The pressure and temperature at the beginning of compression of a cold air-standard Diesel cycle are 100 kPa and 300K, respectively. At the end of the heat addition, the pressure is 7.2 MPa and the temperature is 2250 K. Assume constant specific heats evaluated at 300 K. Determine the cut-off ratio. There is a +/- 5% tolerance.

Answer:

31.1 slug,  453.4 Kg

Explanation:

given,

mass of the beam is 1000 lb

to convert mass of beam into slugs and kilograms.

1 lb is equal to 0.0311 slug

1000 lb = 1000 × 0.0311

             = 31.1 slug

now, conversion of lb into kg

1 lb is equal to 0.4534 kg

so,

1000 lb = 1000 × 0.4534

             = 453.4 Kg

hence, 1000 lb of beam in slugs is equal to 31.1 slugs and in kilo gram is 453.4 Kg.

Answer:

Part 1) Speed at s = 420 meters =12.26 m/s

Part 2) Time required to cover the distance = 26 seconds

Explanation:

This problem can be solved using third equation of kinematics as follows

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

where

'v' is the final velocity

'u' is the initial velocity

'a' is the acceleration of the car

's' is the distance covered between change in velocities

Now during the time at which the car moves it cover's a distance of 420 m-100 m=320 meters

Thus applying values in the above equation we get

[tex]v^2=12^{2}+2\times 0.01\times 320\\\\v^2=150.4\\\\\therefore v=12.26m/s[/tex]

The time to reach this velocity  can be found using first equation of kinematics as

[tex]v=u+at\\\\\therefore t=\frac{v-u}{a}\\\\t=\frac{12.26-12}{0.01}=26seconds[/tex]

Answer:

461 C

862 F

Explanation:

The specific gas constant for CO2 is

R = 189 J/(kg*K)

Using the gas state equation:

p * v = R * T

T = p * v / R

v = 1/δ

T = p  / (R * δ)

T = 9.3*10^6  / (189 * 67) = 734 K

734 - 273 = 461 C

461 C = 862 F

To avoid cavitation for water at 160 °F flowing through a sharp bend, the minimum absolute pressure should be slightly above the vapor pressure of water at this temperature, which is approximately 0.363 psi.

When water flows through a sharp bend, cavitation can occur if the local pressure falls to or below the fluid's vapor pressure. To estimate the minimum absolute pressure without causing cavitation for water at 160 °F, we must consider water's vapor pressure at this temperature. At 160 °F (about 71 °C), the vapor pressure of water is approximately 0.363 psi. Since fluids cannot have a negative absolute pressure, and to avoid cavitation, the absolute pressure must stay above this vapor pressure. Therefore, considering atmospheric pressure to be approximately 14.7 psi, to avoid cavitation, the minimum absolute pressure in the system should be slightly above 0.363 psi to ensure no cavitation occurs.

Answer:

9.7g / cm^3

Explanation:

To calculate a conbined density we must find the ratio  between the sum of the masses and the sum of the volumes remembering that the equation to find the density is α=m/v, taking into account the above the following equation is inferred.

αc=copper density

αs=silver density

Vs=volume of silver

Vc=volume of copper

α= density of alloy

[tex]\alpha =\frac{({\alpha c}{Vc} +{\alpha s }{Vs} )}{Vs+Vc} \\\alpha =\frac{(8.9)(10) +(10.5)(10) }{10+10} \\\\\alpha =9.7g / cm^3[/tex]

the density of the alloy is 9.7g / cm^3

Answer:

The diameter of the test cylinder should be 7.65 meters.

Explanation:

The Hooke's law relation between stress and strain is mathematically represented as

[tex]Stress=E\times strain\\\\\sigma =e\times \epsilon[/tex]

Where 'E' is modulus of elasticity of the material

Now by definition of strain we have

[tex]\epsilon =\frac{\Delta L}{L_{o}}[/tex]

Applying values to obtain strain we get

[tex]\epsilon =\frac{0.5}{380}=0.001316[/tex]

Thus the stress developed in the material equals

[tex]\sigma = 110\times 10^{3}\times 0.001316=144.76N/m^{2}[/tex]

Now by definition of stress we have

[tex]\sigma =\frac{Force}{Area}\\\\\therefore Area=\frac{Force}{\sigma }\\\\\frac{\pi D^{2}}{4}=\frac{6660N}{144.76}=46m^{2}[/tex]

Solving for 'D' we get

[tex]D=\sqrt{\frac{4\times 46}{\pi }}=7.653meters[/tex]

Answer:d

Explanation:

Given

Temperature[tex]=200^{\circ}\approc 473 K[/tex]

Also [tex]\gamma for air=1.4[/tex]

R=287 J/kg

Flow will be In-compressible when Mach no.<0.32

Mach no.[tex]=\frac{V}{\sqrt{\gamma RT}}[/tex]

(a)[tex]1000 km/h\approx 277.78 m/s[/tex]

Mach no.[tex]=\frac{277.78}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.63

(b)[tex]500 km/h\approx 138.89 m/s[/tex]

Mach no.[tex]=\frac{138.89}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.31

(c)[tex]2000 km/h\approx 555.55 m/s[/tex]

Mach no.[tex]=\frac{555.55}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=1.27

(d)[tex]200 km/h\approx 55.55 m/s[/tex]

Mach no.[tex]=\frac{55.55}{\sqrt{1.4\times 287\times 473}}[/tex]

Mach no.=0.127

From above results it is clear that for Flow at velocity 200 km/h ,it will be incompressible.

Answer:

wr = 6.32 rad/s

Explanation:

m = 750 kg

k = 30 kN/m

This system has no dampening, therefore the resonance frequency will simply be the natural frequency of the system.

[tex]wr = w0 = \sqrt{\frac{k}{m}}[/tex]

[tex]wr = \sqrt{\frac{30000}{750}} = 6.32 rad/s[/tex]

In this case the force applied doesn't matter. because we are calculating the resonance frequency.

Answer:

Please, see the attachment.

Explanation:

First, we have to create two input boxes that allows the user to write the current year in one of them and his/her birth year in the another one. Also, we have to create a label that will show the result of the desired variable. We can write a message "Your age is:" and it will be attached to the result.

For the algorithm, let's call the variables as follows:

CY = Current Year

BY = Birth Year

X = Age of user

When the user inserts the current year and his/her birth year, the program will do the following operation:

X = CY - BY; this operation will give us the age of the user

After this the user will see something like "Your age is:" X.

Answer:

[tex]\eta = 91.7[/tex]%

Explanation:

Determine the initial velocity

[tex]v_1 = \frac{\dot v}{A_1}[/tex]

    [tex] = \frac{0.1}{\pi}{4} 0.08^2[/tex]

     = 19.89 m/s

final velocity

[tex]v_2 =\frac{\dot v}{A_2}[/tex]

      [tex]= \frac{0.1}{\frac{\pi}{4} 0.12^2}[/tex]

      =8.84 m/s

total mechanical energy is given as

[tex]E_{mech} = \dot m (P_2v_2 -P_1v_1) + \dot m \frac{v_2^2 - v_1^2}{2}[/tex]

[tex]\dot v = \dot m v[/tex]                       [tex]( v =v_1 =v_2)[/tex]

[tex]E_{mech} = \dot mv (P_2 -P_1) + \dot m \frac{v_2^2 - v_1^2}{2}[/tex]

                [tex] = mv\Delta P + \dot m  \frac{v_2^2 -v_1^2}{2}[/tex]

                 [tex]= \dot v \Delta P  + \dot v \rho \frac{v_2^2 -v_1^2}{2}[/tex]

              [tex]   = 0.1\times 500 + 0.1\times 860\frac{8.84^2 -19.89^2}{2}\times \frac{1}{1000}[/tex]

[tex]E_{mech} = 36.34 W[/tex]

Shaft power

[tex]W = \eta_[motar} W_{elec}[/tex]

    [tex]=0.9\times 44 =39.6[/tex]

mechanical efficiency

[tex]\eta{pump} =\frac{ E_{mech}}{W}[/tex]

[tex]=\frac{36.34}{39.6} = 0.917  = 91.7[/tex]%

Answer:

a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

Answer:

9 cm

Explanation:

The liquid on the bend will be affected by two accelerations: gravity and centripetal force.

Gravity will be of 9.81 m/s^2 pointing down at all points.

The centripetal acceleration will be of

ac = v^2/r

Pointing to the center of the bend (perpendicular to gravity).

The velocity will depend on the radius

v = (1 m^2/s) / r

Replacing:

ac = (1/r)^2 / r

ac = (1 m^4/s^2) / r^3

If we set up a cylindrical reference system with origin at the center of the bend, the total acceleration will be

a = (-1/r^3 * i - 9.81 * j)

The surface of the liquid will be an equipotential surface, this means all points on the surface have the same potential energy.

The potential energy of the gravity field is:

pg = g * h

The potential energy of the centripetal force is:

pc = ac * r

Then the potential field is:

p = -1/r^2 * - 9.81*h

Points on the surface at r = 1 m and r = 3 m have the same potential.

-1/1^2 * - 9.81*h1 = -1/3^2 * - 9.81*h2

-1 - 9.81*h1 = -1/9 - 9.81*h2

-1 + 1/9 = 9.81 * (h1 - h2)

h1 - h2 = (-8/9) / 9.81

h2 - h1 = 0.09 m

The outer part will be 9 cm higher than the inner part.

Answer:

 inside dimension  [tex]= 142 mm \times 42 mm[/tex]

cross section area [tex]= 7.5\times 10^{-3} m^2[/tex]

mass of 1.2 meter log steel  [tex] = 1.843\times 10^{-3} \rho[/tex]

Explanation:

given data:

Outside dimension of steel rectangular [tex]= 150 mm\times 50mm[/tex]

Thickness = 4 mm

Long = 1.22 m

inside dimension will be [tex]= (150- 8)mm \times ( 50-8)mm[/tex]

                                         [tex]= 142 mm \times 42 mm[/tex]

cross section area [tex]= 150\times 50 mm^2[/tex]

                               [tex]= 7500\times 10^{-6} m^2[/tex]

                               [tex]= 7.5\times 10^{-3} m^2[/tex]

let the density be assumed as \rho

mass of 1.2 meter log steel will be [tex]= 1.2  \times (7.5\times 10^{-3} -  0.142\times 0.048)\times \rho[/tex]

                                                       [tex] = 1.843\times 10^{-3} \rho[/tex]

Answer:

No, is not a violation of the first law of thermodynamics.

Explanation:

The first law of thermodynamics states that energy is neither created nor destroyed in an isolated system (a system without mass or energy transfer with ambient).

In a heat pump work is used to transfer heat from a cold body to a hot body (see figure). In a free system heat would go from the heat source to the cold one, that is why you need work. Work and heat are energy in transit.

W + Qin = Qout

In your case

2 kW + Qin =  7 kW

Qin =5 kW

It would be a violation to the first law of thermodynamics if Qout is less than  2 kW.

Answer:

Water needs to be added into the pool at the rate of 1064.814 Liters/min.

Explanation:

The concentration in mg/L of the chlorine in the pool that is induced by dumping the whole container of chlorine into the pool equals

[tex]\frac{500\times 10^{3}}{200\times 10^{3}}=2.5mg/l[/tex]

Since this is in excess to the required concentration of 0.20 mg/L

Let the amount of pure water we need to add be equal to 'V' liters now since pure water does not have any chlorine thus

By the conservation of mass principle we have

[tex]0\times V+2.5\times 200\times 10^{3}=0.20\times (V+200\times 10^{3})[/tex]

Solving for V we get

[tex]V=\frac{2.5\times 200\times 10^{3}-0.20\times 200\times 10^{3}}{0.20}=2300000Liters[/tex]

Thus 2300000 Liters of water needs to be further added in 36 hours

Thus the rate of flow in L/min equals

[tex]Q=\frac{2300000}{36\times 60}=1064.814Liters/min[/tex]

Answer:

1) Acceleration of player is 7.8125 [tex]m/s^{2}[/tex].

2) Constant speed of player is 12.5 m/s.

Explanation:

Let the acceleration of the player by 'a' and let he complete the initial 10 meters in [tex]t_1[/tex] time

The initial 10 meters are case of uniformly accelerated motion and hence we can relate the above quantities using second equation of kinematics as

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

now since the player starts from rest hence u = 0 thus the equation can be written as

[tex]10=\frac{1}{2}at_1^2\\\\at_{1}^{2}=20...........(i)[/tex]

The speed the player reaches after [tex]t_{1}[/tex] time is obtained using first equation of kinematics as

[tex]v=u+at\\v=at_{1}[/tex]

Since the total distance traveled by the player is 40 meters hence the total time of trip is 4 seconds hence we infer that he covers 30 meters of distance at a constant speed in time of [tex]4-t_{1}[/tex] seconds

Hence we can write

[tex](4-t_{1})\cdot at_{1}=30.......(ii)\\\\[/tex]

Solving equation i and ii we get

from equation 'i' we obtain [tex]a=\frac{10}{t_{1}^{2}}[/tex]

Using this in equation 'ii' we get

[tex](4-t_{1})\cdot \frac{20}{t_{1}^{2}}\cdot t_{1}=30\\\\30t_{1}=80-20t_{1}\\\\\therefore t_{1}=\frac{80}{50}=1.6seconds\\\\\therefore a=\frac{20}{1.6^2}=7.8125m/s^{2}[/tex]

Thus constant speed equals [tex]v=7.8125\times 1.6=12.5m/s[/tex]

Answer:

6243.6 lbs

Explanation:

We have given weight = 2838 kg

We have to convert this weight into lbs

Both kg and lbs are unit of measuring the weight of the body so they are changeable

We know that 1 lbs = 2.20 kg

So for converting kg into lbs we have to multiply with 2.20

So 2838 kg = 2838×2.2 = 6243.6 pound

So weight of 2838 kg will be equivalent to 6243.6 lbs

Answer:

Infinite slow thermodynamic process is Quasi-equilibrium process. All the thermodynamic model or equation is based on Quasi-equilibrium process. So, Quasi-equilibrium process is very important process in engineering.

Explanation:

Step1

Quasi-equilibrium process is the thermodynamic process which is infinitely slow. All the thermodynamic variables or properties are taken as uniform. Pressure or temperature is uniform throughout the process. Quasi-equilibrium process is represented by complete joint line in thermodynamics not with the dash line. So, this process has infinite equilibrium points near to each other.  

Step2

All the thermodynamic analysis or equations are based on Quasi-equilibrium process. This gives estimation of heat, work, enthalpy, entropy etc. Quasi-equilibrium process gives maximum power output in power producing devices like turbine or engine. The entire thermodynamic engineering model is designed on the basis of Quasi-equilibrium process. Thus, this process is very important in terms of engineering.  

Answer:

V=33.66 m/s

[tex]Re=448.8\times 10^6[/tex]

Re>4000, The flow is turbulent flow.

Explanation:

Given that

Pressure difference  = 50 mm of Hg

We know that density of Hg=136000[tex]Kg/m^3[/tex]

ΔP= 13.6 x 1000 x 0.05 Pa

ΔP=680 Pa

Diameter of tunnel = 200 mm

Property of air at 25°C

ρ=1.2[tex]Kg/m^3[/tex]

Dynamic viscosity

[tex]\mu =1.8\times 10^{-8}\ Pa.s[/tex]

Velocity of fluid given as

[tex]V=\sqrt{\dfrac{2\Delta P}{\rho_{air}}}[/tex]

[tex]V=\sqrt{\dfrac{2\times 680}{1.2}}[/tex]

V=33.66 m/s

Reynolds number

[tex]Re=\dfrac{\rho _{air}Vd}{\mu }[/tex]

[tex]Re=\dfrac{1.2\times 33.66\times 0.2}{1.8\times 10^{-8}}[/tex]

[tex]Re=448.8\times 10^6[/tex]

Re>4000,So the flow is turbulent flow.

Answer:

1113kN

Explanation:

The ouside diameter OD of the pipe is 61cm and the thickness T is 0.9cm, so the inside diameter ID will be:

Inside Diameter = Outside Diameter - Thickness

Inside Diameter = 61cm - 0.9cm = 60.1cm

Converting this diameter to meters, we have:

[tex]60.1cm*\frac{1m}{100cm}=0.601m[/tex]

This inside diameter is useful to calculate the volume V of water inside the pipe, that is the volume of a cylinder:

[tex]V_{water}=\pi  r^{2}h[/tex]

[tex]V_{water}=\pi (\frac{0.601m}{2})^{2}*120m[/tex]

[tex]V_{water}=113.28m^{3}[/tex]

The problem gives you the water density d as 1.0kg/L, but we need to convert it to proper units, so:

[tex]d_{water}=1.0\frac{Kg}{L}*\frac{1L}{1000cm^{3}}*(\frac{100cm}{1m})^{3}[/tex]

[tex]d_{water}=1000\frac{Kg}{m^{3}}[/tex]

Now, water density is given by the equation [tex]d=\frac{m}{V}[/tex], where m is the water mass and V is the water volume. Solving the equation for water mass and replacing the values we have:

[tex]m_{water}=d_{water}.V_{water}[/tex]

[tex]m_{water}=1000\frac{Kg}{mx^{3}}*113.28m^{3}[/tex]

[tex]m_{water}=113280Kg[/tex]

With the water mass we can find the weight of water:

[tex]w_{water}=m_{water} *g[/tex]

[tex]w_{water}=113280kg*9.8\frac{m}{s^{2}}[/tex]

[tex]w_{water}=1110144N[/tex]

Finally we find the total weight add up the weight of the water and the weight of the pipe,so:

[tex]w_{total}=w_{water}+w_{pipe}[/tex]

[tex]w_{total}=1110144N+2500N[/tex]

[tex]w_{total}=1112644N[/tex]

Converting this total weight to kN, we have:

[tex]1112644N*\frac{0.001kN}{1N}=1113kN[/tex]

Answer:

4.29%

Explanation:

Given:

Density of water, ρ = 1000 kg/m³

elevation change of Δz = 800 m

work per unit mass delivered, Wp = 8,200 J/kg

Now,

The percent irreversibility = [tex](1-n_p)\times100[/tex]

where,

[tex]n_p\frac{W_{actual}}{W_{theoretical}}[/tex]

also,

[tex]W_{actual}=\frac{g\Delta z}{1000}[/tex]

Where, g is the acceleration due to the gravity

on substituting the values, we get

[tex]W_{actual}=\frac{9.81\times800}{1000}[/tex]

or

[tex]W_{actual}=7.848\ KJ/kg[/tex]

or

[tex]W_{actual}=7848\ J/kg[/tex]

Therefore,

The percent irreversibility = [tex](1-n_p)\times100[/tex]

on substituting the values, we get

The percent irreversibility = [tex](1-\frac{7848}{8,200})\times100[/tex]

or

The percent irreversibility = 4.29%

Answer:

Point to point control motion system:

  In point to point control motion system tool perform specific task at a particular location.Point to point control motion system is also called positioning system.It perform intermittent operation.

Ex:  Drilling operation is a point to point motion control system.

Continuous path control system:

 Continuous path control system is continuous operation that perform by tool.The program used in  continuous path control system is more complex than point to point motion control system.

Ex :Milling operation is a  continuous path control system.

Answer:

Wind

Hydro electric

Explanation:

Energy are of two types

1.Renewable energy

 These have unlimited source of energy.

Ex: Solar energy,wind energy,geothermal energy,hydro power ,biomass etc

2.Non renewable energy

These have limited source of energy.

Ex:  Petroleum,Coal

Wind is the fastest renewable source of energy.This energy is produce by using the natural velocity of air.

Hydro electric power plant is the mostly used renewable source of energy to produce electricity.

Answer with Explanation:

The growth of renewable sources of energy depend on various factor's and their grown is also a regional dependent process. Many different countries use different renewable sources of energy and the growth of the renewable sources of energy depend on the location of the place. As an example in India the solar energy is the most widely growing source of energy due to the location of the place. while as in European union Bio energy is the source of renewable energy that has shown the most growth.

Water is the renewable source which is  used to produce the most electricity in the world accounting for 16.3% of the total global electricity production.

Answer:

The gravitational force between the masses is [tex]1.0\times 10^{-8}Newtons[/tex]

Explanation:

For 2 masses 'm' and 'M' separated by a distance 'd' the gravitational force between them is given by Newton as

[tex]F=G\cdot \frac{mM}{d^{2}}[/tex]

where

'G' is universal gravitational constant whose value is [tex]6.67\times 10^{-11}m^3kg^{-1}s^{-2}[/tex]

Applying the values in the above relation we get

[tex]F=6.67\times 10^{-11}\times \frac{8\times 12}{(800\times 10^{-3})^{2}}=1.0\times 10^{-8}Newtons[/tex]

Weight of 8 kg mass =[tex]8\times 9.81=78.45Newtons[/tex]

Weight of 12 kg mass =[tex]12\times 9.81=117.72Newtons[/tex]

thus we see that gravitational force between the masses is completely negligible as compared to the weight of the masses.

Answer:

4.186 × 10⁷ J

Explanation:

Heat gain by water = Q

Thus,    

[tex]m_{water}\times C_{water}\times \Delta T=Q[/tex]

For water:  

Volume = 1 m³ = 1000 L ( as 1 m³ = 1000 L)

Density of water= 1 kg/L

So, mass of the water:  

[tex]Mass\ of\ water=Density \times {Volume\ of\ water}[/tex]  

[tex]Mass\ of\ water=1 kg/L \times {1000\ L}[/tex]  

Mass of water  = 1000 kg

Specific heat of water = 4.186 kJ/kg K

ΔT = 10 K

So,

[tex]1000\times 4.186\times 10=Q[/tex]  

Q = 41860 kJ

Also, 1 kJ = 1000 J

So, Q = 4.186 × 10⁷ J

Explanation:

Step1

Volume flow rate is the rate of change of volume of fluid that is flowing in the duct of pipe per unit time. It is measure in m³/s or l/s. Volume flow rate is very important parameter in fluid analysis.

Step2

For the given duct, the volume flow rate is the product of average velocity to the cross section area of duct.

Expression for volume flow rate is given as follows:

Q=AV

Here, Q is the flow rate, A is area of the duct and V is the average velocity of flowing fluid.

Answer:

22.90 × 10⁸ kg

Explanation:

Given:

Diameter, d = 0.02 m

ωₙ = 0.95 rad/sec

Time period, T = 0.35 sec

Now, we know

T= [tex]2\pi\sqrt{\frac{L}{g}}[/tex]

where, L is the length of the steel cable

g is the acceleration due to gravity

0.35= [tex]2\pi\sqrt{\frac{L}{9.81}}[/tex]

or

L = 0.0304 m

Now,

The stiffness, K is given as:

K = [tex]\frac{\textup{AE}}{\textup{L}}[/tex]

Where, A is the area

E is the elastic modulus of the steel = 2 × 10¹¹ N/m²

or

K = [tex]\frac{\frac{\pi}{4}d^2\times2\times10^11}{0.0304}[/tex]

or

K = 20.66 × 10⁸ N

Also,

Natural frequency, ωₙ = [tex]\sqrt{\frac{K}{m}}[/tex]

or

mass, m = [tex]\sqrt{\frac{K}{\omega_n^2}}[/tex]

or

mass, m = [tex]\sqrt{\frac{20.66\times10^8}{0.95^2}}[/tex]

mass, m = 22.90 × 10⁸ kg

Answer:43.70 MPa

Explanation:

Given

mass of engine [tex] 700 lb \approx 317.515 kg[/tex]

diameter of cable [tex]0.375 in.\approx 9.525 mm[/tex]

[tex]A=\frac{\pi d^2}{4}=71.26 mm^2[/tex]

we know stress([tex]\sigma [/tex])[tex]=\frac{load\ applied}{area\ of\ cross-section}[/tex]

[tex]\sigma =\frac{317.515\times 9.81}{71.26\times 10^{-6}}=43.70 MPa[/tex]

Answer:

The force between the charges when the separation decreases to 0.7 meters equals 126.955 Newtons

Explanation:

We know that for two point charges of magnitude [tex]q_{1},q_{2}[/tex] the magnitude of force between them is given by

[tex]F=\frac{k_{e}q_{1}\cdot q_{2}}{r^{2}}[/tex]

where

[tex]k_{e}[/tex] is constant

[tex]r[/tex] is the separation between the charges

Initially when the charges are separated by 2.4 meters the force can be calculated as

[tex]F_{1}=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\10.8=\frac{k_{e}\cdot q_{1}q_{2}}{2.4^{2}}\\\\\therefore k_{e}\cdot q_{1}q_{2}=10.8\times 2.4^{2}=62.208[/tex]

Now when the separation is reduced to 0.7 meters the force is similarly calculated as

[tex]F_{2}=\frac{k_{e}\cdot q_{1}q_{2}}{0.7^{2}}[/tex]

Applying value of the constant we get

[tex]F_{1}=\frac{62.208}{0.7^{2}}[/tex]

Thus [tex]F_{2}=126.955Newtons[/tex]

Answer:

1) Velocity at t =6 =32m/s

2) Position of particle at t = 6 secs = 67 meters

3) Distance covered in 6 seconds equals 72 meters.

Explanation:

By definition of acceleration we have

[tex]a=\frac{dv}{dt}\\\\dv=a(t)dt\\\\\int dv=\int a(t)dt\\\\v(t)=\int (2t-1)dt\\\\v(t)=t^{2}-t+c_{1}[/tex]

Now at t = 0 v = 2 m/s thus the value of constant is obtained as

[tex]2=0-0+c_{1}\\\\\therefore c_{1}=2[/tex]

thus velocity as a function of time is given by

[tex]v(t)=t^{2}-t+2[/tex]

Similarly position can be found by

[tex]x(t)=\int v(t)dt\\\\x(t)=\int (t^{2}-t+2)dt\\\\x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+c_{2}[/tex]

The value of constant can be obtained by noting that at time t = 0 x = 1.

Thus we get

[tex]1=0+0+0+c_{2}\\\\\therefore c_{2}=1[/tex]

thus position as a function of time is given by

[tex]x(t)=\frac{t^{3}}{3}-\frac{t^{2}}{2}+2t+1[/tex]

Thus at  t = 6 seconds we have

[tex]v(6)=6^{2}-6+2=32m/s[/tex]

[tex]x(6)=\frac{6^{3}}{3}-\frac{6^{2}}{2}+2\times 6 +1=67m[/tex]

The path length can be obtained by evaluating the integral

[tex]s=\int_{0}^{6}\sqrt{1+(t^{2}-t+2)^{2}}\cdot dt\\\\s=72meters[/tex]