The point where the electric field is zero is (0, 3/5).
How to find the point where the electric field is zero?The electric field due to the 6.0 uC charge given by:
\(E1 = k * q1 / r1^2\\r1 = \sqrt{x} (x^2 + y^2)\)
The electric field due to the 4.0 uC charge is given by:
\(E2 = k * q2 / r2^2\\r2 = \sqrt{x} ((x-0)^2 + (y-1)^2) = \sqrt{x} (x^2 + (y-1)^2)\)
Since the electric field is zero, the vector sum of E1 and E2 must be zero:
E1 + E2 = 0
Substituting the expressions for, we get:
\(k * q1 / r1^2 + k * q2 / r2^2 = 0\)
Simplifying and substituting the values, we get:
\((9.0 x 10^9) * (6.0 x 10^-6) / (x^2 + y^2) + (9.0 x 10^9) * (4.0 x 10^-6) / (x^2 + (y-1)^2) = 0\)
Multiplying both sides by (x^2 + y^2) * (x^2 + (y-1)^2), we get:
\((9.0 x 10^9) * (6.0 x 10^-6) * (x^2 + (y-1)^2) + (9.0 x 10^9) * (4.0 x 10^-6) * (x^2 + y^2) = 0\)
Simplifying further, we get:
\(90x^2 + 90y^2 - 108y + 54 = 0\)
This equation represents a circle centered at (0, 3/5) with a radius of r = \(\sqrt{x} (3/5).\)
Therefore, the point where the electric field is zero is (0, 3/5).
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12 A car travels in a straight line at speed v along a horizontal road. The car moves
against a resistive force F given by the equation
F = 400+kv²
where F is in newtons, v in ms-1 and k is a constant.
At speed v = 15ms-1, the resistive force F is 1100 N.
a
Calculate, for this car:
i the power necessary to maintain the speed of 15ms-¹,
ii the total resistive force at a speed of 30 ms-¹,
iii the power required to maintain the speed of 30ms-¹.
Answer:
i) Power = Force * Velocity = 1100 * 15 = 16500 W = 16.5 kW(ii) Find the value of k first: F = 400 + k(15^2) k = 28/9 F = 400 +(28/9)(30^2) = 320
Explanation:
the force measuring instrument is called
Answer:
A force gauge (also called a force meter) is a measuring instrument used to measure forces. Applications exist in research and development, laboratory, quality, production and field environment. There are two kinds of force gauges today: mechanical and digital force gauges.
A roller coaster is traveling at 80m/hr until the emergency breaks turn on, causing it to come to a complete stop in 10s. What is the acceleration of the roller coaster?
This means that the roller coaster is decelerating at a rate of 8m/s² until it comes to a complete stop.
What is decelerating?Decelerating is the process of slowing down or decreasing the speed of a vehicle, person, or object. It is the opposite of accelerating, which is the process of increasing speed. Deceleration can be caused by the force of friction from the road or other surfaces, or by reducing the amount of engine power or braking. Deceleration is necessary to safely slow down and stop a vehicle and to negotiate turns or curves in the road.
The acceleration of the roller coaster can be calculated using the equation
a = (Vf - Vi)/t, where Vf is the final velocity, Vi is the initial velocity, and t is the time taken for the change in velocity to occur.
In this case, Vf = 0m/hr, Vi = 80m/hr, and t = 10s.
Therefore, the acceleration of the roller coaster is -8m/s².
This means that the roller coaster is decelerating at a rate of 8m/s² until it comes to a complete stop.
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a ball rolls from rest down an incline with a uniform acceleration of 4m/s². what is it speed after 8 seconds
witch of the following questoons would you expect to see on an interest inventory
Answer: Do you like science experiments ? 1)
Would You enjoy reading a fashion magazine? 2)
Explanation:
how does an elevator model Newton’s First Law of Motion?
When the elevator is going up, though, you are accelerating, which adds more force to the scale and increases your apparent weight. When the elevator is going down, the same is true, but the acceleration is negative, subtracting force from the scale and decreasing your apparent weight.
please help! you are amazing! brainliest!
The NGC 3603 region of space is shown below. It is a cloud of gas and dust. Stars are forming in this cloud. How would NGC 3603 be classified?
A. As a nebula
B. As a galaxy
C. As a planet
D. As a solar system
The NGC 3603 region of space is shown below. It is a cloud of gas and dust. Stars are forming in this cloud. The NGC 3603 be classified as a nebula. The correct option is A.
What is a nebula?A tiny, ball-shaped nebula is known as a planetary nebula. It generates an ionized gas that is luminous. Large and typically erratic clouds are produced by diffuse nebulae.
Due to their big stars, they typically form a cluster of stars. a cloud of gas in space that can be seen in the night sky as a vaguely discernible brilliant patch, as something black, or as a shadow against other luminous objects.
Dust and gases, primarily hydrogen and helium, make up nebulae. The mass of gas and dust eventually grows so large that it is forced to collapse by gravity. The material in the cloud's core heats up as a result of the collapse, and this hot core is the start of a star.
Therefore, the correct option is A. As a nebula.
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Can sum one pls help me with this assignment I will mark brainlieniest for whoever helps me (:
Answer:
Explanation:
R r
R RR Rr
r Rr rr
homozygous: 2 (out of 4)
heterozygous: 2 (out of 4)
t t
T Tt Tt
T Tt Tt
Tall: 4 (all)
short: 0 (none)
When balancing a chemical equation, can a coefficient within a chemical equation be adjusted to
balance the chemical equation?
AS WE CLIMB UP, THE PRESSURE INSIDE OUR BODY WILL BE.................. *
1 )low
2)high
3)very high
4)same
how would you walk on red-hot coals without getting your feet burned?
Walking on red-hot coals is a traditional practice that has been performed for centuries by certain cultures for ceremonial or religious purposes.
While it may seem like a dangerous or painful activity, it is possible to walk on hot coals without getting your feet burned by following some basic guidelines:
Prepare the fire: The fire should be built with small, dry pieces of wood, and the coals should be allowed to burn down until they are glowing red and no longer producing flames.
Test the coals: Before walking on the coals, test the temperature by walking around the edge of the bed of coals. This will give you an idea of the heat level and whether it is safe to walk on.
Keep your feet clean and dry: Dirt, oil, or sweat on your feet can make them more susceptible to burning. Before walking on the coals, make sure your feet are clean and dry.
Walk quickly: Walking quickly and confidently over the coals can reduce the amount of time your feet are in contact with the hot surface, which can help prevent burns.
Use a shuffling motion: Rather than taking long strides, use a shuffling motion to walk over the coals. This helps distribute your weight more evenly and reduces the pressure on any one part of your foot.
Stay focused: Keep your mind focused on your goal of walking across the coals. Avoid distractions and maintain a positive attitude.
It's important to note that walking on hot coals can still be dangerous and should only be attempted with proper guidance and training from experienced practitioners. It's not recommended for everyone and should be approached with caution.
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Marie curie was a scientist who won two nobel prizes in the early 1900s for her work with.
She remains the only person to be honored for accomplishments in two separate sciences. Curie received the Nobel Prize in Physics in 1903, along with her husband and Henri Becquerel, for their work on radioactivity.
What did Marie Curie discover?Relentless regardless of a vocation of truly requesting and at last lethal work, she found polonium and radium, supported the utilization of radiation in medication and essentially changed how we might interpret radioactivity. Curie was conceived Marya Skłodowska in 1867 in Warsaw.
Curie was the first person to win two Nobel Prizes.She managed it all without a fancy lab.Nobel Prizes were a family affair.Curie was the first female professor at Sorbonne University.Curie is buried in the Panthéon in Paris.To learn more about Marie Curie from the given link
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Distinguish between force and friction. class 8
Answer: Friction is the resistance to motion of one object moving relative to another. For example, when you try to push a book along the floor, friction makes this difficult. Force: Force is essentially a push, or a pull action, that can lead to certain outcomes.
Twice a year the Sun’s rays strike Earth north or south of the....
Answer:
Tropic of Cancer in the north and the Tropic of Capricorn in the south.
Explanation:
Element E is in group 1A and element D is in group 6A they form an ionic compound. What is the formula for this compound
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A mass of 3.6 kg oscillates on a horizontal spring with a spring constant of 160 N/m. When the mass passes the equilibrium point its speed is 5.2 m/s. d. What is the maximum displacement of the mass?
Answer:
x = 0.78 m = 78 cm
Explanation:
Applying the law of conservation of energy to the spring-mass system, we can write the following equation:
\(Kinetic\ Energy\ at\ Mean\ Position= Potential\ Energy\ at\ extreme\ position\\\frac{1}{2}mv^2 = \frac{1}{2}kx^2\\\\x^2 = \frac{mv^2}{k}\\\\x = \sqrt{\frac{mv^2}{k} }\)
where,
x = maximum displacement = ?
m = mass = 3.6 kg
v = speed at mean position = 5.2 m/s
k = spring constant = 160 N/m
Therefore,
\(x = \sqrt{\frac{(3.6\ kg)(5.2\ m/s)^2}{160\ N/m} }\\\\\)
x = 0.78 m = 78 cm
1.Under normal atmospheric pressure what temperature does:
i.Ice turn to water?
ii.water turn to steam?
2.In which order do the following take place during water purification?
a)chlorination b)screening c)filtration d)sedimentation
b.Which of these processes ensures that the water contains no harmful microorganisms
Answer:
1.i)heat
ii.)heat
2.a.)b,c,d,a
b.)a.chlorination
Answer:
B
Explanation:
B
g what best describes the camera motion in the scenes directly prior to the old man's defiant hand gesture?
The action that best describes the camera motion in the scenes directly prior to the Old Man's Defiant hand gesture is: "Zooming in and out" (Option C)
What are the various types of Camera Motions and why are they important?Some interesting camera motions are:
Static shot - The camera is stationary and does not move during the shot. This can create a sense of stability and calm.Panning shot - The camera moves horizontally from left to right or vice versa, following the action or moving to reveal a new part of the scene.Tilt shot - The camera moves vertically up or down, following the action or revealing a new part of the scene.Tracking shot - The camera moves horizontally, vertically, or diagonally while following the action or a moving object.Note that the different camera motions described above are important because they allow filmmakers to convey mood and emotion, create visual interest, and guide the audience's attention within a scene.
For example, a static shot can create a sense of calm or stability, while a handheld shot can create a sense of intimacy or instability. A panning shot can follow the action and keep the audience engaged, while a zoom shot can draw attention to a specific part of the scene.
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Full Question:
What best describes the camera motion in the scenes directly prior to the old man's defiant hand gesture?
O Static and secured on a tripod
O Moving chaotically
O Zooming in and out
O Filming at 96 frames a second
A distant star system is discovered in which a planet with twice the radius of the earth and rotating 3.0 times as fast as the earth orbits a star with a total power output of 5.8 x 1029 W. If the star's radius is 6.0 times that of the sun, what is the electromagnetic wave intensity at the surface? Astronomers call this the surface flux. Mean radius of the Sun is 6.96 x 109 m. Express your answer with the appropriate units. HA ? I Value Units Submit Request Answer Every planet-day (one rotation), the planet receives 9.4 x 1022 J of energy. What is the planetos distance from its star? Give your answer in astronomical units (AU), where 1 AU is the distance of the earth from the sun ΤΟ ΑΣφ ?
The electromagnetic wave intensity at the surface of the planet in the distant star system is approximately X W/m². The planet is located at a distance of Y astronomical units (AU) from its star.
To find the electromagnetic wave intensity at the surface of the planet, we need to consider the star's total power output and its radius compared to that of the Sun. The surface flux, or electromagnetic wave intensity, can be calculated using the following formula:
Surface flux = (Star's power output) / (4π × (Star's radius)²)
Given that the star's power output is 5.8 x 10^29 W and the star's radius is 6.0 times the radius of the Sun (6.0 × 6.96 x 10^9 m), we can substitute these values into the formula:
Surface flux = (5.8 x 10^29 W) / (4π × (6.0 × 6.96 x 10^9 m)²)
Simplifying the expression, we can calculate the surface flux in W/m². This gives us the electromagnetic wave intensity at the surface of the planet.
To find the planet's distance from its star, we can use the fact that the planet receives 9.4 x 10^22 J of energy every planet-day (one rotation). This energy is equal to the total energy emitted by the star over a period of one planet-day. The energy received from the star decreases with distance, following an inverse square law.
By equating the energy received from the star to the energy emitted by the star and solving for the distance, we can determine the planet's distance from its star in astronomical units (AU). The formula to calculate the distance is:
Distance = sqrt((Star's power output × Planet-day duration) / (4π × Surface flux))
Substituting the given values, we can calculate the distance in AU.
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what do you add to distilled water for making it to conduct electricity
Answer:
we add salt to conduct electricity
REEEEEEEEEEE ion know but, Explain how body heat is produced.
A car drives 16 miles south and then 12 miles west. What is the magnitude of the car’s displacement?
Answer:20 miles
Explanation:
the end of a wrench that is tilted θθ = 30 ∘∘ above the horizontal and is rrr = 31 cmcm long, what force must he apply to exert a torque of -17 n⋅mn⋅m ?
The force the person must apply to exert a torque of -17 N·m is approximately 110.84 N. Torque (τ) is calculated as the product of the force applied (F) and the lever arm (r).
The force the person must apply to exert a torque of -17 N·m can be calculated using the given parameters: the angle θ = 30° and the length of the wrench r = 31 cm.
Torque (τ) is calculated as the product of the force applied (F) and the lever arm (r), where the lever arm is the perpendicular distance from the axis of rotation to the line of action of the force. In this case, the torque is given as -17 N·m, which indicates that the direction of the torque is opposite to the direction of the force.
Using the equation τ = r * F * sin(θ), we can rearrange it to solve for the force F:
F = τ / (r * sin(θ))
Substituting the values, we have:
F = -17 N·m / (0.31 m * sin(30°))
Evaluating this expression, the force the person must apply to exert a torque of -17 N·m is approximately 110.84 N.
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steady flow devices that result in a drop in the working fluid pressure from inlet to exit are
Steady flow devices that result in a drop in the working fluid pressure from inlet to exit are known as pressure drop devices. These devices are commonly used in various applications to regulate and control fluid flow.
Pressure drop devices are designed to create a pressure difference between the inlet and exit of the fluid flow. This pressure drop can be achieved through various mechanisms, such as constricting the flow area, introducing resistance, or utilizing fluid dynamics principles.
Examples of pressure drop devices include valves, nozzles, orifices, venturis, and diffusers. These devices are used in various industries and applications, such as plumbing systems, hydraulic systems, HVAC systems, and fluid flow control systems.
The pressure drop across these devices is essential for controlling flow rates, regulating fluid pressures, and achieving specific fluid behavior or performance. By creating a pressure drop, these devices can control the speed, volume, or direction of fluid flow, ensuring optimal operation and functionality in different applications.
In summary, steady flow devices that result in a drop in the working fluid pressure from inlet to exit are referred to as pressure drop devices, which play a crucial role in regulating and controlling fluid flow in a wide range of applications.
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7. A ship can carry heavier load in salty water than that in fresh water, why?
Because of the buoyancy of the salt particles
Neutrinos from Supernova 1987A In 1987, 13 neutrinos were detected on Earth at two different neutrino detection experiments, nearly simultaneously with the light observed from an exploding massive star, or core-collapse supernova, which astronomers dubbed SN 1987A. During a core collapse supernova, the central densities and temperatures become so great that electrons (e) can capture onto protons (p) to form neutrons (n) and electron neutrinos (%) Neutrinos interact very weakly with matter, so the neutrinos formed in this process may scatter briefly through the expanding supernova envelope, before freely streaming across the gulf of interstellar space. This process, known as neutronization, is responsible for the formation of neutron stars as remnants of core collapse supernovae. In this problem, we will use the spread of neutrino event detection times following SN 1987A to place an upper limit on the mass of the electron neutrino a. Show that if a neutrino has mass m and total energy E measured in some reference frame, the velocity u of the neutrino in that reference frame is given by the expression This result demonstrates that if neutrinos move at speeds very close to the speed of light (u c), their rest mess energy must be much less than their total energy E (mc2 《 E) In other words, nearly all of the energy of a highly-relativistic particle is in the form of relativistic kinetic energy, and not the rest mess energy. b. Show that for me 《 E, the time for a neutrino to reach Earth from SN 1987A is 2 2E where L is the distance to SN1987A c. The observed energies of the neutrinos measured on Earth ranged from 8- 41 MeV. If the distance L 168000 light-years, and the observed spread in arrival times of the neutrinos was 12 seconds, what is the upper bound on the neutrino mass placed by SN 1987A?
The question discusses the detection of neutrinos from the supernova SN 1987A and uses the spread of neutrino event detection times to derive an upper limit on the mass of the electron neutrino. It demonstrates that if neutrinos move at speeds close to the speed of light, their rest mass energy must be much less than their total energy. It further shows that the time for a neutrino to reach Earth from SN 1987A is proportional to the square of the neutrino energy. Using the given distance to SN 1987A and the observed spread in arrival times of the neutrinos, we can determine an upper bound on the neutrino mass.
The velocity (u) of a neutrino with mass (m) and total energy (E) in a reference frame is given by the expression u = c * (1 - (m^2c^4 / E^2))^(1/2), where c is the speed of light. This equation demonstrates that if the neutrinos move at speeds close to the speed of light (u ≈ c), their rest mass energy (mc^2) must be much less than their total energy (E).
For neutrinos with me << E, the time (t) for a neutrino to reach Earth from SN 1987A is given by t = L / u, where L is the distance to SN 1987A. Substituting the expression for u, we have t = L / (c * (1 - (m^2c^4 / E^2))^(1/2)).
Given that the observed spread in arrival times of the neutrinos is 12 seconds and the distance L is 168000 light-years, we can use these values to determine an upper bound on the neutrino mass. Rearranging the equation for t, we have (1 - (m^2c^4 / E^2))^(1/2) = L / (ct), and squaring both sides, we get 1 - (m^2c^4 / E^2) = (L^2 / c^2t^2).
Since we are looking for an upper bound, we can assume that m^2c^4 is negligible compared to E^2. Therefore, the equation simplifies to 1 ≈ (L^2 / c^2t^2), and rearranging further, m^2c^4 ≤ E^2 - (E^2 * L^2 / c^2t^2).
Substituting the given values of L, t, and the observed energy range of the neutrinos (8-41 MeV) into the equation, we can calculate the upper bound on the neutrino mass placed by SN 1987A.
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A kid pushes a stationary
merry-go-round, creating an
acceleration of 0.135 rad/s^2.
How much time does it take the
merry-go-round to complete
2.00 rotations?
(Unit = s)
Remember: CCW is +, CW is. 1 rev= 2*pi rad
The merry-go-round takes approximately 29.41 seconds to complete 2.00 rotations.
Given data:
Acceleration (α) = 0.135 rad/\(s^2\)
Number of rotations (θ) = 2.00
To find the time taken (t) for 2.00 rotations, we need to use the formula:
θ = 0.5 * α * \(t^2\)
Rearranging the formula, we get:
\(t^2\) = (2 * θ) / α
Plugging in the given values, we have:
\(t^2\) = (2 * 2.00) / 0.135
\(t^2\) = 29.63
Taking the square root of both sides, we find:
t ≈ √29.63
t ≈ 5.439
Therefore, the time taken for the merry-go-round to complete 2.00 rotations is approximately 5.439 seconds.
Note: It's important to round the final answer to an appropriate number of significant figures, considering the given data. In this case, we have used four significant figures in the final answer.
However, if we want to adhere to the given significant figures in the acceleration (0.135 rad/\(s^2\)), the answer should be rounded to three significant figures. In that case, the final answer would be approximately 5.44 seconds.
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The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due... (i need help rn lol)
Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
kid is bouncing on a pogo stick. he oscillates 22.0 times in 14.9 s. What is his period?
Answer:
Period = 0.68 seconds
Explanation:
Given the following data;
Number of oscillation = 22
Time = 14.9 seconds
To find the period;
Method I.
Period = time/number of oscillation
Period = 14.9/22
Period = 0.68 seconds.
Method II.
We would find the frequency of the wave;
Frequency = time/number of oscillation
Frequency = 22/14.9
Frequency = 1.48 Hertz
Next, we find the period;
Period = 1/frequency
Period = 1/1.48
Period = 0.68 seconds
consider building a system of two positive charges: is the work done by an external agent positive or negative?
By considering building a system of two positive charges: The work done by an external agent is negative.
Any attempt by an outside force to combine two positive charges would be harmful. This is done in order to overcome the electrical force that now opposes the movement of the two positive charges toward one another. Negative work is defined as an effort made to control a disagreeable factor. However, if the two positive charges are already close to one another and the external agent operates to separate them, the work done would be beneficial. This is due to how weak the electrical force that holds the two positive charges together becomes when they are separated, necessitating the aid of an outside force. When work done to separate charges, it is viewed favorably. It's important to keep in mind that the concept of work includes all kinds of forces, not only electrical ones. The sign of the work generated by an external source is determined by the force's direction and the charges' displacement. If both the force and the displacement are travelling in the same direction, the work is positive. If the force and displacement are moving in the opposite directions, the work is negative.
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