An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton

Answers

Answer 1

5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get

[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]

Taking the square root of r^2, we then get the distance as

[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]

The values are given as follows:

[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]

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

[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]

[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]

Putting in all of these values in our equation for r,

[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]


Related Questions

HOW DO U FEEL WHEN U PLAY OR WATCH BADMINTON?

Answers

Answer:

I feel exited and happy I enjoy it with my friend

Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+

Answers

Answer:

Explanation:

I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was

Conservation of momentum

14000(8.1) + m(0) = (14000 + m)(3.6)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

The mass of the second car is 17500 kg.

What is law of conservation of momentum?

The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.

Newton's third law of motion has a direct impact on the idea of momentum conservation.

mass of the first car = 14000 kg.

Initial speed of first car = 8.1 m/s.

Final speed of the two cars = 3.6 m/s.

From law of conservation of momentum; we can write:

Total initial momentum = total final momentum

14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)

14000(8.1) = 14000(3.6) + m(3.6)

m = 14000(8.1 - 3.6) / 3.6

m = 17500 kg

Hence, the mass of the second car is 17500 kg.

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3. What can you infer about the position of the galaxies 100 million years before this telescope photo was taken

Answers

The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.

A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.

Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.

According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.

Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.

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Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer​

Answers

Answer:

thx for the points

Explanation:

no need brainliest

A car of mass 5000 kg was initially moving at 100 km/h and stops at a distance of 55 m. Find the
magnitude of the net force (in N) acting to stop the car.

Answers

Answer:

 |F| = 35 kN

Explanation:

a = F/m

100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s

v² = u² + 2as

   a = (v² - u²) / 2s

F/m = (v² - u²) / 2s

   F = m(v² - u²) / 2s

   F = 5000(0² - 27.8²) / 2(55)

   F = - 35,072.9517...

tính các mô men quán tính chính trung tâm của tiết diện, cho biết a = 10cm

Answers

Answer:

i dont know need points

Explanation:

I need the answer to this question

Answers

Answer:

I don't see anything so you should repost

How does climate change?

Answers

Answer:

Con el cielo del agua

Explanation:

espero que te ayude

(wheather cycle)

hope it helps

. If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change

Answers

Answer:

a1 = -w^2 A sin w t        where w is the angular frequency

a2 =  - (2 w)^2 A sin 2 w t       where w2 = 2 w1

a2 / a1 = 4 sin 2 w t / sin  w t

Since sin 2 w t / sin w t = 2     sin w t will change twice as fast

a2 / a1 = 8      the max acceleration must change 8 tims as fast

Explain different layers of atmosphere and the pressure in each layer. Draw diagram​

Answers

Answer:

Our atmosphere has five different layers. They are:

1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.

2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.

3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.

4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.

5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.

In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead to be 5 N/m. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective "interatomic spring stiffness" for an oscillator is 4*5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms).
What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?

Answers

Answer:

8.01e-22

Explanation:

The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

Extension produced by one mole of lead

The extension produced by one mole of lead atom is calculated by applying Hooke's law;

F = kx

mg = kx

x = mg/k

x = (0.207 x 9.8) / (20)

x = 0.101 m

Energy stored in the lead block

The Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;

E = ¹/₂kx²

E = ¹/₂ (20)(0.101)²

E = 0.102 J

Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.

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i mark u brainliest answer please answer me​

Answers

Answer:

bathing and wear clean clothes 2 pollute 3throwing waste in land air water pollution

A camera takes a properly exposed photo at f/5.6 and 1/250 s. What shutter speed should be used if the lens is changed to f/4.0?

a.1/65 s
b.1/125 s
c.1/250 s
d.1/500 s
e.1/1000 s

Answers

The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.

The relationship between the shutter speed and the focal ratio is an inverse relationship.

Ratio of Areas = 5.6²/4²

A1/A2 = 1.96 = 2

Since; T1A1 = T2A2

T2 = T1(A1/A2)

T2 = 250 × 2 = 500 seconds

In essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.

Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.

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If you were told an atom was an ion, you would know the atom must have a?
A neutral charge
B charge
C negative charge
D positive charge

Answers

Answer:

its b it must be charge

Explanation:

I need to choose a theme for my physics assignment My experiment is finding g

Answers

Question:

How to find g (acceleration due to gravity)

Solution:

We know,

Acceleration due to gravity (g)

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

where, G = Gravitational constant

[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]

M = Mass of the earth

[tex] = 6 \times {10}^{24} \: kg[/tex]

R = Radius of the earth

[tex] = 6.4 \times {10}^{6} m[/tex]

Putting these values of G, M and R in the above formula, we get

[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]

So, the value of acceleration due to gravity is

[tex]9.8m/s ^{2} [/tex]

Hope it helps.

Do comment if you have any query.

Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:

Answers

The internal energy after heating in terms of U is 100U.

The given parameters;

initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰C

Assuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.

[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]

Thus, we can conclude that the internal energy after heating in terms of U is 100U.

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Consider 5.00 mol of liquid water. (a) What volume is occupied by this amount of water? The molar mass of water is 18.0 g/mol. (b) Imagine the molecules to be, on average, uniformly spaced, with each molecule at the center of a sma11 cube. What is the length of an edge of each sma11 cube if adjacent cubes touch but don't overlap? (c) How does this distance compare with the diameter of a molecule?

Answers

This question involves the concepts of density, volume, and mass.

(a) The volume occupied by this amount of water is "90 m³".

(b) The length of an edge of each small cube is "83 μm".

(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.

(a)

The volume can be found using the following formula:

[tex]V = \frac{nM}{\rho}[/tex]

where,

V = volume occoupied = ?

n = no. of moles = 5 mol

M = molar mass = 18 g/mol

[tex]\rho[/tex] = density of water = 1 g/m³

Therefore,

[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]

V = 90 m³

(b)

First, we will find the volume of an individual molecule:

[tex]V_i =\frac{V}{nN_A}[/tex]

where,

[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol

Therefore,

[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]

Vi = 3 x 10⁻²³ m³

This volume can be given as a volume of the sphere:

[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]

r = 4.15 x 10⁻⁷ m

Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)

d = 8.3 x 10⁻⁷ m = 83 μm

since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.

Edge Length = L = d

L = 8.3 x 10⁻⁷ m = 83 μm

(c)

The edge length is equal to the diameter of the molecule.

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49. A particle starts from rest at time t=0 and movies along the x axis. if the net force on is proportional to t its kinetic energy is proportional to? ​

Answers

Answer:

F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]

Explanation:

In pic

_________________

(hopet his helps can I pls have brainlist (crown)☺️)

HELPP

When two forces are in opposite directions, and they are the exact same magnitude, the forces will _______.

a. subtract from each other
b. cancel out
c. go on infinitely
d. eventually reach equilibrium

Answers

Answer:

i think the correct answer is B. cancel out

name two examples where the cohesive force dominates over the adhesive force and vice versa​

Answers

Attractive forces between molecules of the same type are called cohesive forces. ... Attractive forces between molecules of different types are called adhesive forces. Such forces cause liquid drops to cling to window panes, for example.

Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.

What is the difference between cohesive force and adhesive force?

Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.

The force that draws molecules of the same substance together is called the cohesive force. The force that holds molecules of various substances together is known as the adhesive force.

Between molecules of the same substance, there are cohesive forces. There is a natural tendency to resist separation due to these intermolecular forces between like elements. Conversely, adhesive forces draw disparate molecules together.

In physics, cohesion refers to the intermolecular attraction that exists between two adjacent parts of a substance, especially one that is solid or liquid. A piece of matter is held together by this force. Adhesion is a term for the intermolecular forces that act when two dissimilar substances come into contact.

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can someone help me answer this question? A student connected an ammeter as shown in this picture. Did the student connect the ammeter correctly? Explain

Answers

Answer:

No - It is connected in parallel instead of series

Explanation:

The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.

What is Ammeter?

An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.

The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).

The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.

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During a picnic, you and two of your friends decide to have a three-way-tug-of-war, with three ropes in the middle tied into a knot. Roberta pulls to the west with 277 N of force; Michael pulls to the south with 603 N. a) With what force should you pull to keep the knot from moving

Answers

Answer:

Explanation:

I'm not 10% sure but i think that if you were to do:

603 + 277 = 880

1000 - 880 = 120

So to keep it from moving, i think you would need to pull with a force of 120N

The force with which it should be pulled to keep the knot from moving is 663.58 N.

What is Newton's third law ?

Newton's third law states that, for every action, there is an equal and opposite reaction.

Here,

Force exerted by Roberta, F₁ = 277 N

Force exerted by Michael, F₂ = 603 N

Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.

The force needed to keep the knot from moving is the resultant force of these two forces.

Therefore, the resultant force,

Fₙ = √F₁² + F₂²

Fₙ = √(277)² + (603)²

Fₙ = 663.58 N

Hence,

The force with which it should be pulled to keep the knot from moving is 663.58 N.

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As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air acts as an ideal gas, calculate the work done per mol of the air.
A. -80.35 J
B. 80.35 J
C. -811.93 J
D. 811.93 J

(please show calculation)
can use this formula W=nRT ln(p1/p2)​

Answers

Answer:

B.-80.35 J

i dont know the calculation

A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.

Answers

Answer:

  1.5 m/s

Explanation:

Conservation of momentum means the momentum of the system before the collision is the same as after.

The before, after momentum of each ball is ...

  5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)

  10 kg ball: (10 kg)(0 m/s), (10 kg)(v)

The sum of the "before" products is the same as the sum of the "after" products:

  (5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v

  (10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides

  v = (15 kg·m/s)/(10 kg) = 1.5 m/s

The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.

the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?​

Answers

Answer:

Explanation:

350 cm / 25 cm = 14 radians

14 rad / 2π rad/rev = 2.23 revolutions

An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.

Answers

The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is

F = -f = ma = (4.00 kg) a

Assuming constant acceleration a , the acceleration applied by friction is such that

(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)

Solve for the acceleration :

a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²

Then the frictional force exerted a magnitude of

-f = (4.00 kg) (-14.8 m/s²)

f59.4 N

and was directed opposite the block's motion.

calculate the change in gravitational potential energy when a 10kg object falls from 6m on earth

Answers

Answer:

600

Explanation:

GPE=mgh (mass*gravitational force*height)

m=10 kg

g=10

h=6m

10*10*6= 600 Joules

What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3

Answers

Answer:

25.0 cm3

Explanation:

The volume is 25.0 cm3 .

2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=

Answers

Answer:

Explanation:

ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface

          n₁sinθ₁ = n₂sinθ₂

          1 sin52 = 1.33sinθ₂

                  θ₂ = arcsin(sin52 / 1.33)

                  θ₂ = 36°

as measured from the perpendicular to the surface

Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples

Answers

The motion of the rope which is perpendicular to the direction of the

propagation of the wave is a transverse wave motion.

The mass of the box is approximately 9.93 kg

Reasons:

The given function for the wave speed is presented as follows;

[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]

Where;

[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]

Taking the mass of the rope as, m = 2.00 kg

The length of the rope, L = 80.0 m

The mass hanging on the rope, M = 20.0 kg

We have;

T = 20.0 kg × 9.81 m/s² = 196.2 N

[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]

Therefore;

Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;

v = f × λ

Therefore;

v = 7.9 Hz × 7.9 m = 62.41 m/s

Which gives;

[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]

T = 62.41² × 0.025 = 97.3752025

[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]

Where;

g = The acceleration due to gravity which is approximately 9.81 m/s²

[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]

Therefore;

The mass of the box, m ≈ 9.93 kg

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The parameters obtained from a similar question online are;

[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]

Length of the rope, L = 80.0 m

Mass of the rope, m = 2.0 kg

Frequency of a point on the rope, f = 20 Hz

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