11. A tight guitar string has a frequency of 540 Hz as its third harmonic. What will be its fundamental frequency if it is fingered at a length of only 70% of its original length

Answer: 3217.79 hours.

Explanation:

Given, A 140 lb. climber saved her potential energy as she descended from Mt. Everest (Elev. 29,029 ft) to Kathmandu (Elev. 4,600 ft).

Power = 0.4 watt

Mass of climber = 140 lb

= 140 x 0.4535 kg  [∵ 1 lb= 0.4535 kg]

⇒ Mass of climber (m) = 63.50 kg

Let [tex]h_1=29,029\ ft= 8848.04\ m\ \ \ \ [ 1 ft=0.3048\ m ][/tex] and [tex]h_2= 4,600 ft = 1402.08\ m[/tex]

Now, Energy saved =[tex]mg(h_1-h_2)=(63.50)(9.8)(8848.04-1402.08)=4633620.91\ J[/tex]

[tex]\text{Power}=\dfrac{\text{energy}}{\text{time}}\\\\\Rightarrow 0.4=\dfrac{4633620.91}{\text{time}}\\\\\Rightarrow\ \text{time}=\dfrac{4633620.91}{0.4}\approx11584052.28\text{ seconds}\\\\=\dfrac{11584052.28}{3600}\text{ hours}\ \ \ [\text{1 hour = 3600 seconds}]\\\\=3217.79\text{ hours}[/tex]

Hence, she can power her 0.4 watt flashlight for 3217.79 hours.

Answer:

A.16.5x10^-6C

B. 47.5x10^-6C

C.77x10^-6C

Explanation:

Pls see attached file

Answer:

The value for  A  is A= 0.6

The angular acceleration is  [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]

Explanation:

From the question we are told that  

    The radius of the disk is  [tex]r = 25.0 \ cm = 0.25 \ m[/tex]

     The linear acceleration is  [tex]a(t) = At[/tex]

     At time   [tex]t = 3 \ s[/tex]

     [tex]a(3) = 1.80 \ m/s^2[/tex]

Generally angular acceleration is  mathematically represented as  

         [tex]\alpha(t) = \frac{a(t)}{r}[/tex]

Now  at t = 3 seconds  

         a(3) =  A *  3

=>      1.80 =  A  *  3  

=.>       A =  0.6

So  therefore

             a(t) =  0.6 t  

Now  substituting this into formula for angular acceleration

        [tex]\alpha (t) = \frac{0.6 t }{R}[/tex]

substituting for  r  

         [tex]\alpha (t) = \frac{0.6 t }{0.25}[/tex]

         [tex]\alpha (t) = 2.4 \ t \ m/s^2[/tex]

Answer:

the greatest intensity is obtained from   c

Explanation:

An electromagnetic wave stagnant by the expression

           E = E₀ sin (kx -wt)

when two waves meet their electric fields add up

           E_total = E₁ + E₂

the intensity is

           I = E_total . E_total

           I = E₁² + E₂² + 2E₁ E₂ cos θ

where θ  is the phase angle between the two rays

Let's examine the two waves

in this case E₁ = E₂ = E₀

          I = Eo2 + Eo2 + 2 E₀ E₀ coasts

         I = E₀² (2 + 2 cos θ )

         I = 2 I₀ (1 + cos θ )

     let's apply this expression to different cases

a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀

b) cos π = -1     this implies that     I_total = 0

c) the cosine is  1,

         I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ

         I = E₀² (5 +4 cos θ)

         I = E₀² 9

         I = 9 Io

d) in this case the cos pi = -1

          I = E₀² (5 -4)

          I = I₀

e) we rewrite the equation

         I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ

         I = Eo2 (10 + 6 cos θ)

         cos π = -1

         I = E₀² (10-6)

         I = 4 I₀

the greatest intensity is obtained from   c

The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.

An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.

In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.

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Answer:

v = 73.75 m/s

Explanation:

It is given that,

A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.

Let us assume we need to find the velocity of the rocket when it runs out of fuel.

Let v is the final speed. Using the third equation of kinematics as :

[tex]v^2-u^2=2as[/tex]

u = 0

[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]

So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s

Answer:

450°C

Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .

η = 1 - T2 / T1

η = 1/6 initially

when T2 is reduced by 65°C then η becomes 1/3

Solution

η = 1/6

1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)

When η = 1/3 :

η = 1 - ( T2 - 75 ) / T1

1/3 = 1 - (T2 - 75)/T1.........................(2)

T2 - T1 = -75 [ because T2 is reduced by 75°C ]

T2 = T1 - 75...........................(3)

Put this in (2) :

> 1/3 = 1 - ( T1 - 75 - 75 ) / T1

> 1/3 = 1 - (T1 - 150 ) /T1

> (T1 - 150) / T1 = 1 - 1/3

> ( T1 -150 ) / T1 = 2/3

> 3 ( T1 - 150 ) = 2 T1

> 3 T1 - 450 = 2 T1

Collecting the like terms

3 T1- 2 T1 = 450

T1 = 450

The temperature initially was 450°C

Answer:

F' = F

Hence, the magnitude of the attractive force remains same.

Explanation:

The force of attraction between two bodies is given by Newton's Gravitational Law:

F = Gm₁m₂/r²   --------------- equation 1

where,

F = Force of attraction between balls

G = Universal Gravitational Constant

m₁ = mass of first ball

m₂ = mass of 2nd ball

r = distance between balls

Now, we double the masses of both balls and the separation between them. So, the force of attraction becomes:

F' = Gm₁'m₂'/r'²

here,

m₁' = 2 m₁

m₂' = 2 m₂

r' = 2 r

Therefore,

F' = G(2 m₁)(2 m₂)/(2 r)²

F' = Gm₁m₂/r²

using equation 1:

F' = F

Hence, the magnitude of the attractive force remains same.

Answer:

The induced current through resistor R will

b) flow from a to b

Explanation:

The image is shown below, and the full question is written down as

The two solenoids in the figure are coaxial and fairly close to each other. While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor R will

a) Flow from b to a

b) flow from a to b

c) be zero because the rate is constant.

From the image, the current in the left hand solenoid flows from the positive terminal of the battery to the negative terminal in an anticlockwise direction by convention.

Varying a rheostat causes a change in the resistance of electricity through the solenoid, and a changing current through a solenoid will induce current to flow through another solenoid placed nearby. Therefore, the left-hand solenoid induces a current flow on the right-hand solenoid.

Since the current in the left-hand solenoid flows in an anticlockwise direction, then it will have an equivalent magnetic polarity of a north pole on a magnet.

Also remember that Lenz law states that the induce current acts in such a way as to oppose the motion, or action producing it.

In this case, the induced current in the right-hand solenoid will act as to repel the left-hand solenoid away from itself. The only way is by the right-hand solenoid also having a north pole equivalent magnetic pole on it since like poles repel each other. This means that the induced current in the right-hand solenoid will flow in an anticlockwise manner too, from a to b.

Answer:

48 i believe

Explanation:

Answer:

The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

Explanation:

Given;

first object with mass, m₁ = 285 kg

second object with mass, m₂ = 585 kg

distance between the two objects, r = 4.3 m

The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m

Gravitational force between the first object and the 42 kg object;

[tex]F = \frac{GMm}{r^2}[/tex]

where;

G = 6.67 x 10⁻¹¹ Nm²kg⁻²

[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]

Gravitational force between the second object and the 42 kg object

[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]

Magnitude of net gravitational force exerted on 42kg object;

F = 3.545x 10⁻⁷ N  -  1.727 x 10⁻⁷ N

F = 1.818 x 10⁻⁷ N

Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N

Answer:

The frequency will be reduced by a factor of √2/2

Explanation:

Pls see attached file

The new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

Let the initial mass of block be m.

And new mass is, 4m.

The frequency of oscillating motion is defined as the number of complete oscillation made during the time interval of 1 second. The mathematical expression for the frequency of oscillation of block-spring system is given as,

[tex]f = \dfrac{1}{2 \pi}\sqrt{\dfrac{k}{m}}[/tex]

Here,

k is the spring constant.

If the mass of block increased to 4m, then the new frequency of oscillation of spring will be,

[tex]f' = \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{4m}}\\\\\\f' =\dfrac{1}{2} \times \dfrac{1}{2 \pi} \sqrt{\dfrac{k}{m}}\\\\\\f' =\dfrac{1}{2} \times f[/tex]

Thus, we can conclude that the new frequency of oscillation will be half the original frequency of oscillation of spring-block system.

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Answer:

Explanation:

For circular path in magnetic field

mv² / R = Bqv ,

m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .

a )

R = mv / Bq

If v is changed  to 2v , keeping other factors unchanged , R will be doubled

b )

magnitude of acceleration inside field

= v² / R

= Bqv / m

As v is doubled , acceleration will also be doubled

c )

If T be the time inside the magnetic field

T = π R / v

=  π  / v x  mv / Bq

= π m / Bq

As is does not contain v that means T  remains unchanged .

d )

Net force acting on electron

= m v² / R = Bqv

Net force = Bqv

As v becomes twice force too becomes twice .

So a . b , d are correct answer.

Answer: B. The copper wire (resistance 0.1)

Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law

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

So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.

Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:

P = [tex]i^{2}*R[/tex]

As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.

In conclusion, the copper wire will dissipate more heat when connected in parallel.

Answer:

The  wavelength is  [tex]\lambda = 1029 nm[/tex]

Explanation:

From the question we are told that

    The  wavelength of the left light is  [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]

      The  voltage across A  and  B is  [tex]V_{AB } = 1.2775 \ V[/tex]

Let the stopping potential  at A  be [tex]V_A[/tex] and the electric potential at B  be  [tex]V_B[/tex]

The voltage across A and B is mathematically represented as

      [tex]V_{AB} = V_A - V_B[/tex]

Now  According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side  in terms of electron volt is mathematically represented as

        [tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]

Where  W is the work function of the metal

             h is the Planck constant with values  [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]

             c  is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]

And  the stopping potential at B for the ejected electron from the right side  in terms of electron volt is mathematically represented as

          [tex]eV_B = \frac{h * c}{\lambda } - W[/tex]

So  

      [tex]eV_{AB} = eV_A - eV_B[/tex]

=>    [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]

=>   [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]

=>   [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]

=>  [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]

Where e is the charge on an electron with the value  [tex]e = 1.60 *10^{-19} \ C[/tex]

=>   [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]      

=>  [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]  

=>   [tex]\lambda = 1.029 *10^{-6} \ m[/tex]

=>   [tex]\lambda = 1029 nm[/tex]

Answer:

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

Explanation:

The formula of electric power is given as follows:

Power = (Voltage)(Current)

Current = Power/Voltage

In this question, we have:

Power = 45 KW = 45000 W

Voltage of Battery Pack = 340 V

Current needed to be drawn = ?

Therefore,

Current = 45000 W/340 V

Current = 132.35 A

The motor needs to draw 132.35 Amperes current from the battery.

Answer:

The correct answer is A    

Explanation:

In this exercise we are given a graph of temperature versus time.

In calorimeter processes there are two types

* one that when giving thermal energy to the system its temperature increases, this fundamentally due to the greater kinetic energy of the molecular ones, this process observes in the graphs as a straight line of constant slope

* A process donates all the thermal energy that is introduced is cracked in breaking the molecular bonds, taking matter from one thermodynamic state to another, for example: liquid to gas.

This process in curves as a horizontal line, that is, there is no temperature change,

When analyzing the graph shown, parts C and D are the one that show a change in temperature with thermal energy. The correct answer is A

Answer:

C and D

Explanation:

Just took the quiz

Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]

Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:

y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]

where:

|a| is initil displacement

[tex]\frac{2.\pi}{\omega}[/tex] is period

For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]

For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:

[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]

period = [tex]\frac{2.\pi}{\omega}[/tex]

12 = [tex]\frac{2.\pi}{\omega}[/tex]

ω = [tex]\frac{\pi}{6}[/tex]

Replacing values:

[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]

The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].

Answer:

The rotor of the gas turbine rotates 12250 revolutions before coming to rest.

Explanation:

Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:

[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]

Where:

[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.

[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.

[tex]t[/tex] - Time, measured in minutes.

[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.

The angular acceleration is now cleared:

[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]

If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:

[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]

[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]

Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:

[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]

Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.

The change in angular position is cleared herein:

[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]

If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:

[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]

[tex]n-n_{o} = 12250\,rev[/tex]

The rotor of the gas turbine rotates 12250 revolutions before coming to rest.

Answer:

The critical angle is  [tex]i = 41.84 ^o[/tex]

Explanation:

From the  question we are told that

    The index of refraction of the sugar solution is  [tex]n_s = 1.5[/tex]

   The  index of refraction of air is  [tex]n_a = 1[/tex]

Generally from Snell's  law

      [tex]\frac{sin i }{sin r } = \frac{n_a }{n_s }[/tex]

Note that the angle of incidence in this case is equal to the critical angle

Now for total internal reflection the angle of reflection is [tex]r = 90^o[/tex]

So  

      [tex]\frac{sin i }{sin (90) } = \frac{1 }{1.5 }[/tex]

      [tex]i = sin ^{-1} [\frac{ (sin (90)) * 1 }{1.5} ][/tex]

      [tex]i = 41.84 ^o[/tex]

Answer:

  R = v √(2h / g)

Explanation:

This exercise can be solved using the concepts of science, projectile launching

let's calculate the time it takes to get to the water

           y = y₀ +[tex]v_{oy}[/tex] t - ½ g t²

as the stone is skipped the vertical speed is zero

           y = y₀ - ½ g t²

for y=0

           t = √ (2y₀ / g)

the horizontal distance it covers in this time is

           R = v₀ₓ t

            R = v₀ₓ √(2 y₀ / g)

let's call the horizontal velocity as v and the height is h

            R = v √(2h / g)

Correct answer is 2.7 x 10^-3 J/m3

I hope that helps ! <33

The maximum energy density of the capacitor is closest to: 2.7 x 10^-3 J/m3.

Energy density is defined as the total energy per unit volume of the capacitor. Since, Now, for a parallel plate capacitor, A × d = Volume of space between plates to which electric field E = V / d is confined. Therefore, Energy is stored per unit volume.

All Answers (14) Energy density is equal to 1/2*C*V2/weight, where C is the capacitance you computed and V should be your nominal voltage (i.e 2.7 V). Power Density is V2/4/ESR/weight, where ESR is the equivalent series resistance.

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Answer:

The  magnitude of the  electric field intensity is  [tex]E = 7.89 *10^{6} \ V/m[/tex]

Explanation:

From the question we are told that

    The  voltage is  [tex]\epsilon = 72.7 \ mV = 72.7 *10^{-3} V[/tex]

    The  thickness of the membrane is  [tex]t = 9.22 \ nm = 9.22 *10^{-9} \ m[/tex]

Generally the electric field intensity is mathematically represented as

                [tex]E = \frac{\epsilon }{t}[/tex]

 substituting values

                [tex]E = \frac{72.7 *10^{-3} }{9.22 *10^{-9}}[/tex]

                [tex]E = 7.89 *10^{6} \ V/m[/tex]

Answer:

The acceleration is  [tex]a = 2.190 *10^{-7} \ m/s^2[/tex]

Explanation:

From the question we are told that

      The  distance of separation of the ship is  [tex]r= 109 \ m[/tex]

       The mass of each ship is  [tex]M = 39,000 \ metric\ tons =39,000 * 1000 = 3.9 *10^{7}\ kg[/tex]

The gravitational force of attraction exerted on each other is mathematically represented as

            [tex]F_g = \frac{ GMM}{r^2}[/tex]

Where G is the gravitational  constant with value

substituting values

          [tex]F_g = \frac{ 6.674 30 * 10^{-11} (3.9 *10^{7})^2}{109^2}[/tex]

         [tex]F_g = 8.54 \ N[/tex]

This force can also be mathematically represented as

        [tex]F_g = M * a[/tex]

=>   [tex]a = \frac{F_g}{M}[/tex]

substituting values

     [tex]a = \frac{8.544}{3.9 *10^{7}}[/tex]

     [tex]a = 2.190 *10^{-7} \ m/s^2[/tex]

Answer:

No point on the x-axis

Pls see attached file

Answer:

Diamagnetic

Explanation:

Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).

For vanadium V ion, there are 18 electrons which will be arranged as follows;

1s2 2s2 2p6 3s2 3p6.

All the electrons present are spin paired hence the ion is expected to be diamagnetic.

Answer:

its paramagnetic

Explanation:

i took this quiz

Answer:

0.055 kg

Explanation:

According to the given situation the solution of the mass of the string is shown below:-

Speed of the wave is

[tex]v = \sqrt{\frac{F_T\times Length\ of\ string}{Mass\ of\ string}}[/tex]

[tex]30.0 m/s = \sqrt{\frac{10 kg m/s^2\times 5.00 m}{Mass\ of\ string}[/tex]

Mass of string is

[tex]= \sqrt{\frac{10 kg m^2/s^2\times 5.00 m}{900 m^2 s^2}[/tex]

After solving the above equation we will get the result that is

= 0.055 kg

Therefore for calculating the mass of the string we simply applied the above formula.

Answer:

a. 24 m

Explanation:

Destructive interference occurs when two waves arrive at a point, out of phase. In a completely destructive interference, the two waves cancel out, but in a partially destructive interference, they produce a wave with a time varying amplitude, but maintain a wavelength the wavelength of one of the original waves. Since the two waves does not undergo complete destructive interference, then the possible value of the new wave formed can only be 24 m, from the options given.

Answer:

W = 1413.75 J

Explanation:

It is given that,

Mass of car, m = 650 kg

Initial speed of the car, u = 0.7 m/s

Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s

Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.

[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]

So, the work done by the car is 1413.75 J.

Explanation:

It is given that, in the first trial, the initial velocity is [tex]v_o[/tex] and in the second it is [tex]4v_o[/tex].

The total energy of the system remains constant. So,

[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}kx^2=\text{constant}[/tex] ....(1)

x is amplitude

It means that the amplitude is directly proportional to velocity. If velcoity increases to four times, then the amplitude also becomes 4 times.

Differentiating equation (1) we get :

[tex]mv\dfrac{dv}{dt}+kx\dfrac{dx}{dt}=0[/tex]

Since,

[tex]\dfrac{dv}{dt}=a,\ \text{acceleration}[/tex] and [tex]\dfrac{dx}{dt}=v,\ \text{velocity}[/tex]

So,

[tex]mva+kxv=0[/tex]

It means that the acceleration is also proportional to the amplitude. So, acceleration also becomes 4 times.

Hence, the correct option is (B) "both the amplitude and the maximum acceleration are four times as great"

Answer:

Acceleration(a) = 2.588 m/s²

TL = 230.784 N

TR = 220.5 N

Explanation:

Given:

M = 7.95 kg

mL = 32 kg

mR = 17.8 kg

g = 9.8 m/s²

Find:

Acceleration(a)

TL

TR

Computation:

Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]

Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]

Acceleration(a) = [139.16] / [53.775]

Acceleration(a) = 2.588 m/s²

TL = mL(g-a)

TL = 32(9.8-2.588)

TL = 230.784 N

TR = mR(g+a)

TR = 17.8(9.8+2.588)

TR = 220.5 N