The magnitude of the Poynting vector of a planar electromagnetic wave has an average value of 0.724 W/m^2. What is the maximum value of the magnetic field in the wave?

a. 77.9 nT
b. 55.1 nT
c. 38.9 nT
d. 108 nT

Answers

Answer 1

Answer:

a. 77.9 nT

Explanation:

Given;

average value of the electromagnetic wave, E = 0.724 W/m^2

The average value of a Poynting vector is given by;

[tex]S = \frac{1}{2\mu_o} *E_oB_o\\\\But, E_o = cB_o\\\\S = \frac{1}{2\mu_o} *cB_o^2\\\\cB_o ^2= 2\mu_o S\\\\B_o^2 = \frac{2\mu_o S}{c}\\\\B_o = \sqrt{ \frac{2\mu_o S}{c}} \\\\B_o = \sqrt{ \frac{2(4\pi*10^{-7}) (0.724)}{3*10^8}}\\\\B_o = 7.79*10^{-8} \ T\\\\B_o = 77.9*10^{-9} \ T\\\\B_o = 77.9 \ nT[/tex]

Therefore, the maximum value of the magnetic field in the wave is 77.9 nT.

Answer 2

Answer:

A I think : )

Explanation:


Related Questions

A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .

Answers

Complete Question

A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex]  to  [tex]t_2 = 6.0s[/tex]

What is its average velocity over this time interval?

Answer:

The velocity is  [tex]v = 3.903 \ m/s[/tex]

Explanation:

From the question we are told that

    The first position of the ball is  [tex]x_1 = 8.0 \ cm[/tex]

    The second position of the ball is  [tex]x_2 = - 4.1 \ cm[/tex]

Generally the average velocity is mathematically represented as

       [tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]

=>    [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]

=>    [tex]v = 3.903 \ m/s[/tex]

differences between a resultant and a equilibrium​

Answers

Answer:

I found "As nouns the difference between resultant and equilibrium is that resultant is anything that results from something else; an outcome while equilibrium is the condition of a system in which competing influences are balanced, resulting in no net change. is following as a result or consequence of something."

Explanation:

someone help please


waves disturb ____, but do not transmit it.

a. energy
b. matter
c. sound
d. none of the above

Answers

Answer:

b. matter

Explanation:

Waves disturb matter but do not transmit it.

Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.

Energy is propagated by a wave. When for example, sound waves are produced, the disturbance is propagated via particle - particle interaction But after the wave train moves, the particles remain.

An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?

Answers

[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]

We have:

Mass of the object = 3 kgForce on the object = 9 N

We need to find:

Resulting accleration of the object?

Solution:

According to Newton's 2nd law of motion, or quantitative measure of Force:

Force = Mass × Accleration

Using this,

➝ F = ma

➝ 9N = 3 kg × a

➝ a = 9/3 m/s²

➝ a = 3 m/s²

Hence,

The resulting accleration of the object is 3 m/s². And we are done! :D

⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]

If mass (3.0 kg) multiplying (*) acceleration gives you force (newtons), force dividing mass gives you acceleration; 9/3 =


3 m/s^2

List at least 4 aspects to evaluate the quality of an internet site

Answers

Answer:

authority, accuracy, objectivity, currency, coverage, and appearance.

Explanation:

There are six (6) criteria that should be applied when evaluating any Web site. or each criterion, there are several questions to be asked. The more questions you can answer "yes", the more likely the Web site is one of quality.

authority, accuracy, objectivity, currency, coverage, and appearance

How far out from the sun is the rock line?

Answers

Answer:

5 million miles

Explanation:

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.

Answers

Complete Question

A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?

Answer:

The position of the object at t = 10s is  [tex]X = 38.3 \ m[/tex]

Explanation:

From the question we are told that

The acceleration along the x axis is  [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]

  The position of the object at t = 0 is  x = -14.0 m

  The velocity at t = 0 s is  [tex]v_{0}x = 7.10 m/s[/tex]

Generally from the equation for acceleration along x axis we have that

     [tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]

=>   [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]

=>   [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

At  t =0  s   and  [tex]v_{0}x = 7.10 m/s[/tex]

=>   [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]

=>   [tex]K_1 = 7.10[/tex]      

So  

      [tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]

=>  [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]

=>  [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]

At  t =0  s   and   x = -14.0 m

  [tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]

=>   [tex]K_2 = -14[/tex]

So

     [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]

At  t = 10.0 s

      [tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]

=>   [tex]X = 38.3 \ m[/tex]

             

     

A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

(a) Neglecting air resistance, how long does it take the skier to reach the ground?

(b) How far horizontally does the skier travel in the air before landing?

m

Answers

Explanation:

Given

Velocity v = 23.0m/s

Distance S = 3.45m

Required

Time it will take the skier to reach the ground;

Using the equation of motion;

S = ut + 1/2gt²

3.45 = 23t + 1/2(9.8)t²

3.45 = 23t + 4.9t²

4.9t²+23t-3.45 = 0

Factorize;

t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)

t = -23 ±√529+67.62/9.8

t = -23±√596.62/9.8

t = -23±24.43/9.8

t = 1.43/9.8

t = 0.146 secs

Hence take the skier 0.146 secs to reach the ground.

b) Horizontal distance covered is the range;

Range = U√2H/g

Range = 23√2(3.45)/9.8

Range = 23√6.9/9.8

Range = 23√0.7041

Range = 23(0.8391)

Range = 19.29m

Hence the horizontal distance travelled in air is 19.29m

(a).The time taken by the skier to reach the ground is 0.145 second.

(b).The skier travel in the air before landing is 19.29 meter.

a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.

Using equation of motion.

       [tex]S=ut+\frac{1}{2}gt^{2}[/tex]

Where S is vertical distance , u is initial velocity and g is gravitational acceleration.

Substitute S = 3.45 m, u = 23m/s  and g = 9.8 in above equation.

        [tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]

Since, time can not be negative.

So that,  [tex]t=0.145s[/tex]

b. The horizontal distance travel before landing is known as Range.

           Horizontal distance  ,

                                             [tex]=v*\sqrt{\frac{2S}{g} }[/tex]

Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.

                         [tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]

Thus, The skier travel in the air before landing is 19.29 meter.

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If a dog has a mass of 2.5 kg, what is its weight and what is the normal force that it feels.
I

Answers

Answer:

Weight = normal force = 24.5 N

Explanation:

Given that,

Mass of a dog, m = 2.5 kg

We need to find its weight and the normal force that it feels.

The weight of an object is given by :

W = mg

Where g is the acceleration due to gravity

[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]

The normal force is balanced by the weight of an object. So,

Weight = normal force = 24.5 N

a bus with a mass of 5000kg is uniformly acceleration from rest. The net force acting on the bus is

Answers

Answer:

50,000N

Explanation:

According to Newton's second law of motion;

Net Force = Mass * acceleration

Given

Mass = 5000kg

Let the acceleration = 10m/s²

Net force = 5000 * 10

Net force = 50,000N

Hence the net force acting on the bus is 50000N

a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?

Answers

Answer:

5,000

Explanation:

Vf = Vi + a * t

You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. The design specifications call for it to have a rotational kinetic energy of 330 J after it has turned through 30.0 revolutions.

What should be the moment of inertia of the flywheel about its rotation axis?
Express your answer with the appropriate units.

Answers

Answer: 1.14 kg*m/s

Explanation:

The first person explained everything right, they just forgot to convert the angular acceleration to rads/sec^2 from revs/sec^2. Once that is converted, your answer should come out right.

Another small thing, the answer there has an extra unnecessary step. It tells you to find the square root of w^2 to find w but that is unnecessary since the final equation asked for w^2. Hope this helps! :)

The moment of inertia I of the flywheel about its rotation axis is

[tex]1.39Kgm^2[/tex]

Given

Angular displacement,

[tex]\theta = 30rev \\\\\theta = (30) * 2\pi rad \\\\\theta = 188.495rad[/tex]

Therefore, Final angular velocity (w) will be:

[tex]w^2 = 2\alpha\theta\\\\w^2 = 2 * (0.200 * 2\pi) * (188.49)\\\\w^2 = 473.73\\\\w = 21.76 rad/s[/tex]

Therefore,

moment of inertia

[tex]I = 2 * K / w^2[/tex]

[tex]I = 2 * 330 / 473.73[/tex]

[tex]I = 1.39kgm^2[/tex]

For more information on moment of inertia

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I want to know about the inventions caused due to rain. Like the Benjamin Franklin's Lightning Conductor. I have to make a chart.​

Answers

Franklin had been waiting for an opportunity like this. He wanted to demonstrate the electrical nature of lightning, and to do so, he needed a thunderstorm.

He had his materials at the ready: a simple kite made with a large silk handkerchief, a hemp string, and a silk string. He also had a house key, a Leyden jar (a device that could store an electrical charge for later use), and a sharp length of wire. His son William assisted him.

Franklin had originally planned to conduct the experiment atop a Philadelphia church spire, according to his contemporary, British scientist Joseph Priestley (who, incidentally, is credited with discovering oxygen), but he changed his plans when he realized he could achieve the same goal by using a kite.

A jet airplane with a 75.0 m wingspan is flying at 260 m/s. What emf is induced between the wing tips in V if the vertical component of the Earth’s magnetic field is 3.00 × 10-5 T?

Answers

Answer:

0.585V

Explanation:

Given that:

B = 3.00 × 10-5 T

l = 75.0 m

v = 260 m/s

From Blv = emf between the wing tips

= 3.00 × 10-5 T × 75×260

= 117/200

= 0.585V

Hence, the emf between the wing tips is 0.585V

What is magnet made of

Answers

Answer:

metals like iron or nickel

Explanation:

a current of 200mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the potential drop across the conductor

Answers

Answer:

V = 6.65 [volt]

Explanation:

We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.

[tex]P=\frac{E}{t}[/tex]

where:

P = power [w]

E = energy = 40 [J]

t = time = 30 [s]

[tex]P =40/30\\P = 1.33[w][/tex]

Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.

[tex]P =V*I[/tex]

where:

V = voltage [volts]

I = current = 200mA = 0.2 [A]

[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]

it is the question 12 part okay​

Answers

Answer:

Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.

So if you fall into icy water your gonna freeze that’s a given but you can stop it if you get out fast and get in a real hot bath and then get a towel on and you get Hypothermia because your body physically can’t take the cold so it decides to freeze your insides until your heart eventually just stops because it’s frozen.

Can I pls have brainliest :)

If Earth’s Moon were replaced with a typical neutron star, what would the angular diameter of the neutron star be as seen from Earth?

Answers

Answer:

[tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex]

Explanation:

d = Diameter of typical neutron star = 20 km = 20000 m

D = Distance between Earth and Moon = [tex]384.4\times 10^6\ \text{m}[/tex]

Here, [tex]D>>d[/tex] so we use small angle approximation

[tex]\delta=\dfrac{d}{D}\\\Rightarrow \delta=\dfrac{20000}{384.4\times 10^6}\\\Rightarrow \delta=0.00005202\ \text{rad}=\dfrac{0.00005202\times 180}{\pi}=0.003^{\circ}[/tex]

The angular diameter of the neutron star would be [tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex] from Earth.

Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is traveling in the opposite direction at 5 m/s. They crash into each other and hold onto each other so that they move as one mass. How fast are they traveling?

Answers

Answer:

-7/19

Explanation:

A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days

Answers

Answer:

= 720000 [k]

Explanation:

The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.

[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]

= 720000 [k]

What happens when calcium reacts with chlorine?
A. Electrons become delocalized among the atoms
B. Electrons are shared between the calcium atoms and the chlorine
atoms
C. Electrons move from the chlorine atoms to the calcium atoms,
D. Electrons move from the calcium atoms to the chlorine atoms,

Answers

When calcium reacts with chlorine. Electrons move from the calcium atoms to the chlorine atoms. Option D is correct.

What is a chemical reaction?

When one or more chemicals are transformed into one or more other compounds, a chemical reaction occurs.

In an ionic compound, calcium chloride is present. To build a complete outer shell of electrons, the calcium atom loses two electrons and each chlorine atom obtains one.


When calcium combines with chlorine, it forms calcium chloride. From the calcium atoms to the chlorine atoms, electrons travel.

Hence, option D is correct.

To learn more about the chemical reaction refer to the link;

https://brainly.com/question/6876669

#SPJ1

A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci

Answers

Answer:

The resultant velocity has a magnitude of 38.95 m/s

Explanation:

Vector Addition

Given two vectors defined as:

[tex]\vec v_1=(x_1,y_1)[/tex]

[tex]\vec v_2=(x_2,y_2)[/tex]

The sum of the vectors is:

[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]

The magnitude of a vector can be calculated by

[tex]d=\sqrt{x^2+y^2}[/tex]

Where x and y are the rectangular components of the vector.

We have a plane flying due west at 34 m/s. Its velocity vector is:

[tex]\vec v_1=(-34,0)[/tex]

The wind blows at 19 m/s south, thus:

[tex]\vec v_2=(0,-19)[/tex]

The sum of both velocities gives the resultant velocity:

[tex]\vec v =(-34,-19)[/tex]

The magnitude of this velocity is:

[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]

[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]

d = 38.95 m/s

The resultant velocity has a magnitude of 38.95 m/s

A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?

Answers

Answer:

599.245km/hr

Explanation:

A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?

We solve the above question using vectors

In vector form Air speed is -540i + 0j Wind speed is (-80/√2)i + (80/√2)j

Vector notation wind speed is given as: -56.5685 i + 56.5685j

The vector for the ground speed of the plane =

-540i + 0j -56.5685i + 56.5685j

= -596.56854249i + 56.5685j

The the ground speed of the plane √[(596.56854249)² + (56.5685)²]

= √359094.021081 km/hr

= 599.24454197 km/hr

Approximately

= 599.245km/hr

A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.

Answers

As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.

Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.

The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:

n + (-w) = 0

n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N

The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be

f = 0.67 (51.94 N) ≈ 35 N

so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.

The block starts moving as soon as x = 35 N, at which point f = 35 N.

At any point after the block starts moving, we have

f = 0.48 (51.94 N) ≈ 25 N

so that x = 25 N is the required force to keep the block moving at a constant speed.

As x  is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force

Magnitude of normal force ( object at rest );  n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x  after object start moving   x = 25 N

Given data :

mass of block at rest ( m ) = 5.3 kg

Coefficient of static friction ( μ_s ) =0.67

Coefficient of kinetic friction is ( μ_k ) = 0.48

Horizontal force applied to block = x  

First step : magnitude of normal force ( n ) when object is at rest

n = w            where w = m*g

n - w = 0

n - ( 5.3 * 9.81 ) = 0     ∴  n = 51.94 N

Second step : Required magnitude of x before the movement of object

F =  μ_s * n

F = 0.67 * 51.94  = 34.79 N  ≈ 35 N

∴ The object will start moving once F and x = 35 N

Final step : Magnitude of x  after object start moving

F = μ_k  * n

  = 0.48 * 51.94 = 24.93 N  ≈ 25 N

∴ object will continue to accelerate at a constant speed once F and x = 25N

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Write Radar gun summary

Answers

Answer:

A radar gun is a device for measuring the speed of moving objects. ... The radar gun is a Doppler radar unit that can be static, vehicle-mounted or hand-held. It measures the

Explanation:

True or false. when objects collide , some momentum is lost

Answers

Answer:

It is neither false nor true. When they collide some of one of the objects goes to the other object.

Explanation:

Answer: True

Explanation:

A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?

Answers

Answer:

1000x Newton

Explanation:

Step one

given data

acceleration= 1000 m/s²

The question did not specify the mass of the mass of the space ship.

So, let's assume the mass is x kg

Step two:

Required is the force F in Newton

From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.

F=mass x Acceleration

F=ma

Substituting our given data we have

F=1000x Newton

A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?

Answers

Answer:

The displacement of the car after 6s is 43.2 m

Explanation:

Given;

velocity of the car, v = 12 m/s

acceleration of the car, a = -1.6 m/s² (backward acceleration)

time of motion, t = 6 s

The displacement of the car after 6s is given by the following kinematic equation;

d = ut + ¹/₂at²

d = (12 x 6) + ¹/₂(-1.6)(6)²

d = 72 - 28.8

d = 43.2 m

Therefore, the displacement of the car after 6s is 43.2 m

A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s

Answers

Explanation:

Using the formula;

2x = vt

x is the distance up from the ocean floor the submarine is

v is the speed of sound in water

t is the time

Given

t = 2.3s

v = 1490m/s

Required

how high (approx) up from the ocean floor is the submarine x

From the formula;

x = vt/2

x = 1490(2.3)/2

x = 745(2.3)

x = 1,713.5m

Hence the submarine is 1713.5m high up from the ocean floor

. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.

Answers

Answer:

The maximum speed of the car is 35 m/s

The total distance traveled by the car is 658.33 m

Explanation:

Given;

initial velocity of the car, u = 15 m/s

acceleration of the car, a = 2 m/s²

time of car motion, t = 10 s

(i)

Initial distance traveled by the car is given by;

d₁ = ut + ¹/₂at²

d₁ = (15 x 10) + ¹/₂(2)(10)²

d₁ = 150 + 100

d₁ = 250 m

The maximum speed of the car during this is given by;

v² = u² + 2ad₁

v² = (15)² + (2 x 2 x 250)

v² = 1225

v = √1225

v = 35 m/s

(ii)

The final distance cover by the car during the deceleration of 1.5 m/s².

Note: the final or maximum speed of the car becomes the initial velocity during deceleration.

v² = u² + 2ad₂

where;

v is the final speed of the car when it stops = 0

0 = u² + 2ad₂

0 = (35²) + (2 x - 1.5 x d₂)

0 = 1225 - 3d₂

3d₂ = 1225

d₂ = 1225 / 3

d₂ = 408.33 m

The total distance traveled by the car is given by;

d = d₁ + d₂

d = 250 m + 408.33 m

d = 658.33 m

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