Particle q1 has a positive 6 µC charge. Particle q2 has a positive 2 µC charge. They are located 0.1 meters apart.


Recall that k = 8.99 × 109 N•meters squared per Coulomb squared.


What is the force applied between q1 and q2?

Answer:

Explanation:

Diffraction is the term used to describe the bending of a wave around an obstacle. It is one of the general properties of waves.

1. Diffraction of sound is the bending of sound waves around an obstacle which propagates from source to a listener. Two of the daily phenomena that exhibit diffraction of sound are:

i. The voices of people talking outside a building can be heard by those inside.

ii. The sound from the horn of a car can be heard by people at certain distances away.

When sound waves are produced, the surrounding air molecules are required for its transmission. This is because sound wave is a mechanical wave which requires material medium for its propagation. When a source produces a sound, the sound waves bend around obstacles on its path to reach listeners.

2. Light waves are electromagnetic waves which can undergo diffraction. Diffraction of light is the bending of the rays of light around an obstacle. Two of the daily phenomena that exhibit diffraction of light are:

i. The shadow of objects which has the umbra and penumbra regions.

ii. The apparent color of the sky.

A ray of light is the path taken by light, and the combination of two or more rays is called a beam. A ray or beam of light travels in a straight line, so any obstacle on its path would subject the light to bending around it during propagation. These are major applications in pin-hole cameras, shadows, rings of light around the sun etc.

Some significant differences between diffraction of light and that of the sound are:

i. Diffraction of light is not as common as that of sound.

ii. Sound propagates through a wider region than light waves.

iii. Sounds are longitudinal waves, while lights are transverse waves.

Answer:2 volts

Explanation:

Power=current x voltage

6=3 x voltage

Divide both sides by 3

6/3=(3 x voltage)/3

2=voltage

Voltage=2volts

If part of an electric circuit dissipates energy at 6 Watts when it draws a current of 3 Amperes, then the voltage is impressed across it would be 2 volts .

The rate of doing work is known as power. The Si unit of power is the watt .

The power dissipated in the circuit  = work / time

As given in the problem we have to find out what voltage is impressed across the circuit If part of an electric circuit dissipates energy at 6 Watts  when it draws a current of 3 Amperes,

The power dissipated across the circuit  = Voltage × current

                                              6 watts  = Voltage × 3 ampere

                                              Voltage  = 6 / 3

                                              Voltage = 2 volts

Thus , the voltage across the circuit would be  2 volts .

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Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

Answer:

a) W = 1815 J

b) W = 135 J

Explanation:

We need to model the force using Hooke's law

We first get the elastic constant, k

A force of 60 N causes an extension of 0.5 m

F = 60 N, e = 0.5 m

F = k * e

60 = 0.5 k

k = 60/0.5

k = 120 N/m

Therefore the force in terms of the extension, x is:

F(x) = 120x

a) Work done in stretching the spring 5.5 m from its equilibrium position

b = 0 m, a = 5.5 m

[tex]W = \int\limits^a_b {F(x)} \, dx[/tex]

[tex]W = \int\limits^a_b {120x} \, dx \\a =0, b = 5.5[/tex]

[tex]W = 60x^{2} \\W = 60 [5.5^{2} -0^{2} ][/tex]

[tex]W = 1815 J[/tex]

b) Work required to compress the spring 1.5 m from its equilibrium position

[tex]W = \int\limits^a_b {F(x)} \, dx[/tex] b = 0, a = -1.5

[tex]W = 60x^{2} \\W = 60 [(-1.5)^{2} -0^{2} ]\\[/tex]

W = 135 J

The work done to compress the spring 1.5 m from its equilibrium position is 135 J

Answer:

(a). [tex]\dfrac{f_3'}{f_3} =\sqrt{5}.[/tex]

(b). The wavelength remains unchanged.

Explanation:

The speed [tex]v[/tex] of the waves on the string with tension [tex]T_i[/tex] is given by

[tex]v = \sqrt{\dfrac{T_iL}{m} }[/tex]

And if the string is vibrating, its fundamental wavelength is [tex]2L[/tex], and since the frequency [tex]f[/tex] is related to the wave speed and wavelength by

[tex]f = v/\lambda[/tex]

the fundamental frequency [tex]f_1[/tex] is

[tex]f_1 =\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}[/tex]

and since the frequency of the third harmonic is

[tex]f_3 = 3f_1[/tex]

[tex]f_3 = 3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L},[/tex]

and the wavelength is

[tex]\lambda_3 = \dfrac{2L}{3}.[/tex]

(a).

Now, if we increase to the string tension to

[tex]T_f = 5.0T_i[/tex]

the third harmonic frequency becomes

[tex]f_3' = 3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L},[/tex]

The ratio of this new frequency to the old frequency is

[tex]\dfrac{f_3'}{f_3} = \dfrac{3\sqrt{\dfrac{5T_iL}{m} }*\dfrac{1}{2L}}{3\sqrt{\dfrac{T_iL}{m} }*\dfrac{1}{2L}}[/tex]

[tex]\boxed{\dfrac{f_3'}{f_3} =\sqrt{5}.}[/tex]

(b).

The wavelength of the third harmonic remains unchanged because [tex]\lambda_3 = \dfrac{2L}{3}.[/tex] depends only on the length of the string

Answer: 4.821m/s

Explanation:

initial momentum = final momentum 

(80 x 4.1) + (85 x 5.5) = (80 + 85)u

328 + 467.5 = 165u  =  795.5 / 165 = 4.821m/s

Answer:

Explanation:

We shall solve this question with the help of Ampere's circuital law.

Ampere's ,law

∫ B dl = μ₀ I , B is magnetic field at distance x from the axis within wire

we shall find magnetic field at distance x . current enclosed in the area of circle of radius x

=  I x π x²  / π R²

= I x²  /  R²

B x 2π x = μ₀  x current enclosed

B x 2π x = μ₀  x  I x²  /  R²

B =  μ₀   I x  / 2π R²

Maximum magnetic B₀ field  will be when x = R

B₀ = μ₀I   / 2π R

Given

B = B₀ / 3

μ₀   I x  / 2π R² = μ₀I   / 2π R x 3

x = R / 3

b ) The largest value of magnetic field is on the surface of wire

B₀ = μ₀I   / 2π R

At distance x outside , let magnetic field be B

Applying Ampere's circuital law

∫ B dl = μ₀ I

B x 2π x = μ₀ I

B = μ₀ I / 2π x

Given B = B₀ / 3

μ₀ I / 2π x = μ₀I   / 2π R x 3

x = 3R .

The location inside the wire where the magnetic field equals a third of its maximum value is at r/R = 1/3. There is no point outside the wire where the magnetic field reaches a third of its maximum value because it monotonically decreases.

To determine at what location inside the wire the magnetic field produced by the current is equal to a third of its largest value, we need to apply Ampère's Law. Inside a conductor carrying uniform current, the magnetic field B increases linearly with the distance r from the center of the wire due to the proportion of current enclosed. Thus, the magnetic field is given by B = μ_0 J r / 2, where J is the current density and μ_0 is the permeability of free space. Since the value at the surface (r=R) will be maximum (B_max), for B to be a third of B_max, we must have (1/3)B_max = (μ_0 J r / 2). Solving for r, we find r/R = 1/3.

For locations outside the wire, Biot-Savart Law or Ampère's Law show that the magnetic field decreases with 1/r. However, since the field is maximum at the surface, and we do not have an equation that varies outside the wire, we cannot directly calculate when the field will be a third without additional information on how the field varies with r beyond the wire's surface. Normally, the largest value of B is at the surface, and it decreases monotonically outside, never increasing again to allow for a location at which it is a third of its largest value.

The magnetic field for points inside the wire is proportional to the distance from the center, and for points outside the wire, it is inversely proportional to the distance from the center. However, the magnetic field does not reach a third of its largest value at any given point outside the cylindrical wire since it monotonically decreases.

Answer:

T = 1.07 s

a = 1.034 m/s2

Explanation:

Metric unit conversion

m = 145 g = 0.145 kg

x = 3 cm = 0.03 m

Suppose this is a simple harmonic motion, then its period T can be calculated using the following equation

[tex]T = 2\pi\sqrt{\frac{m}{k}}[/tex]

[tex]T = 2\pi\sqrt{\frac{0.145}{5}} = 1.07 s[/tex]

The maximum acceleration would occurs when spring is at maximum stretching length, aka 0.03m

The spring force at that point would be

[tex]F_s = kx = 5*0.03 = 0.15 N[/tex]

According to Newton's 2nd law, the acceleration at this point would be

[tex]a = F_s/m = 0.15 / 0.145 = 1.034 m/s^2[/tex]

Answer:

Between R and S

Explanation:

Answer:

C) Between R and S

Explanation:

Answer:

Explanation:

find the solution below

Final answer:

Calculating voltage, current at maximum power, and fill factor for a PV cell involves applying the Shockley diode equation under illumination. These calculations help in identifying the Maximum Power Point (MPP), crucial for optimizing the cell's performance. Detailed calculations were not provided, but the importance of plotting the I-V characteristics to validate the MPP was emphasized.

Explanation:

The task requires calculating the voltage and current at maximum power and the fill factor for a photovoltaic (PV) cell, given its light current density and reverse saturation current density, at an ambient temperature of 32°C. This involves understanding the behavior of PV cells under varying conditions and applying theoretical models to determine the operating characteristics that yield maximum power output. The fill factor (FF), the maximum power point (MPP), and the open-circuit voltage (Voc) are crucial aspects in assessing the performance and efficiency of PV cells. To derive an accurate answer, one would typically employ the Shockley diode equation adjusted for light conditions and solve for the power at different voltages to find the maximum. However, the calculation details are beyond this response scope due to the required complexity and dependency on specific model parameters not provided. Generally, these calculations involve detailed semiconductor physics and numerical methods to solve nonlinear equations reflective of the cell's I-V characteristic in illuminated conditions. Plotting current and power as functions of voltage from V=0 to Voc, helps validate the maximum power point approximation. This plot demonstrates the balance between voltage and current that a PV cell must achieve to optimize power output, usually found using methods like the perturb and observe algorithm in practical MPPT (Maximum Power Point Tracking) systems.

a) 10 kg

b) 60 N

c) 98 N

d) 0 N

Explanation:

a)

In this problem, there are two forces acting on the backpack:

- The restoring force from the spring, upward, of magnitude [tex]F=60 N[/tex]

- The weight of the backpack, downward, of magnitude [tex]mg[/tex], where

m = mass of the backpack

[tex]g=9.8 m/s^2[/tex] acceleration due to gravity

So the net force on the backpack is (taking upward as positive direction)

[tex]F_{net}=F-mg[/tex]

According to Newton's second law of motion, the net force must be equal to the product between the mass of the backpack and its acceleration, so

[tex]F-mg=ma[/tex]

where

[tex]a=-3.8 m/s^2[/tex] is the acceleration of the backpack and the elevator, downward (so, negative)

If we solve the formula for m, we can find the mass of the backpack:

[tex]F=m(a+g)\\m=\frac{F}{a+g}=\frac{60}{-3.8+9.8}=10 kg[/tex]

b)

In this case, the elevator is moving upward, and it is slowing down at a rate of [tex]3.8 m/s^2[/tex].

Since the elevator is slowing down, it means that the direction of the acceleration is opposite to the direction of motion: and since the elevator is moving upward, this means that the direction of the acceleration is downward: so the acceleration is negative,

[tex]a=-3.8 m/s^2[/tex]

The net force acting on the backpack is still:

[tex]F_{net}=F-mg[/tex]

where

F is the restoring force in the spring, which this time is unknown (it corresponds to the reading on the scale)

Using again Newton's second law of motion,

[tex]F-mg=ma[/tex]

Therefore in this case, the reading on the scale will be:

[tex]F=m(g+a)=(10)(9.8-3.8)=60 N[/tex]

So the reading is the same as in part a).

c)

In this case, the elevator is moving at a constant velocity.

The net force on the backpack is still:

[tex]F_{net}=F-mg[/tex]

However, since here the elevator is moving at constant velocity, and acceleration is the rate of change of velocity, this means that the acceleration of the elevator is zero:

[tex]a=0[/tex]

So Newton's second law of motion can be written as:

[tex]F_{net}=0[/tex]

So

[tex]F-mg=0[/tex]

Which means that the reading on the scale is equal to the weight of the backpack:

[tex]F=mg=(10)(9.8)=98 N[/tex]

d)

In this  case, the elevator had no brakes and the cable supporting it breaks loose.

This means that the elevator is now in free fall.

So its acceleration is simply the acceleration due to gravity (which is the acceleration of an object in free fall):

[tex]a=-9.8 m/s^2[/tex]

And the direction is downward, so it has a negative  sign.

The net force on the backpack is still

[tex]F_{net}=F-mg[/tex]

So Newton's second law can be rewritten as

[tex]F-mg=ma[/tex]

Therefore, we can re-arrange the equation to find F, the reading on the scale, and we find:

[tex]F=m(g+a)=(10)(9.8-9.8)=0 N[/tex]

So, the reading on the scale is 0 N.

Answer:

nail will stop after traveling 1000 m    

Explanation:

We have given mass m = 50 gram = 0.05 kg

Frictional force which is used to stop the mass F = 62.5 kN

Initial velocity is given u = 50 m/sec

From newton's law force is equal to F = ma, here m is mass and a is acceleration

So [tex]a=\frac{F}{m}=\frac{62.5\times 10^3}{0.05}=1.25\times 10^6m/sec^2[/tex]

As finally nail stops so final velocity v = 0 m/sec

From third equation of motion [tex]v^2=u^2+2as[/tex]

So [tex]0^2=50^2-2\times 1.25\times 10^6\times s[/tex]

s = 1000 m

So nail will stop after traveling 1000 m

Answer:

A) = 0.63 m

B)This is approximately 0.21

Explanation:

Part A)

Now if spring is connected to the block then again we can use energy conservation

so we will have

[tex]\frac{1}{2}kx^2 = mg(x + x')sin\theta + \frac{1}{2}kx'^2[/tex]

so we will have

[tex]\frac{1}{2}(70)(0.5^2) = 2(9.81) (0.50 + x') sin41 + \frac{1}{2}(70)x'^2[/tex]

[tex]8.75 = 6.43 + 12.87 x' + 35 x'^2[/tex]

[tex]x' = 0.13 m[/tex]

so total distance moved upwards is

[tex]L = 0.5 + 0.13 = 0.63 m[/tex]

Part B

Now the incline has a coefficient of kinetic friction μk. If the block, attached to the spring, is observed to stop just as it reaches the spring's equilibrium position, what is the coefficient of friction μk

KE = 8.75 – 10.78 * sin 41

This is approximately 1.678 J. This is the kinetic energy at the equilibrium position. For the block to stop moving at this position, this must be equal to the work that is done by the friction force.

Ff = μ * 10.78 * cos 41

Work = 0.5 * μ * 10.71 * cos 41

μ * 10.78 * cos 41 = 8.75 – 10.78 * sin 41

μ = (10 – 8.82 * sin 41) ÷ 8.82 * cos 41

μ = 1.678/8.136

μ = 0.206

This is approximately 0.21

Answer:

A) d = 0.596m

B) μ = 0.206

Explanation:

A) The potential energy stored in a spring when it is compressed is given as;

U = (1/2)kx² - - - - - - (eq1)

Where

U is potential energy stored in spring

k is spring constant

x is compressed length of the spring

Let the height the mass moved before coming to rest be h.

Thus, the potential energy at this height is;

U = mgh - - - - - - (eq2)

Since the mass is connected to the spring, according to the principle of conservation of energy, initial potential energy of the spring is equal to the sum of the final potential energy in the spring and the potential energy of the mass. Thus, we have;

(1/2)K(x1)² = (1/2)K(x2)² + mgh

Where x1 is the initial compressed length and x2 is the final compressed length.

Now, h will be dsin41 while x2 will be d - x1

Where d is the distance the mass moves up before coming to rest

Thus, we now have;

(1/2)K(x1)² = (1/2)K(d - x1)² + mg(dsin41)

(1/2)K(x1)² = (1/2)K(d² - 2dx1 + (x1)²) + mg(dsin41)

(1/2)K(x1)² = (1/2)Kd² - Kdx1 + (1/2)K(x1)² + mg(dsin41)

(1/2)Kd² - Kdx1 + mg(dsin41) = 0

(1/2)Kd² + d[mg(sin41) - Kx1] = 0

From the question,

k = 70 N/m

m = 2.2 kg

x1 = 0.5m

g = 9.8 m/s²

Thus, plugging in these values, we now have;

(1/2)(70)d² + d[(2.2•9.8•0.6561) - (70•0.5)] = 0

35d² - 20.8545d = 0

35d² = 20.8545d

Divide both sides by d to get;

35d = 20.8545

d = 20.8545/35

d = 0.596m

B) Here we are looking for the coefficient of friction.

First of all, let's find the kinetic energy at the equilibrium position;

The potential energy of the spring;

P.E = (1/2)K(x1)² = (1/2)(70)(0.5)² = 8.75J

The energy that will cause the block to decelerate = mg(x1)sin41 = 2.2 x 9.8 x 0.5 x 0.6561 = 7.073

So,

Net KE = 8.75 – 7.073 = 1.677J

Now, for the block to stop moving at this equilibrium position, the work done by the frictional force must be equal to KE of 1.677J

Thus,

F_f(x1) = 1.677J

F_f = μmgcos 41

Where, μ is the coefficient of friction;

So, F_f = μ(2.2 x 9.8 x 0.7547)

Ff = 16.271μ

Thus,

16.271μ x 0.5 = 1.677J

8.136μ = 1.677

μ = 1.677/8.136

μ = 0.206

The position of the mass after t seconds can be found using the equation of motion for a mass-spring system. We can calculate the angular frequency, determine the phase angle, and substitute the values into the equation to find the position.

The position of the mass after t seconds can be found using the equation of motion for a mass-spring system. The equation is given by:

x(t) = A * cos(ω * t + φ)

Where A is the maximum displacement, ω is the angular frequency, and φ is the phase angle. In this problem, the maximum displacement is 0.2 meters, the angular frequency can be calculated using ω = sqrt(k/m) (where k is the spring constant and m is the mass), and the phase angle can be determined using the initial conditions.

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The problem is a case of simple harmonic motion which is solved by first identifying the spring constant using Hooke's Law, then finding the equation for the motion, resulting in x(t) = 0.5 cos(sqrt(41.67) * t + π/2), which gives the position of the mass as a function of time.

The problem presented is a classical physics problem dealing with the motion of a mass attached to a spring. This is a scenario of simple harmonic motion which is characterized by an oscillating motion about a stable equilibrium.

In simple harmonic motion, the spring force is given by Hooke's law, F = -kx, where k is the spring constant and x is the displacement of the spring from its equilibrium. In this case, we are given that the force needed to stretch the spring 0.2m beyond its natural length is 50N. So we can find the spring constant k by rearranging Hooke's law: k = F/x = 50N / 0.2m = 250 N/m.

Next, we know that the equation for the position of the mass in simple harmonic motion is given by x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency and φ is the phase.

Here, the amplitude A is the initial position, which is 0. The initial velocity is given as 0.5 m/s, and ω can be found using the relation ω = sqrt(k/m) = sqrt(250 / 6) = sqrt(41.67) rad/s. The phase can be found using the fact that the initial velocity v0 is v0 = ωA sin(φ), and since A=0 and v0=0.5, φ = arcsin(v0/ωA) = arcsin((0.5 m/s) / (sqrt(41.67) rad/s * 0) = π/2.

Therefore, the equation for the motion becomes x(t) = 0.5 cos(sqrt(41.67) * t + π/2). This equation gives the position x of the mass as a function of time t.

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Answer:

[tex]u(t)=1.15 \sin (8.68t)cm[/tex]

0.3619sec

Explanation:

Given that

Mass,m=148 g

Length,L=13 cm

Velocity,u'(0)=10 cm/s

We have to find the position u of the mass at any time t

We know that

[tex]\omega_0=\sqrt{\frac{g}{L}}\\\\=\sqrt{\frac{980}{13}}\\\\=8.68 rad/s[/tex]

Where [tex]g=980 cm/s^2[/tex]

[tex]u(t)=Acos8.68 t+Bsin 8.68t[/tex]

u(0)=0

Substitute the value

[tex]A=0\\u'(t)=-8.68Asin8.68t+8.68 Bcos8.86 t[/tex]

Substitute u'(0)=10

[tex]8.68B=10[/tex]

[tex]B=\frac{10}{8.68}=1.15[/tex]

Substitute the values

[tex]u(t)=1.15 \sin (8.68t)cm[/tex]

Period =T = 2π/8.68

After half period

π/8.68 it returns to equilibruim

π/8.68 = 0.3619sec

Answer:

a

The potential of the earth surface is  [tex]V_E= - 9.6*10^9 V[/tex]

b

when the potential of the earth is zero the charge is choose to be negative this implies that  the potential at infinity would be [tex]V_\infty=+9.6*10^9V[/tex] and (Ignore the fact that positive charge in the ionosphere approximately cancels the Earth's net charge the potential at the surface increase and the electric field remains the same

Explanation:

From the question we are told that

      The magnitude  electric field is  [tex]E = 150V/m[/tex]

The potential of the earth surface when  V = 0 and  [tex]r = \infty[/tex] is mathematically represented as

                [tex]V_E = - \frac{q}{4 \pi \epsilon_o r} = - E *R[/tex]  

Where q is the charge on the surface of the earth  which is negative

           R is the radius of the earth

            [tex]V_E = 150 * 64 *10^6[/tex]

                 [tex]V_E= - 9.6*10^9 V[/tex]

when the potential of the earth is zero the charge is choose to be negative

Then the potential at infinity would be [tex]+9.6*10^9V[/tex]

Options:

a) more than 0.8 m .

b) equal to 0.8 m .

c) between 0.5 m and 0.8 m .

d) less than 0.5 m .

Answer:

b) equal to 0.8 m .

Explanation:

Note:

An upside down image = Inverted Image

An image that appears in front of the mirror = Real image

An image that appears behind the mirror = Virtual image

Let the object distance from the pole of the mirror be u

When the child stands 1.2 m from the mirror:

u = 1.2 m ( Real and Inverted image of the child is formed)

When the child stands about 0.8 m from the mirror:

u = 0.8 m (Virtual, erect and magnified image of the child is formed)

When the child stands about 0.5 m from the mirror:

u = 0.5 m ( Virtual and erect image of the child is formed)

Note: All objects positioned behind the focal length of a concave mirror are always real. Objects start becoming virtual when they are placed on the focal length or in front of it (Close to the pole of the mirror), although objects placed on the focus has its image formed at infinity.

Since the nature of the image formed changed from real to virtual when the child stands about 0.8 m from the mirror, then the focal length is approximately equal to 0.8 m

The approximate focal length of the mirror will be "0.8 m". A complete solution is provided below.

According to the question,

When a child stands 1.2 m, the real as well as inverted image will be formed, then

When a child stands 0.8 m, the virtual, erect as well as magnified image will be formed, then

When a child stands 0.5 m, the virtual as well as erect image will be formed, then

Since,

Whenever the image is formed from real to virtual, the focal length will become 0.8 m.

Thus the answer above is right.

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Answer:

The velocity of mass 2m is  [tex]v_B = 0.67 m/s[/tex]

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

     The initial speed  of the billiard ball A = [tex]v_1[/tex] =1 m/s

    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = [tex]v_A[/tex]

Let The finial speed  of the billiard ball  B = [tex]v_B[/tex]

      According to the law of conservation of Energy

                 [tex]\frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]

              Substituting values  

                [tex]\frac{1}{2} m (1)^2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2[/tex]

Multiplying through by [tex]\frac{1}{2}m[/tex]

                [tex]1 =v_A^2 + 2 v_B ^2 ---(1)[/tex]

    According to the law of conservation of Momentum

            [tex]mv_1 + 2m(0) = mv_A + 2m v_B[/tex]

    Substituting values

            [tex]m(1) = mv_A + 2mv_B[/tex]

Multiplying through by [tex]m[/tex]

           [tex]1 = v_A + 2v_B ---(2)[/tex]

making [tex]v_A[/tex] subject of the equation 2

            [tex]v_A = 1 - 2v_B[/tex]

Substituting this into equation 1

         [tex](1 -2v_B)^2 + 2v_B^2 = 1[/tex]

         [tex]1 - 4v_B + 4v_B^2 + 2v_B^2 =1[/tex]

          [tex]6v_B^2 -4v_B +1 =1[/tex]

          [tex]6v_B^2 -4v_B =0[/tex]

Multiplying through by [tex]\frac{1}{v_B}[/tex]

          [tex]6v_B -4 = 0[/tex]

            [tex]v_B = \frac{4}{6}[/tex]

            [tex]v_B = 0.67 m/s[/tex]

The stationary billiard ball of mass 2m will have a final velocity of 1/3 m/s after the head-on collision with a ball of mass m moving at 1 m/s due to conservation of momentum.

In this problem, we are dealing with a one-dimensional collision. The principle of conservation of momentum is key in solving this problem. Momentum before the collision is the same as momentum after the collision. The initial momentum is given by mass * velocity of moving ball = m * v1, while the stationary ball has 0 momentum.

So, we can write this as m * 1 = (m + 2m) * v', where v' is the final velocity of the stationary ball after the collision. Solving for v', we get v' = m / 3m = 1/3 m/s. Hence, the stationary ball (of mass 2m) has a final velocity of 1/3 m/s after the collision.

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Answer:

The resulting time constant will be [tex]\bf{5.09~ms}[/tex].

Explanation:

Given:

the time constant of the RC circuit, [tex]\tau = 8~ms[/tex]

The value of the capacitor in the circuit, [tex]C_{1} = 4~\mu F[/tex]

The value of addition capacitor added to the circuit, [tex]C_{2} = 7~\mu F[/tex]

The value of the time constant for a series RC circuit is give by

[tex]\tau = RC[/tex]

So the value of the resistance in the circuit is

[tex]R &=& \dfrac{\tau}{C}\\&=& \dfrac{8 \times 10^{-3}~s}{4 \times 10^-6~F}\\&=& 2000~\Omega[/tex]

When the capacitor [tex]C_{2}[/tex] is added to the circuit, the net value of the capacitance in the circuit is

[tex]C &=& \dfrac{C_{1}C_{2}}{C_{1} + C_{2}}\\&=& \dfrac{4 \times 7}{4 + 7}~\mu F\\&=& \dfrac{28}{11}~\mu F[/tex]

So the new time constant will be

[tex]\tau_{n} &=& (2000~\Omega)(\dfrac{28}{11} \times 10^{-6})~s\\&=& 5.09~ms[/tex]

The question is not complete, so i have attached an image with the complete question.

Answer:

A) speed of block a before it hits block B;v = √2gR

B) speed of combined blocks after collision;V_ab = ½√2gR

C) kinetic energy lost = ½MgR

D) temperature change; ∆t = ½gR/c

E) Additional thermal energy;W_f = 2μMgL

Explanation:

A) From conservation of energy, potential energy = kinetic energy.

Thus; P.E = K.E

So, mgr = ½mv²

m will cancel out to give;

gr = v²

So, v = √2gR

B) from conservation od momentum, momentum before collision = momentum after collision.

Thus;

M_a•V_a = M_ab•V_ab

Where;

M_a is the mass of block A

V_a is speed of block A

M_ab is the mass after collision

V_ab is speed after collision

So, making V_ab the subject, we have;

V_ab = M_a•V_a/M_ab

Now from answer in a above,

V_a = V = √2gR

Also, M_a = M and M_ab = 2M because it's the sum of 2 masses after collision.

Thus;

V_ab = (M√2gR)/(2M)

M will cancel out to give;

V_ab = ½√2gR

C) From the work-kinetic energy theorem, the net work done on the object is equal to the change in the kinetic energy of the object.

Thus;

W_net = K_f - K_i = ½m_ab•v_ab²- ½m_a•v_a²) = ∆K.

We have seen that:

M_a = M and M_ab = 2M

V_a = √2gR and V_ab = ½√2gR

Thus;

∆K = ½(2M•(½√2gR)²) - ½(M•(√2gR)²)

∆K = ½MgR - MgE

∆K = -½MgR

Negative sign means loss of energy.

Thus kinetic energy lost = ½MgR

D) The formula for heat energy is given by;

Q = m•c•∆t

Thus, change in temperature is;

∆t = Q/Mc

Q is the heat energy, thus Q = ½MgR

Thus;∆t = ½MgR/(Mc)

M will cancel out to give;

∆t = ½gR/c

E) Additional thermal energy is gotten from;

Work done by friction;W_f = F_f x d

Where;

F_f is frictional force given by μF_n. F_n is normal force

d is distance moved

Thus;

W_f = μF_n*d

Mass after collision was 2M,thus, F_n = 2Mg

We are told to express distance in terms of L

Thus;

W_f = μ2Mg*L

W_f = 2μMgL

The problem involves principles of energy conservation and kinetic friction. As the block slides down, it converts potential energy to kinetic, collides with another block, and moves together along a rough surface. The work done against friction equals energy at the collision point yielding the expression for velocity after collision V' = sqrt(2µgl).

The important concept here is conservation of energy. As block A slides down, it is converting potential energy into kinetic energy. At point X, the potential energy is maximum because it's at the highest point. The potential energy is given by M*g*R. By the time it reaches point Y, the potential energy has turned into kinetic energy which results in the block's speed (½MV²) at point Y.

After the collision, the two blocks move together with a new mass of 2M and new velocity V': MV = 2MV'. In the horizontal section YZ, due to kinetic friction, the blocks lose energy as work done against friction which causes them to stop.

The work done against friction is: F *d = µ*2M*g*l. Equating the energy at point Y after collision to work done against friction, we get: 2*½MV'² = µ*2M*g*l. Resolving gives the expression V' = sqrt(2µgl).

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Answer:

The object will continue spinning as it has been.

Explanation:

When an object is spinning in a closed system with no unbalance

forces and no external force or torques are applied to it, it will have no change in angular momentum.

According to the law of conservation of angular momentum which states that when no external torque acts on an object, no change of angular momentum will occur

If the change in angular momentum is zero, then the angular momentum is constant; therefore, the object will continue spinning as it has been.

The same scenario is applicable to the rotation and the spinning of the earth.

Answer: The object will continue spinning as it has been.

Explanation: Unbalanced forces refers a force that changes the position, speed or direction of the object to which it is applied. While torque refers to otational or twisting effect of a force; a moment of force, defined for measurement purposes as an equivalent straight line force multiplied by the distance from the axis of rotation.

Any object that spins in place with no unbalanced forces or torques acting upon it will continue spinning as it has been. An example can be seen in the rotation of planets.

Answer:

24.609375 kgm²

Explanation:

I = Moment of inertia of the merry go round

[tex]N_1[/tex] = Initial speed = 51 rpm

[tex]N_2[/tex] = Final speed = 17 rpm

m = Mass of child = 31.5 kg

r = Radius = 1.25 m

In this system the angular momentum is conserved

[tex]IN_1=(I+mr^2)N_2\\\Rightarrow I\times 51=(I+31.5\times 1.25^2)17\\\Rightarrow I\dfrac{51}{17}-I=31.5\times 1.25^2\\\Rightarrow 2I=49.21875\\\Rightarrow I=\dfrac{49.21875}{2}\\\Rightarrow I=24.609375\ kgm^2[/tex]

The moment of inertia of the merry go round is 24.609375 kgm²

Answer:

37.5 N Hard

Explanation:

Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.

Using the expression for hook's law,

F = ke.............. Equation 1

F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.

Given: k = 750 N/m, e = 5.0 cm = 0.05 m

Substitute into equation 1

F = 750(0.05)

F = 37.5 N

Hence the athlete is pushing 37.5 N hard

Answer:

37.5N

Explanation:

According to Hooke's law, the load or force, F, applied on an elastic material (e.g a spring) is directly proportional to the extension or compression, e, caused by the load. i.e

F ∝ e

F = k x e         -------------------------(i)

where;

k = proportionality constant known as the spring constant.

From the question;

k = 750N/m

e = 5.0cm = 0.05m

Substitute these values into equation (i) as follows;

F = 750 x 0.05

F = 37.5N

Therefore, the load applied which is a measure of how hard the athlete is pushing is 37.5N

Complete question:

A 45-mH ideal inductor is connected in series with a 60-Ω resistor through an ideal 15-V DC power supply and an open switch. If the switch is closed at time t = 0 s, what is the current 7.0 ms later?

Answer:

The current in the circuit 7 ms later is 0.2499 A

Explanation:

Given;

Ideal inductor, L = 45-mH

Resistor, R =  60-Ω

Ideal voltage supply, V = 15-V

Initial current at t = 0 seconds:

I₀ = V/R

I₀  = 15/60 = 0.25 A

Time constant, is given as:

T = L/R

T = (45 x 10⁻³) / (60)

T = 7.5 x 10⁻⁴ s

Change in current with respect to time, is given as;

[tex]I(t) = I_o(1-e^{-\frac{t}{T}})[/tex]

Current in the circuit after 7 ms later:

t = 7 ms = 7 x 10⁻³ s

[tex]I(t) = I_o(1-e^{-\frac{t}{T}})\\\\I =0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}}})\\\\I = 0.25(0.9999)\\\\I = 0.2499 \ A[/tex]

Therefore, the current in the circuit 7 ms later is 0.2499 A

The current later after 7 ms is 0.2499 A

The current after a time (t) is given by:

[tex]I(t)=I_o(1-e^{-\frac{t}{\tau} })\\\\I_o=initial\ current\ at\ t=0,\tau=time \ constant\\\\Given\ that\ L=45mH=45*10^{-3}H,R=60\ ohm,V = 15V,t=7\ ms\\\\\tau=\frac{L}{R}=\frac{45*10^{-3}}{60} =7.5*10^{-4} \ s\\\\I_o=\frac{V}{R}=\frac{15}{60}=0.25\ A \\\\\\I(t)=I_o(1-e^{-\frac{t}{\tau} })=0.25(1-e^{-\frac{7*10^{-3}}{7.5*10^{-4}} })\\\\\\I(t)=0.2499\ A[/tex]

The current later after 7 ms is 0.2499 A

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Answer:

a) The shear stress is 0.012

b) The shear stress is 0.0082

c) The total friction drag is 0.329 lbf

Explanation:

Given by the problem:

Length y plate = 2 ft

Width y plate = 10 ft

p = density = 1.938 slug/ft³

v = kinematic viscosity = 1.217x10⁻⁵ft²/s

Absolute viscosity = 2.359x10⁻⁵lbfs/ft²

a) The Reynold number is equal to:

[tex]Re=\frac{1*3}{1.217x10^{-5} } =246507, laminar[/tex]

The boundary layer thickness is equal to:

[tex]\delta=\frac{4.91*1}{Re^{0.5} } =\frac{4.91*1}{246507^{0.5} } =0.0098[/tex] ft

The shear stress is equal to:

[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{1} )(246507)^{0.5} =0.012[/tex]

b) If the railing edge is 2 ft, the Reynold number is:

[tex]Re=\frac{2*3}{1.215x10^{-5} } =493015.6,laminar[/tex]

The boundary layer is equal to:

[tex]\delta=\frac{4.91*2}{493015.6^{0.5} } =0.000019ft[/tex]

The sear stress is equal to:

[tex]\tau=0.332(\frac{2.359x10^{-5}*3 }{2} )(493015.6^{0.5} )=0.0082[/tex]

c) The drag coefficient is equal to:

[tex]C=\frac{1.328}{\sqrt{Re} } =\frac{1.328}{\sqrt{493015.6} } ==0.0019[/tex]

The friction drag is equal to:

[tex]F=Cp\frac{v^{2} }{2} wL=0.0019*1.938*(\frac{3^{2} }{2} )(10*2)=0.329lbf[/tex]

Answer:  4.0024 x 10^ -11 m or 0.040024 nm

Explanation:

λ = h c/ΔE

λ = wave lenght

h = 6.626 x 10 ^ -34  m² kg /s  = planck constant

ΔE = 31 keV potential ( 1 keV = 1.6021 x 10^-16J)

c = velocity of light = 3 x 10⁸ m/s

substitute gives

λ  =    6.626 x 10 ^ -34  m² kg /s x 3 x 10⁸ m/s  = 4.0024 x 10^ -11 m

                 31 x 1.6021x10^-16 J

Answer:

the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

the internal energy generated in crossing the rough patch is 820.26 J

Explanation:

Given that,

Mass, m = 62 kg,

Initial speed,  = 6.90 m/s

Length of rough patch, L = 4.50 m,

coefficient of friction,  = 0.3

Height of inclined plane, h = 2.50 m

[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity

According to energy conservation equation,

Part (b)

The internal energy generated when crossing the rough patch is equal (in magnitude) to the work done by the friction force on the skier.

The magnitude of the friction force is:

[tex]F_f=\mu mg[/tex]

Therefore, the work done by friction is:

[tex]W=-F_f d =-\mu mg d[/tex]

The negative sign is due to the fact that the friction force is opposite to the direction of motion of the skier

Substituting,

[tex]W=-(0.300)(62.0)(9.8)(4.50)=-820.26 J[/tex]

So, the internal energy generated in crossing the rough patch is 820.26 J

Part (a)

If we take the bottom of the hill as reference level, the initial mechanical energy of the skier is sum of his kinetic energy + potential energy:

[tex]E=K_i +U_i = \frac{1}{2}mu^2 + mgh[/tex]

After crossing the rough patch, the new mechanical energy is

[tex]E'=E+W[/tex]

where

W = -820.26 J is the work done by friction

At the bottom of the hill, the final energy is just kinetic energy,

[tex]E' = K_f = \frac{1}{2}mv^2[/tex]

where v is the final speed.

According to the law of conservation of energy, we can write:

[tex]E+W=E'[/tex]

So we find v:

[tex]\frac{1}{2}mu^2 +mgh+W = \frac{1}{2}mv^2\\\\v=\sqrt{u^2+2gh+\frac{2W}{m}}\\\\=\sqrt{6.9^2+2(9.8)(2.50)+\frac{2(-820.26)}{62.0}}\\\\=8.38 m/s[/tex]

Thus, the skier is moving at a speed of 8.38 m/s when she gets to the bottom of the hill.

The question repeated the units attached the the values

So in order of their appearance in the question, the correct units are;

62kg, 6.9m/s, 4.5m, 0.30, 2.5m

Answer:

A) Speed at bottom of hill = 8.38 m/s

B) Internal energy generated in crossing the patch = 820.26J

Explanation:

From the question, it's clear that there are 2 distinct stages of of her skis. The first stage is when she is on the rough patch while the second stage is when she skis down the hill.

A) For the first stage when she is on the rough patch;

If we apply Newton's second law of motion to the skis along the vertical direction, we obtain;

ΣF_y = N - mg = 0

Thus,N = mg

m = 62kg,thus,N = 62 x 9.8 = 607.6N

Now, let's find the kinetic friction. It's given by; f_k = μN

Where μ is coefficient of kinetic friction

Thus, f_k = 0.3 x 607.6 = 182.28N

Now, let's calculate the work done by this frictional force,

So, W_fk = f_k x distance = 182.28 x

4.5 = 820.26J

This work is done in a direction opposite to the displacement and it will have a negative sign. Thus,

W_fk = - 820.26J

Now, since the skier skies horizontally and perpendicular to the gravitational force, the work done due to gravity is zero. Thus,

W_grav = 0

Let's calculate the kinetic energy at the beginning of the rough patch.

K1 = (1/2)m(v1)²

K1 = (1/2) x 62 x (6.9)²= 1475.91J

Also,let's calculate the kinetic energy at end of rough patch.

K2 = (1/2)m(v2)²

where v2 is final velocity at end of rough patch

K2 = (1/2)(62)(v2)² = 31(v2)²

Now, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K2 - K1

In this case, W_other is W_fk

Thus,

- 820.26J + 0 = 31(v2)² - 1475.91J

31(v2)² = 1475.91J - 820.26J

31(v2)² = 655.65

(v2)² = 655.65/31

v2 =√21.15 = 4.6 m/s

Now, for the second stage when she skis down the hill;

In this case the only force acting is gravity, thus, W_other = 0

Work done by gravity = mgh

W_grav = 62 x 9.8 x 2.5 = 1519 J

Now K2 will be kinetic energy at top of hill while K3 will be kinetic energy at bottom of hill.

Thus,

K2 = (1/2)m(v2)²

K2 = (1/2)(62)(4.6)²

K2 = 655.96J

Similarly,

K3 = (1/2)m(v3)²

Where v3 is velocity at bottom of hill.

Thus,

K3 = (1/2)(62)(v3)²

K3 = 31(v3)²

Again, the total work done when other forces other than gravity do work, is given by ;

W_total = W_other + W_grav = K3 - K1

So,

W_other is zero.

Thus,

1519J = 31(v3)² - 655.96J

31(v3)² = 1519J + 655.96J

31(v3)² = 2174.96J

(v3)² = 2174.96/31

v3 = √70.16

v3 = 8.38 m/s

B) Workdone by non-conservative forces manifests itself as changes in the internal energy of bodies. Now, frictional force experienced in the first stage in crossing the patch is a non - conservative force. Thus,

Internal energy = W_fk = 820.26J

Complete question:

A rectangular loop of wire with dimensions 2.0 cm by 10.0 cm and resistance 1.0 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude of 2.0 T and is directed into the plane. At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

Check the image uploaded

Answer:

The magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

Explanation:

Given;

resistance of the wire; R = 1.0 Ω

magnitude of magnetic field strength, B = 2.0 T

speed of the loop, v = 3.00 m/s

Induced emf is given as;

ε = IR

[tex]I = \frac{emf}{R} = \frac{VBL}{R}[/tex]

magnitude of the force that the magnetic field exerts on the loop:

F = BIL

Substitute in the value of I

[tex]F = \frac{VB^2L^2}{R}[/tex]

where;

L is the displacement vector between the initial and final end of the portion of the wire inside the field = 2.0 cm

Substitute the given values and solve for F

[tex]F = \frac{3*2^2*(2*10^{-2})^2}{1} \\\\F = 4.8 *10^{-3} \ N[/tex]

Therefore, the magnitude of the force that the magnetic field exerts on the loop 4.8 x 10⁻³ N

Complete Question:

A rectangular loop of wire with dimensions 1.50 cm by 8.00 cm and resistance 0.700 Ω is being pulled to the right out of a region of uniform magnetic field. The magnetic field has magnitude 2.20 T and is directed into the plane of.

At the instant when the speed of the loop is 3.00 m/s and it is still partially in the field region, what is the magnitude of the force that the magnetic field exerts on the loop?

Answer:

F = 0.133 N

Explanation:

Magnitude of the magnetic field, B = 2.20 T

Length of the loop = 1.5 cm = 0.015 m

The speed of the loop, v = 3.00 m/s

The emf induced in the loop , e = Blv

e = 2.20 * 0.015 * 3

e = 0.099 V

Current induced in the loop, I = e/R

I = 0.099/0.7

I = 0.1414 A

The magnitude of the force is given by, F = I *l *B sin90

F = 0.1414 * 0.015 * 2.20

F = 0.00467 N

Answer:

Using equation 2dsinФ=n*λ

given d=2.41*10^-6m

λ=512*10^-12m

θ=52.64 degrees

Answer:

a

The angle between the center and the third side bright fringe  is

       [tex]\theta = 39.60^o[/tex]

b

The third side bright fringe move away from the pattern's center

c

The distance by which it moves away is [tex]\Delta z=0.3906 m[/tex]

Explanation:

From the question the

           The wavelength is [tex]\lambda = 512nm[/tex]

In the first question we a asked to obtain the angle between the center and the third side bright fringe

  since we are considering the third side of the bright fringe the wavelength of light on  the three sides would be  evaluated as

                 [tex]\lambda_{3} = 3 * 512nm[/tex]

The slit separation is given as  [tex]d = 2.41 \mu m[/tex]

    The angle between the center and the third side bright fringe is

              [tex]\theta = sin^{-1} (\frac{\lambda_3}{d} )[/tex]

              [tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.24*10^{-6}} )[/tex]

                [tex]= sin^{-1} (0.6374)[/tex]

                [tex]\theta = 39.60^o[/tex]

When the frequency of the light is reduced the wavelength is increased

                i.e [tex]f = \frac{c}{\lambda}[/tex]

and this increase would cause the third side bright to move away from the pattern's center

        Now from the question frequency is reduce to 94.5% this mean that the  wavelength would also increase by the same as mathematically represented below

                  [tex]\lambda_{new} = \frac{512 *10^{-9}}{0.945}[/tex]

                          [tex]= 0.542 \mu m[/tex]

The angle between the center and the third side bright fringe is for new wavelength

                   [tex]\theta = sin^{-1} (\frac{3 *512*10^{-9}}{2.41*10^{-6}} )[/tex]

                     [tex]= 42.46^o[/tex]

 The distance traveled away from the pattern's center is mathematically represented as

                [tex]z = A tan \theta[/tex]

 Where A is the separation between the slits and the screen

              [tex]\Delta z = 4.45(tan 42.46 - tan39.60 )[/tex]

                    [tex]\Delta z=0.3906 m[/tex]

Answer:

Yes it would

Explanation:

The submerged portion of the pen would be affected by water buoyancy force which equals to the water weight that it displaces. By Newton's 3rd law, the pen would then make a reaction force on the water itself, which in turn affect the whole reading on the balance.

Yes, it would: The submerged part of the pen would be influenced by water buoyancy force which equals the water weight that it replaces.

The submerged part of the pen would be influenced by water buoyancy force which equals the water weight that it replaces. According to Newton's 3rd law, the pen would then make a reaction force on the water itself, which in turn impacts the whole reading on the balance.

Newton's Third Law refers to There are two forces resulting from this interaction. A force on the chair and also a force on your body. These two forces are called action and reaction forces and also are the subject of Newton's third law of motion.

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Final answer:

To balance the rod with the attached objects, you need to place a support at a certain distance from the rod's center. The support should be placed at (2/3) times the length of the rod to the right of the rod's center.

Explanation:

To balance the rod with the attached objects, you need to place a support at a certain distance from the rod's center. Let's call this distance 'x'. The torque on the rod is balanced when the clockwise torque is equal to the counterclockwise torque.

The torque on the left end of the rod is given by the force on the left object times the distance from the left end of the rod to the support, which is (m * x). The torque on the right end of the rod is given by the force on the right object times the distance from the right end of the rod to the support, which is ((2m) * (L - x)). Since the torque must be balanced, we set the two torques equal to each other:

(m * x) = ((2m) * (L - x))

Simplifying this equation gives us:

x = (2/3) * L

Therefore, you should place the support at a distance of (2/3) times the length of the rod to the right of the rod's center.