Point charge A is on the x-axis at x = -3.00 cm. At x= 1.00 cm on the x-axis its electric field is 2800 N/C. Point charge B is also on the x-axis, at x= 5.00 cm. The absolute magnitude of charge B is twice that of A.
A) Find the magnitude and direction of the total electric field at the origin if both A and B are positive.
B) Find the magnitude and direction of the total electric field at the origin if both A and B are negative.
C) Find the magnitude and direction of the total electric field at the origin if A is positive and B is negative.
D)Find the magnitude and direction of the total electric field at the origin if A is negative and B is positive

Answers

Answer 1

Answer:

We can use the electric field equation to find the electric field at the origin due to each point charge, and then add them vectorially to find the total electric field. The electric field due to a point charge is given by:

Explanation:

E = k * q / r^2

where k is Coulomb's constant, q is the magnitude of the point charge, and r is the distance from the charge to the point where we want to find the electric field.

A) Both charges are positive. The electric field at the origin due to charge A is:

E_A = k * q_A / r_A^2

where q_A is the magnitude of charge A, r_A is the distance from A to the origin, and we have:

q_A > 0

r_A = 3.00 cm = 0.03 m

The electric field at the origin due to charge B is:

E_B = k * q_B / r_B^2

where q_B = 2 * q_A is the magnitude of charge B, r_B is the distance from B to the origin, and we have:

q_B > 0

r_B = 5.00 cm = 0.05 m

Using Coulomb's constant k = 9.00 x 10^9 N*m^2/C^2, we get:

E_A = (9.00 x 10^9 Nm^2/C^2) * q_A / r_A^2

E_A = (9.00 x 10^9 Nm^2/C^2) * q_A / (0.03 m)^2

E_A = 1.00 x 10^12 q_A N/C

E_B = (9.00 x 10^9 Nm^2/C^2) * q_B / r_B^2

E_B = (9.00 x 10^9 Nm^2/C^2) * 2q_A / (0.05 m)^2

E_B = 7.20 x 10^11 q_A N/C

The total electric field at the origin is the vector sum of E_A and E_B. Since the charges are on the x-axis and the origin is also on the x-axis, the total electric field will be along the x-axis. Therefore, we only need to add the magnitudes of E_A and E_B to get the total electric field:

E_total = |E_A| + |E_B|

E_total = 1.00 x 10^12 q_A N/C + 7.20 x 10^11 q_A N/C

E_total = 1.72 x 10^12 q_A N/C

Substituting q_A = 2800 N/C * (0.01 m)^2 / (9.00 x 10^9 N*m^2/C^2) = 3.11 x 10^-6 C, we get:

E_total = 5.39 N/C

Therefore, the magnitude of the total electric field at the origin is 5.39 N/C, and it is directed along the positive x-axis.

B) Both charges are negative. The only difference from part A is that the charges now have negative signs, so we have:

q_A < 0

q_B = -2 * |q_A| = -2q_A < 0

Substituting these signs into the equations for E_A and E_B, we get:

E_A = -1.00 x 10^12 |q_A| N/C

E_B = -7.20 x 10^11 |q_A| N/C

The total electric field at

not sure if this solves your answer but hope it somewhat helps.

Answer 2
Answer:

A) If both charges A and B are positive, the electric field vectors point away from the charges, and we can use the principle of superposition to find the total electric field at the origin. The electric field at the origin due to charge A is:

E_A = k*q_A/r_A^2

where k is the Coulomb constant, q_A is the charge of A, and r_A is the distance between A and the origin. Since A is on the x-axis, r_A is simply the distance between A and the origin:

r_A = |-3.00 cm| = 3.00 cm = 0.03 m

Using the given electric field at x=1.00 cm, we can solve for the charge of A:

2800 N/C = k*q_A/(0.01 m)^2

q_A = (2800 N/C)(0.01 m)^2/k = 2.5210^-8 C

The electric field at the origin due to charge B is:

E_B = k*q_B/r_B^2

where q_B is the charge of B, and r_B is the distance between B and the origin:

r_B = |5.00 cm| = 5.00 cm = 0.05 m

Since the absolute magnitude of charge B is twice that of A, q_B = 2q_A = 5.0410^-8 C. Using the principle of superposition, the total electric field at the origin is the vector sum of E_A and E_B. Since both electric fields are along the x-axis, the total electric field at the origin will also be along the x-axis:

E_total = E_A + E_B = kq_A/r_A^2 + kq_B/r_B^2

E_total = (910^9 Nm^2/C^2)(2.5210^-8 C)/(0.03 m)^2 + (910^9 Nm^2/C^2)(5.0410^-8 C)/(0.05 m)^2

E_total = 1.87*10^6 N/C

The direction of the electric field is to the right (positive x-axis direction), since the positive charge B is farther to the right than the positive charge A.

Therefore, the magnitude of the total electric field at the origin if both A and B are positive is 1.87*10^6 N/C to the right.

B) If both charges A and B are negative, the electric field vectors point towards the charges, and we can use the same method as above to find the total electric field at the origin. The electric field at the origin due to charge A is:

E_A = -k*q_A/r_A^2

where the negative sign indicates that the electric field is directed towards charge A. The electric field at the origin due to charge B is:

E_B = -k*q_B/r_B^2

Using the same values for q_A and q_B as in part A, we get:

E_total = E_A + E_B = -kq_A/r_A^2 - kq_B/r_B^2

E_total = -(910^9 Nm^2/C^2)(2.5210^-8 C)/(0.03 m)^2 - (910^9 Nm^2/C^2)(5.0410^-8 C)/(0.05 m)^2

E_total = -2.23*10^6 N/C

The magnitude of the total electric field at the origin is 2.23*10^6 N/C, and the direction is to the left (negative x-axis direction), since the negative charges A and B are both to the right of the origin.

Therefore, the magnitude of the total electric field at the origin if both A and B
are negative is 2.23*10^6 N/C to the left.

C) When A is positive and B is negative, the electric field vectors produced by each point charge will be in opposite directions. To find the total electric field at the origin, we need to add the electric field vectors produced by each charge.

The magnitude of the electric field due to charge A at the origin is given by the formula:

E = k * Q / r^2

where k is the Coulomb constant (9.0 x 10^9 N m^2 / C^2), Q is the magnitude of the charge, and r is the distance between the charge and the origin.

Using this formula, the magnitude of the electric field due to charge A at the origin is:

E_A = k * Q_A / r_A^2
= (9.0 x 10^9 N m^2 / C^2) * Q_A / (0.03 m)^2
= (9.0 x 10^9 N m^2 / C^2) * (|Q_A|) / (0.03 m)^2
= (3.2 x 10^6) * |Q_A| N/C

where |Q_A| is the absolute magnitude of charge A.

Similarly, the magnitude of the electric field due to charge B at the origin is:

E_B = k * Q_B / r_B^2
= (9.0 x 10^9 N m^2 / C^2) * (-2|Q_A|) / (0.05 m)^2
= - (1.4 x 10^6) * |Q_A| N/C

where |Q_B| is the absolute magnitude of charge B.

The total electric field at the origin is the vector sum of E_A and E_B:

E_total = E_A + E_B

Since E_A is positive and E_B is negative, the direction of the total electric field is towards the negative x-axis.

The magnitude of the total electric field is:

|E_total| = |E_A + E_B|
= |E_A| - |E_B|
= (3.2 x 10^6) * |Q_A| - (1.4 x 10^6) * |Q_A|
= (1.8 x 10^6) * |Q_A| N/C

Therefore, the magnitude of the total electric field at the origin is (1.8 x 10^6) times the magnitude of charge A, and the direction of the electric field is towards the negative x-axis.


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Answer:

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Answers

first of all you need to transform the value of the charge to the SI system which means

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[tex]\eta =\frac{ 1 - T_C}{T_H}[/tex]

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I am very pleased to be here today to welcome all of you to the laboratory.

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A uniform sphere with mass 28.0 kg and radius 0.380 m is rotating at constant angular velocity about a stationary axis that lies along a diameter of the sphere. If the kinetic energy of the sphere is 236 J, what is the tangential velocity of a point on the rim of the sphere?

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The tangential speed of a point on the sphere's rim is 2.31 m/s.

Where does a cycle wheel with a radius of 0.5 metres rotate at a constant angular speed of 10 metres per second?

An area of 0.1 T magnetic field lies perpendicular to the plane of a cycle wheel with a radius of 0.5 m, rotating at a constant rate of 10 rad/s. Between its centre and rim, there is a 0.5 V. zero EMF created.

I = (2/5) * m * r²

where m is the mass of the sphere and r is the radius.

Substituting the given values, we get:

I = (2/5) * 28.0 kg * (0.380 m)² = 1.35 kg m²

K = (1/2) * I * ω²

Substituting the given value of K and the calculated value of I, we get:

236 J = (1/2) * 1.35 kg m² * ω²

Solving for ω, we get:

ω = √(2 * 236 J / (1.35 kg m²)) = 6.62 rad/s

v = r * ω

Substituting the given value of r and the calculated value of ω, we get:

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The wave given in the diagram is a third harmonic wave

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A.1
B.2
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The amplitude of the graph is 4. Option D.

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Olympic swimmer, Micheal Phelps, swam a 200 meter race in 1 minute and 54 seconds. What would his velocity be in meters/seconds

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During the 200-meter race, Michael Phelps moved at a speed of 1.75 metres per second.

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More than any other swimmer in history, he won Olympic medals, world titles, US national titles, and set world records. Phelps has made it a priority to support the growth of swimming at all levels during his career. With a total of 28 medals, he is the most successful and decorated Olympian of all time.

Velocity = Distance / Time

In this case, the distance swum is 200 meters and the time taken is 1 minute and 54 seconds, or 114 seconds.

Velocity = 200 meters / 114 seconds

Velocity = 1.75 meters/second (rounded to two decimal places)

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Mica is investigating the speed of waves. She hypothesizes that waves travel at different speeds depending on the media. She sets up an experiment where she sends two different waves through air, water, and glass. She keeps the temperature of eah medium constant and measure the speed of the waves as they travel through each medium. What would be the most likely result if Mica performed the same experiment again, but sent both waves through a vacuum instead of air, water, and glass?

Answers

Answer:

If Mica sends both waves through a vacuum instead of air, water, and glass, then she would find that the waves travel at the same speed. This is because a vacuum is a completely empty space with no matter or medium to interact with, and the speed of light is constant in a vacuum. Therefore, any electromagnetic waves, including light waves, radio waves, and microwaves, would travel at the same speed in a vacuum.

Today, which measures are not helping people survive or protect property from earthquakes?
A. building codes that set standards for earthquake-resistant construction
B. long-term predictions about where earthquakes will strike
C. short-term predictions about when earthquakes will strike
D. safety procedures to follow when earthquakes strike

Answers

The practices don't assist people survive earthquakes or safeguard property safety precautions to take when an earthquake occurs.

What are the basic plans for buildings that can withstand earthquakes?

The ability of something like a building to bend, wobble, and deform again without collapsing is referred to as ductility, and it is frequently incorporated into earthquake-resistant designs. When subjected to the either vertical or horizontal shear stresses of an earthquake, a ductile structure can bend and flex.

Do structures withstand earthquakes?

Based just on seismic dangers in the area, buildings are made to endure a particular amount of shaking. For instance, a structure in Los Angeles would be constructed to withstand a stronger earthquakes than one in Nyc. But, seismologists do not often know the precise size.

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state two difference between mains electcity and that supplied by 1.5 cell

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The  two differences between mains electricity and that supplied by a 1.5V cell:

Voltage:Power output

What is power output?

Power Output means the average rate of electric energy delivery during one Metering Interval.

Power output: Compared to a 1.5V cell, mains electricity has a much greater power output. This is so that more power can be delivered to devices and appliances by mains electricity, which is provided at a higher voltage and current.

Compared to a 1.5V cell, the voltage provided by mains electricity is usually much higher. A 1.5V cell produces DC (direct current) voltage, whereas the mains electricity in the majority of nations is typically supplied at a voltage of 110–120V or 22–240V AC (alternating current). The kinds of devices and appliances that can be powered by mains electricity versus those that can be powered by a 1.5V cell are affected by this difference in voltage levels.

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Spectroscopy has much in common with what other cosmological concepts?
O
redshift and blueshift
gravity and weight
dark matter and dark energy
speed and velocity

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Spectroscopy has much in common with the cosmological concept of redshift and blueshift.

What is Spectroscopy?

Spectroscopy is the study of the interaction between matter and electromagnetic radiation, and it is commonly used to study the properties of stars and galaxies.

Redshift and blueshift, on the other hand, are phenomena that occur when light is emitted from an object that is moving away or towards an observer, respectively. These concepts are commonly used in cosmology to study the expansion of the universe and the motion of galaxies.

Spectroscopy can be used to measure the redshift or blueshift of light emitted from distant objects, providing valuable information about their distance, velocity, and composition.

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remembering and forgetting: crash course psychology #14

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Crash Course Psychology #14 focuses on the concepts of remembering and forgetting. The episode discusses various aspects of memory, such as encoding, storage, and retrieval. Encoding refers to the process of transforming information into a format that can be stored in the brain, while storage is the preservation of that information over time. Retrieval is the act of accessing and recalling stored information when needed.

Forgetting is a natural part of memory processes and can occur due to factors like decay, interference, and retrieval failure. Decay is the gradual fading of memories over time, while interference involves new information disrupting the recall of older memories. Retrieval failure happens when we are unable to access certain memories, even if they still exist in our brains.

In summary, Crash Course Psychology #14 explores the complex processes of remembering and forgetting, highlighting key concepts such as encoding, storage, retrieval, decay, interference, and retrieval failure.

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a satellite moves in a circular orbit around the earth with speed V- 6000mls. Determine the satellite altitude above the earth surface and the period of the satellite's orbit the earth mass and radius are M= 5.97 x 10²⁴kg and R= 6.378 x 206m.​

Answers

The satellite's height above the surface of the planet is roughly 35,800 kilometers, and its orbital period is roughly 1.54 hours.

What is speed?

An object's speed can be defined as how quickly it is going. It measures how quickly an object travels over time. Typically, speed is measured in terms of time/distance.

How do you determine it?

The speed of the satellite must first be converted from miles per second to meters per second:

V= 6000 miles per second = 9656 m/s

The orbital radius of the satellite can then be calculated using the centripetal force equation as follows:

Fc = Fg

Gm M / r² = mv² / r

r = G M / v²

where M is the earth's mass.

By entering the specified values, we obtain:

r = (5.97 x 10²⁴ kg) x (6.67 x 10⁻¹¹N m²/kg²)/ (9656 m/s) 

r= 4.22 x 10⁷ meters

Hence, the satellite's height above the surface of the planet is:

height = r - R.

height= (4.22 x 10⁷ m) - (6.378 x 10⁶ m)

height is 3.58 x 10⁷ meters.

Now we can apply the equation for a circular orbit's period:

T = 2π r / v

When we enter the values we just discovered, we obtain:

T = 2π (4.22 x 10⁷ m) / (9656 m/s)

T = 5.56 x 10³ seconds

Hence, the satellite's orbit has a period of about 1.54 hours.

So, the satellite's height above the surface of the planet is roughly 35,800 kilometers, and its orbital period is roughly 1.54 hours.

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In the scientific method, which of the following should a researcher do after gathering raw data? Perform the experiment to test the hypothesis Analyze the data collected and check to see if they support the hypothesis Propose new directions for further research Report the results obtained through the experiment and list the methodology used

Answers

Answer:

Analyze the data collected and check to see if they support the hypothesis

Explanation: A. if they already have the raw data thaat means they have already conducted the experiment so it can not be the first option

B. once you collect the raw data you need to check if it corresponds to the data so it would have to be the 2 option.

Hopes this helps.

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