Compute the size of the charge necessary for two spheres separated by 1m to be attached with the force of 1N. How many electrons is this charge?

Answers

Answer 1

Answer:

[tex]q\approx 6.6\cdot 10^{13}~electrons[/tex]

Explanation:

Coulomb's Law

The force between two charged particles of charges q1 and q2 separated by a distance d is given by the Coulomb's Law formula:

[tex]\displaystyle F=k\frac{q_1q_2}{d^2}[/tex]

Where:

[tex]k=9\cdot 10^9\ N.m^2/c^2[/tex]

q1, q2 = the objects' charge

d= The distance between the objects

We know both charges are identical, i.e. q1=q2=q. This reduces the formula to:

[tex]\displaystyle F=k\frac{q^2}{d^2}[/tex]

Since we know the force F=1 N and the distance d=1 m, let's find the common charge of the spheres solving for q:

[tex]\displaystyle q=\sqrt{\frac{F}{k}}\cdot d[/tex]

Substituting values:

[tex]\displaystyle q=\sqrt{\frac{1}{9\cdot 10^9}}\cdot 1[/tex]

[tex]q = 1.05\cdot 10^{-5}~c[/tex]

This charge corresponds to a number of electrons given by the elementary charge of the electron:

[tex]q_e=1.6 \cdot 10^{-19}~c[/tex]

Thus, the charge of any of the spheres is:

[tex]\displaystyle q = \frac{1.05\cdot 10^{-5}~c}{1.6 \cdot 10^{-19}~c}[/tex]

[tex]\mathbf{q\approx 6.6\cdot 10^{13}~electrons}[/tex]


Related Questions

A 500-N box is at rest on the floor. Dennis Elbo makes several
attempts to move the box, pushing against the box with varying
amounts of horizontal force. Yet the box never does move. In this
situation, the amount of static friction force experienced by the box
Select all that apply.
-
0 is 500 N
O is equal to the force with which Dennis exerts on the
box
has an upper limit and Dennis O has not yet exceeded the upper limit
Ois always the coefficient of friction multiplied by the normal force value

Answers

Answer:

Select the second and the third options you listed.

Explanation:

Select the answer options:

"is equal to the force with which Dennis exerts on the  box."

and

"has an upper limit and Dennis has not yet exceeded the upper limit."

In fact, this upper limit of the static friction force is the product of the coefficient of static friction ([tex]\mu[/tex]) times the weight of the box.

What is the volume of a box if he has Length=7 cm Width=5cm , Height=10cm ?

Answers

Answer:

Volume of Cuboid = Height*Width*Length

Explanation:

Volume of Cuboid = 10*5*7

= 350 cu² cm

Answer:

Diagram:-

[tex]\setlength{\unitlength}{0.74 cm}\begin{picture}\thicklines\put(5.6,5.4){\bf A}\put(11.1,5.4){\bf B}\put(11.2,9){\bf C}\put(5.3,8.6){\bf D}\put(3.3,10.2){\bf E}\put(3.3,7){\bf F}\put(9.25,10.35){\bf H}\put(9.35,7.35){\bf G}\put(3.5,6.1){\sf 5cm}\put(7.7,6.3){\sf 7cm}\put(11.3,7.45){\sf 10cm}\put(6,6){\line(1,0){5}}\put(6,9){\line(1,0){5}}\put(11,9){\line(0,-1){3}}\put(6,6){\line(0,1){3}}\put(4,7.3){\line(1,0){5}}\put(4,10.3){\line(1,0){5}}\put(9,10.3){\line(0,-1){3}}\put(4,7.3){\line(0,1){3}}\put(6,6){\line(-3,2){2}}\put(6,9){\line(-3,2){2}}\put(11,9){\line(-3,2){2}}\put(11,6){\line(-3,2){2}}\end{picture}[/tex]

Required Answer:-

It is a cuboid

where

length =l=7cmwidth=b=5cmheight =h=10cm

As we know that in a cuboid

[tex]{\boxed{\sf Volume=lbh}}[/tex]

Substitute the values

[tex]{:}\longrightarrow[/tex][tex]\sf Volume =7×5×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=35×10 [/tex]

[tex]{:}\longrightarrow[/tex][tex]\sf Volume=350cm^3 [/tex]

A uniform electric field has a magnitude of 10 N/C and is directed upward. A charge brought into the field experiences a force of 50 N downward. The charge must be:_______.

Answers

Answer:

q = 5 C

Explanation:

The electric field is defined as the force experienced by a unit charge when it is brought into the field. Hence, the formula used to find the electrical field is given as follows:

E = F/q

where,

E = Electric Field Magnitude = 10 N/C

F = Force Experienced by the test charge = 50 N

q = Magnitude of the Charge = ?

Therefore,

10 N/C = 50 N/q

q = 50 N/(10 N/C)

Therefore,

q = 5 C

What physical property does the symbol I_enclosed in problem 5 represent? a. The current along the path in the same direction as the magnetic field b. The current in the path in the opposite direction from the magnetic field c. The total current passing through the loop in either direction d. The net current through the loop

Answers

Answer:

C

Explanation:

Current passing through the loop in either direction

A medicine ball has a mass of 5kg and is thrown with a speed of 3 m/sec what is it's kinetic energy

Answers

KE = 1/2mv^2

KE = 1/2 (5kg)(3m/s)

KE = 22.5 J

when you stir a cup of tea the floating clip collect at the centre of the Cup rather than in the outer Rim why​

Answers

Answer:

hsvshxansjusjsnwjwisks

Explanation:

When we stir a cup of tea we create a force in the center which pulls out all the particles towards it this is the basic reason for collection of tea leaves at the center of the cup rather than at the rim of the cup, it is similar to the the case of tornado where it takes all the particles present on it way to its ..

A parallel-plate capacitor with circular plates of radius 40 mm is being discharged by a current of 6.0 A. At what radius (a) inside and (b) outside the capacitor gap is the magnitude of the induced magnetic field equal to 75% of its maximum value?(c) What is that maximum value?

Answers

Answer:

A) r = 0.03 m

B) r = 0.0533 m

C) B_max = 0.00003 T

Explanation:

Formula for magnetic field inside the capacitor when it is parallel to the length element is;

B_in = (μ_o•I•r/(2πR²)

Formula for maximum magnetic field is;

B_max = (μ_o•I/(2πR)

Formula for magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

A) Magnetic field inside the capacitor is gotten from our first equation above;

B_in = (μ_o•I•r/R²)

Since we want to find the radius at which the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value.

Thus;

B_in = 0.75B_max

(μ_o•I•r/(2πR²) = 0.75((μ_o•I/(2πR))

μ_o•I and 2πR will cancel out to give;

r/R = 0.75

r = 0.75R

We are given R = 40 mm = 0.04 m

r = 0.75 × 0.04

r = 0.03 m

B) magnetic field outside the capacitor is; B_out = (μ_o•I/(2πr)

Thus for the magnitude of the induced magnetic field equal to 75% or 0.75 of its maximum value:

B_out = 0.75B_max

(μ_o•I/(2πr) = 0.75((μ_o•I/(2πR))

μ_o•I and 2π will cancel out to give;

1/r = 0.75/R

r = R/0.75

r = 0.04/0.75

r = 0.0533 m

C) B_max = μ_o•I/(2πR)

μ_o is a constant known as vacuum of permeability with a value of 4π × 10^(-7) T.m/A

Thus;

B_max = (4π × 10^(-7) × 6)/(2π × 0.04)

B_max = 0.00003 T

An electron is accelerated from rest through a potential difference of 300V. it then passes through a uniform 0.001-T magnetic field, oriented perpendicular to the electrons velocity. what is the magnitude of the magnetic force on the electron?

Answers

Given :

Potential difference, V = 300 V.

Magnetic Field, B = 0.001 T.

To Find :

The magnitude of the magnetic force on the electron.

Solution :

We know, for perpendicular orientation, force is given by :

[tex]F = qVB\\\\F = 1.6 \times 10^{-19} \times 300\times 0.001\ N\\\\F = 4.8\times 10^{-20}\ N[/tex]

Hence, this is the required solution.

Approximating Venus's atmosphere as a layer of gas 50 km thick, with uniform density 21 kg/m3, calculate the total mass of the atmosphere.Express your answer using two significant figures.m venus atmosphere = ____ kg

Answers

Answer:

m = 4.9 10⁸ kg

Explanation:

The expression for the density is

           ρ = m / V

           m = ρ V

the volume of the atmosphere is the volume of the sphere of the outer layer of the atmosphere minus the volume of the plant

          V = V_atmosphere - V_planet

           V = 4/3 π R_atmosphere³ - 4/3 π R_venus³

           V = 4/3 π (R_atmosphere³ - R_venus³

)

the radius of the planet is R_venus = 6.06 10⁶ m.

The radius of the outermost layer of the atmosphere

          R_atmosphere = 50 10³ + R_ venus = 50 10³ + 6.06 10⁶

           R_atmosphere = 6.11 10⁶ m

let's find the volume

           V = 4/3 pi [(6,11 10⁶)³ - (6,06 10⁶)³]

            V = 23,265 10⁶ m³

let's calculate the mass

          m = 21  23,265 10⁶

          m = 4.89   10⁸ kg

with two significant figurars is

          m = 4.9 10⁸ kg

A 8.45μC particle with a mass of 6.15 x 10^-5 kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m. How much time will it take for the particle to complete one orbit?

a. 92.7 s
b. 0.0927 s
c. 9.27 s
d. 927 s

Answers

This question is incomplete, the complete question is;

A 8.45μC particle with a mass of 6.15 x 10⁻⁵ kg moves perpendicular to a 0.493-T magnetic field in a circular path of radius 34.1 m.

How much time will it take for the particle to complete one orbit?

a. 92.7 s

b. 0.0927 s

c. 9.27 s

d. 927 s

Answer:

it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

Explanation:

Given that;

mass m =  6.15 x 10⁻⁵ kg

q = 8.45μC = 8.45 × 10⁻⁶ C

B = 0.493

we know that

Time period T = 2πr / V

where r = mv/qB

so T = 2πm/qB

we substitute

T = (2 × 3.14 × 6.15 x 10⁻⁵) / ( 8.45 × 10⁻⁶ × 0.493)

T = 0.0003862 / 0.000004165

T = 92.7 sec

Therefore it will take 92.7 seconds for the particle to complete one orbit.

Option a) 92.7 s is the correct option

An extraterrestrial creature is standing in front of plane mirror. The height of this creature is H and we know that this creature has eyes positioned h below the top of its head. This creature sees its reflection which fit exactly the mirror, it means, this creature can just see the top of head and the bottom of its feet (or whatever it uses for motion). We can conclude that the top of a mirror is exactly:________

a. H/2 above the ground
b. H above the ground
c. (H-h/2) above the ground
d. (H-h) above the ground
e. We can not guess anything without information about the nature of this creature.

Answers

Answer:

c. (H-h/2) above the ground

Explanation:

The mirror must be at least half as tall as the alien, and its base must be located at half of the distance between the alien's eyes and the ground (assuming that the alien doesn't float or levitate).

This question is about the Law of Reflection which states that the angle of reflection = angle of incidence.

I attached an image that can help you understand the concept, although the alien is not included.

What percentage of an iron anchor’s weight will be supported by buoyant force when submerged in salt water?

Answers

Answer:

0.87

Explanation:

To solve this, we use the principle of Archimedes. Archimedes Principal of flotation states that "the buoyant force of an object is equal to the total weight of the fluid it displaces."

In the attachment, I stated the mathemacal formula, of which

F(B) = The buoyant force

w(fl) = The weight of the salt water displaced

p(iron) = density of iron

p(salt) = density of the salt water = 1025 kg/m³

F' = weight of the iron in air

F = weight of the iron in salt water

p(man) = density of man = 7680 kg/m³

The rest are the easy calculations done by substituting the values


Find the mass of an object on planet F if its weight is 650 N (g = 13m/s^2)

Answers

Answer:

the object's mass is 50 kg

Explanation:

We use Newton's second law to solve for the mass:

F = m * a , then   m = F / a

In our case, the acceleration is the gravitational acceleration on the planet, and the force is the weight of the object on the planet. So we get:

m = w / a = 650 N / 13 m/s^2 = 50 kg

Then, the object's mass is 50 kg.

On a surface of a planet of radius R and mass M the acceleration due to gravity is 7m/s?. Consider another planet of radius 2R and mass 0.4M. What would the acceleration due to gravity be on this new planet? Show your calculations.

Answers

Answer:

0.7 m/[tex]s^{2}[/tex]

Explanation:

From Newton's law of universal gravitation,

F = [tex]\frac{GMm}{r^{2} }[/tex]

and from Newton's second law of motion,

F = mg

So that;

mg = [tex]\frac{GMm}{r^{2} }[/tex]

⇒ g = [tex]\frac{GM}{r^{2} }[/tex]

For the first planet,

7 = [tex]\frac{GM}{R^{2} }[/tex]

⇒ G = [tex]\frac{7R^{2} }{M}[/tex] .............. 1

For the second planet,

g = [tex]\frac{0.4GM}{(2R)^{2} }[/tex]

   = [tex]\frac{0.4GM}{4R^{2} }[/tex]

⇒ G = [tex]\frac{4gR^{2} }{0.4M}[/tex] ............. 2

Equating 1 and 2, we have;

[tex]\frac{7R^{2} }{M}[/tex] = [tex]\frac{4gR^{2} }{0.4M}[/tex]

g = [tex]\frac{7R^{2} *0.4M}{4R^{2}M }[/tex]

  = [tex]\frac{7*0.4}{4}[/tex]

  = [tex]\frac{2.8}{4}[/tex]

g = 0.7

Therefore, the acceleration due to gravity on the new planet is 0.7 m/[tex]s^{2}[/tex].

The speed of a space shuttle is 10 / express this in /�

Answers

Answer:

268.22m/s

Explanation:

Given;

    10mile/min to m/s

We need to convert between the two units;

    1 mile  = 1609.34m

     60s  = 1min

Now;

    10 x [tex]\frac{mile}{min}[/tex] x [tex]\frac{1min}{60s}[/tex] x [tex]\frac{1609.34m}{1mile}[/tex]

  = 268.22m/s

explain the relationship among visible light, the electromagnetic spectrum, and sight.

Answers

Explanation:

The electromagnetic spectrum is the name given to the full range of frequencies and/or wavelengths that electromagnetic phenomena may have.

Human eyes respond to a small range of wavelengths in that spectrum. That response is called sight. Because humans can see that electromagnetic energy, it is called visible light.

Any five physics problems

Answers

Explanation:

There are still some questions beyond the Standard Model of physics, such as the strong CP problem, neutrino mass, matter–antimatter asymmetry, and the nature of dark matter and dark energy.

what is gathering and analyzing information about an object without physical contact with the object

Answers

Answer:

Remote Sensing

Explanation:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answers

Complete Question:

A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.

They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet

Answer:

aw = 3 i + 6 j m/s2

Explanation:

Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:

       [tex]a_{c} = \omega^{2} * r (1)[/tex]

Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.

       ∴ ωp = ωw (2)

           ⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]

               [tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]

Dividing (4) by (3), from (2), we have:

        [tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]

Solving for aw, we get:

        [tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]

The components of lifetime fitness include all of the following components except

Answers

Answer:it’s A

Explanation:

because i took the quiz

Answer:

D is the correct answer, not A

Explanation:

Objects falling through air are slowed by the force of air resistance. Which objects were slowed the most by air resistance?

Answers

Answer:

This question is incomplete

Explanation:

This question is incomplete. However, it should be noted that when objects of different sizes fall in absence of air resistance, the objects will get to the ground at the same time. But with the presence of air resistance, the heaviest object gets to the ground first; meaning it has the least air resistance while the lightest object will arrive at the ground last because it has the greatest air resistance and is slowed down the most by the air resistance.

Thus, the lightest object in the completed question is the answer.

A current of 3.75 A in a long, straight wire produces a magnetic field of 2.61 μT at a certain distance from the wire. Find this distance.

Answers

Given :

Current, I = 3.75 A .

Magnetic Field, [tex]B = 2.61\times 10^{-4}\ T[/tex]

To Find :

The distance from the wire.

Solution :

We know,

[tex]B = K\dfrac{2i}{d}\\\\d = 10^{-7}\times \dfrac{2\times 3.75}{2.61\times 10^{-4}}\\\\d = 0.00287\ m \\\\d = 2.87\times 10^{-3}\ m[/tex]

Hence, this is the required solution.

A student, starting from rest, slides down a water slide. On the way down, a kinetic frictional force (a nonconservative force) acts on her. The student has a mass of 71.0 kg, and the height of the water slide is 12.7 m. If the kinetic frictional force does -5.10 103 J of work, how fast is the student going at the bottom of the slide

Answers

Answer:

[tex]10.27m/s[/tex]

Explanation:

Given data

work W= -5.10 10^3 J

mass m= 71kg

final height of slide h2= 12.7m

initial height of slide h1=0m

initial velocity v1= 0m/s

final velocity v2=?

Step two:

required

Final velocity

The work-energy theorem is expressed as'

[tex]W=1/mv_2^2 +mgh_2-(1/mv_1^2+mgh_1)[/tex]

make V2 subject of formula we have final speed

[tex]v_2=\sqrt{\frac{2W}{m}+v_1^2-2g(h_1-h_2) } \\\\[/tex]

substitute our given data we have

[tex]v_2=\sqrt{\frac{2*(-5.1*10^3)}{71}+0^2-2*9.81(12.7) } \\\\v_2=\sqrt{143.66-249.174 } \\\\v_2=\sqrt{105.514 } \\\\v_2=10.27m/s[/tex]

The student going at  10.27m/s

If a ball rolls across a table to the left with constant speed, which of the following is true about the force(s) on the ball? (3a1)

Question 10 options:

There cannot be any forces applied to the ball.


There must be exactly one force applied to the ball.


The net force applied to the ball is zero.


The net force applied to the ball is directed to the right.

Answers

Answer:

C. The net force applied to the ball is zero.

Explanation:

From Newton's second law of motion;

F = ma

Where F is the force on an object, m is its mass and a is its acceleration.

Therefore, the force on an object is a product of its mass and acceleration as it travel from one point to another.

Since acceleration relates to the rate of change in velocity to time. Then when the object moves at uniform velocity (especially along a straight path), its acceleration is zero.

So that;

F = m x 0

  = 0

No force is applied on the object.

Therefore for the ball in the given question, the net force applied to the ball is zero because it rolls with constant speed along a straight path.

A bullet of mass 4.2 g strikes a ballistic pendulum of mass 1.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

Answers

Answer:

449.020m/s

Explanation:

We have been provided with the following information:

Mass of bullet = Mb = 4.2g

Mass of pendulum = 1.0 = Md

Vertical distance = h = 18cm

Gravitational force = g = 9.8

We have kinetic energy converted to potential energy for the entire system

1/2(Mb+Md)V² = (Mb + Md)gh

We have V as the speed of system during collision

Mbv = (Mb+Md)V

We divide through by Mb

v = (Mb+Md/Mb)√2gh

4.2x10^-3 = 0.0042

18cm = 0.18m

(0.0042+1.0/0.0042)√2x9.8x0.18

= 1.0042/0.0042√3.528

= 239.095x1.878

= 449.020m/s

This is the answer to the question

Thank you!

We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?

Answers

Answer:

A) refraction experiment   n = n₁ sin θ₁ / sin θ₂

B)  n_A = 1.19 ,    n_B = 1.53

Explanation:

A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index

            n₁ sin θ₁ = n₂ sinθ₂

            n₂ = n₁ sin θ₁₁ /sin θ₂

If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is

            n = n₁ sin θ₁ / sin θ₂

B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be

         

material A

             n_A = sin 50 / sin 40

             n_A = 1.19

material B

              n_B = sin 50 / sin30

              n_B = 1.53

Sona wants to install room heater in her living room. She had only two options, either to install heater at the top of the window or near the ground level. . Which method of installing room heater would be the effective way . Why or Why not ?

Answers

Installing the heater near ground level is the better option because heat rises so putting it near the bottom will allow the warmth to rise and warm up more of the air in the room. Especially if it’s below the window because it will heat the cool air seeping through the window.

F = 5 Newtons
W = 75 Joules
d = ?

ANSWER

Answers

Substitution: d = 75 J/5 N
Answer with unit of measure: d = 15 m

Which statement best describes a characteristic of gases?

Gases can be compressed, or squeezed together.
The particles of gases are packed close together.
The particles of gases spread our vertically instead of horizontally.
Gases have a definite shape and volume.

Answers

Answer:

A

Explanation:

I'm pretty sure it's A. That's the one that makes the most sense and checks out

Answer:

Just here to confirm that it is A

Explanation:

The sound intensity at 4 m from a source is 100 W/me. What is the intensity of the sound at 12 m away from the source ?

Answers

Answer:

Intensity at 12 meters will be 11.11 W/m^2

Explanation:

Recall that the intensity of sound is inversely proportional to the square of the distance from the source. Therefore, if at 4 m the intensity is 100 W/m^2

we have: 100 W/m^2 = k/16 and therefore, k = 1600 W

Then the intensity (I) at 12 m will be:

I = k/12^2 = 1600/144  W/m^2 = 11.11 W/m^2

Other Questions
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