An electron is projected with an initial speed Vo = 5.35x10^6 m/s into the uniform field between the parallel plates. The direction of the field is vertically downward, and the field is zero except in the space between the two plates. The electron enters the field at a point midway between the plates.
Length of the plates= 2cm
Distance between the plates= 1cm
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.

To solve this problem we will apply the concepts given by the kinematic equations of motion. For this purpose it will be necessary with the given data to obtain the deceleration. With this it will be possible again to apply one of the kinematic equations of motion that does not depend on time, but on distance, to find how far the block would slide with the quadruplicate velocity

Our values are given as,

[tex]\text{Initial speed} =V_i = 2.3 m/s[/tex]

[tex]\text{Final speed}= V_f = 0 m/s[/tex]

[tex]\text{Stopping distance = }d = 1.9 m[/tex]

[tex]a = acceleration[/tex]

[tex]\text{mass} = m = 3.1kg[/tex]

Using the kinematic equation of motion we have

[tex]V_f^2 = V_i^2 + 2 a d[/tex]

[tex]0^2 = 2.3^2 + 2 a (1.9)[/tex]

[tex]a = -1.39211 m/s^2[/tex]

Now if the initial velocity is quadrupled we have that,

[tex]\text{Initial speed} =V_i' = 2.3*4 m/s = 9.2m/s[/tex]

[tex]\text{Final speed}= V_f' = 0 m/s[/tex]

[tex]\text{Stopping distance = }d'[/tex]

[tex]V_f^2' = V_i^2' + 2 a d'[/tex]

Replacing the values

[tex]0^2 = 9.2^2+ 2 (-1.39211) d'[/tex]

[tex]d' = 30.44m[/tex]

Therefore the block would have slipped around 30.44 if its initial velocity quadrupled.

To solve this problem we will apply the concepts related to the kinematic equations of linear motion. From there we will define the distance as the circumference of the earth (approximate as a sphere). With the speed given in the statement we will simply clear the equations below and find the time.

[tex]R= 6370*10^3 m[/tex]

[tex]v = 239m/s[/tex]

[tex]a = 16.5m/s^2[/tex]

The circumference of the earth would be

[tex]\phi = 2\pi R[/tex]

Velocity is defined as,

[tex]v = \frac{x}{t}[/tex]

[tex]t = \frac{x}{v}[/tex]

Here [tex]x = \phi[/tex], then

[tex]t = \frac{\phi}{v} = \frac{2\pi (6370*10^3)}{239}[/tex]

[tex]t = 167463.97s[/tex]

Therefore will take 167463.97 s or 1 day 22 hours 31 minutes and 3.97seconds

Answer:

The center mass (Xcm) of the two mass = (M₁X₁ + M₂X₂)/(M₁ +M₂)

Explanation:

let the first mass = M

let the position of second  = M

The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring is x(t) = A * cos(ωt + φ), where A represents the amplitude, ω is the angular frequency, and φ is the phase constant.

The formula for the trajectory of the center of mass of a two mass oscillator connected by a spring can be derived using the principles of simple harmonic motion (SHM). Let's assume that the masses are m1 and m2, and the spring constant is k. The equation for the trajectory of the center of mass can be written as:

x(t) = A * cos(ωt + φ)

where:

Answer:

a) p=0, b) p=0, c) p= ∞

Explanation:

In quantum mechanics the moment operator is given by

              p = - i h’  d φ / dx

             h’= h / 2π

We apply this equation to the given wave functions

a)  φ = [tex]e^{ikx}[/tex]

        .d φ dx = i k [tex]e^{ikx}[/tex]

We replace

        p = h’ k [tex]e^{ikx}[/tex]

        i i = -1

The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

            p = 0

b) φ = cos kx

           p = h’ k sen kx

The average sine function is zero,

          p = 0

c) φ = [tex]e^{-ax^{2} }[/tex]

         d φ / dx = -a 2x  [tex]e^{-ax^{2} }[/tex]

         .p = i a g ’2x  [tex]e^{-ax^{2} }[/tex]

       The average moment is

         p = (p₂ + p₁) / 2

         p = i a h ’(-∞ + ∞)

         p = ∞

Answer:

A. the indices of refraction matched

Explanation:

The index of refraction, or refractive

index, is a measure of how fast light

rays travel through a given medium.

Alternatively, it could be said that

the refractive index is the measure of

the bending of a light ray when

passing from one medium to

another. Mathematically, it can be

represented as a ratio between two

different velocities – the velocity of

light in vacuum and the velocity of

light in a given medium.

For example, try putting a pencil in a jar full of water. If you look at the pencil from above, it would look as though the pencil has bent in the water. That happens due to the refraction of light. It occurs because as light rays enter water, they slow down, as the speed of light in water is lower than the speed of light in air. The magnitude of how much a medium refracts a light ray is determined by the index

If you place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched here. Thus, the correct option is A.

Refraction is the phenomena of bending of light, which also happens with the rays of sound, water and other waves as it passes from one transparent substance into another substance or medium. This phenomena of bending of light waves by the refraction makes it possible for the people to have lenses, magnifying glasses, prisms, and rainbows. Even, our eyes depend upon this bending of light for the visibility and other effects of light.

If we place a glass cylinder in Wessin Oil, the view of the cylinder nearly disappeared when the cylinder nearly disappeared when the eyedropper was full of Wessin Oil because the indices of refraction matched.

Therefore, the correct option is A.

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Explanation:

(a) Velocity is given by :

[tex]v=\dfrac{ds}{dt}[/tex]

s is the length of the distance

t is the time

The dimension of v will be, [tex][v]=[LT^-1][/tex]      

(b) The acceleration is given by :

[tex]a=\dfrac{dv}{dt}[/tex]

v is the velocity

t is the time

The dimension of a will be, [tex][a]=[LT^{-2}][/tex]

(c) Since, [tex]d=\int\limits{v{\cdot}dt} =[LT^{-1}][T]=[L][/tex]

(d) Since, [tex]v=\int\limits{a{\cdot}dt} =[LT^{-2}][T]=[LT^{-1}][/tex]

(e)

[tex]\dfrac{da}{dt}=\dfrac{[LT^{-2}]}{[T]}[/tex]

[tex]\dfrac{da}{dt}=[LT^{-3}]}[/tex]

Hence, this is the required solution.

Answer:

34.9 kg

Explanation:

Since there are no net external forces on the system, the center of gravity does not move.

Let's say that m is the mass of the man, M is the mass of the boat, and L is the length of the boat.

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m x + M (x + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m x + M (x + L/2)) / (m + M)

m L + M (L/2) = m x + M (x + L/2)

m L + M (L/2) = m x + M x + M (L/2)

m L = m x + M x

m L − m x = M x

m (L − x) = M x

M = m (L − x) / x

Plugging in values:

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

The required mass of boat is 34.9 kg.

The given problem is based on the concept of the center of mass. The point of analysis where the entire mass of the system is known to be concentrated is known as the center of mass.

Given data:

The mass of man is, m = 89.3 kg.

The length of boat is, L = 5.8 m.

The distance away from the pier is, d = 4.17 m.

Since there are no net external forces on the system, the center of gravity does not move. Let's say that m is the mass of the man, M is the mass of the boat

When the man is at the stern, the distance between the center of gravity and the pier is:

CG = (m L + M (L/2)) / (m + M)

When the man reaches the prow, the distance between the center of gravity and the pier is:

CG = (m d + M (d + L/2)) / (m + M)

Since these are equal:

(m L + M (L/2)) / (m + M) = (m d + M (d + L/2)) / (m + M)

m L + M (L/2) = m d + M (d + L/2)

m L + M (L/2) = m d + M d + M (L/2)

Further solving as,

m L = m d + M d

m L − m d = M d

m (L − x) = M x

M = m (L − x) / x

M = 89.3 kg (5.8 m − 4.17 m) / 4.17 m

M = 34.9 kg

Thus, we can conclude that the required mass of boat is 34.9 kg.

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Answer:

No

Explanation:

Although the direction of position, velocity or acceleration of an object in marry-go-round problem changes continuously,however the magnitude of the position, velocity and acceleration do remains the same. Marry-go-round is nothing but a machine found in fairs that turn round in circular motion. So, the laws of circular motion are applicable in it.

Answer:

Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa

Explanation:

given data

density = 1400 kg/m³

frequency = 370 Hz

wavelength = 8.40 m

solution

we get here bulk modulus of the liquid that is

we know Bulk Modulus = [tex]v^2*\rho[/tex]   ...............

here [tex]\rho[/tex] is density i.e 1400 kg/m³

and v is =  frequency × wavelength

v = 370 × 8.40 = 3108 m/s

so here bulk modulus will be as

Bulk modulus = 3108² × 1400

Bulk modulus = 1.35 × [tex]10^{10}[/tex] Pa

Answer:

Development in science:

There a number of individuals in the world around us who has improved the field of science and made it easy for us to know this universe more easily then ever. This universe is still unexplored and there are going to be more then billions of stars which are still not studied and has an immense amount of information regarding the universe.

Stephen Hawking: The scientist who made it easy for each of us to know the universe and more over the cosmos in a more detailed form, as before this no one knew about the different patterns of entities and there properties.

A genius who was handicapped:

He was a genius in cosmology, mathematics, and other subjects of science as he was diagnosed by the amyotrophic lateral sclerosis(ALS), which made it unable for him to live a normal life.But, he communicated through a computer which detected his nerve signals and shared his thoughts about any thing or subject.

Why him?

Stephen hawking was just tremendous in making it more understandable about the black holes, time worms, and space etc.As it was known to very small number of people before he took the stage.

Answer:

a, b) part a and b on diagram attached

c) sf = -35 i + 20 j

35 km West and 20 km North

Explanation:

For part a and b refer to the attached co-ordinate system:

Note: unit vector i is in West/East direction and unit vector j is in North/South direction.

si = -15 i + 25 j

sf-si  = -20 i - 5 j

Hence,

Mark relative position from habitat sf = si + sf/i

sf = ( -15 i + 25 j ) + ( -20 i - 5 j )

sf = -35 i + 20 j

35 km West and 20 km North

Answer:

For Part A:

ΔX=11.8813 m

Part B:

[tex]t=0.8194 s[/tex]

Explanation:

Note: In order to find Part A we first have to find Part B i.e time

Data given:

[tex]V_i=0[/tex]

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]

V_i=0, Equation will become

Δy=[tex]\frac{1}{2}g*t^2[/tex]

[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]

For Part A:

Again:

Second equation of Motion:

ΔX=[tex]V_i*t+\frac{1}{2}a*t^2[/tex]

Since velocity is constant a=0

V_i=14.5m/s, t=0.8194 sec

ΔX=[tex]V_i*t[/tex]

ΔX=[tex]14.5*0.8194[/tex]

ΔX=11.8813 m

Part B: (Calculated above)

Data given:

[tex]V_i=0[/tex]

a=-9.8m/s^2 (-ve is because ranch is falling down)

Δy=-3.29m (-ve is because ranch is falling down)

Second equation of Motion:

Δy=[tex]V_i*t+\frac{1}{2}g*t^2[/tex]

V_i=0, Equation will become

Δy=[tex]\frac{1}{2}g*t^2[/tex]

[tex]t=\sqrt{2Y/g} \\t=\sqrt{2*-3.29/-9.8} \\t=0.8194 s[/tex]

Answer:

(a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

Explanation:

Given that,

Tensile strength = 196 MPa

Maximum strain = 0.380

Diameter = 50.0 μm

Length = 12.0 cm

We need to calculate the area

Using formula of area

[tex]A=\dfrac{\pi}{4}\times d^2[/tex]

Put the value into the formula

[tex]A=\dfrac{\pi}{4}\times(50.0\times10^{-6})^2[/tex]

[tex]A=1.96\times10^{-9}\ m^2[/tex]

We need to calculate the magnitude of the force

Using formula of force

[tex]F=\sigma A[/tex]

Put the value into the formula

[tex]F=196\times10^{6}\times1.96\times10^{-9}[/tex]

[tex]F=0.38416\ N[/tex]

(b). If the length of a strand of the hair is 12.0 cm at its breaking point

We need to calculate the unstressed length

Using formula of strain

[tex]strain=\dfrac{\Delta l}{l_{0}}[/tex]

[tex]\Delta l=strain\times l_{0}[/tex]

Put the value into the formula

[tex]\Delta l=0.380\times l_{0}[/tex]

Length after expansion is 12 cm

We need to calculate the original length

Using formula of length

[tex]l=l_{0}+\Delta l[/tex]

Put the value into the formula

[tex]I=l_{0}+0.380\times l_{0}[/tex]

[tex]l=1.38l_{0}[/tex]

[tex]l_{0}=\dfrac{l}{1.38}[/tex]

[tex]l_{0}=\dfrac{12\times10^{-2}}{1.38}[/tex]

[tex]l_{0}=0.0869\ m[/tex]

Hence, (a). The magnitude of the force is 0.38416 N.

(b). The original length is 0.0869 m.

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

As additional resistors are connected in series to a constant voltage source, the power supplied by the source decreases.

This can be explained by Ohm's Law, which states that power is equal to the voltage squared divided by the resistance. When resistors are connected in series, the total resistance increases, which leads to a decrease in the power supplied by the source.

For example, if you connect two resistors in series to a constant voltage source, the total resistance would be the sum of the resistances of the two resistors. As a result, the power supplied by the source would decrease.

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Adding more resistors in series to a constant voltage source results in an increase in total resistance, which leads to a decrease in current. Because power is proportional to the square of the current, the power supplied by the voltage source decreases.

When additional resistors are connected in series to a constant voltage source, the power supplied by the source is affected in a specific way. Given that power (P) is calculated by P = I^2R for a given resistance (R) and current (I), and by P = V^2/R for a given voltage (V) and resistance, we can understand the impact of adding resistors in series.

In a series circuit, the current remains constant throughout the resistors, but the total resistance (sum of all individual resistances) increases as more resistors are added. Since the voltage is constant, an increase in the total resistance will result in a decrease in current according to Ohm's law (I = V/R). Consequently, if the current decreases, the power supplied by the source also decreases (P = I^2R), because the power is proportional to the square of the current flowing through the circuit. Therefore, the correct answer is (D) The power supplied by the source decreases.

Answer:

n/(FG) = 3.

Explanation:

At the top of the loop-the-loop, the normal force is directed downwards as well as the weight of the car. So, the total net force of the car is

[tex]F_{net} = N + mg[/tex]

By Newton's Second Law, this force is equal to the centripetal force, because the car is making circular motion in the loop.

[tex]F_{net} = ma = \frac{mv^2}{R}\\N + mg = \frac{mv^2}{R}[/tex]

The critical speed is the minimum speed at which the car does not fall. So, at the critical speed the normal force is zero.

[tex]0 + mg = \frac{mv_c^2}{R}\\v_c = \sqrt{gR}[/tex]

If the car is moving twice the critical speed, then

[tex]N + mg = \frac{m(2v_c)^2}{R} = \frac{m4gR}{R} = 4mg\\N = 3mg[/tex]

Finally, the ratio of the normal force to the gravitational force is

[tex]\frac{3mg}{mg} = 3[/tex]

To solve this problem we will start by differentiating the values in each of the states of matter. Subsequently through the thermodynamic tables we will look for the values related to the entropy, enthalpy and respective specific volumes. Through the relationship of Power defined as the product between mass and enthalpy and mass, specific volume and pressure, we will find the energetic values in the two states investigated. We will start defining the states

State 1

[tex]T_1 = 700\°C[/tex]

[tex]P_1 = 4 Mpa[/tex]

From steam table

[tex]h_1 =3906.41 KJ/Kg[/tex]

[tex]s_1 = 7.62 KJ/Kg.K[/tex]

Now

[tex]s_1 = s_2 = 7.62 KJ/Kg.K[/tex] As 1-2 is isentropic

State 2

[tex]P_2 = 20 Kpa[/tex]

[tex]s_2 = 7.62 KJ/Kg \cdot K[/tex]

From steam table

[tex]h_2 = 2513.33 KJ/Kg[/tex]

PART A) The power produced by turbine is the product between the mass and the enthalpy difference, then

[tex]Power = m \times (h_1-h_2)[/tex]

[tex]P = (50)(3906.41 - 2513.33)[/tex]

[tex]P = 69654kW[/tex]

b) Pump Work

State 3

[tex]P_3 = 20 Kpa[/tex]

[tex]\upsilon= 0.001 m^3/kg[/tex]

The Work done by the pump is

[tex]W= m\upsilon \Delta P[/tex]

[tex]W = (50)(0.001)(4000-20)[/tex]

[tex]W = 199kJ[/tex]

a) The equation 1/2 mv² = 1/2 mv0² + mgh is not dimensionally consistent; b)  The equation v = v0 + at² is not dimensionally consistent; c) The equation ma = v² is not dimensionally consistent.

(a) The left-hand side of the equation has dimensions of energy (M L² T⁻²), while the right-hand side has dimensions of energy plus length (M L² T⁻² + ML).

This is because the term mgh represents the potential energy gained by an object of mass m when it is lifted to a height h above the ground. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

(b)The left-hand side of the equation has dimensions of velocity (L T⁻¹), while the right-hand side has dimensions of velocity plus time squared (L T⁻¹ + T²).

This is because the term at² represents the change in velocity over time due to acceleration. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

(c) The left-hand side of the equation has dimensions of force (M L T⁻²), while the right-hand side has dimensions of velocity squared (L² T⁻²).

This is because the term v² represents the square of the velocity of an object. Therefore, the dimensions of the two sides of the equation do not match, and the equation cannot be correct.

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The dimensional analysis shows that equation (a) is correct while equations (b) and (c) are incorrect. Equation (b) is missing a time component, while equation (c) incorrectly equates force to velocity squared.

Let's analyze each equation in terms of their dimensions:

By comparing dimensions on both sides of an equation, we can spot if the equations are incorrect or not.

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Answer:

B

Explanation:

The sailor, the boat and the wrench are all moving at he same constant rate, so the wrench will appear to fall straight down. This due to that fact there is no relative motion among them and all are at rest w.r.t to one another. Hence the correct answer would be B.

The correct option is (B). The wrench will fall straight down relative to the moving boat and will land directly at the base of the mast.

To determine where the wrench will hit the deck, let's analyze the motion of the wrench.

1. Horizontal Motion:

2. Vertical Motion*: The wrench will accelerate downwards due to gravity.

To find the distance between the first-order red and blue fringes formed by a diffraction grating, we calculate the diffraction angles using the grating equation and then use trigonometry to determine the positions of the fringes on the screen. The distance apart is the absolute difference between these positions.

The student is asking about the distance between the first-order red and blue fringes that appear on a screen as a result of a diffraction pattern formed by light from a hydrogen discharge lamp passing through a diffraction grating. To answer this question, we apply the formula for diffraction maxima, d sin( heta) = m ext{ extlambda}, where d is the distance between the grating lines,  ext{ extlambda} is the wavelength of light,  ext{ exttheta} is the diffraction angle, and m is the order of the maximum.

Given that the grating has 500 lines per mm or 500,000 lines per meter, we can calculate d as 1/500,000 meters. Since we are looking for the first-order maxima (m=1), we can find the angles for each color using the formula sin( heta) =  ext{ extlambda}/d. We can then use basic trigonometry to find the positions of the red and blue fringes on the screen, and their distance apart.

For the 656 nm red light:

ext{ exttheta}_{red} = sin^{-1}( ext{ extlambda}_{red}/d)
ext{ exttheta}_{red} = sin^{-1}(656 ext{ exttimes}10^{-9} m / (1/500,000 m))

For the 486 nm blue light:

ext{ exttheta}_{blue} = sin^{-1}( ext{ extlambda}_{blue}/d)
ext{ exttheta}_{blue} = sin^{-1}(486 ext{ exttimes}10^{-9} m / (1/500,000 m))

After calculating the angles, the positions on the screen for first-order maxima are found by L tan( heta), where L is the distance from the grating to the screen (1.50 m in this case).

The final distance between the red and blue fringes is the absolute difference between their positions on the screen.

Answer:

The velocity of the pin is opposite its acceleration on the way up.

(d) option is correct.

Explanation:

when the juggler throws a bowling pin straight in the air, the acceleration working on the pin is in the downward direction due to the gravitational force of the earth.

According to Newton's Universal Law of Gravitation

''The gravitational force is a force that attracts any objects with mass''

Final answer:

The acceleration due to gravity is always downward, so when the pin is thrown up, its velocity is opposite to its acceleration. At the peak, the velocity is zero and then aligns with the direction of gravity on the descent.

Explanation:

When a juggler throws a bowling pin straight up in the air, the acceleration due to gravity is always directed towards the ground, which means it is downwards. As the pin moves upwards, its velocity is in the opposite direction of the acceleration. At the peak of its motion, the velocity of the pin is zero, after which it starts to fall back down, and its velocity is then in the same direction as acceleration. Thus, the correct statement is (d) The velocity of the pin is opposite its acceleration on the way up.

Answer:

[tex]3.26198\times 10^{-12}\ kg[/tex]

Explanation:

E = Electric field = 20 MN/C

q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]

g = Acceleration due to gravity = 9.81 m/s²

m = Mass of drop

The electrical force will balance the weight

[tex]Eq=mg\\\Rightarrow 20\times 10^{6}\times 10\times 1.6\times 10^{-19}=m\times 9.81\\\Rightarrow m=\dfrac{20\times 10^{6}\times 10\times 1.6\times 10^{-19}}{9.81}\\\Rightarrow m=3.26198\times 10^{-12}\ kg[/tex]

The mass that can be suspended is [tex]3.26198\times 10^{-12}\ kg[/tex]

From charge to mass ratio, the mass of the charges is 3.2  × 10^-12 Kg.

The charge to mass ratio experiment was used by R. A. Millikan to accurately determine the charge to mass ratio of the electron. We have the following information from the question;

Field strength = 20 MN/C

Number of charges = 10

Now;

The magnitude of electric field strength is obtained from;

E = F/q

F = Eq

Where;

F = electric force

E = electric field intensity

q = magnitude of charge

F = 10 × 20  × 10^6  × 1.6 × 10^-19 = 3.2  × 10^-11 N

Where the charges fall freely under gravity;

F = mg

m = F/g

m = 3.2  × 10^-11 N/10 ms-2

m = 3.2  × 10^-12 Kg

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By employing Coulomb's Law and additional physical principles like Newton's third law, it becomes possible to determine the characteristics of an unknown electrical charge from the force it experiences and exerts.

The question concerns two charges, one of which is negative and known (-0.510 µC), and the other is unknown. You're asked to find the value of this unknown charge and the force it exerts on the known charge. To do this, we would use Coulomb's Law, which states that the electric force between two charges is directly proportional to the product of their charges, and inversely proportional to the square of the distance separating them.

So, using the formula F = k*q1*q2/r^2, where F is the force, k is Coulomb's constant (approximately 9 * 10^9 N.m^2/C^2), q1 and q2 are the charges, and r is the distance between them, we can solve for the unknown charge(q2) - which gives us q2 = Fr^2/(kq1). Substituting the provided values in the question, we should be able to solve it.

For part (b), the sign of the force that the unknown charge exerts on the known charge would be exactly opposite in direction to the force it experiences, as per Newton's third law (equal and opposite reactions). However, we'd need to know the magnitude of the unknown charge to calculate the actual force.

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Answer:

The distance covered by the lighting flash is 1700 meters.

Explanation:

Given that,

The speed of sound travels in air at a speed of about 340 meter second, v = 340 m/s

The sound of thunder is heard 5 seconds after the flash of lightning is seen, t = 5 s

To find,

The distance far away was the lightning flash.

Solution,

The distance covered by the lighting flash is given by the product of speed and time. It is given by :

[tex]d=v\times t[/tex]

[tex]d=340\ m/s\times 5\ s[/tex]

d = 1700 meters

So, the distance covered by the lighting flash is 1700 meters. Therefore, this is the required solution.

Answer:

6.08 cm

Explanation:

We are given that

Ratio =1:125000

Let x be the distance on a map between two features .

The distance between two features on ground=y=7.6 km

According to question

[tex]\frac{x}{y}=\frac{1}{125000}[/tex]

Substitute the values then we get

[tex]\frac{x}{7.6}=\frac{1}{125000}[/tex]

[tex]x=\frac{7.6}{125000}=0.0000608 km[/tex]

We know that

[tex]1km=100000 cm[/tex]

0.0000608 km=[tex]0.0000608\times 100000=6.08cm[/tex]

Hence, the distance between two features on the map=6.08 cm

The distance on the map between the two features is 6.08 centimeters.

To find the distance on a map in centimeters between two features if they are 7.6 km apart on the ground and the map has a scale of 1:125000, you can use the scale factor formula: Distance on Map = Distance on Ground / Scale

Plugging in the values: Distance on Map = 7.6 km / 125000 = 0.0000608 km

Since 1 km = 100000 centimeters, multiply by 100000 to convert km to cm:

Distance on Map = 0.0000608 km × 100000 cm/km = 6.08 cm

Therefore, the distance on the map between the two features is 6.08 centimeters.

Answer:

277 s

Explanation:

Given data

We can find the elapsed time using the following expression.

I = C/t

t = C/I

t = 56.0 C/(0.2020 C/s)

t = 277 s

It took 277 seconds.

Answer:

D = 72.68 m

Explanation:

given,

R = 100 m

angular speed = 0.1 rad/s

distance she can be pulled before the centripetal acceleration reaches 5g = 49 m/s².

using conservation of Angular momentum

[tex]I_i\omega_i= I_f\omega_f[/tex]

[tex]mr_i^2\omega_i=m r_f^2\omega_f[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

[tex]\omega_f = \dfrac{r_i^2}{r_f^2}\times \omega_i[/tex]

we know,

centripetal acceleration

[tex]a = \dfrac{v^2}{r}[/tex]

v = r ω

[tex]a =\omega_f^2 r_f [/tex]

[tex]a =(\dfrac{r_i^2}{r_f^2}\times \omega_i)^2 r_f [/tex]

[tex]a =\dfrac{r_i^4\times \omega_i^2}{r_f^3}[/tex]

[tex]r_f^3=\dfrac{100^4\times 0.1^2}{5\times 9.8}[/tex]

[tex]r_f^3=20408.1632[/tex]

[tex]r_f = 27.32\ m[/tex]

distance she has reached inward is equal to

D = 100 - 27.23

D = 72.68 m

To solve this problem we will apply Ohm's law. The law establishes that the potential difference V that we apply between the ends of a given conductor is proportional to the intensity of the current I flowing through the said conductor. Ohm completed the law by introducing the notion of electrical resistance R. Mathematically it can be described as

[tex]V = IR \rightarrow R = \frac{V}{I}[/tex]

Our values are

[tex]I = 500mA = 0.5A[/tex]

[tex]V = 37V[/tex]

Replacing,

[tex]R = \frac{V}{I}[/tex]

[tex]R = \frac{37}{0.5}[/tex]

[tex]R = 74 \Omega[/tex]

Therefore the smallest resistance you can measure is [tex]74 \Omega[/tex]

Answer:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

Explanation:

Option A is correct (0) ( The electric field at the center of circular charged ring is zero)

The reason why electric field at the center of circular charged ring is zero because the fields at the center of the circular ring due to any point are cancelled by electric fields of another point which is at 180 degree to that point i.e opposite to that charge.

Answer: a) 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

Explanation:

Given that the ring is perfectly circular with radius b which has a uniform distributed charge q around it.

The electric field experienced at the centre of the ring from opposite directions are given as

E1 = kₑq/b²

E2 = -kₑq/b²

It experience the two electric field E1 and E2 from opposite directions at the centre. So the resultant electric field is given by:

E = E1 + E2

E = kₑq/b² - kₑq/b²

E = 0

The electric field from opposite directions at the centre of the circular ring cancel each other out and give a resultant of zero

To solve this problem we will apply the concepts related to the potential, defined from the Coulomb laws for which it is defined as the product between the Coulomb constant and the load, over the distance that separates the two objects. Mathematically this is

[tex]V = \frac{kq}{r}[/tex]

k = Coulomb's constant

q = Charge

r = Distance between them

[tex]q = 18 pC \rightarrow q = 1.8*10^-11 C[/tex]

[tex]d = 2.4mm \rightarrow r = 1.2 mm = 1.2*10^-3 m[/tex]

Replacing,

[tex]V = \frac{kq}{r}[/tex]

[tex]V = \frac{ (9*10^9)*(1.8*10^{-11})}{(1.2*10^{-3})}[/tex]

[tex]V = 135 V[/tex]

Therefore the potential at the surface of the raindrop is 135 V

The new power is: New P_avg = 2.52 W / 4 ≈ 0.63 W

Average Power Carried by a Wave

To solve this problem, we need the following information:

Mass of piano wire: 2.95 g = 0.00295 kg

Length of wire: 79.0 cm = 0.79 m

Tension: 29.0 N

Frequency: 105 Hz

Amplitude: 1.80 mm = 0.00180 m

First, calculate the linear mass density (μ) of the wire:

μ = mass / length = 0.00295 kg / 0.79 m ≈ 0.00373 kg/m

Next, find the wave speed using the formula for the speed of a wave on a string:

v = [tex]\sqrt{Tension / \mu}[/tex] =[tex]\sqrt{29.0 N / 0.00373 kg/m}[/tex]≈ 88.19 m/s

Now, we calculate the average power (P_avg) carried by the wave using the formula:

P_avg = 0.5 x μ x v x ω² x A²

Where:

ω = 2πf (angular frequency)

ω = 2 x π x 105 ≈ 659.73 rad/s

Therefore,

P_avg = 0.5 x 0.00373 kg/m x 88.19 m/s x (659.73 rad/s)² x (0.00180 m)² ≈ 2.52 W

Average Power if Amplitude is Halved

If the amplitude (A) is halved, the new amplitude is:

New A = 0.00180 m / 2 = 0.00090 m

Since power is proportional to the square of the amplitude (A²), halving the amplitude reduces the power by a factor of 4.

Thus, the new power is:

New P_avg = 2.52 W / 4 ≈ 0.63 W