcrit. answer in terms of k and m.
An eagle that has a mass of 4.5 kg is spotted 5 m above a lake
An Eagle that has a mass of 4.5 kg is spotted 5 m above a lake. It flies up to its nest in a tree at a height of 25 m. How much gravitational potential energy did the eagle gain?
This is the original question
Answer:
882 J = PEg = MGH M = mass in Kg G = 9.6 m/s H = height
Height in this case is the displacement (25m - 5m) = 20m
Hope this helps.
to cook food in the microwave oven converts electrical energy into. energy. a. electricmagnetic. b. heat. c. mechanical d.nulcleae
Fluorine is more electronegative than oxygen because 1. the outer orbitals of fluorine are located further from the nucleus than the outer orbitals of oxygen. 2. the effective nuclear charge of fluorine is greater than that of oxygen. 3. the effective nuclear charge of oxygen is greater than that of fluorine. 4. the outer orbitals of oxygen are located closer to the nucleus than the outer orbitals of fluorine. 5. false; oxygen is more electronegative than fluorine.
Fluorine is more electronegative than oxygen because the effective nuclear charge of fluorine is greater than that of oxygen.
Electronegativity is the tendency of an atom to attract electrons. The effective nuclear charge is the net positive charge experienced by an electron in an atom. Fluorine has a smaller atomic radius than oxygen. This means that the electrons in the outer orbitals of fluorine are closer to the nucleus than the electrons in the outer orbitals of oxygen.
The closer the electrons are to the nucleus, the stronger the attractive force between the electrons and the nucleus. This means that the effective nuclear charge of fluorine is greater than that of oxygen. Therefore, fluorine is more electronegative than oxygen.
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How does the internal circuitry differ among the ohmmeter, voltmeter, and ammeter?
Positive psychologists believe that happiness can be increased by
Positive psychologists believe that happiness can be increased by various strategies and interventions, including practicing gratitude, engaging in acts of kindness, cultivating positive relationships, finding meaning and purpose in life, setting and achieving goals, and focusing on one's strengths and virtues. Research has shown that these activities and mindsets can enhance happiness and well-being. CBT and positive psychology interventions have also been found to be effective in increasing happiness and improving overall well-being.
Explanation:Positive psychologists believe that happiness can be increased by various strategies and interventions. These include practicing gratitude, engaging in acts of kindness, cultivating positive relationships, finding meaning and purpose in life, setting and achieving goals, and focusing on one's strengths and virtues.
Research has shown that these activities and mindsets can enhance happiness and well-being. For example, studies conducted by Seligman and Lyubomirsky have demonstrated that practicing gratitude and performing acts of kindness can boost happiness levels.
Furthermore, interventions such as cognitive-behavioral therapy (CBT) and positive psychology interventions (PPIs) have been found to be effective in increasing happiness and improving overall well-being.
Final answer:
Positive psychologists believe that happiness can be increased by cultivating and focusing on positive affect and emotions.
Explanation:
Positive psychologists believe that happiness can be increased by cultivating and focusing on positive affect and emotions. This means intentionally seeking out activities and experiences that bring joy, gratitude, contentment, and other positive emotions. Research has shown that practicing gratitude, engaging in acts of kindness, forming strong social connections, setting and achieving meaningful goals, and finding purpose and meaning in life are all ways to increase happinessThe equation of motion of a particle is s = t3 â 27t, where s is in meters and t is in seconds. (assume t ⥠0.) (a) find the velocity and acceleration as functions of t.
To find the velocity and acceleration of a particle as functions of time in a given equation of motion, differentiate the equation with respect to time. The velocity function is v = 3t² - 27 and the acceleration function is a = 6t.
Explanation:Velocity: To find the velocity, differentiate the equation of motion with respect to time which gives v = ds/dt. In this case, v = 3t² - 27.
Acceleration: Similarly, differentiate the velocity equation with respect to time to find acceleration, which results in a = dv/dt. Here, a = 6t.
20. A 0.145 kg mass of tungsten at 130.0 °C is placed in a 0.502 kg of water at 22.0 °C. The mixture reaches equilibrium at 28.6 °C. Calculate the specific heat of tungsten. (specific heat of water = 4180 J/kg C)
Givens:
Unknown:
Equation: macaΔTa = - mbcbΔTb
Substitute:
Solve:
A +1.0 c charged object and a +2.0 c charged object are separated by 100 m. where should a -1.0x10-3 c charged object be placed on a line between the positively charged objects so that the net electrical force exerted on the negatively charged object is zero?
The negative charge should be placed 41 m from +1.0 C charged object
Further explanationElectric charge consists of two types i.e. positively electric charge and negatively electric charge.
There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :
[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]
F = electric force (N)
k = electric constant (N m² / C²)
q = electric charge (C)
r = distance between charges (m)
The value of k in a vacuum = 9 x 10⁹ (N m² / C²)
Let's tackle the problem now !
Given:
Q₁ = +1.0 C
Q₂ = +2.0 C
R₁ = x
R₂ = 100 - x
q = -1.0 × 10⁻³ C
Unknown:
x = ?
Solution:
In order to produce a total force = 0, the electric forces acting on this negative charge must be in the opposite direction and equal in magnitude.
We assume the distance of this negative charge to the positive charge +1.0 C is x, then:
[tex]F_1 = F_2[/tex]
[tex]k\frac{Q_1q}{R_1^2} = k\frac{Q_2q}{R_2^2}[/tex]
[tex]\frac{Q_1}{R_1^2} = \frac{Q_2}{R_2^2}[/tex]
[tex]\frac{1}{x^2} = \frac{2}{(100 - x)^2}[/tex]
[tex]\sqrt{\frac{1}{x^2}} = \sqrt{\frac{2}{(100 - x)^2}}[/tex]
[tex]\frac{1}{x} = \frac{\sqrt{2}}{100 - x}[/tex]
[tex]100 - x = \sqrt{2}x[/tex]
[tex]x + \sqrt{2}x = 100[/tex]
[tex]x ( 1 + \sqrt{2} ) = 100[/tex]
[tex]x = \frac{100}{1 + \sqrt{2}}[/tex]
[tex]x = 100(\sqrt{2} - 1) ~ m[/tex]
[tex]x \approx 41 ~ m[/tex]
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Subject: Physics
Chapter: Static Electricity
Keywords: Series , Parallel , Measurement , Absolute , Error , Combination , Resistor , Resistance , Ohm , Charge , Small , Forces