a 14n force is applied for 0.33 seconds, calculate the impulse

Answer:

joule is the answer

Explanation:

I hope it helps you

Answer:

hello your question has a missing information

The other is Δ-connected with an impedance of (60 - j45) ohm per phase.

answer : A) 5A ∠0° ,

               p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

 B) 193.64 v

C) current at load 1 = 2.236 A , current at load 2 = 4.472 A

 D) Load 1 : 450 watts(real power ) , 600 VAR ( reactive power )

      Load 2 : 1200 watts ( real power ), -900 VAR ( reactive power )

Explanation:

First convert the Δ-connection to Y- connection attached below is the conversion and pre-solution

A) determine the current, real power and reactive power delivered by the sending-end source

current power delivered (Is)  =  5A ∠0°

complex power delivered ( s ) = 3vs Is  

                                                  = 3 * 120∠0° * 5∠0° = 1800 + j0 ---- ( 1 )

also s = p + jQ  ------ ( 2 )

comparing equation 1 and 2

p( real power )  = 1800 and  Q ( reactive power ) = 0 VAR

B) determine Line-to-line voltage at the load

Vload = √3 * 111.8

           = 193.64 v

c) Determine current per phase in each load

[tex]I_{l1} = Vl1 / Zl1[/tex]

     = [tex]\frac{111.8<-10.3}{50<53.13}[/tex] = 2.236∠ 63.43° A   hence current at load 1 = 2.236 A

[tex]I_{l2} = V_{l2}/Z_{l2}[/tex]  

     = [tex]\frac{111.8<-10.3}{25<-36.87}[/tex]  = 4.472 ∠ 26.57° A hence current at load 2 = 4.472 A

D) Determine the Total three-phase real and reactive powers absorbed by each load

For load 1

3-phase real power = [tex]3I_{l1} ^{2} R_{l1}[/tex] = 3 * 2.236^2 * 30 = 450 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l1}[/tex] = 3 * 2.236^2 * 40 = 600 VAR

for load 2

3-phase real power = [tex]3I_{l1} ^{2} R_{l2}[/tex]  = 1200 watts

3-phase reactive power = [tex]3I_{l1} ^{2} X_{l2}[/tex] = -900 VAR

The sum of load powers and line losses, 1800 W+ j0 VAR and The line voltage magnitude at the load terminal is 193.64 V.

(a) The impedance per phase of the equivalent Y,

[tex]\bar{Z}_{2}=\frac{60-j 45}{3}=(20-j 15) \Omega[/tex]

The phase voltage,

[tex]\bold { V_{1}=\frac{120 \sqrt{3}}{\sqrt{3}}=120 VV }[/tex]

Total impedance from the input terminals,

[tex]\bold {\begin{aligned}&\bar{Z}=2+j 4+\frac{(30+j 40)(20-j 15)}{(30+j 40)+(20-j 15)}=2+j 4+22-j 4=24 \Omega \\&\bar{I}=\frac{\bar{V}_{1}}{\bar{Z}}=\frac{120 \angle 0^{\circ}}{24}=5 \angle 0^{\circ} A\end{aligned} }[/tex]

The three-phase complex power supplied  [tex]\bold {=\bar{S}=3 \bar{V}_{1} \bar{I}^{*}=1800 W}[/tex]  

P =1800 W and Q = 0 VAR delivered by the sending-end source.

(b) Phase voltage at load terminals will be,  

[tex]\bold {\begin{aligned}\bar{V}_{2} &=120 \angle 0^{\circ}-(2+j 4)\left(5 \angle 0^{\circ}\right) \\&=110-j 20=111.8 \angle-10.3^{\circ} V\end{aligned} }[/tex]  

The line voltage magnitude at the load terminal,  

[tex]\bold{\left(V_{ LOAD }\right)_{L-L}=\sqrt{3} 111.8=193.64 V(V }[/tex]    

(c) The current per phase in the Y-connected load,  

[tex]\bold {\begin{aligned}&\bar{I}_{1}=\frac{\bar{V}_{2}}{\bar{Z}_{1}}=1-j 2=2.236 \angle-63.4^{\circ} A \\&\bar{I}_{2}=\frac{\bar{V}_{2}}{\bar{Z}_{2}}=4+j 2=4.472 \angle 26.56^{\circ} A\end{aligned} ​}[/tex]

The phase current magnitude,  

[tex]\bold {\left(I_{p h}\right)_{\Delta}=\frac{I_{2}}{\sqrt{3}}=\frac{4.472}{\sqrt{3}}=2.582 }[/tex]

(d) The three-phase complex power absorbed by each load,

[tex]\bold {\begin{aligned}&\bar{S}_{1}=3 \bar{V}_{2} \bar{I}_{1}^{*}=430 W +j 600 VAR \\&\bar{S}_{2}=3 \bar{V}_{2} \bar{I}_{2}^{*}=1200 W -j 900 VAR\end{aligned}}[/tex]

The three-phase complex power absorbed by the line is  

[tex]\bold{\bar{S}_{L}=3\left(R_{L}+j X_{L}\right) I^{2}=3(2+j 4)(5)^{2}=150 W +j 300 VAR }[/tex]

Since, the sum of load powers and line losses,  

[tex]\bold {\begin{aligned}\bar{S}_{1}+\bar{S}_{2}+\bar{S}_{L} &=(450+j 600)+(1200-j 900)+(150+j 300) \\&=1800 W +j 0 VAR\end{aligned} }[/tex]

To know more about voltage,

https://brainly.com/question/2364325

​

Answer:

the correct one is the first,   the refractive index of the two materials must be the same

Explanation:

When a beam of light passes through two materials, it must comply with the law of refraction

         n₁ sin θ₁ = n₂ sin θ₂

where n₁ and n₂ are the refractive indices of each medium.

In this case, it indicates that the light does not change direction, so the input and output angle of the interface must be the same,

       θ₁ = θ₂ = θ

substituting

          n₁ = n₂

therefore the refractive index of the two materials must be the same

When reviewing the answers, the correct one is the first

Answer:

5.8 m/s

Explanation:

Let v = velocity of bike relative to Betty = -12.0 m/s (since the bike is moving away from betty).

u = velocity of ball relative to bike = + 17.8 m/s

and V = velocity of ball relative to Betty.

So, by Galilean relativity,

V = v + u

V = -12.0 m/s + 17.8 m/s

V = 5.8 m/s

So, the velocity of the ball as measured by Betty is 5.8 m/s

Answer:

The answer is below

Explanation:

We are going to use Gauss’ law to find the electric field equation. Since electric field is coming from an infinite line of charge, hence it is going out in a radial direction.  

Therefore we use the area of the electric field which passes through, forming a Gaussian cylinder. We neglect the ends of the area.

Hence:

[tex]\int\limits {E} \, dA=\frac{Q_{enc}}{\epsilon_o}\\\\E(2\pi rL)= \frac{\lambda L}{\epsilon_o}\\\\E=\frac{\lambda}{2\pi r\epsilon_o} \\\\Given \ that:\\\\r=a=8.7\ cm=0.087\ m, \lambda=-2.3 \mu C/cm=-2.3*10^{-4}\ C/m,\epsilon_o=8.85*10^{-12}F/m.\\\\Hence:\\\\E=\frac{-2.3*10^{-4}}{2\pi *0.087*8.85*10^{-12}}=-4.75*10^7\ N/C[/tex]

The value of the x-component of the electric field is -475213.968 newtons per coulomb.

In this question we shall apply Gauss' Law to determine the magnitude of the electric field ([tex]E_{x}[/tex]), in newtons per coulomb, rapidly and based on the assumptions of uniform charge distribution and cylindrical symmetry.

[tex]\frac{Q_{enc}}{\epsilon_{o}} = \oint\,\vec E\,\bullet d\vec A[/tex] (1)

Where:

Based on all assumptions, we simplify (1) as follows:

[tex]\frac{\lambda\cdot l}{\epsilon_{o}} = E \cdot (2\pi\cdot r\cdot l)[/tex]

And the equation of the x-component of the electric field is:

[tex]E = \frac{\lambda}{2\pi\cdot \epsilon_{o}\cdot r}[/tex] (2)

Where [tex]\lambda[/tex] is the linear charge density, in coulomb per meter.

If we know that [tex]\lambda = -2.3\times 10^{-6}\,\frac{C}{m}[/tex] and [tex]a = 0.087\,m[/tex], then the electric field produced by the line of charge at point P is:

[tex]E = \frac{\left(-2.3\times 10^{-6}\,\frac{C}{m} \right)}{2\pi\cdot \left(8.854\times 10^{-12}\,\frac{s^{4}\cdot A^{2}}{kg\cdot m^{3}} \right)\cdot (0.087\,m)}[/tex]

[tex]E_{x} = -475213.968 \,\frac{N}{C}[/tex]

The value of the x-component of the electric field is -475213.968 newtons per coulomb. [tex]\blacksquare[/tex]

The figure is missing, we present the corresponding image in the file attached below.

To learn more on electric fields, we kindly invite to check this verified question: https://brainly.com/question/12757739

Answer:

276.12 pC

Explanation:

We are given that

Area,A=[tex]2.6 cm^2=2.6\times 10^{-4}m^2[/tex]

Where [tex]1 cm^2=10^{-4} m^2[/tex]

[tex]d=0.25 mm=0.25\times 10^{-3} m[/tex]

[tex]1mm=10^{-3} m[/tex]

Potential difference, V=30  V

We have to find the charge on the capacitor when the plates are pulled to a separation of 0.55 mm.

We know that

Charge ,[tex]Q=\frac{\epsilon_0 A V}{d}[/tex]

Where [tex]\epsilon_0=8.85\times 10^{-12}[/tex]

Using the formula

[tex]Q=\frac{8.85\times 10^{-12}\times 2.6\times 10^{-4}\times 30}{0.25\times 10^{-3}}[/tex]

[tex]Q=2.7612\times 10^{-10} C[/tex]

[tex]Q=276.12p C[/tex]

[tex]1 pC=10^{-12} C[/tex]

When the plates are now pulled to a separation of 0.55 mm.Then, the charge on the plates remain same because the  battery has been disconnected.

Therefore, charge on the capacitor=276.12 pC

Answer:

Explanation:

Mechanical weathering types

Mechanical weathering is the breaking down of rocks into smaller pieces without changing the composition of the minerals in the rock. This can be divided into four basic types – abrasion, pressure release, thermal expansion and contraction, and crystal growth.

Answer:

pitch and loudnesss

thanks for your question

Answer:

a) k= 3232.30 N / m,  b)  f = 4,410 Hz

Explanation:

In this exercise, the car + spring system is oscillating in the form of a simple harmonic motion, as the four springs are in parallel, the force is the sum of the 4 Hocke forces.

The expression for the angular velocity is

          w = √k/m

the angular velocity is related to the period

          w = 2π / T

we substitute

          T = 2[tex]\pi[/tex]  √m/ k

a) empty car

           k = 4π² m / T²

           k = 4 π² 1310/2 2

           k = 12929.18 N / m

This is the equivalent constant of the short springs

           F1 + F2 + F3 + F4 = k_eq x

           k x + kx + kx + kx = k_eq x

           k_eq = 4 k

           k = k_eq / 4

           k = 12 929.18 / 4

            k= 3232.30 N / m

b) the frequency of oscillation when carrying four passengers.

In this case the plus is the mass of the vehicle plus the masses of the passengers

            m_total = 1360 + 4 70

            m_total = 1640 kg

angular velocity and frequency are related

              w = 2pi f

we substitute

             2 pi f = Ra K / m

in this case the spring constant changes us

             k_eq = 12929.18 N / m

             f = 1 / 2π √ 12929.18 / 1640

             f = π / 2 2.80778

             f = 4,410 Hz

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration ([tex]a[/tex]) can be defined by this ordinary differential equation in terms of distance:

[tex]a = v\cdot \frac{dv}{ds}[/tex] (1)

Where:

[tex]v[/tex] - Speed of the automobile, measured in meters per second.

[tex]s[/tex] - Distance travelled by the automobile, measured in meters.

If we know that [tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex], then the equation of acceleration is:

[tex]a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)[/tex]

[tex]a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)[/tex]

[tex]a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)[/tex]

But distance covered by the vehicle is defined by the following formula:

[tex]s = \theta \cdot r[/tex] (2)

Where:

[tex]\theta[/tex] - Arc angle, measured in radians.

[tex]r[/tex] - Radius, measured in radians.

Then, we expand (1) by means of this result:

[tex]a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)[/tex]

[tex]a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)[/tex]

[tex]a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r[/tex]

And finally we get the following third order polynomial:

[tex]1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0[/tex] (3)

If we know that [tex]r = 50\,m[/tex], [tex]a = 0.6\cdot g[/tex] and [tex]g = 9.807\,\frac{m}{s^{2}}[/tex], then the polynomial becomes into this:

[tex]1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0[/tex] (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

[tex]\theta_{1} \approx 4.365\,rad[/tex], [tex]\theta_{2} \approx 0.818+i\,0.441\,rad[/tex], [tex]\theta_{3}\approx 0.818 -i\,0.441\,rad[/tex], [tex]\theta_{4} \approx 1.563\,rad[/tex]

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid first. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: ([tex]r = 50\,m[/tex], [tex]\theta \approx 1.563\,rad[/tex])

[tex]s = (1.563\,rad)\cdot (50\,m)[/tex]

[tex]s = 78.15\,m[/tex]

Lastly, the speed of the automobile at this location is: ([tex]s = 78.15\,m[/tex])

[tex]v = 0.5\cdot s - 0.0025\cdot s^{2}[/tex] (4)

[tex]v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}[/tex]

[tex]v = 23.806\,\frac{m}{s}[/tex]

The automobile is running at speed of 23.806 meters per second.

Answer:

If she bends forward

Explanation:

because the equilibrium of gravity will not stay the same causing her to move forward

Answer:

While Bohr's atomic model hypothesizes that electrons move in particular energy levels around the nucleus, the electron cloud model suggests that electrons move in an unpredictable pattern but are more likely to be in certain regions than others.

Explanation:

Answer:

Chemical energy

Explanation:

The energy in the torch is stored as chemical energy before the torch gets switch on.

The chemical energy energy in the battery of cell will power the cell and allows it to produce light.

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

Answer:

B. 0.2 kg x m/s

Explanation:

Answer: The correct statement is that the Last Quarter Moon (rises near midnight and sets near midday).

Explanation:

Phases of the moon also called the LUNAR PHASE can be defined as the different shades of illumination on the moon as seen from the earth. The moon is the natural satellite of the earth that illuminates upon reflection of light from the sun. This means it doesn't have power to shine on its own. When carefully observed, there are times it gets dark and beings to glow brighter over a period of time. This occurs because as the moon completes its four weeks lunar cycle round the earth, how much of its face we see illuminated by sunlight depends on the angle the Sun makes with the Moon.

There are 8 main types of the moon phases these includes:

--> New moon: This is when the moon is not visible to the earth because it's between the earth and the sun. It rises at sunrise and sets at sunset.

--> The waxing crescent: At this phase the moon gets brighter and illuminated from the sun that a crescent shape is seen.

--> First quarter: this occurs one week after the new moon. The moon rises at noon and sets at midnight.

--> The waxing gibbous: This occurs after the first quarter phase where more than half of the lit part of the moon is seen.

--> Full moon: This occurs when the moon and the sun are opposite each other. That is way it is said to rise at sunset and sets at sun rise.

--> The waning gibbous: this occurs when more than half of the lit part of the moon gradually becomes darker

--> Third quarter ( Last Quarter): The moon rises at midnight and sets at noon. This occurs a week after the full moon.

--> The waning crescent: This occurs after the last quarter phase where a very thin fading crescent shaped moon is seen, just before the Moon is invisible again at the start of the cycle, the new moon.

Answer:

the diameter of the bright circular spot formed is 0.787 m  

Explanation:

Given that;

Radius of the flat circular mirror = 0.100 m

height of small ight source = 0.920 m

high ceiling = 2.70 m  

now;

Diameter(mirror) = 2×r = 2 × 0.100 = 0.2 m

D(spot) = [Diameter(mirror) × ( 2.70m + 0.920 m)] /  0.920 m

so

D(spot) = 0.2m × 3.62m /  0.920 m

D(spot) = 0.724 m / 0.920 m

D(spot) = 0.787 m  

Therefore, the diameter of the bright circular spot formed is 0.787 m  

Answer:  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface; [tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature; [tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e; [tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

Explanation:  

Given the data in question;

we write the general form of the heat conduction equation equation in cylindrical coordinates with internal heat generation.

1/r[tex]\frac{d}{dr}[/tex]( kr[tex]\frac{dT}{dr}[/tex] ) + 1/r² [tex]\frac{d}{d\beta }[/tex](  ( k[tex]\frac{dT}{dr}[/tex] ) + [tex]\frac{d}{dz}[/tex]( k[tex]\frac{dT}{dr}[/tex]) + q = 0

where radius of cylinder is r, thermal conductivity of the cylinder is k, and q is heat generated in cylinder.

Now, Assume one dimensional heat conduction

lets substitute the condition for conducting rod with steady state condition.

[tex]k_{y}[/tex]/r [tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{y} }{dr}[/tex] ) + q = 0

Apply the conditions for cladding by substituting 0 for q

[tex]\frac{d}{dr}[/tex]( r[tex]\frac{dT_{r} }{dr}[/tex] ) = 0

Apply the following boundary conditions;  

a) For radial heat transfer to be zero along the perfectly insulated adiabatic surface;

[tex]\frac{dT_{y} }{dr}[/tex][tex]|_{r-0}[/tex] = 0

b) For constant temperature

[tex]T_{y}[/tex]([tex]r_{i}[/tex]) = [tex]T_{C}[/tex]([tex]r_{i}[/tex])

c) The heat transfer in the conducting rod and the cladding material is the same, i.e

[tex]k_{r}[/tex][tex]\frac{dT_{y} }{dr}[/tex] [tex]|_{ri}[/tex] =  [tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{ri}[/tex]  

d) The convection surface conduction by cooling fluid will be;

[tex]k_{c}[/tex][tex]\frac{dT_{c} }{dr}[/tex] [tex]|_{r0}[/tex] = h( [tex]T_{c}[/tex]( [tex]r_{0}[/tex] ) - [tex]T_{\infty}[/tex] )

Answer:

The new force becomes 4 times the initial force.

Explanation:

The force of attraction or repulsion is given by the relation as follows :

[tex]F=k\dfrac{q_1q_2}{d^2}[/tex]

Where

d is the distance between the interacting charges

F is inversely proportional to the distance between charges.

If the distance is halved, d'=(d/2), new force is given by :

[tex]F'=k\dfrac{q_1q_2}{d'^2}\\\\=k\dfrac{q_1q_2}{(\dfrac{d}{2})^2}\\\\=k\dfrac{q_1q_2}{\dfrac{d^2}{4}}\\\\=4\times \dfrac{kq_1q_2}{d^2}\\\\F'=4F[/tex]

So, the new force becomes 4 times the initial force.

Answer:

the second time there is a gas between you and the star,

Explanation:

When you observe the star for the first time you do not have a given between you and the star, therefore you observe the emission spectrum of the same that is formed by lines of different intensity and position that indicate the type and percentage of the atoms that make up the star.

 When you observe the same phenomenon for the second time there is a gas between you and the star, this gas absorbs the wavelengths of the star that has the same energies and the atomisms and molecular gas, therefore these lines are not observed by seeing a series of dark bands,

The information obtained from the two spectra is the same, the type of atoms that make up the star

Answer:

i dont know i am right but here Luminous intensity is defined as dI=dΨλ / dΩ, where dΨλ is the luminous flux (light energy flux in watts per m2) emitted within a solid angle dΩ. The light energy flux may be expressed in terms of the incident x-ray energy flux and the x-ray absorption and conversion properties of the scintillator(7,8,9).

Explanation:

Answer:

220 km

Explanation:

3 hours at 40 km/hr

40 km = 1 hour

40 km times 3 = 120 km

2 hours at 50 km/hr

50 km = 1 hour

50 times 2 = 100 km

Total distance

120 km + 100 km = 220 km

Answer:

LAW 1 :  For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation.  

---------------------------------------------

LAW 2: For a given metal, there exists a certain frequency below which the photoelectric emission does not take place. This frequency is called threshold frequency.

-----------------------------------------------

LAW 3: For a frequency greater than the threshold frequency, the kinetic energy of photoelectrons is dependent upon frequency or wavelength but not on the intensity of light.

-----------------------------------------------

LAW 4: Photoelectric emission is an instantaneous process. The time lag between incidence of radiations and emission of electron is 10^-9 seconds.

Explanation:

Answer:

LAW 1 : For a given metal and frequency, the number of photoelectrons emitted is directly proportional to the intensity of the incident radiation. ... LAW 4: Photoelectric emission is an instantaneous process.

To Find :

Which is larger: 65 mph (miles per hour) or 120 kph (kilometers per hour).

Solution :

We know, 1 mph = 1.61 kph

So, 65 mph = 1.61 × 65 kph

65 mph = 104.65 kph

Since, 65 mph is 104.65 kph which is smaller than 120 kph.

Therefore, 120 kph is faster than 65 mph by ( 120 - 104.65 ) = 15.35 kph.

Answer:

a) if the two trains are going to collide

b) x = 538 m

Explanation:

To solve this exercise we use the kinematics relations in one dimension.

We set a reference system at the starting point of the passenger train

Freight train, we use index 1 for this train

          x₁ = x₀ + v₁ t

passenger train

          x₂ = v₂ t - ½ a₂ t²

as we are using the same system to measure the position of the two trains, at the meeting point the position must be the same

           x₁ = x₂

we substitute

          x₀ + v₁ t = v₂ t - ½ a₂ t²

         ½ a₂ t² + t (v₁ -v₂) + x₀ = 0

we subjugate the values

         ½ 0.1 t² + t (15-25) + 200 = 0

         0.05 t² - 10 t + 200 = 0

          t² - 200 t + 4000 = 0

we solve the second degree system

         t = [200 ±[tex]\sqrt{200^2 - 4 \ 4000}[/tex] ]/ 2

         t = [200 ± 154.9] / 2

         t₁ = 177.45 s

         t₂ = 22.55 s

therefore for the smallest time  the two trains must meet  t₂ = 22.55 s

merchandise train

            x = 200+ 15  22.55

            x = 538 m

passenger train

           x = 25 22.55 -1/2  0.100  22.55²

           x = 538 m

we see that the trains meet for this distance

a) if the two trains are going to collide

b) x = 538 m

c) see attachment for schematic graphics

Answer and Explanation: A charge exerts a force over another charge even if they are very far apart. This force is called Electrostatic Force.

If the two charges have the same sign, e.g. both aare positive, the force between them is opposite. If they have opposite sign, the force is towards each other. In other words, for electrostatic force, equal charges repel and different charges attract.

So,

1. If Q2 and Q3 have opposite signs, it is TRUE force in Q2 will go the left;

2. If the 2 are negative, they have the same sign, so it's FALSE force is to the right;

Sentences 3 and 4 are also TRUE due to the reasons described above;

5. If the charges have opposite signs, it means force is towards each other, or, to the right, so the sentence is TRUE;

1. Force is directly proportional to charges in Coulomb [C] and inversely proportional to distance squared in [m]:

[tex]F=\frac{k.q.Q}{r^{2}}[/tex]

where k is a constant that equals 9 x 10⁹ N.m²/C²

Calculating force between 1 and 2:

[tex]F_{12}=\frac{9.10^{9}(1.9.10^{-6})(2.84.10^{-6})}{(0.301)^{2}}[/tex]

[tex]F_{12}=536.02.10^{-3}[/tex] N

Force between 2 and 3:

[tex]F_{23}=\frac{9.10^{9}(2.84.10^{-6})(3.03.10^{-6})}{(0.169)^{2}}[/tex]

[tex]F_{23}=2711.63.10^{-3}[/tex] N

Total force is the net force. Since Q2 is negative and the others are positive, force of 2 related to 1 is to left and related to 3 is to the right. Therefore, total force is the difference between those two forces:

[tex]F_{T}=2711.63.10^{-3}-536.02.10^{-3}[/tex]

[tex]F_{T}=2175.61.10^{-3}[/tex] N

The total force on Q2 is 2175.61 x 10⁻³ N

2. For net force to be 0, [tex]F_{13}=F_{23}[/tex]. Suppose distance from 1 to 3 is x, then from 2 to 3 is [tex]x-0.301[/tex]

Calculating:

[tex]\frac{k(1.90.10^{-6})(3.03.10^{-6})}{x^{2}}=\frac{k(2.84.10^{-6})(3.03.10^{-6})}{(x-0.301)^{2}}[/tex]

[tex]\frac{5.757.10^{-12}}{x^{2}} =\frac{8.6052.10^{-12}}{x^{2}-0.602x+0.090601}[/tex]

[tex]\frac{5.757.10^{-12}}{8.6052.10^{-12}}=\frac{x^{2}}{x^{2}-0.602x+0.090601}[/tex]

[tex]x^{2}=0.67x^{2}-0.40x+0.061[/tex]

[tex]0.33x^{2}+0.40x-0.061=0[/tex]

roots = 0.14 or -1.35

Solving quadratic equation gives 2 roots, but one of the roots is negative. As distance is a measure that cannot be negative, the solution is x = 0.14.

The distance of Q3 relative to Q1 is 0.14 m

Answer:

True.

Explanation:

Welding requires extensive training because welding involves fire and we need to use fire safety measurements. A normal man can't just simply go and weld so a person must require extensive training for welding.

Answer:

1/9

Explanation:

Let A denote the bigger piece and let B denote the smaller piece.

We are told that one with three times the mass of the other.

Therefore, we have;

M_a = 3M_b

Firecracker is placed in the block and it explodes and thus, momentum is conserved.

Thus;

V_ai = V_bi = 0

Where V_ai is initial velocity of piece A and V_bi is initial velocity of piece B.

Since initial momentum equals final momentum, we have;

P_i = P_f

Thus;

0 = (M_a × V_af) + (M_b × V_bf)

Since M_a = 3M_b, we have;

(3M_b × V_af) + (M_b × Vbf) = 0

Making V_af the subject, we have;

V_af = -⅓V_bf

The kinetic energy gained by each block during the explosion will later be lost due to the negative work done by friction. Thus;

W_f = -½M_b•(v_bf)²

Now, let's express the work is in terms of the force and the distance.

Thus;

W_f = F_f × Δx × cos 180°

Frictional force is also expressed as μmg

Thus;

W_f = -μM_b × g × Δx

Earlier, we saw that;

W_f = -½M_b•(v_bf)²

Thus;

-½M_b•(v_bf)²= -μM_b × g × Δx

Δx = (v_bf)²/2μg

Let the distance travelled by block A be Δx_a and that travelled by B be Δx_b

Thus;

Δx_a/Δx_b = ((v_ba)²/2μg)/((v_bf)²/2μg)

Δx_a/Δx_b = ((v_af)²/((v_bf)²)

Δx_a/Δx_b = (-⅓V_bf)²/(V_bf)²

Δx_a/Δx_b = 1/9