A 0.36 kg object, attached to a spring with constant k=10n/m, is moving on a horizontal frictionless surface in simple harmonic motion of amplitude 0.082 m. What is it speed when it’s displacement is 0.041 m

A 0.36 Kg Object, Attached To A Spring With Constant K=10n/m, Is Moving On A Horizontal Frictionless

Answers

Answer 1

Answer:

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.

Explanation:

First, we need to determine the angular frequency of the system ([tex]\omega[/tex]), in radians per second:

[tex]\omega = \sqrt{\frac{k}{m} }[/tex] (1)

Where:

[tex]k[/tex] - Spring constant, in newtons per meter.

[tex]m[/tex] - Mass, in kilograms.

If we know that [tex]k = 10\,\frac{N}{m}[/tex] and [tex]m = 0.36\,kg[/tex], then the angular frequency of the system is:

[tex]\omega = \sqrt{\frac{10\,\frac{N}{m} }{0.36\,kg} }[/tex]

[tex]\omega \approx 5.270\,\frac{rad}{s}[/tex]

The kinematic formulas for the position ([tex]x(t)[/tex]), in meters, velocity ([tex]\dot x(t)[/tex]), in meters per second, and acceleration of the object ([tex]\ddot x(t)[/tex]), in meters per square second, are:

[tex]x(t) = A\cdot \cos \omega t[/tex] (2)

[tex]\dot x(t) = -\omega \cdot A \cdot \sin \omega t[/tex] (3)

[tex]\ddot x(t) = -\omega^{2}\cdot A \cdot \cos \omega t[/tex] (4)

Where [tex]A[/tex] is the amplitude of the motion, in meters.

From (2) we determine the time associated with position [tex]x(t) = 0.041\,m[/tex] ([tex]\omega \approx 5.270\,\frac{rad}{s}[/tex], [tex]A = 0.082\,m[/tex]):

[tex]t = \frac{1}{\omega}\cdot \cos^{-1} \left(\frac{x(t)}{A} \right)[/tex] (5)

[tex]t = \frac{1}{5.270\,\frac{rad}{s} }\cdot \cos^{-1}\left(\frac{0.041\,m}{0.082\,m} \right)[/tex]

[tex]t = 0.199\,s[/tex]

And the speed of the object is:

[tex]\dot x(t) = -\left(5.270\,\frac{rad}{s} \right)\cdot (0.082\,m)\cdot \sin \left[\left(5.270\,\frac{rad}{s} \right)\cdot (0.199\,s)\right][/tex]

[tex]\dot x(t) \approx -0.375\,\frac{m}{s}[/tex]

The speed of the object when displacement is 0.041 meters is 0.375 meters per second.


Related Questions

I let go of a piece of bread from a balcony. A bird flying 5.0 m overhead sees me drop it, and starts to dive straight down towards the bread the instant that I release it. She catches it after it falls 3.0 m. Assuming she accelerates constantly from rest (v0 = 0) at the time I let go of the bread, what is her acceleration? Show your work

Answers

This question can be solved using the equations of motion. There are two scenarios where the equations of motion can be used. The first scenario is the free-fall motion of the piece of bread. The second scenario is the uniformly accelerated motion of the bird.

The acceleration of the bird is  "a = 26.13 m/s²".

First, we will calculate the time taken by the bread to fall 3 m. Using the second equation of motion for this free-fall motion:

[tex]h = v_it + \frac{1}{2}gt^2[/tex]

where,

h = height fall = 3 m

vi = initial velocity = 0 m/s

g = acceleration due to gravity = 9.8 m/s²

t = time taken = ?

Therefore,

[tex]3\ m = (0\ m/s)t+\frac{1}{2}(9.8\ m/s^2)t^2\\t = \sqrt{\frac{(3\ m)(2)}{9.8\ m/s^2}}\\\\t = 0.78\ s[/tex]

The bird took the same time to catch the bread. Now applying the second equation of motion to the bird's motion:

[tex]s = v_it + \frac{1}{2}at^2[/tex]

where,

s = distance covered by the bird = 5 m + 3 m = 8 m

vi = initial velocity of the bird = 0 m/s

a = acceleration of the bird = ?

t = time taken = 0.78 s

Therefore, using these values we get:

[tex]8\ m = (0\ m/s)(0.78\ s)+\frac{1}{2}a(0.78\ s)^2\\\\a = \frac{16\ m}{(0.78\ s)^2}[/tex]

a = 26.13 m/s²

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Which of the following changes would not lead to changes in the efficiency of
a heat engine?
A. Doubling the work done while keeping the heat flow into the
engine the same
B. Doubling the heat flow into the engine while halving the work done
C. Doubling both the work done and the heat flow into the engine
D. Doubling the heat flow into the engine while keeping the work
done the same

Answers

Explanation:

B. Doubling the heat flow into the engine while halving the work done

hope this helps you

have a nice day

any correct ans guess for 30 pts and brainlist​

Answers

Answer:

a) ii

b) i

c)iii

d)iii

e) i

f)iii

Explanation:

i hope this will help

Answer:

the answer of b is N and d is mol

Un cuerpo es saltado desde cierta altura , llegando al piso luego de 7 s. Determine de altura se le soltó . (g=10m/segundo al cuadrado )

Answers

Answer:

Maximum height, H = 35 meters

Explanation:

Given the following data;

Time = 7 seconds

Acceleration due to gravity, g = 10 m/s²

To find the maximum height;

Mathematically, the maximum height is given by the formula;

H = ½gt²

H = ½ * 10 * 7

H = 5 * 7

Maximum height, H = 35 meters

Si la fuerza de fricción cinetica es 250N ¿Qué fuerza se necesita para mantener a la caja deslizándose a una velocidad constante ¿

Answers

Answer:

250N

Explanation:

According to newton second law,

\sumF = ma

Fm - Ff = ma

Since the velocity is constant, a = 0m/s

Frictional force Ff = 250N

Substitute

Fm - 250 = m(0)

Fm - 250 = 0

Fm = 250N

Hence the force to keep the box sliding at constant speed is 250N

What are the uses of magnetic force?​

Answers

Answer:

Computer hard drives use magnetism to store the data on a rotating disk. More complex applications include: televisions, radios, microwave ovens, telephone systems, and computers. An industrial application of magnetic force is an electromagnetic crane that is used for lifting metal objects.

Answer:

Examples of magnetic force is a compass, a motor, the magnets that hold stuff on the refrigerator, train tracks, and new roller coasters. All moving charges give rise to a magnetic field and the charges that move through its regions, experience a force.

I Hope this will help you if not then sorry :)

what is velocity give its SI unit​

Answers

Answer:

Velocity is the speed of something given in a direction. the SI unit of velocity is metre per second or m/s.

Calculate the acceleration of a train of mass 30 000 kg when driven by a force of 15000 N.

Answers

Answer:

Explanation:

F = ma, so filling in:

15000 = 30000a and

a = .50 m/s/s

The acceleration of a train is 0.50 [tex]m/s^{2}[/tex].

What is acceleration?

Acceleration is the rate of change of the velocity of an object with respect to time.

F = ma

a = F/m

a = 15000/30000

a = 0.50 [tex]m/s^{2}[/tex]

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when you go out to your car one cold winter morning you discover a 0.80 cm thick layer of ice on the windshield which has an area of 1.4m square if the temperature of the ice is - 2.0 degrees celsius and its density is 917 kg/m to the third find the heat required to melt all the ice

Answers

Answer:

The total heat required to melt the ice is approximately 3.473 MJ

Explanation:

The given parameters for the layer of ice are;

The thickness of the layer of ice, t = 0.80 cm = 0.008 m

The area of the wind shield, A = 1.4 m²

The initial temperature of the ice, T₁ = -2.0 °C = 271.15 K

The density of the ice, ρ = 917 kg/m

The temperature at which ice melts, T₂ = 0 °C = 273.15 K

We have;

The mass of the ice, m = ρ × t × A

∴ m = 917 kg/m³ × 0.008 m × 1.4 m² = 10.2704 kg

The specific heat capacity of ice, c = 2,090 J/(kg·K)

∴ The equation for the heat capacity of the ice to melt, is given as follows;

ΔQ = m·c·ΔT

Where;

ΔT = T₂ - T₁

∴ ΔT = T₂ - T₁ = 273.15 K - 271.15 K = 2 K

ΔQ = 10.2704 kg × 2,090 J/(kg·K) × 2 K = 42.930272 kJ

The latent heat to melt the ice, Q = The latent heat of fusion of ice, L × Mass of ice, m

The latent heat of fusion of ice, L = 334 kJ/kg

∴ Q = 334 kJ/kg × 10.2704 kg = 3,430.3136 kJ

The total heat required to melt the ice, H = ΔQ + Q

∴ H = 42.930272 kJ + 3,430.3136 kJ = 3,473.243872 kJ ≈ 3.473 MJ.

A cricket ball of mass 800g has momentum of 20 kg m/sec . Calculate velocity of ball

Answers

Answer: 25 m/s

Explanation:

1. rearrange equation p = mv to find v

2. v = p/m - then convert 800 g into kg (= 0.8 kg)

3. plug in p and m - v = 20 / 0.8 = 25 m/s

Answer:

V= 25m/s

Explanation:

Mass=800g

Momentum=20kgm/s

V= ?

Convert 800g to kg

800/1000 = 0.8kg

Momentum=mass× Velocity

Momentum/Mass = Velocity

20/0.8 =25m/s

An object moving north with an initial velocity of 14 m/s accelerates 5 m/s2 for 20 seconds. What is the final velocity of the object?
39 m/s
90 m/s
114 m/s
414 m/s

Answers

Answer:

option C

Explanation:

Final velocity of the object is 114 m / s. Hence, final velocity of the object is 114 m / s.

Here is the answer ! Hope it helps.

Enlight the contribution of science and technology for current pandemic of Corona virus??​

Answers

The vaccine ofcourse. They helped in making the vaccine.

Why aren’t magazine photos a good representation of what a healthy person looks like

Answers

Because it all fake the photo shop so it’s not real
Magazine photos are supposed to be eye catchy! Only then people will invest in the advertised product. They are photoshopped, airbrushed and what not? Even the celebrities admit that they do not look like that in reality.

There is given an ideal capacitor with two plates at a distance of 3 mm. The capacitor is connected to a voltage source with 12 V until it is loaded completely. Then the capacitor is disconnected from the voltage source. After this the two plates of the capacitor are driven apart until their distance is 5 mm. Now a positive test charge of 1 nC is brought from the positively charged plate to the negatively charges plate. How large is the kinetic energy of the test charge? The test charge of 1 nC can be regarded to be so small that it does not influence the electric field between the two plates of the capacitor.

Answers

The kinematic energy of the positive charge is 2 10⁻⁸ J

This electrostatics exercise must be done in parts, the first part: let's start by finding the charge of the capacitor, the capacitance is defined by

        C = [tex]\frac{Q}{\Delta V}[/tex]

        C = ε₀ [tex]\frac{A}{d}[/tex]

we solve for the charge (Q)

        [tex]\frac{Q}{\Delta V} = \epsilon_o \frac{A}{d}[/tex]

indicates that for the initial point d₁ = 3 mm = 0.003 m and the voltage is DV₁ = 12

         Q = [tex]\epsilon_o \ \frac{A \ \Delta V_1 }{d_1}[/tex]

Now the voltage source is disconnected so the charge remains constant across the ideal capacitor.

For the second part, the condenser is separated at d₂ = 5mm = 0.005 m

         Q = \epsilon_o \  \frac{A \ \Delta V_2 }{d_2}

we match the expressions of the charge and look for the voltage

          [tex]\frac{\Delta V_1}{d_1} = \frac{\Delta V_2}{d_2}[/tex]

          ΔV₂ = [tex]\frac{d_2}{d_1 } \ \Delta V_1[/tex]

The third part we use the concepts of conservation of energy

starting point. With the test load (q = 1 nC = 1 10⁻⁹ C) next to the left plate

          Em₀ = U = q DV₂

          Em₀ = q  \frac{d_2}{d_1 } \ \Delta V_1

           

final point. Proof load on the right plate

         Em_f = K

energy is conserved

         Em₀ = em_f

         q  \frac{d_2}{d_1 } \ \Delta V_1 = K

   

we calculate

         K = 1 10⁻⁹  12  [tex]\frac{0.005}{0.003}[/tex]  

         K = 20 10⁻⁹ J

In this exercise, as the conditions at two different points of separation give, the area of ​​the condenser is not necessary and with conservation of energy we find the final kinetic energy of 2 10⁻⁸ J

The number of windings on the primary coil of a transformer
is 1.5 times greater than on the secondary coil. The primary
coil has a current of 3.0 A and a voltage of 12.0 V. Determine
the voltage and current on the secondary coil.

Answers

Answer:

I. Vs = 8.0 Volts.

II. Is = 4.5 Amperes.

Explanation:

Given the following data;

Np = 1.5Ns = [tex] \frac {N_{P}}{N_{S}} = 1.5 [/tex]  ..... equation 1

Ip = 3.0 A

Vp = 12 V

To find the voltage and current on the secondary coil;

I. For the voltage in the secondary coil (Vs), we would use the following formula;

[tex] \frac {V_{P}}{V_{S}} = \frac {N_{P}}{N_{S}} [/tex]  ...... equation 2.

Substituting eqn 1 into eqn 2, we have;

[tex] \frac {V_{P}}{V_{S}} = 1.5 [/tex]

[tex] \frac {12}{V_{S}} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] V_{S} * 1.5 = 12 [/tex]

[tex] V_{S} = \frac {12}{1.5} [/tex]

Vs = 8.0 V

II. For the current in the secondary coil (Is), we would use the following formula;

[tex] \frac {I_{S}}{I_{P}} = \frac {N_{P}}{N_{S}} [/tex]  .... equation 3

Substituting eqn 1 into eqn 3, we have;

[tex] \frac {I_{S}}{I_{P}} = 1.5 [/tex]

[tex] \frac {I_{S}}{3.0} = 1.5 [/tex]

Cross-multiplying, we have;

[tex] I_{S} = 1.5 * 3.0 [/tex]

Is = 4.5 A

3kg of water at 80degree celcius is added to 8 kg of water at 25 degree celcius. find the temperature of final mixture provided there is no loss of heat in the surrounding. the specific heat capacity is 4200j/kg​

Answers

Answer:

hope fully it help s

If 20N force produces an acceleration of 5ms^-2 In a body then the mass of the body will be:
A.4kg
B.5kg
C.1/4kg
D.1/5kg

Answers

F = ma
m = F/a
m = 20N / 5m/s^2
m = 4kg

Which statement about the ocean is true?

No evaporation or precipitation in the water cycle occurs over the ocean.
Most evaporation and precipitation in the water cycle occur over the ocean.
All evaporation and precipitation in the water cycle occur over the ocean.
Evaporation, but not precipitation, in the water cycle occurs over the ocean.

Answers

Answer:

most evaporation and precipitation in the water cycle occus over the ocean

The option Most evaporation and precipitation in the water cycle occur over the ocean. is the correct answer

which statement accurately describes how the acceleration of an object in free fall changes

Answers

Answer:

A ball rolls off the table and free falls to the ground. Which statement accurately describes its acceleration? Its acceleration is downward but remains a constant value as it falls.

You toss an apple across the room to a friend. Which of the following statements is true about the apple at the top of its trajectory?
A. Its acceleration is zero.
B. The horizontal component of its velocity is zero.
C. The vertical component of its velocity is 9.8 m/s down.
D. Its acceleration is 9.8 m/s2 down.​

Answers

B is the correct answer!! The horizontal component of its velocity is zero.

Can someone pls help, thank you in advance!
What is an example of a force applied at an angle to displacement

Answers

Answer:

  an object sliding down hill

Explanation:

On a slope, the force applied is due to gravity. Its direction is straight down. If the object is sliding down the hill, its displacement is at an angle to the applied force. The angle of displacement will depend on the steepness of the hill.

1
An object has a gravitational potential energy of
41772.5J and has a mass of 1550kg. How high is
it above ground?

Answers

Answer:

h = 2.78 m

Explanation:

Potential Energy(PE)

Mass(m)

Gravity(g) which is approximately equal to 10.

Height (h)

PE = mgh

41772.5 = 1500 * 10 * h

41772.5 = 15000h

h = 41772.5/15000

h = 2.78 m

Plz Help me thank uuu...
Which of these is a likely impact of the stronger than normal trade winds on the eastern Pacific ocean?

Warm surface water builds up, causing lower than average temperature.
Warm surface water builds up, causing higher than average temperature.
Warm surface water is reduced, causing colder conditions than normal.
Warm surface water is reduced, causing hotter conditions than normal.

Answers

Answer:

Jet stream would be displaced southwards causing heavy rain and flooding.

Explanation:

The other options of the question were A) Jet stream would be displaced northwards causing drought. B) Jet stream would be displaced southwards causing drought. D) Jet stream would be displaced northwards causing heavy rain and flooding,

The statement that is a likely impact of stronger than normal trade winds in the Pacific Northwest to the United States is "Jet stream would be displaced southwards causing heavy rain and flooding."

We are talking about climate or weather terminology. In this case, we are referring to the "El Niño" (the Children) effect. Its presence affects the weather in North America. This phenomenon combines with the "La Niña) effect and it presents itself every two to seven years, ad they last from 8 to 12 months, affecting the weather conditions of the region.

Answer:

jet stream should be displaced southwards which causes heavy rainfall and flood.

A low-power laser used in a physics lab might have a power of 0.50 mW and a beam diameter of 3.0 mm. Calculate:a.The average light intensity of the laser beam.b. The intensity of a lightbulb producing 100-W light viewed from 2.0 m.c.Compare the intensity of the laser to the intensity of the lightbulb. Is it advisable to look directly at a laser

Answers

Answer:

A) I_laser = 70.74 W/m²

B) I_bulb = 1.989 W/m²

C) it is not advisable to look at the laser beam directly.

Explanation:

We are given;

Power; P = 0.50 mW = 0.5 × 10^(-3) W

Diameter; d = 3 mm = 0.003 m

Radius; r = d/2 = 0.003/2 = 0.0015 m

A) Area of beam; A = πr²

A = 0.0015²π

Now, formula for average intensity is;

I = P/A

I = (0.5 × 10^(-3))/0.0015²π

I = 70.74 W/m²

B) We are told to find the intensity of a lightbulb producing 100-W.

Thus, P = 100 W

A light bulb is spherical in shape. Thus;

Area; A = 4πr²

We are told it's 2 m away.

Thus; r = 2 m

A = 4π(2)²

A = 16π

Thus, I = P/A = 100/16π

I = 1.989 W/m²

C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.

Why are road accidents at high speeds very much worse than road accidents at low speeds?

Answers

Answer:

The momentum makes it worse.

Explanation:

The momentum of vehicles running at faster speeds is very high and causes a lot of damage to the vehicles.

Vehicles with high speeds have high momenta (plural of momentum). When they come to rest in very small time they experience high forces
we know that rate of change of momentum is force, lower the time higher is force. So, slow moving vehicles experience less force. It is important to note that if any one vehicle involving has high speed it can do more damage both the vehicles involving.

3 write the three laws given by kepler.How did they help Newton to arrive at the inverse square law of gravity?​

Answers

Answer:

Kepler's laws apply: First Law: Planetary orbits are elliptical with the sun at a focus. Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times. Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semimajor axis is the same for all planets.

HELP‼️‼️ A car horn creates a 595 Hz tone at rest. Two cars pass on the street, each going 20.0 m/s; the first car honks. What frequency does the other car hear before they pass each other?

Answers

The frequency of the other car is 669HZ

Let start off by writing out the parameters given in the question

frequency= 595 HZ

Speed of sound(v)= 343m/s

Speed of the car(vs)= 20m/s

Speed of the second car(vo) = 20m/s

The frequency of the other car can be calculated as follows

(v+vo/v-vs) f

= (343+20/343-20)595

= (363/323)595

= 1.1238(595)

= 669 HZ

Hence the frequency of the other car is 669 HZ

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which is not a common theory on how the effects of hypnosis occurs

Answers

Answer:

EFFECTS Adverse reactions to hypnosis are rare, but may include: Headache. Drowsiness. Dizziness

A 6.0 kg mass is placed on a 20º incline which has a coefficient of friction of 0.15. What is the acceleration of the mass down the incline? show all work please​

Answers

Answer:

Explanation:

The form of Newton's 2nd Law that we use for this is:

F - f = ma where F is the Force pulling the mass down the ramp forward, f is the friction trying to keep it from moving forward, m is the mass and a is the acceleration (and our unknown).

We know mass and we can find f, but we don't have F. But we can solve for that by rewriting our main equation to reflect F:

[tex]wsin\theta-\mu F_n=ma[/tex] That's everything we need.

w is weight: 6.0(9.8). Filling in:

6.0(9.8)sin20 - .15(6.0)(9.8) = 6.0a and

2.0 × 10¹ - 8.8 = 6.0a and

11 = 6.0a so

a = 1.8 m/s/s

From the description of the question, the acceleration is 1.95 ms-2.

What is friction?

The term friction is defined a the force that seeks to prevent the movement of a surface on another.

Now we know that;

ma =mgsinθ -  ukmgcosθ

ma =( 6 * 9.8 * sin 20) - ( 0.15 * 6 * 9.8 * cos 20)

a =  ( 6 * 9.8 * sin 20) - ( 0.15 * 6 * 9.8 * cos 20) /6

a = 20 - 8.29/6

a = 1.95 ms-2

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cho 11,2 lit hỗn hợp gồm metan và etilen tác dụng với dung dịch brom. sau phản ứng xảy ra toàn thấy có chất khí thoát ra và 40 gam brom tham gia phản ứng .
a) viết phương trình phản ứng xảy ra
b) tính thành phần phần trăm thể tích mỗi chất trong hỗn hợp
c) đốt cháy toàn bộ thể tích khí thoát ra sau đó cho chất khí thu được tác dụng với 100ml dung dịch Ca(OH)2 1,25 M. tính khối lượng chất thu được sau phản ứng . biết metan ko tác dụng với brom , các chất khí đo ở điều kiện tiêu chuẩn

Answers

Yes he was the strongest hahaha
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