Suppose that 4 J of work is needed to stretch a spring from its natural length of 36 cm to a length of 47 cm. (a) How much work is needed to stretch the spring from 41 cm to 45 cm? (Round your answer to two decimal places.) J (b) How far beyond its natural length will a force of 15 N keep the spring stretched? (Round your answer one decimal place.)

Answer:

Ok, the minimal quantity of charge that we can find is on the electron or in the proton (the magnitude is the same, but the sign is different)

Where the charge of a single proton is:

C = 1.6x10^-19 C

Now, you need to remember that when we are working with charges, we are working with discrete math:

What means that?

If the minimum positive is the charge of one proton, then the consecutive charge will be the charge of two protons (there is no somethin in between)

So the consecutive charge will be:

C = 2*1.6x10^-19 C = 3.2x10^-19 C.

So, because we are working in discrete math, we can not have any object that has charge between  1.6x10^-19 C and 3.2x10^-19 C.

Particularly, 2.0x10^-19 C is in that range, so we can conclude that:

No, an object can not carry a charge of 2.0x10^-19 C.

Answer:

Explanation:

 An insulator is a material that impedes the movement of heat or electric current from flowing.

Theoretically good heat insulators stops the movement of heat, while practically this insulation can only be slowed down.

Hence from the options listed the correct answer practically is

Answer:

Option A

Explanation:

Acceleration will be obviously zero when Force = 0

That is how:

Force = Mass * Acceleration

So, If force = 0

0 = Mass * Acceleration.

Dividing both sides by Mass

Acceleration = 0/Mass

Acceleration = 0 m/s²

Answer:

[tex]\boxed{\mathrm{A. \: It \: will \: be \: 0 \: meters \: per \: second \: per \: second. }}[/tex]

Explanation:

[tex]\mathrm{force=mass \times acceleration}[/tex]

The force is given 0 newtons.

[tex]\mathrm{force=0 \: N}[/tex]

Plug force as 0.

[tex]\mathrm{0=mass \times acceleration}[/tex]

Divide both sides by mass.

[tex]\mathrm{\frac{0}{mass} =acceleration}[/tex]

[tex]\mathrm{0 =acceleration}[/tex]

[tex]\mathrm{acceleration= 0\: m/s/s}[/tex]

Answer:

The Independent variable in this experiment is the time taken by the ball to roll down each distance.

The dependent variable is the distance  through which the ball rolls

The control variables are: slope of hill, weight, of the ball, size of ball, wind speed, surface characteristics of the ball.

Explanation:

The complete question is

Jane is collecting data for a ball rolling down a hill. She measures out a set of different distances and then proceeds to use a stop watch to find the time it takes the ball to roll. What are the independent, dependent, and control variables in this experiment?

Independent variable have their values not dependent on any other variable in the scope of the experiment. The time for the ball to roll down the hill is not dependent on any other variable in the experiment. Naturally, some common independent variables are time, space, density, mass, fluid flow rate.

A dependent variable has its value dependent on the independent variable in the experiment. The value of the distance the ball rolls depends on the time it takes to roll down the hill.

The relationship between the dependent and independent variables in an experiment is given as

y = f(x)

where y is the output or the dependent variable,

and x is the independent variable.

Control variables are those variable that if not held constant could greatly affect the results of an experiment. For an experiment to be more accurate, control variables should be confined to a given set of value throughout the experiment.

Answer:

4.9x10^-6T

Explanation:

See attached file

That TIME is called the "period" of the wave.

(It's not one of the choices.)

Answer:

d. if any of the above occurs

Explanation:

That is The number of fringes per unit length on the screen can be doubled if

if the distance between the slits is doubled.

if the wavelength is changed to λ = λ/2.

And if the distance between the slits is quadruple the original distance and the wavelength is changed to λ = 2λ

Answer:

The  inductance is [tex]L = 40\mu H[/tex]

Explanation:

From the question we are told that

    The number of turns is  [tex]N = 163 \ turns[/tex]

    The  diameter is  [tex]D = 6.13 \ mm = 6.13 *10^{-3} \ m[/tex]

    The  length is  [tex]l = 2.49 \ cm = 0.0249 \ m[/tex]

The radius is evaluated as [tex]r = \frac{d}{2}[/tex]

substituting values

        [tex]r = \frac{6.13 *10^{-3}}{2}[/tex]

       [tex]r = 3.065 *10^{-3} \ m[/tex]

The  inductance of the Tarik's solenoid is mathematically represented as

            [tex]L = \frac{\mu_o * N^2 * A }{l }[/tex]

Here [tex]\mu_o[/tex] is the permeability of free space with value  

        [tex]\mu_o = 4\pi *10^{-7} \ N/A^2[/tex]

A is the area which is mathematically evaluated as

         [tex]A = \pi r^2[/tex]

substituting values

       [tex]A = 3.142 * [ 3.065*10^{-3}]^2[/tex]

       [tex]A = 2.952*10^{-5} \ m^2[/tex]

substituting values into formula for L  

      [tex]L = \frac{ 4\pi *10^{-7} * [163]^2 * 2.952*10^{-5} }{0.0249 }[/tex]

     [tex]L = 40\mu H[/tex]

Answer:

D. 0.01 s

Explanation:

The maximum speed is the amplitude times the angular frequency.

v = Aω

4.4 m/s = (0.0070 m) ω

ω = 628.6 rad/s

The period is:

ω = 2π / T

T = 2π / ω

T = 2π / (628.6 rad/s)

T = 0.01 s

Now, there is some information missing to this problem, since generally you will be given a figure to analyze like the one on the attached picture. The whole problem should look something like this:

"Beam AB has a negligible mass and thickness, and supports the 200kg uniform block. It is pinned at A and rests on the top of a post, having a mass of 20 kg and negligible thickness. Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously."

Answer:

[tex]\mu_{sB}=0.126[/tex]

[tex]\mu_{sC}=0.168[/tex]

Explanation:

In order to solve this problem we will need to draw a free body diagram of each of the components of the system (see attached pictures) and analyze each of them. Let's take the free body diagram of the beam, so when analyzing it we get:

Sum of torques:

[tex]\sum \tau_{A}=0[/tex]

[tex]N(3m)-W(1.5m)=0[/tex]

When solving for N we get:

[tex]N=\frac{W(1.5m)}{3m}[/tex]

[tex]N=\frac{(1962N)(1.5m)}{3m}[/tex]

[tex]N=981N[/tex]

Now we can analyze the column. In this case we need to take into account that there will be no P-ycomponent affecting the beam since it's a slider and we'll assume there is no friction between the slider and the column. So when analyzing the column we get the following:

First, the forces in y.

[tex]\sum F_{y}=0[/tex]

[tex]-F_{By}+N_{c}=0[/tex]

[tex]F_{By}=N_{c}[/tex]

Next, the forces in x.

[tex]\sum F_{x}=0[/tex]

[tex]-f_{sB}-f_{sC}+P_{x}=0[/tex]

We can find the x-component of force P like this:

[tex]P_{x}=360N(\frac{4}{5})=288N[/tex]

and finally the torques about C.

[tex]\sum \tau_{C}=0[/tex]

[tex]f_{sB}(1.75m)-P_{x}(0.75m)=0[/tex]

[tex]f_{sB}=\frac{288N(0.75m)}{1.75m}[/tex]

[tex]f_{sB}=123.43N[/tex]

With the static friction force in point B we can find the coefficient of static friction in B:

[tex]\mu_{sB}=\frac{f_{sB}}{N}[/tex]

[tex]\mu_{sB}=\frac{123.43N}{981N}[/tex]

[tex]\mu_{sB}=0.126[/tex]

And now we can find the friction force in C.

[tex]f_{sC}=P_{x}-f_{xB}[/tex]

[tex]f_{sC}=288N-123.43N=164.57N[/tex]

[tex]f_{sC}=N_{c}\mu_{sC}[/tex]

and now we can use this to find static friction coefficient in point C.

[tex]\mu_{sC}=\frac{f_{sC}}{N}[/tex]

[tex]\mu_{sC}=\frac{164.57N}{981N}[/tex]

[tex]\mu_{sB}=0.168[/tex]

Answer:

The electric field just inside the charged conductor is zero.

Explanation:

Electric field is defined as the region where electrical force is experienced by an electric charge usually as a result of the presence of another electric charge. A charged conductor is said to be in electrostatic equilibrium when it is in an electrostatically balanced state. This simply means a state in which the free electrical charges in the charged conductor have stopped moving.

For any charged conductor that has attained electrostatic equilibrium, the electric field at any point below the surface of the charged conductor falls to zero. Hence the electric field just inside the charged conductor is zero.

Answer:

The speed of the police car is 294 m/s

Explanation:

Given;

frequency of the siren in air, f = 280 Hz

speed of sound in air, v = 343 m/s

Determine the wavelength of the sound in air to the stationary car:

v = fλ

where;

λ is wavelength of the sound

λ = v/f

λ = 343 / 280

λ = 1.225 m

Now, determine the speed at which the police car is approaching the stationary car;

The actual frequency of the police car, F = 240 Hz

V = Fλ

Where;

V is speed of the police car

λ is the distance between the police car and the stationary car, (wavelength)

V = 240 x 1.225

V = 294 m/s

Therefore, the speed of the police car is 294 m/s

Answer:

733m/s

Explanation:

Assuming that Just after impact:

given that at lowest point Is

T2 + V2 = T3 + V3

1 /2(4 + 0.002) (vB)²2 + 0 = 0 + (4 + 0.002)(9.81)(1.25)(1 -cos 6°)

(vB)2 = 0.3665

m>s For the system of bullet and block: ( S+ ) Σmv1 = Σmv2 0.002(vB)1 = (4 + 0.002)(0.3665) (vB)1 = 733 m/s

Answer:

A field is a way of mapping forces surrounding any object that can act on another object at a distance without apparent physical connection. The field represents the object generating it. Gravitational fields map gravitational forces, electric fields map electrical forces, and magnetic fields map magnetic forces.

Explanation:

Answer:

The final angular speed is 6.25 rad/s

Explanation:

Given;

initial angular speed, ω₁ = 5 rad/s

initial moment of inertia, I₁ = 2.25 kg.m²

Final moment of inertia, I₂ = 1.8 kg.m²

final angular speed, ω₂ = ?

Based on conservation of angular momentum, we will have the following expression;

ω₁I₁ = ω₂I₂

ω₂ = (ω₁I₁ ) / I₂

ω₂ = (5 x 2.25) / 1.8

ω₂ = 6.25 rad/s

Therefore, the final angular speed is 6.25 rad/s

Answer:

Higher.

Explanation:

The greater the frequency the bigger the amplitude gets and the greater pitch gets.

Think - more energy, bigger waves, more waves, and higher sound

Answer:

The flow rate is  [tex]R =2.357 *10^{-4} \ m^3/s[/tex]

Explanation:

From the question we are told that

    The velocity is [tex]v = 3.0 \ m/s[/tex]

   The  diameter of the pipe is  [tex]d = 1.0 \ cm = 0.01 \ m[/tex]

The  radius of the pipe is mathematically represented as

            [tex]r = \frac{d}{2}[/tex]

substituting values

            [tex]r = \frac{0.01}{2}[/tex]

           [tex]r = 0.005 \ m[/tex]

The flow rate is mathematically represented as

       [tex]R = v * A[/tex]

Where is the cross-sectional area of the pipe which is mathematically evaluated as

      [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * (0.005)^2[/tex]

     [tex]A = 7.855 * 10^{-5} \ m^2[/tex]

So

    [tex]R = 3.0 * 7.855 *10^{-5}[/tex]

    [tex]R = 2.357*10^{-4} \ m^3 /s[/tex]

Answer:

2.64 x 10⁻⁶T

Explanation:

The magnitude of the magnetic field produced by a long straight wire carrying current is given by Biot-Savart law as follows: "The magnetic field strength is directly proportional to the current on the wire and inversely proportional to the distance from the wire". This can be written mathematically as;

B = (μ₀ I) / (2π r)                ----------------(i)

B is magnetic field

I is current through the wire

r is the distance from the wire

μ₀  is the magnetic constant = 4π x 10⁻⁷Hm⁻¹

From the question;

I = 0.7A

r = 0.053m

Substitute these values into equation (i) as follows;

B =  (4π x 10⁻⁷ x 0.7) / (2π x 0.053)

B = 2.64 x 10⁻⁶T

Therefore the approximate magnitude of the magnetic field at that location is 2.64 x 10⁻⁶T

Answer:

E   = α/2∈₀ [ 1 - a²/r² ]

Ф = α/2∈₀

Explanation:

Using Gauss Law:

    ρ(r) = a/r, dA

          = 4 π r²d r

    Ф = [tex]\int\limits^r_a[/tex] ρ(r')dA

    Ф[tex]_{encl}[/tex] = [tex]\int\limits^r_a[/tex] ρ(r')dA

             = 4πα [tex]\int\limits^r_a[/tex] r'dr'

Ф[tex]_{encl}[/tex]     = 4 π α 1/2(r²-a²)

E(4πr²) = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀

           = [tex]2\pi\alpha (r^{2}-a^{2} )/[/tex]∈₀(4πr²)

           = α (r² - a²) / 2 ∈₀ (r²)

           = α/2∈₀ [ r²/r² - a²/r² ]

      E   = α/2∈₀ [ 1 - a²/r² ]

Electric field of the point charge:

E[tex]_{q}[/tex] = q / 4π∈₀r²

[tex]E_{total}[/tex] = α / 2 ∈₀ - (α / 2 ∈₀ )(a² / r²) + q / 4 π ∈₀ r²

For [tex]E_{total}[/tex]  to be constant:

- (αa²/ 2 ∈₀ ) + q / 4 π ∈₀ = 0 and q = 2παa²

-> α / 2 ∈₀ - αa²/ 2 ∈₀ + 2παa² / 4 π ∈₀

= α - αa² + αa² / 2 ∈₀

= α /2 ∈₀

Hence:

Ф = α/2∈₀

Answer:

a)

6t i + t² j m

b)

6 i + 2 t j m/s

c)

x = 24 m and y = 16 m

d)

10 m/s

Explanation:

Explanation is given in the attached document.

Answer:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

Explanation:

1. The electric force is given by the relation

           F = k Q e / r2

where k is the Coulomb constant, Q the charge used, e the charge of the electron and r the distance between the two.

 The strength depends on:

a) true.

b) True

c) False. In the equation above the mass does not appear

d) True

e) False. Mass does not appear in the equation

f) False. The load even when distributed in the space can be considered concentrated in the center

two.

a) True

b) Treu

c) Fail

f) false

For a single electron located at a distance from a positive charge, we have:

1. The force on the electron depends on the distance between it and the positive charge (option a) and the charge of both particles (option b and d).      

2. The electric field at the electron's position depends on the distance between the positive charge and it (option a) and the charge of the positive particle (option d).    

The force on a single electron at a distance from the point charge is given by Coulomb's law:

[tex] F = \frac{Kq_{1}q_{2}}{r^{2}} [/tex]    (1)

Where:

As we can see in equation (1), the force on the electron by the positive charge depends on both charges q₁ and q₂, and the distance, so the correct options are:

a. The distance between the positive charge and the electron

b. The charge on the electron

d. The charge of the positive charge

The other options (c, e, f, and g) are incorrect because the electric force does not depend on the particles' masses or their radii.

The electric field (E) at a distance "r" from a point charge is given by:

[tex] E = \frac{Kq_{1}}{r^{2}} [/tex]   (2)

From equation (2), we can see that the electric field is directly proportional to the charge and inversely proportional to the distance of interest (r).  

The electric field at the electron's position is given by the one produced by the positive charge, so the correct options are:

a. The distance between the positive charge and the electron

d. The charge of the positive charge

The other options (b, c, e, f, and g) are incorrect because the electric field is independent of the mass of the charges involved and their radii.

Therefore, the correct options for part 1 are a, b, and d and for part 2 are a and d.

Learn more about the electric field here:

brainly.com/question/13308086

I hope it helps you!

Answer:

P = cte / V

therefore pressure and volume are inversely proportional

Explanation:

For this exercise we can join the ideal gases equation

        PV = n R T

they indicate that the amount of matter and the temperature are constant, therefore

         PV = cte

        P = cte / V

therefore pressure and volume are inversely proportional

Answer:

The angle between the polarizing axes of the two polarizers is 54°

Explanation:

Given;

intensity of unpolarized light, I₀ = 54.0 W/m²

intensity of light that emerges from second ideal polarizer, I₁ = 19.0 W/m²

The angle between the polarizing axes of the two polarizers is dtermined by applying Malus' law for intensity of a linearly polarized light passing through a polarizer.

I₁ = I₀Cos²θ

Cos²θ = I₁ / I₀

Cos²θ = 19 / 54

Cos²θ =0.3519

Cos θ = √0.3519

Cos θ = 0.5932

θ = Cos⁻¹(0.5932)

θ = 53.6°

θ = 54°

Therefore, the angle between the polarizing axes of the two polarizers is 54°

Answer:

Weight is 83 385 N

Explanation:

Weight is calculated by multiplying the mass by the gravitational acceleration constant

       Weight = mass* gravity

Assuming that the gravitational constant is 9.81 m/s^2

        Weight = mass* gravity

        Weight of crane = (8500 kg)*(9.81 m/s^2)

        Weight = 83 385 kg*m/s^2 or 83 385 N

Answer:

The weight of the rod is 32.87 N

Explanation:

Density of the rod = 7800 kg/m

length of the rod = 81.2 cm = 0.812 m

diameter of rod = 2.60 cm = 0.026 m

acceleration due to gravity = 9.80 m/s^2

The rod can be assumed to be a cylinder.

The volume of the rod can be calculated as that of a cylinder, and can be gotten as

V = [tex]\frac{\pi d^{2} l}{4}[/tex]

where d is the diameter of the rod

l is the length of the rod

V = [tex]\frac{3.142* 0.026^{2}* 0.812}{4}[/tex] = 4.3 x 10^-4 m^3

We know that the mass of a substance is the density times the volume i.e

mass m = ρV

where ρ is the density of the rod

V is the volume of the rod

m = 4.3 x 10^-4 x 7800 = 3.354 kg

The weight of a substance is the mass times the acceleration due to gravity

W = mg

where g is the acceleration due to gravity g = 9.80 m/s^2

The weight of the rod W = 3.354 x 9.80 = 32.87 N

Answer:

Maximum acceleration is 800m/s^2

Explanation:

See attached file

Explanation:

Given that,

Focal length of a converging lens, f = +25 cm

Size of the object, h = 10 cm

Object distance, u = -30 cm

We need to find the image distance and the size of the image.

Using lens formula, [tex]\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}[/tex], v is image distance

[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{25}+\dfrac{1}{(-30)}\\\\v=150\ cm[/tex]

Magnification,

[tex]m=\dfrac{v}{u}=\dfrac{h'}{h}[/tex]

h' is the size of image

[tex]h'=\dfrac{vh}{u}\\\\h'=\dfrac{150\times 10}{(-30)}\\\\h'=-50\ cm[/tex]

So, the image distance is 150 cm the size of image is 50 cm.

Answer:

Equilibrium

Explanation:

Physics terminology I guess? Equilibrium means that an object isn't moving.

Answer:

Balanced Forces

Explanation:

When forces are in balance, acceleration is zero. Velocity is constant and there is no net or unbalanced force. A plane will fly at constant velocity if the acceleration is zero.

5.19 x 10³Hz

The capacitive reactance, [tex]X_{C}[/tex], which is the opposition given to the flow of current through the capacitor is given by;

[tex]X_C = \frac{1}{2\pi fC }[/tex]

Where;

f = frequency of the signal through the capacitor

C = capacitance of the capacitor.

Also, from Ohm's law, the voltage(V) across the capacitor is given by the product of current(I) and the capacitive reactance. i.e;

V = I x [tex]X_{C}[/tex]             [Substitute the value of

=> V = I x [tex]\frac{1}{2\pi fC}[/tex]      [Make f the subject of the formula]

=> f = [tex]\frac{I}{2\pi VC}[/tex]                    ---------------------(i)

From the question;

I = 3.33mA = 0.00333A

C = 8.50nF = 8.50 x 10⁻⁹F

V = 12.0V

Substitute these values into equation (i) as follows;

f = [tex]\frac{0.00333}{2 * 3.142 * 12.0 * 8.50 * 10^{-9}}[/tex]            [Taking [tex]\pi[/tex] = 3.142]

f = 5.19 x 10³Hz

Therefore, the frequency is closest to f = 5.19 x 10³Hz