A student bikes to school by traveling first dN = 0.800 miles north, then dW = 0.300 miles west, and finally dS = 0.100 miles south.
Similarly, let d--> W be the displacement vector corresponding to the second leg of the student's trip. Express d⃗ W in component form.
Express your answer as two numbers separated by a comma. Be careful with your signs.

Answer:

Statics and dynamics

Explanation:

Rigid-body mechanics is divided into two areas: statics deals with the equilibrium of bodies, that is, those that are either at rest or move with a constant velocity; whereas dynamic is concerned with the accelerated motion of bodies.

In rigid-body mechanics, there are two various branches: statics and dynamics. Statics studies the stability of bodies, i.e., those that are at rest or move at a constant speed, whereas dynamics studies the accelerated movement of bodies.

Even if neither a system's energy state nor its phase of motion tends to change over time, the system is said to be in equilibrium.

A simple mechanical body is considered to have been in equilibrium if it neither suffers accelerometers nor angular acceleration. Unless an external force disrupts this state, it will remain in equilibrium indefinitely.

When all forces acting on a single particle are added together and equal to zero, equilibrium results. In addition to the conditions outlined in the article above, a rigid body is said to be in equilibrium if the vector sum of any torques operating on it equals zero, maintaining the body's steady state of rotational motion.

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Answer:

2000Hz and 1500Hz

Explanation:

Using

a) f = f0((c+vr)/(c+vs))

=>>> f0((c)/(c-0.5c))

=>>>1000/0.5 = 2000Hz

b) f = f0((c+vr)/(c+vs))

=>>>f0((c+0.5c)/(c))

=>>>>1000 x 1.5 = 1500Hz

Here we have a problem referring to the Doppler effect, the solutions are:

The Doppler effect is an effect that explains how the perception of waves changes as the source moves or as the listener moves.

The formula, for sound, is:

[tex]f = \frac{v + v_0}{v - v_1}*f_0[/tex]

where:

a) First we have that the source moves towards a stationary listener, then we have:

[tex]f = \frac{v }{v - v/2}*1.0 kHz\\\\f = \frac{v }{v/2}*1.0 kHz = 2.0 kHz[/tex]

b) in this case, the listener moves towards the source, so we have:

[tex]f = \frac{v + v/2 }{v }*1.0 kHz\\\\f = \frac{ (3/2)*v }{v}*1.0 kHz = 1.5 kHz[/tex]

So in this case the perceived frequency is smaller than in the point a.

This is because the waves will move at a fixed rate in air, in one case, the successive waves are emitted from different points in space (each time closer to the listener) while in the other case the waves are emitted from a fixed point, and the listener moves towards them, thus feels that the waves move faster.

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Answer:

The geocentric model of Ptolemy in which the sun and other planets were believed to move round the Earth.

Explanation:

The Copernicus heliocentric model of the solar system was put forward by  Nicolaus Copernicus and was published in 1543. The model proposed by Copernicus puts the Sun near the center of the Universe, motionless, with Earth and the other planets orbiting around the sun in circular paths, modified by epicycles, and at uniform speed. This is contrary to Ptolemy's geocentric solar system that puts the Earth at the center of the solar system, in which the sun and other planets were believed to move round the Earth.

Answer:

1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Explanation:

1km = 10^3 m = 10^5 cm

1h = 60x60 = 3600s

--> 1km/h^2 = 10^5 cm / (3600^2) s^2

<=> 1km/h^2 = 5/648 cm/s^2 [about 7.7 x 10^(-3) /s^2]

Answer:

True

Explanation:

In the foundational topics of physics mechanics consists of motion, energy and forces, theses areas are core concerns in the study of physics hence they are the foundational topics and are often times treated first in all matters.

What is motion?

Simply put, it is alternating from points to points.

the areas of study that involves motion and force together is call dynamics.

while the area of study of motion without any applied force is call statics.

Answer:

1) 5.52 cm , C) 5.5 cm

Explanation:

When a measurement is carried out, in addition to the value of the magnitude, the error or uncertainty of the measurement must occur, in a direct measurement with an instrument the uncertainty is equal to the appreciation of the instrument.

Uniform see the errors by the number of significant figures days, in this cases they are two decimals for which the appreciation of the instrument ± - 0.01

now we can analyze the measurements made

1) 5.52 cm. Validate. It is a valid measurement is within the uncertainty range

2) 6.63 cm. It does not validate. It is out of the error range

3) 5.5 cm Valid. It is within the given error range,

4) 5.93 cm Not Valid. It is out of the error range.

Answer:

6.63 and 5.93

Explanation:

Answer:

The angular acceleration of the top is 0.18 rad/s²

Explanation:

Given;

angular velocity of the top, ω = 16 rad/s

time it takes to come to a complete stop, t = 88.9s

The angular acceleration of the top, is calculated as;

[tex]\alpha = \frac{\omega}{t}[/tex]

α is the angular acceleration

ω is angular acceleration

t is the time

[tex]\alpha = \frac{\omega}{t}\\\\\alpha = \frac{16}{88.9}\\\\\alpha =0.18 \ rad/s^2[/tex]

Therefore, the angular acceleration of the top is 0.18 rad/s²

Answer:

Direction is downwards

Explanation:

This is by employing Flemings Right hand rule which says If you point your pointer finger in the direction the positive charge is moving, and then your middle finger in the direction of the magnetic field, your thumb points in the direction of the magnetic force pushing on the moving charge.

Answer:

1670 kg

Explanation:

Convert the dimensions to cm.

12 ft × (12 in/ft) × (2.54 cm/in) = 365.76 cm

32 in × (2.54 cm/in) = 81.28 cm

Calculate the volume.

V = πr²h

V = π (81.28 cm / 2)² (365.76 cm)

V = 1,897,813.27 cm³

V = 1,897,813.27 mL

V = 1,897.81 L

Density of benzene is 0.88 kg/L.

m = (0.88 kg/L) (1,897.81 L)

m = 1670 kg

Explanation:

Assuming the circular path is horizontal, the sum of forces in the centripetal direction is:

∑F = ma

T = mv²/r

T = (1.0000 kg) (5.00 m/s)² / (0.500 m)

T = 50.0 N

Answer:

Zero

Explanation:

Because using

Deta X= dsinစ x n(lambda)

But we know that for central maxima

n is zero

So after substituting

Deta x = 0

Answer:

The force is  [tex]F = 423.04 \ N[/tex]

Explanation:

From the question we are told that

   The weight is  [tex]W = 35 .0 \ N[/tex]

    The  force constant is  [tex]k = 220 \ N/m[/tex]

    The frequency is  [tex]f = 10.5 \ Hz[/tex]

     The amplitude is  [tex]A = 3.00 \ cm = 0.03 \ m[/tex]

Generally the maximum driving force is mathematically represented as

     [tex]F = m * w^2 A[/tex]

Here  m is the mass of the weight which is  mathematically represented as  

       [tex]m = \frac{ W }{g}[/tex]

=>     [tex]m = \frac{ 35 }{9.8 }[/tex]

=>     [tex]m = 3.571 \ kg[/tex]

  Also   [tex]w[/tex] is the angular frequency of the weight which is mathematically represented as

         [tex]w = 2 \pi * f[/tex]

         [tex]w = 2* 3.142 * 10[/tex]

=>      [tex]w = 62.84 \ rad/s[/tex]

So

    [tex]F = 3.571 * 62.84^2 * 0.03[/tex]

    [tex]F = 423.04 \ N[/tex]

Answer:

force applied by red = 50 N

force applied by blue = - 63 N (since it is in the opposite direction)

net force = force by red + force by blue

net force = 50 + (- 63)

net force = - 13 N  

Complete Question

A wave is described by y(x,t) = 0.1 sin(3x + 10t), where x is in meters, y is in centimetres and t is in seconds. The angular wave  frequency is

Answer:

The  value is [tex]w =  10 \ rad /s[/tex]

Explanation:

From the question we are told that  

    The equation describing the wave is y(x,t) = 0.1 sin(3x + 10t)

Generally the sinusoidal equation representing the motion of a wave is mathematically represented as

         [tex]y(x,t) =  Asin(kx + wt )[/tex]

Where  w  is the  angular frequency

Now comparing this equation  with that given we see that

       [tex]w =  10 \ rad /s[/tex]

Answer:

2260000 J

Explanation:

From the question,

Q = ml.................... Equation 1.

Where Q = Heat, m = mass of water, l = specific latent heat of vaporization.

But we can get the mass o water using the formula of density.

D = m/v........... Equation 2

Where D = density of water, v = volume of water.

make m the subject of the formula in equation 2

m = D×v............ Equation 3

Substitute equation 3 into equation 1

Q = D×v×l................. Equation 4

Given: v = 1.00 L = 0.001 m³.

Constant: l = 226000 J/kg, D = 1000 kg/m³

Substitute these values into equation 4

Q = 0.001(2260000)(1000)

Q = 2260000 J

Question:  Initially, the car travels along a straight road with a speed of 35 m/s. If the brakes are applied and the speed of the car is reduced to 13 m/s in 17 s, determine the constant deceleration of the car.

Answer:

1.29 m/s²

Explanation:

From the question,

a = (v-u)/t............................ Equation 1

Where a = deceleration of the car, v = final velocity of the car, u = initial velocity of the car, t = time.

Given: v = 13 m/s, u = 35 m/s, t = 17 s.

a = (13-35)/17

a = -22/17

a = -1.29 m/s²

Hence the deceleration of the car is 1.29 m/s²

Answer:

height of cliff (h) = 112.38m

Explanation:

The time 5.12 s is the total time it takes for the rock to fall, and for the soundwave to travel back to the top of the cliff before it is heard.

[tex]5.12 = t_f\ +\ t_s - - - - -(1)\\where:\\t_f = time\ of\ fall\ of\ the\ piece\ of\ rock\\t_s = time\ travelled\ by\ the\ return\ sound[/tex]

Let h be the height of the cliff in meters, the time taken for the rock to fall is given by:

[tex]t_f=\sqrt{\frac{2h}{g} } \\where:\\t_f = time\ of\ fall\\h = height\ of\ cliff\\g= acceleration\ due\ to\ gravity= 9.8 m \slash s^2[/tex]

[tex]\therefore t_f = \sqrt{\frac{2h}{9.8}} \\squaring\ both\ sides\\(t_f)^2 = \frac{2h}{9.8}\\ 2h = 9.8 \timess\ (t_f)^2\\h = \frac{9.8 \timess\ (t_f)^2}{2} \\h= 4.9(t_f)^2 - - - - - (2)[/tex]

Next, let us calculate the time taken fot the sound to return

[tex]t_s = \frac{h}{v} \\where:\\t_s = time\ for\ sound\ to\ travel\ up\ the\ cliff\\h= distance\ tavelled\ = height\ of\ cliff\\v= speed\ = 340m \slash s\ (speed\ of\ sound)\\\therefore t_s = \frac{h}{340} - - - - - (3)\\[/tex]

now putting the values of  h from equation 2 into equation (3)

[tex]t_s = \frac{4.9(t_f)^2}{340}[/tex]

Putting the value of [tex]t_s[/tex] into equation (1)

[tex]5.12 = t_f +\frac{4.9(t_f)^2}{340} \\[/tex]

multiplying through by 340

[tex]1740.8 = 340(t_f) + 4.9(t_f)^2\\4.9(t_f)^2 + 340 (t_f) - 1740.8 = 0[/tex]

now let us solve the quadratic equationsss;

[tex]Let\ (t_f) = x[/tex]

[tex]4.9x^2 + 340x - 1740.8 = 0\\using\ quadratic\ formula\\x = \frac{-b \pm\sqrt{b^2 - 4ac} }{2a} \\x = \frac{-340 \pm\sqrt{(340)^2 -\ 4 \times4.9 \times(-1740.8)} }{2\times4.9}\\x = \frac{-340\ \pm\ 386.936}{9.8} \\x =\frac{386.938 - 340}{9.8} \\x = \frac{46.936}{9.8}\\ x = 4.789\\x = t_f\\t_f=4.789s[/tex]

note, time cannot be negative, so we ignored the negative answer

putting the value of  [tex]t_f[/tex]  into equation (2) to find height of cliff (h)

[tex]h= 4.9(t_f)^2\\h = 4.9 \times(4.789)^2\\h = 112.38m[/tex]

Therefore, height of cliff (h) = 112.38m

A cyclist must lean into a turn to prevent tipping over in the other direction.The frictional force provides the centripetal force necessary to turn the cyclist to the left.But the frictional force also produces a clockwise torque that will cause the rider and the bicycle to tip clockwise to the right.The force is provided by the friction of the tires.

Answer:

to become stable

Explanation:

when it bends it body, it moves closer to the center of gravity which makes the bicycle stable and hence the turn can be taken easily

Answer:

We know that the relation between period and frequency is:

T = period.

f = frequency.

T = 1/f.

Then, if the period is 35 μs = 35x10^-9 seconds.

The frequency will be:

f = (1/ 35x10^-9 s) = (1/35s)*10^9

now, 1Hz = 1/s

1KHz = 1/1000s = 10^-3 s

f = (10^9/35)*10^-3 KHz = 10^6/35 KHz = 28,571.429 KHz.

Now we must divide this by 2.5:

28,571.429 KHz/2.5 = 11,428.572 KHz

Now we can use the relation:

T = 1/f

T = 1/11,428.572 KHz = (1/11,428.572 Hz)*10^-3

T = 8.750x10^-8 seconds.

And we want this expressed in  μs = 10`-9 seconds, we have:

T = 8.750x10^-8 s = (10/10)8.750x10^-8 s = 87.50x10`-9s = 87.50 μs

So as expected, if the frequency is reduced by a factor of 2.5, the  period will increase by a factor of 2.5

As the initial period was 35 μs, and:

2.5*35 μs = 87.5 μs

Answer:

t = 0.25 seconds

Explanation:

Given that,

Initial speed of a bullet, v = 800 m/s

Distance from the target is 200 m

We need to find the time required for the bullet to reach the target. Time is simply calculated by the definition of velocity i.e.

[tex]t=\dfrac{d}{v}\\\\t=\dfrac{200\ m}{800\ m/s}\\\\t=0.25\ s[/tex]

So, it will take 0.25 seconds to reach the target.

Answer:

a

     [tex]l_o  =52 \  m[/tex]

b

      [tex]l = 37.13 \ LY[/tex]

Explanation:

From the question we are told that

    The  speed of the spaceship is  [tex]v  =  0.800c[/tex]

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

    The  length is  [tex]l = 31.2 \  m[/tex]

     The  distance of the star for earth is [tex]d = 145 \  light \  years[/tex]

     The  speed is [tex]v_s = 2.90 *10^{8}[/tex]

Generally the from the length contraction equation we have that

       [tex]l  =  l_o  \sqrt{1 -[\frac{v}{c } ]}[/tex]

Now the when at rest the length is  [tex]l_o[/tex]

So  

      [tex]l_o =\frac{l}{\sqrt{ 1 - \frac{v^2}{c^2 } } }[/tex]

      [tex]l_o =\frac{ 31.2 }{ \sqrt{1 - \frac{(0.800c ) ^2}{c^2} } }[/tex]

      [tex]l_o=52 \  m[/tex]

Considering b  

  Applying above equation

            [tex]l  =l_o \sqrt{1 -  [\frac{v}{c } ]}[/tex]

Here [tex]l_o  =145 \  LY(light \ years )[/tex]

So

           [tex]l=145 *  \sqrt{1 -  \frac{v_s^2}{c^2 } }[/tex]

            [tex]l =145 *  \sqrt{ 1 - \frac{2.9 *10^{8}}{3.0*10^{8}} }[/tex]

            [tex]l = 37.13 \ LY[/tex]

Answer:

The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

Explanation:

Given that,

Initial position of particle [tex]r_{1}=4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k\ m[/tex]

Final position of particle [tex]r_{2}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k\ m[/tex]

We need to calculate the magnitude of the position vector

Using formula of position vector

[tex]\vec{r}=\vec{r_{2}}-\vec{r_{1}}[/tex]

Put the value into the formula

[tex]\vec{r}=-2.00\times10^{-9}i+4.00\times10^{-9}j+ 3.00\times10^{-9}k-(4.00\times10^{-9}i-5.00\times10^{-9}j-2.00\times10^{-9}k)[/tex]

[tex]\vec{r}=-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k[/tex]

[tex]\vec{|r|}=\sqrt{(9^2+9^2+5^2)\times10^{-18}}[/tex]

[tex]\vec{|r|}=13.6\times10^{-9}\ m[/tex]

We need to calculate the unit vector along electric field direction

Using formula of unit vector

[tex]\hat{r}=\dfrac{\vec{r}}{|\vec{r}|}[/tex]

Put the value into the formula

[tex]\hat{r}=\dfrac{-9\times10^{-9}i+9\times10^{-9}j+5\times10^{-9}k}{13.6\times10^{-9}}[/tex]

[tex]\hat{r}=-0.66i+0.66j+0.36k[/tex]

We need to calculate the electric field due to charge at distance r

Using formula of electric field

[tex]E=\dfrac{kq}{r^2}\hat{r}[/tex]

Put the value into the formula

[tex]E=\dfrac{9\times10^{9}\times1.6\times10^{-19}}{(13.6\times10^{-9})^2}\times(-0.66i+0.66j+0.36k)[/tex]

[tex]E=7.78\times10^{6}\times(-0.66i+0.66j+0.36k)[/tex]

[tex]E=-0.8169\times10^{6}i+0.8169\times10^{6}j+2.800\times10^{6}k\ N/C[/tex]

[tex]E=(-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

Hence, The electric field due to charge at distance r is [tex](-0.8169\times10^{6},0.8169\times10^{6}, 2.800\times10^{6})\ N/C[/tex]

Answer:

Radio waves have longer wavelength

Explanation:

Radio wave is an electromagnetic frequency that has the ability to travel through long distance. They have frequencies shuttling been the range of 10^4 hz and a frequency of 10^12 hz

Light wave is also called visible light. This is because it is visible to the naked eye, despite it being in the electromagnetic spectrum. It's frequency is usually between 4*10^-7 hz and a frequency of 7*10^-7 hz.

As can be seen from both, the radio waves length are quite far stronger than that of the light waves.

Answer:

The antenna which is a transmitting and receiving device emits energy from current as radio waves, it does this by attracting the radio waves which are a form of EMWs and converts it to small voltages which are amplified to the final voltage signal which hear

Complete question:

The length of nylon rope from which a mountain climber is suspended has an effective force constant of 1.40 ×10⁴ N/m.

What is the frequency (in Hz) at which he bounces, given that his mass plus the mass of his equipment is 84.0 kg?

Answer:

The frequency (in Hz) at which he bounce is 2.054 Hz

Explanation:

Given;

effective force constant, K = 1.40 ×10⁴ N/m.

The total mass = his mass plus the mass of his equipment, m = 84 kg

The frequency (in Hz) at which he bounce is given by;

[tex]f = \frac{1}{2\pi} \sqrt{\frac{k}{m}}\\\\f = \frac{1}{2\pi} \sqrt{\frac{1.4*10^4}{84}}\\\\f = 2.054 \ Hz[/tex]

Therefore, the frequency (in Hz) at which he bounce is 2.054 Hz

Answer:

The Sun is known to emit almost all wavelengths of electromagnetic radiation but 99% of the radiation emitted by the sun lie in the ultraviolet, visible, and infrared regions.

Ultraviolet (UV) is a form of electromagnetic radiation with wavelength from 10 nm to 400 nm (750 THz). The wavelength is shorter than that of visible light but longer than X-rays. UV rays make up about 10% of the e-m waves from the sun. UV radiation is carcinogenic to the skin, and is absorbed by the melanin pigment in the skin.

Visible light is the only e-m wave that our eyes can pick up, i.e the only e-m wave we can see. The frequency of this spectrum corresponds to a band in the vicinity of 405–790 THz. It can further be separated into different colors. It makes up a large portion of the e-m waves coming from the sun.

Infrared wave is an e-m radiation whose wavelengths longer than those of visible light. It is is generally invisible to the human eye. The wavelength of an infrared wave extend from the red edge of the visible spectrum at 700 nanometers (frequency 430 THz), to 1 millimeter (300 GHz). Most of the thermal radiation emitted by objects near room temperature is infrared. Most of the heat from the sun reach us as infrared radiation. As with all e-m radiation, infrared radiation carries radiant energy and behaves both like a wave and like a photon.

In each case, the larger unit is the one that represents a greater quantity or magnitude compared to the other unit. For example, a centimeter is larger than a millimeter because one centimeter is equal to 10 millimeters. Similarly, a megagram is larger than a kilogram as one megagram is equivalent to 1,000 kilograms. The pattern continues for each pair, where the larger unit represents a higher order of magnitude or a greater number of the smaller units.

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Complete Question

The human eye can readily detect wavelengths from about 400 nm to 700 nm. If white light illuminates a diffraction grating having 710 lines/mm, over what range of angles does the visible m = 1 spectrum extend

Answer:

The  value  [tex]\theta = 16.5 ^ o[/tex]

Explanation:

From the question we are told that

   The  diffraction grating haves [tex]a =  710 \  lines /mm[/tex]

Generally the separation of the slit is mathematically represented  as

      [tex]d = \frac{1}{710} \ mm =0.001408 \ mm =   1.408 *10^{-6} \  m[/tex]

 Generally the condition for constructive interference is mathematically represented as  

           [tex]dsin(\theta ) = n \lambda[/tex]

So

          [tex]\theta =  sin ^{-1} [\frac{n * \lambda }{d} ][/tex]

=>       [tex]\theta  =  sin^{-1}[\frac{1 *  400 *10^{-9}}{ 1.408*10^{-6}} ][/tex]

=>         [tex]\theta = 16.5 ^ o[/tex]

Answer:

soccer ball

Explanation:

Because it's round in shape and therefore cannot be stopped as both moves at same speed. There's no edges or corners to bounce up and slow the ball.

Answer:

They will both be since they are going the same speed

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Answer:

1)   α = 2.2 rad / s² , 2)   t = 7.068 s , 3) in this interval   s = 23.096 m

total distance    s = 57.4 m

Explanation:

For this exercise we use the angular kinematic relations, before starting the problem we reduce all the magnitude to the SI system

        v₀ = 88 km / h (1000 m / 1km) 1h / 3600s) = 24.44 m / s

        v = 56.0 km / h = 15.55 m / s

        θ = 90 rev (2pi rad / 1 rev) = 565,487 rad

        d = 0.84 m

          r = d / 2

         r = 0.84 / 2

         r = 0.42 m

1) ask for angular acceleration

        w² = w₀² - 2 α Δθ

        α = (w₀² -w²) / 2 Δθ

To find the angular velocities we use the acceleration between the linear and angular velocity

         v = w r

         w = v / r

         w₀ = 24.44 / 0.42

         w₀ = 58.20 rad / s

         w = 15.55 / 0.42

         w = 30.037 rad / s

we calculate

        α = (58.20² - 30.037²) / (2   565.487)

        α = 2.2 rad / s²

2) how much longer does it take to stop

         w₂ = 15.55 rad / s

         w = 0

         w = w₂ - α t

         t = (w₂ -0) / α

         t = 15.55 / 2.2

         t = 7.068 s

3) the distance that the car travels from the beginning of the movement, we can find it by looking for the number of revolutions until it stops and then using the relationship between the angular and linear variable

         w² = w₀² - 2 α θ

at the end of the movement speed is zero

         0 = w₀² - 2 α θ

         θ = w₀² / 2 α

         θ = 24.44² / (2 2.2)

          θ = 135.80 rad

If the angles are measured in radians, we can apply the relation

         θ = s / R

         s = R ttea

         s = 0.42 135

         s = 57.4 m

this is the distance from when the movement starts

the distance for the final part of the movement is

          w = 15.55 rad / s

          θ = w² / 2 α

          θ = 15.55 2 / (2 2.2)

          θ = 54.99 rad

the distance in this interval is

          s = 0.42 54.99

          s = 23.096 m