Quizlet A supermassive black hole is in the center of many galaxies, and a huge amount of electromagnetic radiation is emitted from a region near to that black hole. The typical mass of that black hole is:

Answer:

36m would be the correct answer

Answer:

The period will decrease.

T (period) = 2 π (L / g)^1/2      period of simple pendulum

As the length of a simple pendulum decreases its period of motion will also decrease (it will swing faster)

The centripetal force is the force that keeps a body moving in a circular path.

The centripetal force is given by; F = mv^2/r

1) We have;

F = 5,600 N

v = 20 m/s

r =?

m = 700 kg

Making r the subject of the formula;

r =mv^2/F

r = 700 × (20)^2/5,600

r = 50 m

2) F = mv^2/r

F = 900 × (5)^2/30

F = 750 N

3) Making r the subject of the formula;

r =mv^2/F

r = 800 × (40)^2/ 16,000

r = 80 m

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Answer:

The trees would remove more carbon from the atmosphere.

Explanation:

I just completed the test, and got a 100%.

The time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

In rock, S waves generally travel about 60% the speed of P waves, and the S wave always arrives after the P wave.

v ∝ 1/t

v₁t₁ = v₂t₂

t₁/t₂ = v₂/v₁

t₁/t₂ = 0.6v₁/v₁

t₁/t₂ =  0.6

t₁ = 0.6t₂

t₁ = 0.6 x 22 mins

t₁ = 13.2 mins

Thus, the time taken for the first p-wave to reach the same seismic station is approximately 13 minutes.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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Given :

A 1200 kg truck traveling at 20 m/s towards the east.

To Find :

The magnitude and direction of the momentum.

Solution :

We know, momentum is given by :

Momentum, P = mv

P = 1200 × 20 kg m/s

P = 24000 kg m/s

Now, we know, direction of momentum is same as the direction of velocity.

Therefore, momentum of truck is 24000 kg m/s and direction is towards the east.

Answer:

EQUAL

Explanation:

Energy entering the Earth needs to _____EQUAL_____________ the energy leaving the Earth.

Answer:

Equal

Explanation:

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a) The linear charge density of the insulating shell is \(\lambda_2\)=-4.058 μC/m.

b) The value of the x-component of the electric field at point P, located a distance 7.9 cm along the y-axis from the line of charge is 0 N/C.

c) The value of the y-component of the electric field at point P, located a distance 7.9 cm along the y-axis from the line of charge is 4.842 N/C.

d) The value of the x-component of the electric field at point R, located a distance 1.1 cm along a line that makes an angle of 30 degrees with the x-axis is 3.926 N/C.

e) The value of the y-component of the electric field at point R, located a distance 1.1 cm along a line that makes an angle of 30 degrees with the x-axis is 3.437 N/C.

f) The value of r for which the magnitude of the electric field is equal to 0 is more than one.

g) If we were to double λ₁, the magnitude of the electric field at point P, that is, E, would double.

h) In order to produce an electric field of zero at some point r > 4.1 cm, λ₁ would have to change its sign and decrease its magnitude.

a) The total charge enclosed by the insulating shell is equal to the volume charge density times the volume of the shell: \(Q = \rho*(4/3)*\pi*(b^3-a^3)\).

Therefore, the linear charge density of the insulating shell is

\(\lambda_2 = Q/(2*\pi*(b-a)) = (3*\rho*(b^2+a*b+a^2))/(2*(b-a))\)

= -4.058 μC/m.

b) The x-component of the electric field at point P is zero since it lies on the y-axis which is perpendicular to the line of charge.

c) The y-component of the electric field at point P can be found using the formula for the electric field of an infinite line of charge:

\(E = (\lambda/(2*\pi*\epsilon*r))\),

where r is the distance from the line of charge.

Thus, \(E = (\lambda_1/(2*\pi*\epsilon*\sqrt{r^2+d^2}))\)

= \((7.2/(2*\pi*8.85*10^{-12}*\sqrt{7.9^2+(2.2*10^{-2})^2}))\)

= 4.842 N/C,

where d is the distance from the line of charge to the point P along the y-axis.

d) The x-component of the electric field at point R can be found by first finding the distance between the line of charge and point R along the x-axis, which is r*cos(30°) = 0.55 cm.

Then, \(E = (\lambda_1/(2*\pi*\epsilon*r))*cos(30^{\circ})\)

= \((7.2/(2*\pi*8.85*10^{-12}*0.55))*cos(30^{\circ})\)

= 3.926 N/C.

e) The y-component of the electric field at point R can be found by first finding the distance between the line of charge and point R along the y-axis, which is r*sin(30°) = 0.55 cm.

Then, \(E = (\lambda_1/(2*\pi*\epsilon*r))*sin(30^{\circ})\)

= \((7.2/(2*\pi*8.85*10^{-12}*0.55))*sin(30^{\circ})\)

= 3.437 N/C.

f) The magnitude of the electric field is equal to zero at all points inside the insulating shell since the shell is uniformly charged and the electric fields from each infinitesimal element of charge cancel each other out. Therefore, there are an infinite number of values of r where the magnitude of the electric field is zero.

g) The magnitude of the electric field at point P is proportional to λ₁. Thus, if we double λ₁, E will also double.

h) In order to produce an electric field of zero at some point r > 4.1 cm, λ₁ would have to be negative and equal to

\(-\rho*(4/3)*\pi*(r^3-b^3)/(2*\pi*(r-b))\).

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Answer: STP is 273.15 K and 0.986 92 atm.

Explanation:

STP is defined as 273.15 K and a pressure of exactly 100 kPa (1 bar).

Hope this helps! :)

Answer:

it is correct

Explanation:

Though no rounded numbers can be defined as accurate, if we were going by people's discovery, and research, we can define that the number, g = 9.8m/s^2, is accurate

The period of one vibration of the tone will be 0.00390625 seconds.

The period (T) of a waveform is the time it takes for one complete cycle or vibration. It is calculated as the inverse of the frequency (f) of the waveform.

So, if the frequency of middle C is 256.0 Hz, then the period of one vibration of this tone is:

T = 1/f

T = 1/256.0 Hz

T = 0.00390625 seconds

Therefore, the period of one vibration of middle C is approximately 0.00390625 seconds.

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The force diagram on the desk consists of normal reaction and weight of the pencil.

The net force on the desk in vertical direction is 1 N.

The net force on the desk in horizontal direction is 0.

The given parameters;

The force diagram of the desk is sketched as follows;

                                             ↑ N

                                             ⊕

                                              ↓ W

Where;

The net force on the desk in vertical direction = 1 N

The net force on the desk in horizontal direction = 0

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The y-component of the vector is 4.97 m.

A vector is a quantity that has both magnitude and direction.

To calculate the y-component of the vector in the diagram above, we use the formula below.

Formula:

From the question,

Given:

Substitute these values into equation 1

Hence, the y-component of the vector is 4.97 m.

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Answer:

i) Hooke's Law works for some materials

ii) Hooke's Law only applies up to a certain force

iii) The force at which Hooke's Law stops working for springs is lower than for most materials

Explanation:

Hooke's Law of elasticity states that the deformation of an elastic material is proportional to the applied for small deformations

The law is applicable to elastic materials whose values of deformation or extension as well as the applied load or stress are expressible by a single real number

Hooke's Law is applicable when the force and the extension of the elastic material are proportional, at larger force, the elastic material is observed to expand more than as expected based on Hooke's Law

The nature of springs is such that its elastic limit is reached by a much lower force than for most materials. Therefore, Hooke's law stops working for springs at a lower force than for most materials.

Answer:

1.63366

Explanation:

I got this answer from calculator soups physics calculators. I really recommend their website for formulas.  

Answer:

Using the range formula   S = V^2 * sin 2 theta / g

So the cannonball should travel

S = 35^2 * sin 90 / 9.8 = 125 m

Then the cannon should be placed 100 m from the wall

H = Vy t - 1/2 g t^2   height of cannonball after time t

Vy = 35 * sin 45 = 24.7 m/s     vertical speed of cannonball

Vx = 35 * cos 45 = 24.7 m/s

t = 100 m / 24,7 m/s = 4.04 sec    time for cannonball to reach wall

H = Vy t - 1/2 g t^2       height of cannonball after time t

H = 24.7 * 4.04 - 1/2 * 9.8 * 4.04^2 = 19.8 m

19.8 m - 15 m = 4.8 m     cannonball clears wall by 4.8 m

The projectile launch ratios allow to find the results on the cannonball launch questions are:

 a) The canyon is at x = 100 m from the wall.

 b) The bullet hits the wall at a height of 4.8 m above the height of the wall.

Projectile launching is an application of kinematics where there is no acceleration on the x-axis and the y-axis is gravity acceleration.

A) Indicate that the initial speed of the bullet is v = 35 m/s with an angle  θ=45º, which is the distance from the wall. The target is 25 m behind the wall, in the attachment we have a scheme of the system.

Let's find the range that is the horizontal distance for which the height is  zero.

       \(R = \frac{v_p^2 \ sin \ 2\theta}{g}\)

Let's calculate.

       R = \(\frac{35^2 \ sin ( 2 45(}{9.8}\)  

       R = 125 m

Therefore the canonl must be 100 m from the wall.

B) The height of the wall is y = 15m, the blah manages to pass it and by how much.

We use trigonometry to find the components of velocity.

         v₀ₓ = v₀ cos θ

         \(v_{oy}\) = v₀ sin θ

          v₀ₓ = 34 cos 45 = 24.7 m / s

          \(v_{oy}\) = vo sin 45 = 24.7 m / s

Let's find the time it takes for the cannonball to travel the x = 100 m

        \(v_x = \frac{x}{t} \\ t = \frac{x}{v_x}\)

        t = \(\frac{100}{35 \ cos 45 }\)  

        t = 4.04 s

Let's find the height for this time.

        y = \(v_[oy}\) t - ½ g t²

Let's calculate

        y = 24.7 4.04 - ½ 9.8 4.04²

        y = 19.8 m

This height is greater than the height of the wall h = 15m, therefore the cannoball passes the wall, by a difference in height.

        Δy = y -h

        Δy = 19.8-15

        Δy = 4.8 m

In conclusion, using the projectile launch ratios we can find the results on the cannonball launch questions are:

 a) The canyon is at x = 100 m from the wall.

 b) The cannonball hits the wall at a height of 4.8m above the height of the wall.

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The distance at which the package is dropped at the target is equal to 636m.

\(s= ut+\frac{1}{2} at^2\)

\(88m= 0+\frac{1}{2} (9.8m/s^2)(t^2)\)

\(t=\sqrt{\frac{2*88}{9.8m/s^2}} = 4.24s\)

\(s=vt\\s= 150*4.24=636m\)

In normal use and in kinematics, the rate of an object is the importance of the change of its role over time or the significance of the alternate of its position in keeping with a unit of time; its miles accordingly a scalar amount.

The price of change of function of an object in any path. the pace is measured as the ratio of distance to the time wherein the gap becomes included. the pace is a scalar amount because it has the handiest path and no value. speed is defined because of the rate of trade of distance with time. It has the size of distance by means of time. hence, the SI unit of speed is given because of the aggregate of the primary unit of distance and the primary unit of Time. for this reason, the SI unit of pace is a meter in keeping with the second.

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The higher the energy, the faster the particles vibrate (which means a higher frequency).

option D

Waves like sound  wave is produced when an object vibrates, creating a pressure wave.

This pressure wave causes particles in the surrounding medium to have vibrational motion. As the particles vibrate, they move nearby particles, transferring energy between neighboring particles, creating a faster vibration as the energy transmitted increases.

Thus, we can conclude that, waves have energy that causes particles in matter to vibrate. So, it stands to reason that the higher the energy, the faster the particles vibrate (which means a higher frequency).

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A noise of 1 decibel is subaudible. Option D.

When it comes to human hearing noise below 70 dB is considered safe. 1 decibel is below this limit so this level is considered safe for human hearing. Noise above 85 dB is considered potentially dangerous with prolonged exposure. The human hearing threshold is approximately 0 decibels.

Above this threshold sounds with higher sound pressure levels are perceived as louder. To make the noise twice as audible the noise must be increased by 10 dB. For example, 10 violins can only be heard twice as loud as 1 violin. The risk of hearing loss from noise increases with sound intensity, not loudness. Almost all firearms produce noise in excess of 140 dB small. 22 caliber rifles can make about 140 dB of noise while large caliber rifles and pistols can make more than 175 dB of noise.

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In an  electron (mass = 9.1 x 101 kg) is released at a speed of 1.5 x 10 m/s in a direction perpendicular to a uniform magnetic field,the magnitude of the magnetic field required for the electron to move in a circular path of radius 4.4 cm is approximately 7.98 Tesla.

To find the magnitude of the magnetic field required for an electron with given mass, speed, and radius to move in a circular path, we can use the formula for the centripetal force:

F = mv² / r

Where:

F is the centripetal force,

m is the mass of the electron,

v is the speed of the electron, and

r is the radius of the circular path.

The centripetal force is provided by the magnetic force in this case, which is given by:

F = |q|.v.B

Where:

|q| is the magnitude of the charge (absolute value of electron charge, 1.6 x 10⁻¹⁹ C),

v is the speed of the electron, and

B is the magnitude of the magnetic field.

Equating the two expressions for the centripetal force, we have:

mv² / r = |q|v B

Simplifying the equation:

B = (m .v) / (|q|r)

Given:

Mass of the electron (m) = 9.1 x 10⁻³¹ kg

Speed of the electron (v) = 1.5 x 10⁶ m/s

Radius of the circular path (r) = 4.4 cm = 0.044 m

Charge of the electron (|q|) = 1.6 x 10⁻¹⁹ C

Plugging in the values:

B = ((9.1 x 10⁻³¹ kg) * (1.5 x 10⁶ m/s)) / ((1.6 x 10⁻¹⁹ C) * (0.044 m))

Simplifying the expression:

B ≈ 7.98 T

Therefore, the magnitude of the magnetic field required for the electron to move in a circular path of radius 4.4 cm is approximately 7.98 Tesla.

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Explanation:

Plastics with a density less than or equal to 1.025 g/mL will float in seawater, while plastics with a density greater than 1.025 g/mL will sink.

Here are some examples of plastic densities:

Polyethylene terephthalate (PET) has a density of approximately 1.38 g/mL, so it will sink in seawater.

Polypropylene (PP) has a density of approximately 0.9 g/mL, so it will float in seawater.

High-density polyethylene (HDPE) has a density of approximately 0.95 g/mL, so it will float in seawater.

Polystyrene (PS) has a density of approximately 1.05 g/mL, so it will sink in seawater.

Based on the above examples, we can see that plastics such as polypropylene and high-density polyethylene will definitely float in seawater, while others such as polyethylene terephthalate and polystyrene will definitely sink. Other types of plastics with densities close to 1.025 g/mL may float or sink depending on their exact density and the conditions of the water, such as temperature and salinity.

Answer:

1. 571.43m/s

2. 142.9m and 342.9m

Explanation:

1.Take the difference in time.

1.2-0.7=0.7 seconds

Take the distance between them and divide with differnce in time.

400÷0.7=571.43 seconds.

2.Take the time of the two men and divide by two.

0.5÷2= 0.25 secs

1.2÷2= 0.6 secs

multiply each with the velocity.

0.25×571.43=142.9m

0.6×571.43=342.9m

The magnitude of the magnetic field is 2.80 T, directed in the negative y-direction.

When a charged particle moves through a magnetic field, it experiences a force known as the Lorentz force. This force can be expressed using the equation F = q(v × B), where F is the force, q is the charge of the particle, v is its velocity, and B is the magnetic field.

In this case, the proton is moving perpendicular to the magnetic field B, with a velocity in the positive z-direction. The acceleration experienced by the proton is given as 3.00 × 10¹³ m/s²  in the positive x-direction.

We know that the force acting on the proton is given by the equation F = m × a, where m is the mass of the proton and a is its acceleration. Since we have the acceleration value, we can calculate the force acting on the proton.

Next, we can use the equation for the Lorentz force to relate the magnetic field, velocity, and force acting on the proton. Since the proton experiences an acceleration in the positive x-direction, we can conclude that the Lorentz force must act in the negative x-direction to cause this acceleration.

The magnitude of the Lorentz force can be found by equating it to the force calculated earlier. From this equation, we can isolate the magnitude of the magnetic field B.

Finally, by substituting the given values into the equation, we find that the magnitude of the magnetic field B is 2.80 T, directed in the negative y-direction.

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Cart attached to a spring is displaced from equilibrium and then released from rest, The spring's potential energy is negative between points A and I.

The spring's potential energy is negative when the displacement of the cart from equilibrium is in the opposite direction of the displacement at equilibrium. This occurs when the displacement is to the left of point A and to the right of the point I.

At point A, the displacement is at its maximum to the left, indicating that the cart has moved the furthest distance away from equilibrium in the left direction. At this point, the spring is stretched to its maximum length, which means that it has stored the most potential energy. However, this potential energy is still positive because the displacement is still in the same direction as at equilibrium. As the cart moves further to the left, the displacement becomes negative, and the potential energy of the spring becomes negative as well. This continues until the cart reaches point I, where the displacement is at its maximum to the right and the potential energy of the spring is once again positive.

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Answer:

Braking distance 126 meters

Explanation:

Given:

V₁ = 43 km/h = 43 000 m /  3 600 s ≈ 12 m/s

D₁ =14 m

V₂ = 129 km/h = 129 000 m / 3 600  s ≈ 36 m/s

____________________

D₂ - ?

D₁ = V₁² / (2·a)

2·a·D₁ = V₁²

a = V₁² / (2·D₁)

D₂ = V₂² / (2·a) = V₂²·2·D₁ / (2·V₁²) =

= V₂²·D₁ / V₁² = D₁·(V₂ / V₁)²

D₂ = 14·(36 / 12)² = 14·3² =  126 m

The braking distance has increased 9 times!

Answer:

Distance, Mass