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Tom [10]
3 years ago
11

A particle of mass m moves under the influence of a force given by F = (−kx + kx3/α2) where k and α are positive constants. a) F

ind the potential energy function U(x) (Take U(x=0) = 0). b) Find any equilibrium points and determine if they are stable or unstable. c) Sketch the potential as a function of position. d)What is the maximum total energy, Emax a particle may have if its motion is to remain bounded? e) For the case of bounded motion with E
Physics
1 answer:
Romashka-Z-Leto [24]3 years ago
4 0

Answer:

Explanation:

a ) F = (-kx + kx³/a²)

intensity of field

I = F / m

=  (-kx + kx³/a²) / m

If U be potential function

- dU / dx =  (-kx + kx³/a²) / m

U(x)  = ∫  (kx - kx³/a²) / m dx

= k/m ( x²/2 - x⁴/4a²)

b )

For equilibrium points , U is either maximum or minimum .

dU / dx = x - 4x³/4a² = 0

x = ± a.

dU / dx = x - x³/a²

Again differentiating

d²U / dx² = 1 - 3x² / a²

Put the value of x = ± a.

we get

d²U / dx²  = -2 ( negative )

So at x = ± a , potential energy U is maximum.

c )

U =  k/m ( x²/2 - x⁴/4a²)

When x =0 , U = 0

When x=  ± a.

U is maximum

So the shape of the U-x curve is like a bowl centered at x = 0

d ) Maximum potential energy

put x = a or -a in

U(max)  =  k/m ( x²/2 - x⁴/4a²)

= k/m ( a² / 2 - a⁴/4a²)

= k/m ( a² / 2 - a²/4)

a²k / 4m

This is the maximum total energy where kinetic energy is zero.

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In the design of a rapid transit system, it is necessary to balance the average speed of a train against the distance between st
bekas [8.4K]

Answer:

a) t = 746 s

b) t = 666 s

Explanation:

a)

  • Total time will be the sum of the partial times between stations plus the time stopped at the stations.
  • Due to the distance between stations is the same, and the time between stations must be the same (Because the train starts from rest in each station) we can find total time, finding the time for any of the distance between two stations, and then multiply it times the number of distances.
  • At any station, the train starts from rest, and then accelerates at 1.1m/s2 till it reaches to a speed of 95 km/h.
  • In order to simplify things, let's first to convert this speed from km/h to m/s, as follows:

       v_{1} = 95 km/h *\frac{1h}{3600s}*\frac{1000m}{1 km} = 26.4 m/s  (1)

  • Applying the definition of acceleration, we can find the time traveled by the train before reaching to this speed, as follows:

       t_{1} = \frac{v_{1} }{a_{1} } = \frac{26.4m/s}{1.1m/s2} = 24 s (2)

  • Next, we can find the distance traveled during this time, assuming that the acceleration is constant, using the following kinematic equation:

       x_{1} = \frac{1}{2} *a_{1} *t_{1} ^{2} = \frac{1}{2} * 1.1m/s2*(24s)^{2} = 316.8 m  (3)

  • In the same way, we can find the time needed to reach to a complete stop at the next station, applying the definition of acceleration, as follows:

       t_{3} = \frac{-v_{1} }{a_{2} } = \frac{-26.4m/s}{-2.2m/s2} = 12 s (4)

  • We can find the distance traveled while the train was decelerating as follows:

       x_{3} = (v_{1} * t_{3})   + \frac{1}{2} *a_{2} *t_{3} ^{2} \\ = (26.4m/s*12s) - \frac{1}{2} * 2.2m/s2*(12s)^{2} = 316.8 m - 158.4 m = 158.4m  (5)

  • Finally, we need to know the time traveled at constant speed.
  • So, we need to find first the distance traveled at the constant speed of 26.4m/s.
  • This distance is just the total distance between stations (3.0 km) minus the distance used for acceleration (x₁) and the distance for deceleration (x₃), as follows:
  • x₂ = L - (x₁+x₃) = 3000 m - (316.8 m + 158.4 m) = 2525 m (6)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{2525m}{26.4m/s} = 95.6 s   (7)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 95.6 s + 12 s = 131.6 s (8)
  • Due to we have six stations (including those at the ends) the total time traveled while the train was moving, is just t times 5, as follows:
  • tm = t*5 = 131.6 * 5 = 658.2 s (9)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 4 intermediate stops, we need to add to total time 22s * 4 = 88 s, as follows:
  • Ttotal = tm + 88 s = 658.2 s + 88 s = 746 s (10)

b)

  • Using all the same premises that for a) we know that the only  difference, in order to find the time between stations, will be due to the time traveled at constant speed, because the distance traveled at a constant speed will be different.
  • Since t₁ and t₃ will be the same, x₁ and x₃, will be the same too.
  • We can find the distance traveled at constant speed, rewriting (6) as follows:
  • x₂ = L - (x₁+x₃) = 5000 m - (316.8 m + 158.4 m) = 4525 m (11)
  • The time traveled at constant speed (t₂), can be found from the definition of average velocity, as follows:

       t_{2} = \frac{x_{2} }{v_{1} } = \frac{4525m}{26.4m/s} = 171.4 s   (12)

  • Total time between two stations is simply the sum of the three times we have just found:
  • t = t₁ +t₂+t₃ = 24 s + 171.4 s + 12 s = 207.4 s (13)
  • Due to we have four stations (including those at the ends) the total time traveled while the train was moving, is just t times 3, as follows:
  • tm = t*3 = 207.4 * 3 = 622.2 s (14)
  • Since we know that the train was stopped at each intermediate station for 22s, and we have 2 intermediate stops, we need to add to total time 22s * 2 = 44 s, as follows:
  • Ttotal = tm + 44 s = 622.2 s + 44 s = 666 s (15)
7 0
2 years ago
Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

       The  between the first and second dark fringes is  y =  4.2 \ mm  =  4.2 *10^{-3} \ m

Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

   Hence

        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

7 0
3 years ago
How is sound intensity measured?
muminat

Answer:

Sound intensity is measured with a sound level meter or Sound pressure Level (SPL) Meter.

Explanation:-

It measure sound intensity is the sound pressure level. The unit of measurement is decibels.

7 0
2 years ago
a. What proportion of resistors have resistances less than 90 Ω? b. Find the mean resistance. c. Find the standard deviation of
Anettt [7]

Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

c) 9.43 Ω

d) 1

Explanation:

The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

4 0
3 years ago
A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad
dalvyx [7]
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2 
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration 
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1 
w1 = wo + ãt 
w1 - final angular velocity 
wo - initial angular velocity 
18 = 0 + 4.5t t = 4 s
3 0
3 years ago
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