A 2.8 kg block slides with a speed of 2.4 m/s on a frictionless horizontal surface until it encounters a spring. Part A If the block compresses the spring 5.6 cm before coming to rest, what is the force constant of the spring

Answer:

D

Explanation:

Now the net force is the applied force minus the frictional force; this is expressed mathematically as:

Fnet= Fappplied - Ffrictional

Now the frictional force is given as ;

Coefficient of friction × normal reaction

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction of the human is ;

35 × 9.8 = 343N { note that weight = m× g and g= 9.8m/S2, a known standard }

Hence the Frictional force =343×0.4 =137.20N

Hence Fnet = 200-137.20 = 62.8N

Fnet = 63N to the nearest whole

The net force on the couch as it slides is  63N.

What is frictional force?

When an object is moving on a rough surface, it experiences opposition. This opposing force is called the friction force.

The friction force is given by

f = coefficient of friction x Normal force

Given, the coefficient of kinetic friction between a couch and the floor is 0.4. If the couch has a mass of 35 kg and you push it with a force of 200 N.

Normal reaction is the weight of the human acting in opposite direction.

Normal reaction N =35 × 9.8 = 343N

Frictional force f =0.4 x 343

                          f =137.20N

The net force will be

Fnet= Fappplied - Ffrictional

Fnet = 200-137.20 = 62.8N

Fnet = 63N

Thus,  the net force on the couch as it slides is  63N.

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Answer:

The length of organ pipe A is [tex]L = 0.3611 \ m[/tex]

The length of organ pipe B is  [tex]L_b = 0.2708 \ m[/tex]

Explanation:

From the question we are told that

    The fundamental frequency is  [tex]f = 475 Hz[/tex]

     The speed of sound is  [tex]v_s = 343 \ m/s[/tex]

The fundamental frequency of the organ pipe A  is mathematically represented as

        [tex]f= \frac{v_s}{2 L}[/tex]

Where L is the length of  organ pipe

   Now  making L the subject

        [tex]L = \frac{v_s}{2f}[/tex]

substituting values

        [tex]L = \frac{343}{2 *475}[/tex]

        [tex]L = 0.3611 \ m[/tex]

The second harmonic frequency of the  organ pipe A is mathematically represented as

       [tex]f_2 = \frac{v_2}{L}[/tex]

The third harmonic frequency of the  organ pipe B is mathematically represented as      

      [tex]f_3 = \frac{3 v_s}{4 L_b }[/tex]

So from the question

       [tex]f_2 = f_3[/tex]

So

    [tex]\frac{v_2}{L} = \frac{3 v_s}{4 L_b }[/tex]

Making  [tex]L_b[/tex] the subject

     [tex]L_b = \frac{3}{4} L[/tex]

substituting values

    [tex]L_b = \frac{3}{4} (0.3611)[/tex]

    [tex]L_b = 0.2708 \ m[/tex]

Complete question is;

Assuming 100% efficient energy conversion how much water stored behind a 50 centimeter high hydroelectric dam would be required to charge the battery with power rating, 12 V, 50 Ampere-minutes.

Answer:

Amount of water required to charge the battery = 7.35 m³

Explanation:

The formula for Potential energy of the water at that height = mgh

Where;

m = mass of the water

g = acceleration due to gravity = 9.8 m/s²

h = height of water = 50 cm = 0.5 m

We know that in density, m = ρV

Where;

ρ = density of water = 1000 kg/m³

V = volume of water

So, potential energy is now given as;

Potential energy = ρVgH = 1000 × V × 9.8 × 0.5 = (4900V) J

Now, formula for energy of the battery is given as;

E = qV

We are given;

q = 50 A.min = 50 × 60 = 3,000 C

V = 12 V

Thus;

qV = 3,000 × 12 = 36,000 J

E = 36,000 J

At a 100% conversion rate, the energy of the water totally powers the battery.

Thus;

(4900V) = (36,000)

4900V = 36,000

V = 36,000/4900

V = 7.35 m³

Answer:

a) 463.29 K

b) 8065.65 Pa

c) 0 J

Explanation:

The parameters given are;

Volume of the tank, V = 3.72 m³

Number of moles of gas present in the tank, n = 22.1 moles

Temperature of the gas before heating, T₁ = 300 k

Heat added to the gas, ΔQ = 4.5 × 10⁴ J

Specific heat capacity at constant volume, [tex]c_v[/tex], for monatomic gas = 12.47 J/K/mole

Avogadro's number = 6.022 × 10²³ particles per mole

a) ΔQ = n × [tex]c_v[/tex] × ΔT

Where:

ΔT = T₂ - T₁

T₂ = Final temperature of the gas

Hence, by plugging in the values, we have;

4.5 × 10⁴ = 22.1 × 12.47 × (T₂ - 300)

[tex]T_{2} - 300 = \frac{4.5\times 10^{4}}{22.1\times 12.47}[/tex]

T₂ = 300 + 163.29 = 463.29 K

b) The pressure of the gas is found from the relation;

P×V = n×R×T

[tex]P = \dfrac{n \times R \times T}{V}[/tex]

Where:

P = Pressure of the gas

R = Universal gas constant = 8.3145 J/(mol·K)

T = Temperature of the gas

V = Volume of the gas = 3.72 ³ (constant)

n = Number of moles of gas present = 22.1 moles (constant)

Hence the change in pressure is given by the relation;

[tex]\Delta P = \dfrac{n \times R \times (T_2 - T_1)}{V} = \dfrac{n \times R \times \Delta T}{V}[/tex]

Plugging in the values, we have;

[tex]\Delta P = \dfrac{22.1 \times 8.3145 \times 163.29}{3.72} = 8065.65 \, Pa[/tex]

c) Work done, W, by the gas is given by the area under the pressure to volume graph which gives;

W = f(P) × ΔV

The volume given in the question is constant

∴ ΔV = 0

Hence, W =  f(P) × 0 = 0 J

No work done by the gas during the process.

Answer:

normal force = 10 N

Explanation:

Given data

frictional force = 0.400 N

coefficient of kinetic friction = 0.04

Solution

we get here normal force that is express as

normal force = [tex]\frac{Frictional\ force}{coefficient\ of\ friction}[/tex]        ............1

put here value and we will get value

normal force = [tex]\frac{0.400}{0.04}[/tex]  

solve it we get

normal force = 10 N

Answer: it's weight decreases

Explanation:

Answer:

24 Newtons

Explanation:

The force exerted in the middle cube needs to be enough to move the middle cube and the right cube with an acceleration of 2 m/s2.

The mass of those two cubes combined is 6 + 6 = 12 kg

So, using the following equation, we can find the force:

Force = mass * acceleration

Force = 12 * 2

Force = 24 Newtons

Answer:

v = 7.67 m/s

Explanation:

The equation for apparent weight in the situation of weightlessness is given as:

Apparent Weight = m(g - a)

where,

Apparent Weight = 360 N

m = mass passenger = 61.2 kg

a = acceleration of roller coaster

g = acceleration due to gravity = 9.8 m/s²

Therefore,

360 N = (61.2 kg)(9.8 m/s² - a)

9.8 m/s² - a = 360 N/61.2 kg

a = 9.8 m/s² - 5.88 m/s²

a = 3.92 m/s²

Since, this acceleration is due to the change in direction of velocity on a circular path. Therefore, it can b represented by centripetal acceleration and its formula is given as:

a = v²/r

where,

a = centripetal acceleration = 3.92 m/s²

v = speed of roller coaster = ?

r = radius of circular rise = 15 m

Therefore,

3.92 m/s² = v²/15 m

v² = (3.92 m.s²)(15 m)

v = √(58.8 m²/s²)

v = 7.67 m/s

Answer:

☟

Explanation:

Fuels made from oil mixtures containing large hydrocarbon molecules are not efficient as they do not flow easily and are difficult to ignite. Crude oil often contains too many large hydrocarbon molecules and not enough small hydrocarbon molecules to meet demand. This is where cracking comes in.

Cracking allows large hydrocarbon molecules to be broken down into smaller, more useful hydrocarbon molecules. Fractions containing large hydrocarbon molecules are heated to vaporise them. They are then either:

heated to 600-700°C

passed over a catalyst of silica or alumina

These processes break covalent bonds in the molecules, causing thermal decompositionreactions. Cracking produces smaller alkanesand alkenes (hydrocarbons that contain carbon-carbon double bonds). For example:

hexane → butane + ethene

C6H14 → C4H10 + C2H4

Some of the smaller hydrocarbons formed by cracking are used as fuels, and the alkenes are used to make polymers in plastics manufacture. Sometimes, hydrogen is also produced during cracking.

Fractional distillation of crude oil

Fractional distillation separates a mixture into a number of different parts, called fractions.

A tall fractionating column is fitted above the mixture, with several condensers coming off at different heights. The column is hot at the bottom and cool at the top. Substances with high boiling points condense at the bottom and substances with lower boiling points condense on the way to the top.

Crude oil is a mixture of hydrocarbons. The crude oil is evaporated and its vapours condense at different temperatures in the fractionating column. Each fraction contains hydrocarbon molecules with a similar number of carbon atoms and a similar range of boiling points.

Oil fractions

The diagram below summarises the main fractions from crude oil and their uses, and the trends in properties. Note that the gases leave at the top of the column, the liquids condense in the middle and the solids stay at the bottom.

As you go up the fractionating column, the hydrocarbons have:

lower boiling points

lower viscosity (they flow more easily)

higher flammability (they ignite more easily).

Other fossil fuels

Crude oil is not the only fossil fuel.

Natural gas mainly consists of methane. It is used in domestic boilers, cookers and Bunsen burners, as well as in some power stations.

Coal was formed from the remains of ancient forests. It can be burned in power stations. Coal is mainly carbon but it may also contain sulfur compounds, which produce sulfur dioxide when the coal is burned. This gas is a cause of acid rain. Also, as all fossil fuels contain carbon, the burning of any fossil fuel will contribute to global warming due to the production of carbon dioxide.

In fractional distillation, the crude oil is added to the chamber and heated. The components with the highest boiling point will condense in the lower part of the column and the components with the lower boiling point will condense at the top of the column. Petrol with a low boiling point is collected from the top of the column.

Fractional distillation can be described as the separation of a mixture into its component fractions. The chemical compound is separated by heating them to a temperature at which fractions of the mixture will vaporize.

Generally, the components have boiling points that differ by less than 25 °C  from each other under one atmosphere. When the mixture is heated, the component with the lower boiling point boils and changes to vapours.

The more volatile component remains in a vapour state and repeated distillations are used in the process, and the mixture is separated into component parts.

Therefore, the petrol from the crude oil can easily be separated as it has a boiling point of about 25-60°C.

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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  [tex]Q =2.094 C[/tex]

Explanation:

From the question we are told that

    The diameter of the wire is  [tex]d = 0.205cm = 0.00205 \ m[/tex]

     The radius of  the wire is  [tex]r = \frac{0.00205}{2} = 0.001025 \ m[/tex]

     The resistivity of aluminum is [tex]2.75*10^{-8} \ ohm-meters.[/tex]

       The electric field change is mathematically defied as

         [tex]E (t) = 0.0004t^2 - 0.0001 +0.0004[/tex]

Generally the charge is  mathematically represented as

       [tex]Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt[/tex]

Where A is the area which is mathematically represented as

       [tex]A = \pi r^2 = (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2[/tex]

 So

       [tex]\frac{A}{\rho} = \frac{3.3 *10^{-6}}{2.75 *10^{-8}} = 120.03 \ m / \Omega[/tex]

Therefore

      [tex]Q = 120 \int\limits^{t}_{0} { E(t) } \, dt[/tex]

substituting values

      [tex]Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt[/tex]

     [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | t} \atop {0}} \right.[/tex]

From the question we are told that t =  5 sec

           [tex]Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] } \left | 5} \atop {0}} \right.[/tex]

          [tex]Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }[/tex]

         [tex]Q =2.094 C[/tex]

The charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds is 2.094 Coulomb.

Given the following data:

Conversion:

Diameter of wire = 0.205 cm to m = 0.00205 meter.

Radius = [tex]\frac{Diameter}{2} =\frac{0.00205}{2} =0.001025\;meter[/tex]

To determine the charge (Q) passing through a cross-section of the conductor between time 0 seconds and time 5 seconds, we would apply Gauss's law in an electric field for a surface charge:

First of all, we would find the area of the wire.

[tex]Area = \pi r^2\\\\Area = 3.142 \times 0.001025^2\\\\Area = 3.3 \times 10^{-6}\;m^2[/tex]

Mathematically, Gauss's law in an electric field for a surface charge is given by the formula:

[tex]Q = \int\limits^t_0 {\frac{A}{\rho } E(t)} \, dt[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]Q= \int\limits^t_0 {\frac{3.3 \times 10^{-6}}{2.75\times 10^{-8} } (0.0004t^2-0.0001t+0.0004)} \, dt\\\\Q=120\int\limits^t_0 1{ (0.0004t^2-0.0001t+0.0004)} \, dt[/tex]

[tex]Q=120(\frac{0.0004t^3}{3} -\frac{0.0001t^2}{2} +0.0004t |\left{5} \atop {0} \right[/tex]

When t = 5 seconds:

[tex]Q=120(\frac{0.0004[5]^3}{3} -\frac{0.0001[5]^2}{2} +0.0004[5])\\\\Q=120(\frac{0.03}{3} -\frac{0.0025}{2} +0.002)\\\\Q=120(0.0167-0.00125+0.002)\\\\Q=120(0.01745)[/tex]

Q = 2.094 Coulomb.

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Question:

A particle moving along the x-axis has a position given by x=(24t - 2.0t³)m, where t is measured in s. What is the magnitude of the acceleration of the particle at the instant when its velocity is zero

Answer:

24 m/s

Explanation:

Given:

x=(24t - 2.0t³)m

First find velocity function v(t):

v(t) = ẋ(t) = 24 - 2*3t²

v(t) = ẋ(t) = 24 - 6t²

Find the acceleration function a(t):

a(t) = Ẍ(t) = V(t) = -6*2t

a(t) = Ẍ(t) = V(t) = -12t

At acceleration = 0, take time as T in velocity function.

0 =v(T) = 24 - 6T²

Solve for T

[tex] T = \sqrt{\frac{-24}{6}} = \sqrt{-4} = -2 [/tex]

Substitute -2 for t in acceleration function:

a(t) = a(T) = a(-2) = -12(-2) = 24 m/s

Acceleration = 24m/s

Answer:

a. 200 m

b. 100 m

Explanation:

Solution:-

- We will first draw three points marked A,B and Q from left most to right most.

- We are told that the antennas at A and B radiate in phase. This means the radio-waves emitted by each antenna are synchronous in terms of ( frequency and wavelength ).

- We will denote the common wavelength of coherent sources of radio-waves ( A and B ) with λ.

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of destructive interference is:

                             AQ - BQ = n*λ/2

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ/2

- To determine the longest wavelength ( λ ) to meet destructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ/2

                           λ = 200 m  .... Answer

- The relation between the wavelength ( λ ) and the path difference between the source and observation point ( Q ) for the case of constructive interference is:

                             AQ - BQ = n*λ

Where,

             n: The order of wavelength

             AQ: The distance between antenna A and point Q

             BQ: The distance between antenna B and point Q

- The point Q is positioned ( 100 + 50 ) m away from antenna A and 50 m from antenna B. Hence,

                            150 - 50 = n*λ

- To determine the longest wavelength ( λ ) to meet constructively at point Q with the given path difference. The order of wavelength ( n ) must be minimum ( 1 ). Therefore,

                           100 = λ

                           λ = 100 m  .... Answer

Answer:

a. 900 J

b. 0.383

Explanation:

According to the question, the given data is as follows

Horizontal force = 150 N

Packing crate = 40.0 kg

Distance = 6.00 m

Based on the above information

a. The work done by the 150-N force is

[tex]W = F x = \mu N x = \mu\ m\ g\ x[/tex]

[tex]W = 150 \times 6[/tex]

= 900 J

b. Now the coefficient of kinetic friction between the crate and surface is

[tex]\mu = \frac {F}{m\timesg}[/tex]

[tex]= \frac{150}{40\times 9.8}[/tex]

= .383

We simply applied the above formulas so that each one part could calculate

We want to find the work and kinetic friction for the given situation. The solutions are:

Here we have a horizontal force of 150N pushing a 40.0 kg packing crate a distance of 6.00m at a constant speed.

a) First we want to find the work, it is given by the force applied times the distance moved, so the work is just:

W = 150N*6.00m = 900 N*m

b) Now we want to find the coefficient of kinetic friction, it must be such that the kinetic friction force is equal to the pushing force, in this way there is no net force, and then there is no acceleration.

Remember that the friction force is:

F = m*g*μ

Where:

Then we must solve:

150N = 40kg*(9.8 m/s^2)*μ = 392N*μ

150N/392N = 0.38

So the coefficient of kinetic friction between the crate and the surface is 0.38

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Answer:

1A,2D,3B

Explanation:

hope this helps

Answer:

SO THAT

Answer:

To determine wavelength of a wave on a string.

Explanation:

The Standing Waves lab study the parameters that affect standing waves in various strings. The effects of string tension and density on wavelength and frequency will be studied.

Answer:

The lifetime of the particle is  [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Explanation:

From the question we are told that

    The average rest energy is [tex]E = 91.19 \ GeV = 91.19GeV * \frac{1.60 *10^{-10} J }{1GeV} = 1.46 *10^{-8}J[/tex]

    The intrinsic width is  [tex]\Delta E =2.5eV = 2.5GeV * \frac{1.60 *10^{-10}J }{1GeV} = 4*10^{-10} J[/tex]

The lifetime is mathematically represented as

     [tex]\Delta t = \frac{h}{\Delta E}[/tex]

Where h is the Planck's constant with a value of  [tex]1.055*10^{-34} \ J\cdot s[/tex]  

substituting values

    [tex]\Delta t = \frac{1.055*10^{-34}}{4 *10^{-10}}[/tex]

     [tex]\Delta t = 2.6*10^{-25} \ s[/tex]

Answer:

since the bed is a cuboid, we find the volume by L×W×H

4.50 × 2.50 × 1.50 = 16.875m³

HOPE THIS HELPS

The average acceleration [tex]\vec a_{\rm ave}[/tex] over some time interval [tex][t_1,t_2][/tex] is equal to the ratio of the change in velocity [tex]\vec v_2-\vec v_1[/tex] over the duration of the interval [tex]t_2-t_1[/tex], or

[tex]\vec a_{\rm ave}=\dfrac{\Delta\vec v}{\Delta t}=\dfrac{\vec v_2-\vec v_1}{t_2-t_1}[/tex]

which can be split into the [tex]x[/tex] and [tex]y[/tex] components as

[tex]a_{\rm{ave},x}=\dfrac{v_{2,x}-v_{1,x}}{t_2-t_1}=\dfrac{-170\frac{\rm m}{\rm s}-90\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-8.67\dfrac{\rm m}{\mathrm s^2}[/tex]

[tex]a_{\rm{ave},y}=\dfrac{v_{2,y}-v_{1,y}}{t_2-t_1}=\dfrac{40\frac{\rm m}{\rm s}-110\frac{\rm m}{\rm s}}{30.0\,\mathrm s-0}\approx-2.33\dfrac{\rm m}{\mathrm s^2}[/tex]

The magnitude of this average acceleration is

[tex]\left\|\vec a_{\rm ave}\right\|=\sqrt{{a_{\rm{ave},x}}^2+{a_{\rm{ave},y}}^2}\approx8.98\dfrac{\rm m}{\mathrm s^2}[/tex]

and its direction is [tex]\theta[/tex] such that

[tex]\tan\theta=\dfrac{a_{\rm{ave},y}}{a_{\rm{ave},x}}\implies\theta\approx-164.9^\circ[/tex]

which corresponds to a direction of about 15.1º South of West.

Answer:

b and e

Explanation:

r x F is the formula for torque.

The "turning effect" or torque happens when concentric forces rotate an object along said center.

a) False because T = Fr = Ia (a = angular acceleration)

b) True

c) False. L = Iw (w = angular velocity), which does not equal Ia

d) False. It is torque, not the product of torque and something else

e) True.

Answer:

Its called PSY

Explanation: I so do not know why they named it this way but, hope i helped.

Answer:  arrows

Explanation:

A vector quantity is usually represented by an arrow, with the direction of the vector being the direction in which the arrow points and the length of the arrow representing the vector's magnitude.

What is the vector quantity unit?

The meter is the only fundamental SI unit that is a vector. The rest are all scalars. Derived quantities might be scalar or vector, but all vector quantities require meters as part of their definition and measurement.

The term "vector quantities" refers to physical quantities whose magnitude and direction are well specified.

Arrows are used to depict vectors. An arrow has a direction and a magnitude (how long it is) (the direction in which it points).

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Explanation:

The first equation of motion in kinematics is given by :

[tex]v=u+at[/tex] .....(1)

u is initial speed

a is acceleration

v is final speed

t is time

Equation (1) is valid when the object is moving with constant acceleration. This equation gives relation between velocity and time.

Answer:

Explanation:

 Time taken by sound waves to cover distance between politician and reporter = time taken by em waves to travel distance between politician and the television viewer.

5.5 / 343 = d / 3 x 10⁸ + 2.7 / 343

d is distance between politician and television set + time taken by sound to travel distance between television and its viewer.

.0160349 = d / 3 x 10⁸  + .0078717

d / 3 x 10⁸  = .0081632

d = 2448960 m

= 2448.96 km

= 2449 km .

Answer:

4.09 MeV

Explanation:

Find the given attachment

Answer:

ΔE = ‒0.271 kJ

Explanation:

Let's begin by listing out the given variables:

q = -0.653 kJ, w = 0.389 kJ

Using the formula ΔE = q + w

ΔE = -0.653 + 0.388

ΔE = (‒0.655 + 0.382) kJ

ΔE = ‒0.271 kJ

Therefore, the change in internal energy is -271 J or -0.271 kJ which implies that the system is exothermic

Answer:

It's true

Explanation:

Because the ship is mafe up of aluminium, which is a light metal.

Answer:

False

Explanation:

Took The Quiz

Answer:

Thus, the energy stored by a 50 Ampere minute battery is found to be  36 KJ.

Explanation:

The power delivered by a battery is given by the formula:

P = VI

where,

P = Power Delivered by battery in 1 second

V = Voltage of battery = 12 volt

I = Current stored in battery

But, if we multiply both sides of equation by time (t), then:

Pt = VIt

where,

Pt = Power x Time = E = Energy Stored = ?

It = Rating of Battery = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Thus, the energy stored by a 50 Ampere minute battery is found to be  36 KJ.

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

Velocity is the rate of change of displacement. During their periods of travel, the cars definitely had the same velocity.

Velocity is the directional speed of a moving object as an indicator of its rate of change in location as perceived from a certain frame of reference and measured by a specific time standard.

Given that the first car travels 40 meters due east in 5 seconds. Therefore, we can write,

Distance = 40 meters

Time = 5 seconds

Velocity = Distance / Time = 40 meter/ 5 sec = 40 m/sec

Also, given that the second car travels 64 meters due west in 8 seconds. Therefore, we can write,

Distance = 64 meters

Time = 8 seconds

Velocity = Distance / Time = 64 meter/ 8 sec = 8 m/sec

Hence, During their periods of travel, the cars definitely had the same velocity.

Learn more about Velocity here:

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