An electric dipole consists of a positive and a negative charge of equal magnitude. Consider an electric dipole with each charge having a magnitude of 1 × 10−6 C. The negative charge is located at (3 cm, 0) and the positive charge is located at (−3 cm, 0). Calculate the electric field from each charge at the points A through E, described below. Use symmetry as much as possible! Using the scale 1 cm = 105 N/C, draw the vector to represent the magnitude and direction of the electric field from each charge. (When entering angle values, enter a number greater than or equal to 0° and less than 360° measured counterclockwise from the +x-axis.) • A = (−13 cm, 0) • B = (−3 cm, 10 cm) • C = (0, 10 cm ) • D = (3 cm, 10 cm) • E = (13 cm, 0) For the negative charge:

Answer:

Explanation:

given equation

aA + bB + C = 0

Putting the given values

a(6i-8j) +b (-8i-3j) +(26i-19j) = 0

i ( 6a - 8b ) - j ( 8a + 3 b ) = - 26 i + 19 j

comparing the coefficients of i and j

6a - 8b = -26

8a + 3b = -19.

multiplying first equation by 4 and second equation by 3  

24a - 32 b = - 104

24a + 9b = -57

9b + 32b = -57 + 104

41 b = 47

b = 1.41

6 a - 8 x 1.41 = -26

6a = -14.72

a = - 2.45  

Answer:

So the induced current opposes the motion that induced it (from Lenz's Law). When we pull the magnet out, the left hand end of the coil becomes a south pole (to try and hold the magnet back). Therefore the induced current must be flowing clockwise.

hope this helps u...

Answer:

option C is correct

................

Answer:

C- in an inelastic collision, both momentum & kinetic energy are conserved

Explanation:

Took the test

Answer:

Explanation:

To convert gram / centimeter³ to kg / m³

gram / centimeter³

= 10⁻³ kg / centimeter³

= 10⁻³  / (10⁻²)³ kg / m³

= 10⁻³ / 10⁻⁶ kg / m³

= 10⁻³⁺⁶ kg / m³

= 10³ kg / m³

So we shall have to multiply be 10³ with amount in gm / cm³ to convert it into kg/m³

2.33 gram / cm³

= 2.33 x 10³ kg / m³ .

Answer:

a)  μ = 0.475 , b)   μ = 0.433

Explanation:

a) For this exercise of Newton's second law, we create a reference system with the x-axis parallel to the plane and the y-axis perpendicular to it

X axis

     Wₓ - fr = m a

the friction force has the expression

     fr = μ N

y Axis

     N - [tex]W_{y}[/tex] = 0

let's use trigonometry for the components the weight

     sin 27 = Wₓ / W

     Wₓ = W sin 27

     cos 27 = W_{y} / W

     W_{y} = W cos 27

     N = W cos 27

     W sin 27 - μ W cos 27 = m a

     mg sin 27 - μ mg cos 27 = m a

      μ = (g sin 27 - a) / (g cos 27)

      very = tan 27 - a / g sec 27

      μ = 0.510 - 0.0344

      μ = 0.475

b) now the block starts with an initial speed of 3m / s. In Newton's second law velocity does not appear, so this term does not affect the result, the change in slope does affect the result

         μ = tan 25 - 0.3 / 9.8 sec 25

         μ = 0.466 -0.03378

         μ = 0.433

Answer:B

Explanation:

Power=1000 watts

Time=5 seconds

Distance=5 meters

Force=(power x time) ➗ distance

Force=(1000 x 5) ➗ 5

Force=5000 ➗ 5

Force=1000

Force=1000N

Answer:1,000

Explanation:

ape.x

Answer:

Here's your answer :

hope it helps!

Answer:

a) 1.248 rad/s

b) 13.728 m/s

c) 0.52 rad/s^2

d) 17.132m/s^2

Explanation:

You have that the angles described by a astronaut is given by:

[tex]\theta=0.260t^2[/tex]

(a) To find the angular velocity of the astronaut you use the derivative og the angle respect to time:

[tex]\omega=\frac{d\theta}{dt}=\frac{d}{dt}[0.260t^2]=0.52t[/tex]

Then, you evaluate for t=2.40 s:

[tex]\omega=0.52(2.40)=1.248\frac{rad}{s}[/tex]

(b) The linear velocity is calculated by using the following formula:

[tex]v=\omega r[/tex]

r: radius if the trajectory of the astronaut = 11.0m

You replace r and w and obtain:

[tex]v=(1.248\frac{rad}{s})(11.0m)=13.728\frac{m}{s}[/tex]

(c) The tangential acceleration is:

[tex]a_T=\alpha r\\\\\alpha=\frac{\omega^2}{2\theta}=\frac{(1.248rad/s)^2}{2(0.260(2.40s)^2)}=0.52\frac{rad}{s^2}[/tex]

(d) The radial acceleration is:

[tex]a_r=\frac{v^2}{r}=\frac{(13.728m/s)^2}{11.0m}=17.132\frac{m}{s^2}[/tex]

Answer:

Q= NBA/R

Explanation:

Check attachment for derivation

The equation relating the total charge, magnitude, turns, time will be "[tex]\frac{NBA}{R}[/tex]".

According to the question,

Resistance = R

Total charge = Q

Current = I

Number of turns = N

Time = Δt

and,

Q = IΔt ...(equation 1)

We know the flux,

→ [tex]\Phi[/tex] = NBA

Emf induced,

   ε = [tex]\frac{- \Delta \Phi}{\Delta t}[/tex]

Δ[tex]\Phi[/tex] = [tex]\Phi_2 - \Phi_1[/tex]

then,

   ε = [tex]\frac{NBA}{\Delta t}[/tex]

As we know, Voltage (V) = iR

then, ε = [tex]\frac{NBA}{\Delta t}[/tex] = iR

         i = [tex]\frac{NBA}{R \Delta t}[/tex]

Hence, by applying the values in "equation 1"    

→ Q = iΔt

      = [tex]\frac{NBA}{R \Delta t}[/tex] × Δt

      = [tex]\frac{NBA}{R}[/tex]

Thus the response above is correct.

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Answer:

Explanation:

In magnetic field , charged particle will have circular path . Let the radius of their circular path be r₁ and r₂ . Let their velocity at the time of entering magnetic field be v₁ and v₂ .

The velocity with which they will come out of electric field can be measured from following equation

Eq = 1/2 m v²  , E is electric field , q is charge on the particle , m is mass and v is velocity .

v² = 2Eq / m

radius of circular path can be measured by the following expression

m v² / r = Bqv

2Eq / r = Bqv

r = 2Eq / Bqv

= 2E / Bv

r² = 4E² / B²v²

= 4E²m / B²x 2Eq

since E , B and q are constant

r² = K . m

r₂² / r₁² = m₂ / m₁

1.5²

m₂ / m₁ = 1.5²

= 2.25

Answer:

Gold, silver, platinum. Gold is the most malleable and ductile.

Explanation:

The elements which are malleable and ductile include the following:

Malleability is the ability of a substance to be hammered into thin sheets

while ductility involves the deformation of a substance without any

breakage occurring in it.

Transition metals are the group of elements which have both

characteristics and examples are listed above.

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Answer:

Explanation:

Given that,

h(t) = -9.8t² / 2 + 125t + 500

for t > 0.

At t = 0, the rocket is at height h = 500m, at a velocity of Vo = 125m/s.

We want to find the maximum height reached by rocket

Using mathematics maxima and minima

let find the turning point when dh/dt = 0

dh/dt = -9.8t + 125

dh / dt = 0 = -9.8t + 125

9.8t = 125

t = 125 / 9.8

t = 12.76s

Let find the turning point to know if this time t = 12.76 is maximum or minimum point

Let find d²h / dt²

d²h / dt² = -9.8

Since, d²h/dt² < 0, then, at t = 12.76s is the maximum points.

Then, the maximum height reached is

h =  -9.8t² / 2 + 125t + 500

h =  -9.8(12.76)² / 2 + 125(12.76) + 500

h = -797.80 + 1595 + 500

h = 1297.2 m

The maximum height reached is 1297.2 m

From the attachment, the maximum height is 1297.2m at t = 12.76sec.

Comment, the result are the same for both the analysis aspect and the graphical aspect.

Answer:

a) Velocity of the block of mass 3.6 kg after collision = 1.13 m/s

Velocity of the block of mass 2.1 kg after collision = 5.43 m/s

b) Initial energy of the 3.6 kg block = 33.282 J

Final energy of the 3.6 kg block = 2.3 J

Initial energy of the 2.1 kg block = 0J

Final energy of the 2.1 kg block = 30.96 J

The two total kinetic energies are the same = 33.30 J

Explanation:

Check the attached files for the complete solution and explanations.

Answer:30

Explanation:

Current=60 milliamps

Current=(voltage)/(resistance)

60=(voltage)/(resistance)

Doubling the resistance means multiplying both sides by 1/2

60x1/2=(voltage)/(resistance) x 1/2

30=(voltage)/2(resistance)

Therefore the resistance would be 30 milliamp if we double the resistance

Answer:

i) Mr. Dunn arrives to home first.

ii) 3 min

Explanation:

i. To find who arrives first to home you calculate the time, by using the following formula:

[tex]t=\frac{x}{v}[/tex]

x: distance

v: velocity

Mr. Dunn:

[tex]t=\frac{64.8km}{48km/h}=1.35h[/tex]

Mrs. Dunn:

[tex]t=\frac{81.2km}{58km/h}=1.4h[/tex]

Hence, Mr. Dunn arrives to home first.

ii. To calculate the difference in minutes, you convert hours to minutes:

[tex]1.35h*\frac{60min}{1h}=81min\\\\1.40h*\frac{60min}{1h}=84min\\\\\Delta\ t=(84-81)min=3min[/tex]

the difference between the times is 3min

(i) Mr. Dunn takes less time so he arrives at home first.

(ii) The second person arrives 3 min late.

(i) We have to calculate the time taken to reach home by Mr. Dunn and Mrs. Dunn.

t = x/v

where x is the distance

and v is the velocity

Time taken by Mr. Dunn:

distance x = 64.8 km

speed v  = 48 km/h

t = 64.8 / 48

t = 1.35 h

Time taken by Mrs. Dunn:

distance x = 81.2 km

speed v  = 58 km/h

t' = 81.2 / 58

t' = 1.4 h

Hence, Mr. Dunn arrives at home first.

(ii) To calculate the difference in minutes, you convert hours to minutes:

The time taken by Mr. Dunn in minutes is:

t = 1.35×60 = 81 minutes

The time taken by Mrs. Dunn in minutes is:

t' = 1.4×60 = 84 minutes

the difference between the times is 3min

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Answer:

Explanation:

The two pictures attached here shows the solution to the two questions from the problem. thank you and I hope it helps you

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

We can assume that the radar wave penetrates the layer and is reflected in the inner part of it, giving rise to an interference phenomenon of the two reflected rays, we must be careful that the ray has a phase change when

* the wave passes from the air to the film with a higher refractive index

* the wavelength inside the film changes by the refractive index

         λ = λ₀ / n

so the ratio for destructive interference is

            2 n t = m λ

            t = m λ / 2n

indicate that the wavelength λ = 2 cm, suppose that the interference occurs for m = 1, therefore it is thickness

            t = 1 2/2 n

            t = 1 / n

where n is the index of refraction of the anti-reflective layer. As they tell us not to take into account the change in wavelength when penetrating the film n = 1

            t = 1 cm

So the thickness of the film for destructive interference is 1 cm

Answer:

the thickness of the film for destructive interference is 1 cm

Explanation:

Answer:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element. Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.Explanation:

An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

Every solid, liquid, gas, and plasma is composed of neutral or ionized atoms. Atoms are extremely small; typical sizes are around 100 picometers.

Each atom is made up of a nucleus and one or more electrons that are linked to it. One or more protons and a significant number of neutrons make up the nucleus. Only the most prevalent type of hydrogen is neutron-free.

Atoms that are neutral or ionized make up every solid, liquid, gas, and form of plasma. Atoms are incredibly tiny, measuring typically 100 picometers across. The nucleus of an atom contains more than 99.94% of its mass.

Therefore, An atom is the smallest constituent unit of ordinary matter that constitutes a chemical element.

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Answer:

[tex]W=5.16 J[/tex]  

Explanation:

Using the Hooke's law we can find the elasticity constant:

[tex]F=-k\Delta x[/tex]

[tex]30.8=-k*0.177[/tex]

[tex]k=|-\frac{30.8}{0.177}|[/tex]

[tex]k=174 N/m[/tex]

Now, we know that the work done is equal to the elastic energy, so we will have:

[tex]W=\frac{1}{2}k(x_{2}^{2}-x_{1}^{2})[/tex]

x2 is the final distance (x2 = 0.177+0.124 = 0.301 m)

x1 is the initial distance (x1 = 0.177 m)

[tex]W=\frac{1}{2}*174(0.301^{2}-0.177^{2})[/tex]

[tex]W=5.16 J[/tex]    

I hope it helps you!

Answer:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Explanation:

To know what is the answer you first take into account the magnetic field generated by each current, for a distance of d:

[tex]B_1=\frac{\mu_oI_1}{2\pi d}=\frac{\mu_o}{2\pi d}(1A)\\\\B_2=\frac{\mu_oI_2}{2\pi d}=\frac{\mu_o}{2\pi d}(2A)=2B_1\\\\[/tex]

Next, you use the formula for the magnetic force produced by the wires:

[tex]\vec{F_B}=I\vec{L}\ X \vec{B}[/tex]

if the direction of the L vector is in +k direction, the first wire produced a magnetic field with direction +y, that is, +j and the second wire produced magnetic field with direction -y, that is, -j (this because the direction of the magnetic field is obtained by suing the right hand rule). Hence, the direction of the magnetic force on each wire, produced by the other one is:

[tex]\vec{F_{B1}}=I_1L\hat{k}\ X\ B_2(-\hat{j})=I_1LB_2\hat{i}=(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}\\\\\vec{F_{B2}}=I_2L\hat{k}\ X\ B_2(\hat{j})=I_2LB_1\hat{i}=-(2A^2)\frac{L\mu_o}{2\pi d}\hat{i}[/tex]

Hence, due to this result you have that:

C. The left wire attracts the right wire and exerts as much force as the right wire does.

Answer:Convex lens spectacles is required for reading purpose..

Explanation:

I don't say you have to mark my ans as brainliest but if it has really helped you please don't forget to thank me...

Answer:

Explanation:

The package leaves the chute with a speed of 3 m/s, and lands in the cart after falling for 0.75 seconds . During .75 second duration . package undergoes free fall due to which additional vertical velocity is added

velocity added = a x t

= 9.8 x .75

= 7.35 m /s

Total vertical velocity

= 3 sin27 + 7.35

= 8.71 m /s

Horizontal component = 3 cos 27

= 2.67 m /s

If v be the resultant velocity of these components

v² = 2.67² + 8.71²

v² = 7.13 + 75.86

v = 9.11 m /s .

Answer:

1.47*10^{-8}s

Explanation:

You first calculate the distance traveled by the electron:

[tex]x=vt\\\\x=(0.941(3*10^8m/s))(15.7*10^{-9}s)=4.43m[/tex]

Next, you calculate the relative speed as measure by an observer in the positron, of the electron:

[tex]u'=\frac{u+v}{1+\frac{uv}{c^2}}\\\\u'=\frac{0.941c+0.941c}{1+\frac{(0.941)^2c^2}{c^2}}\\\\u'=0.99c[/tex]

with this relative velocity you calculate the time:

[tex]t=\frac{x}{u'}\\\\t=\frac{4.43m}{0.99c}=1.47*10^{-8}s[/tex]

Answer:

.7934[tex]m/s^{2}[/tex]

Explanation:

Acceleration = change in velocity / change in time

A = 10.98[tex]m/s[/tex] / 13.84[tex]s[/tex]

A = .7934[tex]m/s^{2}[/tex]

Answer:0.8 m/s^2

Explanation:

initial velocity(u)=18.54m/s

Final velocity(v)=29.52m/s

Time(t)=13.84 sec

Acceleration =(v-u)/t

acceleration =(29.52-18.54)/13.84

Acceleration =10.98/13.34

Acceleration=0.8 m/s^2

Answer:

F = (913.14 , 274.87 )

|F| = 953.61 direction 16.71°

Explanation:

To calculate the resultant force you take into account both x and y component of the implied forces:

[tex]\Sigma F_x=480N+513Ncos(32.4\°)=913.14N\\\\\Sigma F_y=513sin(32.4\°)=274.87N[/tex]

Thus, the net force over the body is:

[tex]F=(913.14N)\hat{i}+(274.87N)\hat{j}[/tex]

Next, you calculate the magnitude of the force:

[tex]F=\sqrt{(913.14N)+(274.87N)^2}=953.61N[/tex]

and the direction is:

[tex]\theta=tan^{-1}(\frac{274.14N}{913.14N})=16.71\°[/tex]

Answer:

Explanation:

Initial volume of gas V₁ = 9.30 x 10⁻⁴ m³

final volume V₂ = 1 / 6 x  9.30 x 10⁻⁴

= 1.55 x 10⁻⁴ m³

Atmospheric pressure P = 1.013 x 10⁵ Pa .

temperature T .

PV = n RT

nRT = 1.013 x 10⁵ x 9.3 x 10⁻⁴

= 94.21

work done in isothermal process

= 2.303 nRT log V₁ / V₂

= 2.303 x 94.21 log 6

= 168.83 J .

Answer:

A i. E = 9.62 × 10⁻⁷ J/s

ii. The absorbed dose is 4.81 × 10⁻⁶ Gy

iii. The equivalent dose is  3.37 × 10⁻⁴ rem/s

iv.  t = 593471.81 seconds

B. i. 4.025 × 10¹⁵/s

ii. 0.512 mW

C. 7218092.2 seconds

D. i. 6.3 × 10⁻¹ J

ii. 1.4 × 10⁻² W

iii. 1.57 × 10³ Curie

E. 0.129 Ω

Explanation:

The given parameters are;

Mass of tumor = 0.20 kg

Activity of Cobalt-60 = 2.60 × 10⁻⁴ Ci

Photon energy = 1.25 MeV

(i) The energy, E, delivered to the tumor is given by the relation;

[tex]E = \frac{1}{2}\left (Number \, of \, decay / seconds \right )\times \left (Energy \, of \, photon \right )[/tex]

[tex]E = \frac{1}{2}\left (2.6\times 10^{-4}Ci )\times \left (\frac{3.70\times 10^{10}decays/s}{1 Ci} \right )\times 1.25\times 10^{6}eV\times \frac{1.6\times 10^{-19}J}{1eV}[/tex]

E = 9.62 × 10⁻⁷ J/s

(ii) The equation for absorbed dose is given as follows;

Absorbed dose, D, in Grays Gy = (Energy Absorbed Joules J)/Mass kg

Therefore, absorbed dose = (9.62 × 10⁻⁷ J/s)/( kg) = 4.81 × 10⁻⁶ Gy

1 Gray = 100 rad

4.81 × 10⁻⁷ Gy = 100 × 4.81 × 10⁻⁶ = 4.81 × 10⁻⁴ rad/s

(iii) Equivalent dose, H, is  given by the relation;

H = D × Radiation factor, [tex]w_R[/tex]

∴ H = 0.7 × 4.81 × 10⁻⁴ rad/s = 3.37 × 10⁻⁴ Sv = 3.37 × 10⁻⁴ rem/s

(iv) The exposure time required for an equivalent dose of 200 rem is given as follows;

[tex]\dot{H} = \dfrac{H}{t}[/tex]

Therefore;

[tex]t= \dfrac{200}{{3.37 \times 10^{-4}} } = 593471.81 \, s[/tex]

∴ t = 6.9 days

B. The number of electrons ejected is given by the relation;

[tex]N = \frac{P}{E} = \frac{P \times \lambda}{hc}[/tex]

[tex]N = \dfrac{2.0 \times 10^{-3} \times 400 \times 10^{-9}}{6.626 \times 10^{-34} \times 3 \times 10^8} = 4.025 \times 10^{15}/s[/tex]

(ii) The power carried by the electron

The energy carried away by the electrons is given by the relation;

[tex]KE_e = hv - \Phi[/tex]

[tex]KE_e = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^{-9}} - 2.31 \times \frac{1.6 \times 10 ^{-19} }{1}[/tex]

[tex]KE_e = 4.9695 \times 10^{-19} - 3.696 \times 10 ^{-19} = 1.2735 \times 10^{-19} J[/tex]

Power, P[tex]_e[/tex], carried away by the electron = 4.025 × 10¹⁵ × 1.2735 × 10⁻¹⁹ = 0.512 mW

C. The given parameters are;

d = 1.19 mm, ∴ r = 1.19/2 = 0.595 × 10⁻³ m

l = 50 mm = 5 × 10⁻³ m

V = 500 ml = 5 × 10⁻⁴ m³

η = 0.0027 Pa

p = 1,900 Pa.

[tex]\dfrac{V}{t} = \dfrac{\pi }{8} \times \dfrac{P/l}{\eta } \times r^4[/tex]

[tex]t = \dfrac{8\times \eta\times V\times l }{\pi \times P \times r^4}[/tex]

[tex]t = \dfrac{8\times 0.0027 \times 5 \times 10^{-4} \times 5 \times 10^{-2} }{\pi \times 1900 \times (0.595 \times 10^{-4} )^4}[/tex]

t = 7218092.2 seconds

D) i. Energy absorbed is given by the relation;

E = m×D

Where:

D = 35 Gray = 35 J/kg

m = 18 g = 18 × 10⁻³ kg

∴ E = 35 × 18 × 10⁻³ = 6.3 × 10⁻¹ J

ii. Total time for treatment = 15 × 5 = 75 minutes

Energy absorbed = 6.3 × 10⁻¹ × 100 = 63 J

Power = Energy(in Joules)/Time (in seconds)

∴ Power = 63/(75×60) = 1.4 × 10⁻² W

iii. Whereby the power is provided by 0.5% of the photons emitted by the source, we have;

[tex]P_{source}= \frac{P_{beam}}{0.005} =\frac{0.0014}{0.005} =0.28 \, W[/tex]

1 MeV = 1.60218 × 10⁻¹³ J

0.03 MeV = 0.03 × 1.60218 × 10⁻¹³ J = 4.80654 × 10⁻¹⁵ J/photon

Therefore, the number of disintegration per second = 0.28 J/s ÷  4.80654 × 10⁻¹⁵ J/photon = 5.83 × 10¹³ disintegrations per second

1 Curie = 3.7 × 10¹⁰  disintegrations per second

Hence, 5.83 × 10¹³ disintegrations per second = (5.83 × 10¹³)/(3.7 × 10¹⁰) Curie

= 1.57 × 10³ Curie

E. The parameters given are;

Density of water = 1000 kg/m³

Volume of water = 250 ml = 0.00025 m³

Initial temperature, T₁, = 25°C

Final temperature, T₂, = 100°C

Change in temperature, ΔT = 100 - 25 = 75°

Specific heat capacity of the water = 4200 J/kg/°C

Mass of water = Density × Volume = 1000 × 0.00025 = 0.25 kg

∴ Heat supplied = 4200 × 0.25 × 75 = 78,750 J

Time to heat the water = 45.0 sec

Therefore, power = Energy/time = 78750/45 = 1750 W

The formula for electrical power = I²R =VI = V²/R

Therefore, where V = 15.0 V, we have;

15²/R = 1750

R = 15²/1750 = 0.129 Ω.

The resistance of the heater = 0.129 Ω.

Answer:

The minimum running time is 319.47 s.

Explanation:

First we find the distance covered and time taken by the train to reach its maximum speed:

We have:

Initial Speed = Vi = 0 m/s    (Since, train is initially at rest)

Final Speed = Vf = 29.17 m/s

Acceleration = a = 0.25 m/s²

Distance Covered to reach maximum speed = s₁

Time taken to reach maximum speed = t₁

Using 1st equation of motion:

Vf = Vi + at₁

t₁ = (Vf - Vi)/a

t₁ = (29.17 m/s - 0 m/s)/(0.25 m/s²)

t₁ = 116.68 s

Using 2nd equation of motion:

s₁ = (Vi)(t₁) + (0.5)(a)(t₁)²

s₁ = (0 m/s)(116.68 s) + (0.5)(0.25 m/s²)(116.68 s)²

s₁ = 1701.78 m = 1.7 km

Now, we shall calculate the end time and distance covered by train, when it comes to rest on next station.

We have:

Final Speed = Vf = 0 m/s    (Since, train is finally stops)

Initial Speed = Vi = 29.17 m/s     (The train must maintain max. speed for min time)

Deceleration = a = - 0.7 m/s²

Distance Covered to stop = s₂

Time taken to stop = t₂

Using 1st equation of motion:

Vf = Vi + at₂

t₂ = (Vf - Vi)/a

t₂ = (0 m/s - 29.17 m/s)/(- 0.7 m/s²)

t₂ = 41.67 s

Using 2nd equation of motion:

s₂ = (Vi)(t₂) + (0.5)(a)(t₂)²

s₂ = (29.17 m/s)(41.67 s) + (0.5)(- 0.7 m/s²)(41.67 s)²

s₂ = 607.78 m = 0.6 km

Since, we know that the rest of 7 km, the train must maintain the maximum speed to get to the next station in minimum time.

The remaining distance is:

s₃ = 7 km - s₂ - s₁

s₃ = 7 km - 0.6 km - 1.7 km

s₃ = 4.7 km

Now, for uniform speed we use the relation:

s₃ = vt₃

t₃ = s₃/v

t₃ = (4700 m)/(29.17 m/s)

t₃ = 161.12 s

So, the minimum running time will be:

t = t₁ + t₂ + t₃

t = 116.68 s + 41.67 s + 161.12 s

t = 319.47 s