Two objects are identical and small enough that their sizes can be ignored relative to the distance between them, which is 0.189 m. In a vacuum, each object carries a different charge, and they attract each other with a force of 1.39 N. The objects are brought into contact, so the net charge is shared equally, and then they are returned to their initial positions. Now it is found that the objects repel one another with a force whose magnitude is equal to that of the initial attractive force. What is the initial charge on each object, part (a) being the one with the greater (and positive) value and part (b) being the other value?

Answers

Answer 1

Answer:

The charges are + 74.3 μC and - 74.3 μC

Explanation:

Let the charges be q and q'.

Since the charges initially attract each other with a force of 1.39 N, the force of attraction is given by

F = kqq'/r² where k = 9 × 10⁹ Nm²/C² and r = distance between the charges = 0.189 m

When the charges are brought together, they share their charge equally and have a net charge of (q + q')/2 each.

They now repel each other.

So, the magnitude of the force of repulsion is given by

F' = k[(q + q')/2][(q + q')/2]/r²

F' = k[(q + q')²/4r²

Since the magnitude of the force of attraction and repulsion are the same, we have that

F = F'

kqq'/r² = k[(q + q')²/4r²

qq' = (q + q')²/4

(q + q')² = 4qq'

q² + 2qq' + q'² = 4qq'

q² + 2qq' - 4qq' + q'² = 0

q² - 2qq' + q'² = 0

(q - q')² = 0

q - q' = 0

q = q'

Substituting q = q' into F, we have

F = kqq'/r²

F = kq²/r²

making q subject of the formula, we have

q² = Fr²/k

q = √(Fr²/k)

q = r√(F/k)

Substituting the values of the variables into the equation, we have

q = 0.189 m√(1.39 N/9 × 10⁹ Nm²/C²)

q = 0.189 m√(0.15444 × 10⁻⁹ Nm²/C²)

q = 0.189 m(0.3923 × 10⁻³ C/m)

q = 0.0743 × 10⁻³ C

q = 74.3 × 10⁻³ × 10⁻³ C

q = 74.3 × 10⁻⁶ C

q = 74.3 μC

Since q and q' initially attract, it implies that they initially had opposite charges.

So, q = 74.3 μC and q' = -74.3 μC

So, the charges are + 74.3 μC and - 74.3 μC


Related Questions

A cello and an organ are playing together. The organ plays a pitch of C with a frequency of 65.4 Hz in a pipe open at both ends. The cello plays its C string (a string fixed at both ends), and a beat frequency of 2.5 Hz is heard. The cellist loosens the string until no beat frequency is heard. what is the length of the pipe

Answers

Answer:

λ/2 = length of pipe

The pipe is open at both ends having wavelength of A-N-A or antinode-node-antinode which is 1/2 wavelength

λ = 331 m/s / (65.4 / s) = 5.06 m     wavelength of sound

L = 5.06 m / 2 = 2.53 m      length of pipe

Where do inherited traits come from​

Answers

Answer:

Your parents or anyone in your ancestry.

Explanation:

A pendulum with a length of 2 m has a period of 2.8 s. What is the period of a pendulum with a length of 8 m

Answers

Answer:

P = 2 pi (L / g)^1/2

P2 / P1 = (8 / 2)^1/2 = 2

The period would be twice as long or 5.6 sec.

Draw the graphs for exothermic and endothermic reactions Label: axes, reactants, products, Energy of reaction, heat energy change​

Answers

Answer:

axes, reactants, products, Energy of reaction, heat energy

Explanation:

axes, reactants, products, Energy of reaction, heat energyaxes, reactants, products, Energy of reaction,grreactantsaphs heat energy

An object, initially traveling at a velocity of 73 m/s, experiences an acceleration of -9.8 m/s^2. How much time will it take it to come to rest?

Answers

7.4 s

Explanation:

Given:

[tex]v_0 = 73\:\text{m/s}[/tex]

[tex]v = 0[/tex]

[tex]a = -9.8\:\text{m/s}^2[/tex]

[tex]t = ?[/tex]

To solve the time it takes for the object to come to a stop, we are going to use the equation below:

[tex]v = v_0 + at \Rightarrow t = \dfrac{v - v_0}{a}[/tex]

Using the given values above, we get

[tex]t = \dfrac{0 - 73\:\text{m/s}}{-9.8\:\text{m/s}^2}[/tex]

[tex]\;\;\;\;= 7.4\:\text{s}[/tex]

. Prior to Remy's trip to Cleveland, his uncle tells him about this amazing barbecue restaurant there and raves about
the food and live music. Remy looks it up once he gets to Cleveland, and he finds that there are a number of
negative reviews about it. Nevertheless, he focuses on the few positive reviews he comes across, convinces himself
they are probably more valid than the negative ones, and decides to go to the restaurant after all. Remy has likely
been influenced by:
the availability heuristic.
confirmation bias.
O the representativeness heuristic.
oming

Answers

whats the restraunt called

Based on the scenario, Remy has likely been influenced by framing effect.

What is framing effect?

In Psychology, framing effect can be defined as a cognitive bias wherein a person's choice from a group of options is primarily influenced by the positive or negative connotations pertaining to the options.

In this scenario, Remy did a research on the barbecue restaurant and finds that there are a number of both negative and positive reviews about it. However, he convinced himself by relying on the positive reviews and decides to go to the restaurant.

In conclusion, we can deduce that Remy has likely been influenced by framing effect.

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A car travels a certain distance from A to B with a speed of 60km/hr and then returns along the same path to the starting point with a speed of 40km/hr. Find the average speed and average velocity.
a) Km/hr
b) m/s

wrong answers will be reported!

Answers

Answer:

Explanation:

Speed is total distance traveled over time taken to do so.

If AB is measured in kilometers, time (t) for the whole trip is

t = AB/60 + AB/40  

t = 2AB/120 + 3AB/120

t = 5AB/120 hrs

Average speed is distance over time

s = 2AB / (5AB/120)

s = 2(120)/5

s = 48 km/hr

s = 48(1000 m/km / 3600 s/hr) = 13.333333.... 13 m/s

Velocity is displacement over time.

As displacement is zero, velocity is zero

v = 0 km/hr = 0 m/s

Pretty harsh reporting answers just because they are wrong.

If a person climbed Mt. Everest has a mass of 105 kg and a weight of 625 N what would be the acceleration due to gravity?

Answers

The acceleration due to gravity would be 5.95 m/s²

A force is known to be a push or pull and it is the change in momentum per time. It can be expressed by using the relation.

Force = mass × acceleration.

From the parameters given:

Mass = 105 kgForce = 625 N

By replacing the given values into the above equation, we can determine the acceleration.

625 N = 105 kg × acceleration.

[tex]\mathbf{acceleration = \dfrac{625 \ N}{105 \ kg}}[/tex]

acceleration = 5.95 N/kg

Since 1 N/kg = 1 m/²

acceleration = 5.95 m/s²

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When a hot metal cylinder is dropped into a sample of water, the water molecules

Answers

Answer:

I believe the answer is speed up.

Explanation:

this is because when water heats up the molecules move father apart from each other they speed up, eventually causing the water to boll

12) A horizontal force of 200 N is applied to move a 55-kg cart (initially at rest) across a 10 m level surface. What is the final speed of the cart? [hint: use work – energy principle] [3 marks]

Answers

Hi there!

We can use the following:

W = ΔKE = F · d

Find the work done on the cart:

W = 200 · 10 = 2000 J

Now, this is equal to the change in kinetic energy of the object. Its initial kinetic energy is 0 J since it starts from rest, so:

2000J = KEf - KEi

KE is given as:

[tex]KE = \frac{1}{2}mv^2[/tex]

2000J = 1/2(55)v²

4000 = 55v²

√(4000/55) = 8.53 m/s


Scenario 3: You are driving through a rain storm talking to your family but you can only hear every other word. What 2 medias are the waves passing in this scenario?

Answers

The sound waves are passing through the sound of the rain and the air around you

48.36
g.
MgSO4 to motes

Answers

Answer:120.3676

Explanation: using the molecular calculator and molar mass of MgSO4. hope this helps!

A car move at an initial velocity of 240m and reach at the final velocity of 540m in 8hours. calculate its acceleration.​

Answers

Answer:

a = 0.01m/s²

Explanation:

V_f = V_0+a*t

V_f = Velocity final

V_0 = Velocity initial

a = acceleration

t = time

a = (V_f-V_0)/t

a = (540m/s-240m/s)/((8hr)*(60min/1hr)*(60s/1min))

a = 0.01m/s²

WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!

A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.

Please show work.​

Answers

Answer:

Explanation:

a) the acceleration of the puck on the rough ice.

a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²

 (comes from μ = F/N = ma/mg = a/g)

b) the distance from the end boards the puck is when it comes to a stop.

v² = u² + 2as

0² = 12.0² + 2(-5.40)s

s = 13.3 ft

so distance from the boards is

15.7 - 13.3 = 2.4 m

by the way...that's some VERY rough ice...more like sand.

in a compoumd are atoms physically or chemically combined

Answers

Answer:

They are...if I'm correct Chemically combined, sorry if I'm wrong.

if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?

Answers

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

PLEASE HELP FOR PHYSICS!
All objects exert a gravitational force on all other objects. This force is given by, F = GMm r2 , where the value of G = 6.673 × 10–11 N–m2/kg2 , M is the mass of the heavier object, m is the mass of the lighter object, and r is the distance between the two objects.
What is the force of gravity between two balls of mass 50 kg each if the distance between them is 25 m. Assume that there is no interference from any other gravitational field.

Answers

Hi there!

Recall Newton's Law of Universal Gravitation:

[tex]\large\boxed{F_g = G\frac{m_1m_2}{r^2}}[/tex]

Where:

Fg = Force of gravity (N)

G = Gravitational Constant

m1, m2 = masses of objects (kg)

r = distance between objects (m)

Plug in the given values stated in the problem:

[tex]F_g = (6.673*10^{-11})\frac{50 * 50}{25^2} = \boxed{2.669 * 10^{-10} N}[/tex]

If a 35 kg box collides with a stationary 120 kg box with a force of 90 N, what must be true of the magnitude of the reaction force?

Answers

Newton's third law allows to find the result for the value of the reaction force during the collision is:

The reaction force is F = 90 N and is applied to the lighter body.

Newton's third law stable that the forces appear in pairs or ea that when two bodies interact, the interaction forces appear in the two bodies simultaneously, in general they are called action and reaction forces.

These furas are of the same magnitude, but in the opposite direction, each one applied to one of the bodies.

They indicate that the most lighter body collides with the one with the greatest mass with a force of F = 90 N. If we call this the action, the larger body must react with a force of equal magnitude on the lighter body.

Consequently, the reaction force is F = 90 N directed towards the lighter body.

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a 4kg sample of water absorbs 500 joules of energy.How much did the water temp change.The specific heat of water is 4200 J/(kg C)

Answers

Answer:

0.02976°C

Explanation:

Heat Supplied = mcø

ø = 500/(4*4200)

ø = 0.02976 °C

why does gas have the most energy but moves the slowest

Answers

Gases have heavier molecules. Since all gases have the same average kinetic energy at the same temperature, lighter molecules move faster and heavier molecules move slower on average.

The qualitative equivalent of external validity is:
A- Credibility

B- Dependability

C- Transformability

D- Confirmability

Answers

c transformability i think

What is an ellipse?

a plane that slices between orbits

an oval-shaped orbit

a circular orbit

the center of gravity between orbiting objects​

Answers

Answer:

i think it's C thx correct me if wrong

What is the approximate value of k when 30 = e^5k?

Answers

Answer:

Explanation:

30 = e^5k

ln30 = lne^5k

ln30 = 5k

k = ln30/5

k = 0.68023947...

round to your heart's content.


A wheel with radius 41.5 cm rotates 5.13 times every second.
Find the period of this motion.
What is the tangential speed of a wad of chewing gum stick to the rim of the wheel?

Answers

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

Let suppose that the wheel rotates at constant angular speed ([tex]\omega[/tex]), in radians per second, the tangential speed of a wad of chewing gum to the rim of the wheel ([tex]v[/tex]), in centimeters per second, is:

[tex]v = 2\pi\cdot r\cdot f[/tex] (1)

Where:

[tex]r[/tex] - Radius of the wheel, in centimeters[tex]f[/tex] - Frequency, in hertz

If we know that [tex]f = 5.13\,hz[/tex] and [tex]r = 41.5\,cm[/tex], then the tangential speed of the chewing gum is:

[tex]v = 2\pi\cdot (41.5\,cm)\cdot (5.13\,hz)[/tex]

[tex]v \approx 1337.659\,\frac{cm}{s}[/tex]

The tangential speed of a wad of chewing gum to the rim of the wheel is approximately 1337.659 centimeters per second.

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In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block

Answers

Answer:

Here we use the conservation of momentum theorem.

m stands for mass, and v stands for velocity. The numbers refer to the respective objects.

m1v1 + m2v2 = m1vf1 + m2vf2

Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.

m1v1 + m2v2 = vf(m1 + m2)

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

Please help me.............................

Answers

Answer:

[tex]a[/tex]

Explanation:

is a corect anser

i just want an answer please

Answers

Answer: An answer on what? I’ll never ignore you!

Explanation:

Answer:

an answer on what?

Explanation: Im here to help!!

20 . A car of mass 2000 kg is moving with a constant velocity of 10 m/s due east. What is the momentum of the car

Answers

Answer:

P=mv

Explanation:

m = 2000kg

v = 10m/s

2000×10=20000

Answer: 2000kgm/s

The momentum will be =  2 * [tex]10^{4}[/tex] kg m/s

What is momentum ?

Momentum is a property of a moving body that the body has by virtue of its mass and motion and that is equal to the product of the body's mass and velocity

momentum = mass * velocity

mass = 2000 kg

velocity = 10 m/s

momentum = 2000 * 10 = 2 * [tex]10^{4}[/tex] kg m/s

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Marks: 1
How much does a man weigh if it takes 60J of energy to climb onto a
table that is 75cm tall?
4500 N
45 N
80 N
0.8 N

Answers

Answer:

Explanation:

E = mgh

mg = E/h

mg = 60/0.75 = 80 N

A 30-cm-tall, 4.0-cm-diameter plastic tube has a sealed bottom. 250 g of lead pellets are poured into the bottom of the tube, whose mass is 30 g, then the tube is lowered into a liquid. The tube floats with 5.0 cm extending above the surface. What is the density of the liquid

Answers

The density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

What is density?

The density of an object is defined as the ratio of the mass of an object to the volume of the object.

Volume of tube = 2^2 * pi * 30 = 377 cm^3

Volume of tube submerged = 25* 377 / 30 = 314 cm^3

Buoyancy = weight of liquid displaced

Volume of liquid displaced = 314 cm^3

Mass of tube and lead = 250 + 30 = 280 g

Now from the mass density by definition

[tex]\rho = \dfrac{m}{v}[/tex]

[tex]m=\rho \times v[/tex]

Mass of liquid displaced = Mass being supported

[tex]314 \times \rho = 280 g[/tex]

[tex]\rho= \dfrac{280}{ 314 } = .892 \frac{g}{cm^3}[/tex]

Thus the density of the liquid will be equal to   [tex]\rho=0.892 \ \dfrac{g}{cm^3}[/tex]

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The density of the liquid is 1.67 g/[tex]cm^3[/tex].

The volume of the tube is

30 * 4 * 3.14 * 0.25 = 94.2 [tex]cm^3[/tex].

The mass of the lead pellets and the plastic tube is

30 + 250 = 280 g.

The volume of the lead pellets is

250 / 11.34 = 22 [tex]cm^3[/tex].

The volume of the liquid that the tube displaces is

94.2 - 22 = 72 [tex]cm^3[/tex].

The density of the liquid is

280 / 72 = 1.67 g/ [tex]cm^3[/tex].

Therefore, the density of the liquid is 1.67 g/ [tex]cm^3[/tex].

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