Because most particles fired at metal foil passed straight through, rutherford concluded that.

The strength of the electrostatic force is inversely proportional to the square of the distance

between the charges. If the distance is doubled, the force is 1/4.

The new force is 0.80/4 = 0.20 N.

Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.

The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

[tex]F=K |\frac{q_{1} q_{2}}{r^{2} } |[/tex]

In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:

Therefore, with q₁ and q₂ values ​​of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = [tex]9 \times 10^9[/tex] N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is [tex]8.854\times10^{-12} C^{2} N^{-1} m^{-2}[/tex].

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The resultant wave is calculated by adding the magnitude of both the waves traveling in same direction.The resulting wave is of the amplitude 5.66cm or 5.66 x 10⁻²m

Let us take A be the amplitude of both the waves that are travelling

We are given that waves are moving 90 degrees out of the phase

Let the equations of the waves be

y1= Asin(ωt-kx)

y2=Acos(ωt-kx)

Since the waves are ninety degrees travelling out of phase ,

The resultant wave is

y=y1+y2

  = Asin(ωt-kx) +Acos(ωt-kx)

  =√2 A sin(ωt-kx+π/4)

So, we can conclude resultant wave is √2A = √2 *4 =5.66cm

=5.66* 10 ⁻²m

Hence the resultant wave is 5.66cm

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The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width.

There is said to be a wandering black hole that can be seen in an approximately  5,000 light-years away from earth.

It is one that can be seen in the Carina-Sagittarius spiral aspect of our galaxy.

Note that, its discovery gives room for astronomers to state that the nearest form of isolated stellar-mass black hole to Earth can be as close as about 80 light-years away.

Therefore, The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width because it has no width, no depth and no height.

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mass of one sheet of paper is 0.00504 kg or 5.04g.

First we will find the surface area of a single sheet of paper

surface area = length ×width

surface area =0.3×0.21

surface area =0.063m²

now 0.063m² is the surface area of a single sheet of paper .

so now we will find number of papers required to form 1m² of area

which will be given by dividing 1m² by surface area of a single sheet

i.e.

number of sheets required = 1/0.063m²

number of sheets required =15.87

mass of one sheet = mass of 1 m^2 of paper / number of times narrower

mass of one sheet = 0.080 kg / 15.873

mass of one sheet = 0.00504 kg

mass of one sheet of paper is 0.00504 kg or 5.04g.

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The speed of light u is equal to c.

Given,

The motion of a transparent medium influences the speed of light.

The water moves with speed v in a horizontal pipe.

Assume the light travels in the same direction as the water moves.

The speed of light with respect to the water is c /n.

The index of refraction of water, n = 1.33

Let us assume u' be the speed of light in water.

u' is related to the refractive index of water,as u' = c/n

where, c is the speed of light.

Let, u be the speed of light in water in the lab frame.

Now, u and u' are related as : u = ( u'+v) /(1+u'v/c^2)

Here, v is the speed of water in the horizontal pipe.

We know the value of u'. so by substituting the value, we will get ,

u = (c/n+v) /(1+cv/nc^2)

u = c/n (1+nv/c) /(1+v/nc)

u in the limit as the speed of the water approaches c.

Thus,

u = lim v-->c [c/n (1+nv/c) /(1+v/nc) ]

u= c/n (1+n) /(1+1/n)

u= c .

Hence, the speed of light u is equal to c.

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Disclaimer: incomplete question. here is the complete question.

Question:The motion of a transparent medium influences the speed of light. This effect was first observed by Fizeau in1851. Consider a light beam in water. The water moves with speed v in a horizontal pipe. Assume the light travels in the same direction as the water moves. The speed of light with respect to the water is c / n , where n = 1.33 is the index of refraction of water.

(a) Use the velocity transformation equation to show that the speed of the light measured in the laboratory frame isu = c/n (1 + nv/c / 1+ v/nc).

(d) Evaluate u in the limit as the speed of the water approaches c .

Answer:

See below

Explanation:

[tex]\displaystyle {\textsf {Acceleration is the rate of change of velocity with time:}}[/tex]

[tex]\boxed {g = \frac{dv}{dt}}[/tex]

[tex]\sf {where \;dv = \;change \;in \;velocity\; and \;dt \;the \;change \;in \;time}[/tex]

here  g is the gravitational force

When a body is thrown vertically upward, it has an initial speed. As it keeps going up, it speed keeps decreasing until it becomes zero, then the ball starts dropping down

On the upward movement, the change in velocity is negative while the change in time is always positive

So dv/dt = a negative number divided by a positive number which is a negative value for g in the equation for g

Answer: 1) D & 2) H

Explanation:

D. Tropical rainy &

H. Long-term average temperature  and precipitation for the area

Answer:

3.42 + 4 + 5.2 + 12= 24.62
So since we have 24.62 it will be 4 sig figs

Explanation:

The electric field on the axis of the ring at the center of the ring at 100.00cm is 6.64×〖10〗^5N/C

Where a positive charge (q) is given, always note that the charges are pointing away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,

E=(K_e×b)/(r^3×Q)

Given that;

E = Electric field

Ke = electric field constant

a = radius of the charge q from the center of the circle

Q = electric charge = 75.0μC

The radius of the charge is a = 10cm= 0.1m

Distance from the electric field is 100cm = 1m

Ke is a constant given as 8.99×〖10〗^(9) Nm^2/c^2

E=(8.99×〖10〗^(9  ) Nm^2/c^2×1)/(〖0.1〗^3×75×〖10〗^(-6) )

E=6.64×〖10〗^5N/C

Therefore, the electric field away from the center of the ring is 6.64×〖10〗^5N/C away from the center of the ring.

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Initial speed for the toys snake will be 122.88 m/s

Given,

Height from where it has been dropped, h = 115 m

Height where the toy lands, s = 595 m

As we know that [tex]t = \sqrt{\frac{2h}{g} }[/tex]

here 'g' is acceleration due tp gravity = 9.81 m/s²

therefore  [tex]t = \sqrt{\frac{2 * 115}{9.81\\ }[/tex]

[tex]t = \sqrt{23.445}[/tex]

t = 4.842 s

Now, According to the Second Equation of the Motion which relates distance with the acceleration and time, we can say that

[tex]S = ut + \frac{1}{2}at^{2}[/tex]

The toy snake is thrown horizontally, therefore acceleration acting on the toy will be zero (0).

s = 595 m

t = 4.842 s

therefore [tex]u = \frac{s}{t}[/tex]

[tex]u = \frac{595}{4.842}[/tex]

u = 122.88 m/s

The initial speed of the toy snake will be 122.88 m/s.

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Answer:

An object at rest remains at rest

Explanation:

An object at rest stays at rest and an object in motion stays in motion with the same speed

The magnitude of the force on the test charge is 1.

Greatness is the quantitative worth of seismic energy. It is a particular worth having no connection with distance and heading of the focal point. The idea of extents traces all the way back to the Ancient Greeks, when stars overhead was sorted into six sizes. Extent alludes to the evaluating of the splendor of stars, the first being the most splendid. It has been moved to different issues since basically the seventeenth 100 years. In Physics, greatness is characterized as the most extreme degree of size and the course of an item. Greatness is utilized as a typical consider vector and scalar amounts. By definition, we realize that scalar amounts are those amounts that have size as it were.

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In the electron-transport chain, as electrons move along a series of carriers, they release energy that is used to pump protons across a membrane.

In plants and animals, the electron transport chain refers to the consistent streaming of electrons between cells where electron donors donate electrons and electron acceptors accept electrons through a redox reaction.

An electrochemical gradient is created as a result of the redox reaction and ATPs are produced.

In this chain, there are 4 major protein complexes that are transferred across the plasma membrane.

This cellular process occurs in mitochondria and is responsible for photosynthesis and cellular respiration. Its primary steps are Krebs' cycle, glycolysis, oxidative phosphorylation, and pyruvate oxidation.

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The distance should be 1.27 m from the wire in order to get magnetic field one-tenth as large .

Let [tex]B_1[/tex] be inside the plane and [tex]B_2[/tex] be outside the plane . It is required to calculate magnetic field at point A .

Direction of magnetic field at A is calculated using Right hand rule.

[tex]B_{net}=B_2-B_1\\\\B_{net}=\frac{u_0I}{2\pi (0.4-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(0.4+\frac{3\times10^{-3}}{2})} \\\B_{net}=7.5\times 10^{-9}T[/tex]....(1)

Let  a distance "R"  from the wire so that the magnetic field is [tex]\frac{B_{net}}{10}[/tex]

Using the magnetic field equation , we get

[tex]\frac{B_{net}}{10} =B_2-B_1\\\\\\frac{7.5\times 10^{-9}}{10} =\frac{u_0I}{2\pi (R-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(R+\frac{3\times10^{-3}}{2})} \\\\\R^2-2.25\TIMES10^{-6}=1.6\\\\R=1.27m[/tex]

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The total voltage across resistor- three in the circuit is 24V

acc to ohm's law

the electrical current flowing through any circuit is directly proportional to the voltage applied across it.

i.e. V ∝ I

V= IR

I = current

v = voltage

R = resistance (proportionality constant)

R in parallel

(1/Rp = 1/R1 + 1/R2 + 1/R3 +1/R4)  

on rearranging the above equation (V=IR)

I = V/R

I = 24/6 = 4A

I is the current in each resistor.

in the parallel circuit voltage across the component is same and total current is the sum of the current  flowing through each component.

the total voltage across resistor-three (vr3) in the circuit is

V= IR₃ = 4×6= 24 volt

the total voltage across the resistor three (vr3) in the circuit is 24volt.

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The acceleration of the robot is 1.43 m/s2 in magnitude.

The acceleration of an object is defined as the variation of the velocity with respect to time. The acceleration is calculated by finding the ratio of the change in velocity to the change in time.

The acceleration is also calculated by using the force formula. The force is actually defined as the acceleration produced in the body by applying the force on an object of mass m.

On a non-stick surface, a toy robot and doll are standing side by side. The robot weighs 0. 30 kg, while the doll weighs 0. 20 kg. With a force of 0. 30 n, the robot pushes against the doll.

According to Newton's second law.

F = mass x acceleration

Given that

Force applied = 1N

Acceleration = Force/mass

Substitute the values and get acceleration.

Acceleration = 1/0.7

Acceleration = 1.43m/s²

Hence, the acceleration of the robot is 1.43 m/s2 in magnitude.

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The speed at which Matt ran is 10 miles per hour. Therefore, option (B) is the correct answer.

Matt runs at a speed of x miles per hour, whereas Doug does it at a speed of 2 miles per hour faster. Doug's speed will thereafter be equal to x + 2 miles per hour. Since there are four quarter-mile laps, Doug completes 1.5 miles in the time it requires Matt to complete 1.25 miles.

Matt:

Speed = x

Distance = 1.25

Time = [tex]\frac{1.25}{x}[/tex]

Doug:

Speed = x+2

Distance = 1.5

Time = [tex]\frac{1.5}{x+2}[/tex]

Making an equation by putting the two timings on the chart equal to one another and then solving for x will reveal that the two boys took the same amount of time from the moment Doug started.

[tex]\frac{1.5}{x+2}[/tex] = [tex]\frac{1.25}{x}[/tex]

1.5x = 1.25 (x+2)

0.25x = 2.5

x = 10

Therefore, Matt’s speed is 10 miles per hour.

The complete question is:

Doug, who runs track for his high school, was challenged to a race by his younger brother, Matt. Matt started running first, and Doug didn’t start running until Matt had finished a quarter-mile lap on the school track. Doug passed matt as they both finished their sixth lap. If both boys ran at a constant speed, with Doug running 2 miles an hour faster than Matt, what was Matt’s speed?

A. 10.5 miles per hour

B. 10 miles per hour

C. 9 miles per hour

D. 8 miles per hour

E. 7.5 miles per hour

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While current is the same in series and parallel, voltage is not. Therefore, because these three resistors are connected in parallel, their voltage is the same, or 12 volts. Using the equation V=IR, 12V=I*10Ω gives you a current of 1.2A.

A resistor is a passive two-terminal electrical component used in circuits to implement electrical resistance. Resistors have a variety of purposes in electronic circuits, reducing current flow, modifying signal levels, dividing voltages, biassing active components, and terminating transmission lines are a few examples. High-power resistors that can generate many watts of heat instead of electrical energy can be utilised as test loads for generators, power distribution systems, and motor controls. Fixed resistors' resistances only sporadically vary as a result of changes in operating voltage, time, or temperature. Variable resistors can be utilised as force sensors, heat sensors, light sensors, volume controls, lamp dimmers, humidity sensors, and chemical activity sensors.

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For a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin is the conclusion which can be made from Gay-Lussac’s law and is denoted as option D.

This is referred to as the perpendicular force applied on a body per unit area and the unit is Pascal.

Gay-Lussac was a scientist who discovered through numerous experiments and observations that at a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin.

This is denoted as the following equation below:

P ∝ T

P = kT where p is pressure, k is constant and t is temperature.

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The options are:

Explanation:

E = hc / wavelength

Given that:

h = 6.626 × 10-34 Js

c = 2.998 x 10^8 m/s

wavelength = 263 x 10^-9 m

Given:

[tex]speed,\;v=449\;m/s[/tex]

We know that,

[tex]adiabatic index, \gamma = 1.667[/tex]

[tex]gas constant,\;R=8.314\;J[/tex]

The form speed of the sound in a monoatomic gas is,

[tex]v=\sqrt{\frac{\gamma\;R\;T}{M} }[/tex]

Substitute the known values in the above equation,

[tex]449=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{M}[/tex]

[tex]M=\sqrt{x} \frac{1.667 \times 8.314 \times 293}{449}[/tex]

[tex]M=0.02014 \times (\frac{1000}{1})[/tex]

[tex]M=20.14\;g/mol[/tex]

Therefore, monoatomic gas is Neon gas (Ne)

Monatomic gas, which differs from diatomic, triatomic, or generally polyatomic gases in that it contains particles (molecules) made up of just one atom, includes gases like helium or sodium vapor. Because a monatomic gas lacks the rotational and vibrational energy components that characterize polyatomic gases, its thermodynamic behavior in the normal temperature range is incredibly straightforward. As a result, its heat capacity is independent of temperature and molecular weight (in this case, atomic weight), and its entropy (a measure of disorder) only depends on temperature and molecular weight.

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When the variable name has been assigned with the value 33, then the value of the name after the assignment name will be 66.

In the programming language the sign ' = ', is referred to as the assignment operator which is used to assign values to a particular variable, where the variable referred to the name of the memory location. As it is given the assignment name = name * 2, it means that name on the right-hand side is the assignment name, so whatever value it gets is replaced by twice that previous value according to the operation.

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The electric flux is 0 N.m²/C.

We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as

Φ = E . S = E . Scosθ

where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.

From the question above, we know that

E = 6.20 × 10⁵ N/C

θ = 90⁰

S = 3.20 m²

By substituting the following parameter, we get

Φ = E . Scosθ

Φ = 6.20 × 10⁵ . 3.20cos(90⁰)

Φ = 0 N.m²/C

Hence, the electric flux is 0 N.m²/C

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The ball is on the 50-yard line. The ball travels west 5 yards. The distance ball traveled was 60 yards.

Distance can be defined as the length of any path between any two locations. Displacement is the direct distance between any two points when calculated via the shortest path between them. When computing distance, the direction is disregarded. The displacement computation takes the direction into consideration.

The 5 yard gain would bring the ball to the 45 yard line (ball was tackled  at the 50 yard line, so 50 - 5 = 45), but the 15 yards lost due to the tackle and push back would bring ball to the 60 yard line (45 + 15 = 60).

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By the charge density, The charge throughout te plastic is 0.059 nC.

We need to know about charge density to solve this problem. The charge density can be determined as

λ = Q / V

where λ is charge density, Q is charge and V is volume.

The parameter given is the charge density and the solid cylinder shape which are :

λ = 500nC/m³

r = 2.5 cm = 0.025 m

L = 6.0 cm = 0.06 m

Find the volume of solid cylinder

V = π .  r² . L

V = π .  0.025² . 0.06

V = 1.18 x 10¯⁴ m³

Find the charges

Q = λ x V

Q = 500 x  1.18 x 10¯⁴

Q = 0.059 nC

Hence, the charge throughout the plastic is 0.059 nC.

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Starting from rest, a boulder rolls down a hill with constant acceleration and travels 4.00 m during the first second, 8 m first it is moving.

Speed increase is the name we provide for any cycle where the speed changes. Since speed is a speed and a heading, there are simply two different ways for you to speed up: change your speed or shift your course — or change both. In mechanics, speed increase is the pace of progress of the speed of an item as for time. Speed increases are vector amounts (in that they have size and heading). The direction of an article's speed increase is given by the direction of the net power following up on that item. The extent of an articles. An item's typical speed increase throughout some undefined time frame is its adjustment of speed, partitioned by the span of the period.  Numerically, Momentary speed increase, in the meantime, is the constraint of the typical speed increase over a microscopic time period. In the terms of analytics, immediate speed increase is the subsidiary of the speed vector regarding time.

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The force on the wire will be

F = 2.88 Newton

We have a current carrying wire in a uniform magnetic field.

We have to determine the magnitude of the magnetic force on this section of wire.

The force on the current carrying wire will be -

F = IBL sinθ

According to the question, we have -

Length (L) = 0.75 m (along +x - axis)

Current (I) = 2.4 A (along + x - axis)

Magnetic Field (B) = 0.39 k T (along +z axis)

Therefore, the angle between Magnetic field and Length = θ = 90°.

Therefore -

F = IBL sin(90)

F = 2.4 x 1.6 x 0.75 x sin(90)

F = 2.88 x sin(90)

F = 2.88 Newton

Hence, the force on the wire will be

F = 2.88 Newton

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Answer:

Mass

No matter where an object is in the universe, the more matter an object has, the more mass it will have as well.

Answer: partially decomposed

Explanation:

The distance should be 4m from the wire in order to get the magnetic field of 0.100μ .

It is given that magnetic field 40.0 cm away from a straight wire is  1.00μT having  current 2.00 A .

From equation (1)  magnetic field 40.0 cm = 0.4m away from a straight wire is 1.00μT which is given by    [tex]1.00=\frac{u_0I}{2\pi \times0.4}[/tex]      .....(2)

From equation (1)  magnetic field 'r' m away from a straight wire is 0.100μT which is given by    [tex]0.100=\frac{u_0I}{2\pi \times r}[/tex]       ...(3)

On dividing equation (2) by (3) , we get

             [tex]\frac{1}{0.1} =\frac{r}{0.4} \\\\r=4m[/tex]

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