NEED HELP ASAP!!!

If 100 newtons of force are applied over a distance of 50 meters for 5 seconds, how much power is used
2,500 watts
1,000 joules
10 watts
1,000 watts

If the maximum tractive effort of a vehicle is 450 lb and the engine generated tractive effort is 425 lb, then the available tractive effort is 450 lb.

The maximum tractive effort of a vehicle is the maximum amount of force that the vehicle can apply to the ground to generate traction and move forward. The engine generated tractive effort is the amount of force that the vehicle's engine can produce to move the vehicle. The available tractive effort is the amount of force that the vehicle can actually use to move forward, and it is determined by the maximum tractive effort of the vehicle. In this case, the available tractive effort is 450 lb, which is the maximum tractive effort of the vehicle.

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Answer: D

Explanation:

It is at organ level because the kidney is an organ

By measuring the terminal velocity of the ball bearing and knowing the properties of the liquid and the ball bearing, the viscosity of the liquid can be calculated using Stoke's law.

The experiment you are referring to is commonly known as the Stoke's law experiment or the falling ball viscometer experiment.

In this experiment, a ball bearing (usually made of steel or glass) is dropped into a vertical tube filled with a liquid of known viscosity. The ball bearing falls through the liquid under the influence of gravity, but also experiences a viscous drag force due to the resistance of the liquid.

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Frankness and communication are not vital to family harmony so it will be false.

While frankness and communication are indeed important factors in maintaining family harmony, it would be inaccurate to claim that they are vital in every situation. While open and honest communication can foster understanding, trust, and strong relationships within a family, there are other elements that contribute to family harmony as well. Factors such as empathy, respect, compromise, and shared values also play significant roles in maintaining a harmonious family environment.

While frankness and communication can help address conflicts, express needs and concerns, and promote healthy relationships, they alone may not guarantee family harmony. Each family is unique, and what works for one may not work for another. Different families may have their own dynamics, communication styles, and values that contribute to their overall harmony.

It is important to recognize that family harmony is a complex and multifaceted concept, influenced by various factors such as individual personalities, family dynamics, cultural backgrounds, and external circumstances. Therefore, it is essential to consider a holistic approach and take into account the broader context of family dynamics when striving for family harmony.

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Answer:

water vapor is a type of gas

Explanation:

all gas are in air state of form

hope this helps

Answer:

chemical potential energy

Explanation:

When the sun's radiant energy falls on Earth's oceans, it causes water to change state by evaporating. The form of energy water vapor have is chemical potential energy. when heat energy of sun falls on water surface, its state change from liquid to gas. This gas rises above in the form of water vapor. Its a chemical change. If this vapor get condense/ cool, it will turn into liquid again. We can say that vapor has potential to transform into water again. Thus, the energy stored in water vapor is called chemical potential energy.

The amount of energy a battery can hold is known as its power capacity. Watt-hours are a common way to measure this power (the symbol Wh).

Current times voltage equals power. This relationship demonstrates how voltage and current are crucial in determining what a battery is good for. The higher the power, the quicker the rate at which a battery can perform work.

P = E/t, where P stands for power, E for energy, and t for time in seconds, is the formula. According to this equation, power is the amount of energy consumed in a given amount of time.

Read the colour codes on the resistor while holding it with the gold or silver band to the right. From the bands on the resistor, choose the appropriate colour code. From left to right, read the colours. Now the resistance value depending on the supplied colour code is visible.

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The velocity of a car is a vector quantity, meaning it has both magnitude and direction. The magnitude of velocity is the speed of the car, while the direction of velocity is the direction in which the car is moving.

If the car is moving in a straight line, then its velocity component in a particular direction can be equal to its magnitude if the car is moving directly towards or away from the observer. In other words, if the direction of motion and the direction in which the velocity is being measured are the same, then the velocity component and the magnitude of velocity will be equal.

For example, if a car is moving due north and its velocity is being measured in the north direction, then the velocity component in the north direction will be equal to the magnitude of the velocity. However, if the car is moving due north and the velocity is being measured in the east direction, then the velocity component and the magnitude will be different.

In general, the direction of the car's motion does affect the relationship between the velocity component and the magnitude. The velocity component in a particular direction will be equal to the magnitude only when the car is moving directly towards or away from the observer in that direction. If the car is moving at an angle to the observer, then the velocity component and the magnitude will be different.

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From the calculations and the data provided, the speed is obtained as 15 m/s

The term momentum refers to the product of mass and velocity. Given that we can now proceed with the answer.

Since;

p = mv

m = 750kg

v = 5 km/hr or 1.4 m/s

p = 750kg *  1.4 m/s

p = 1050 kgm/s

If the mass of the body is 70kg, the speed is

v = p/m

v = 1050 kgm/s/70 Kg

= 15 m/s

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We do not expect to see tidal disruption of sun-like stars by black holes larger than about 10^8 M☉ due to their weaker gravitational tidal forces.

Tidal disruption occurs when a star gets too close to a black hole, and the gravitational forces from the black hole pull on the star more strongly than the internal forces holding it together. This causes the star to be torn apart and accreted onto the black hole. The tidal disruption radius, which is the distance from the black hole at which this happens, depends on the mass and size of the star as well as the mass of the black hole. For a sun-like star, the tidal disruption radius is proportional to the black hole mass. However, once the black hole mass exceeds about 10^8 M☉, the tidal disruption radius becomes larger than the size of the star, making tidal disruption less likely to occur. Therefore, we do not expect to see tidal disruption of sun-like stars by black holes larger than about 10^8 M☉.

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The element that has 4 electrons, 3 protons and 6 neutrons is lithium ion.

An element is one of the simplest chemical substances that cannot be decomposed in a chemical reaction or by any chemical means and made up of atoms all having the same number of protons.

The atom of an element is made up of three subatomic particles namely;

In a neutral element, the number of protons and electrons are the same i.e. equal. According to this question, an element has 4 electrons, 3 protons and 6 neutrons.

The element in the periodic table with 3 protons is Lithium. This means that the element is lithium ion because it has gained one electron.

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Answer:

0.0231 m/s

Explanation:

Given the question :

A small toddler is playing in his yard. Over the course of 29.01 s, he walks 2.099 m S before turning N and walking 1.429 m. What is his average velocity?

Distance walked south = 2.099 metre

Time taken = 29.01s

Distance walked north = 1.429 metre

Average Velocity = Displacement / time

The Displacement is hence,

(2.099 m - 1.429m) = 0.67m

Average Velocity = 0.67m / 29.01s

Average Velocity = 0.02309 m/s

Average Velocity = 0.0231 m/s

The acceleration of the skater is: 1.97 m/s²

Any procedure where the velocity varies is referred to as acceleration. There are only two ways to accelerate: changing your speed or changing your direction, or changing both. This is because velocity is both a speed and a direction.

Time, (t) = 10 sec

Frequency (F) = 1/t = 0.1 Hz.

Velocity, (V) = r × ω

V = 10 x (2 Π/10)

V = 6.28 m/s

Now, Centripetal acceleration,

a = v²/r

or, a = (6.28)²/20

or, a = 1.97 m/s².

Now, the acceleration of the skater is: 1.97 m/s².

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The quality of early caregiver-child relationships has been the most consistent predictor of a secure attachment.
Sensitivity has been the most consistent predictor of a secure attachment.

The most reliable indicator of a stable attachment has historically been the standard of parenting.

According to attachment theory research, the most crucial aspect in determining whether a child will establish a secure attachment style is the quality of parenting, particularly the responsiveness and sensitivity of the carer. A kid who has a secure attachment has faith in their caregiver's availability and responsiveness, which fosters a sense of safety and security that allows them to explore their surroundings and form long-lasting, healthy relationships with others.

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The amount of energy radiated by a star is directly proportional to its surface area and the fourth power of its temperature. Therefore, the more luminous star radiates 81 times more energy than the less luminous star.

If two stars have the same surface area but one has 3 times the temperature of the other, we can calculate the ratio of the energy radiated by the more luminous star compared to the less luminous star.

Let's assume the surface area of both stars is A. The energy radiated by the less luminous star can be represented as E1, and the energy radiated by the more luminous star can be represented as E2.

Since both stars have the same surface area, we can say that \(E1 ∝ T1^4\) and\(E2 ∝ T2^4\), where T1 is the temperature of the less luminous star and T2 is the temperature of the more luminous star.

Given that T2 = 3T1, we can substitute this value into the equations:

\(E1 ∝ T1^4\)
\(E2 ∝ (3T1)^4\)

Simplifying, we get:
\(E1 ∝ T1^4\)
\(E2 ∝ 81T1^4\)

Therefore, the more luminous star radiates 81 times more energy than the less luminous star.

In conclusion, if two stars have the same surface area but one has 3 times the temperature of the other, the more luminous star will radiate 81 times more energy than the less luminous star.

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The magnitude of the second charge is approximately 0.1111 Coulombs.

To determine the magnitude of the second charge, we can use Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

The formula for Coulomb's law is:

\(F = (k \times |q1 \times q2|) / r^2\)

Where:

F is the force between the charges,

k is the Coulomb's constant (\(k = 9.0\times 10^9 Nm^2/C^2\)),

q1 and q2 are the magnitudes of the charges, and

r is the distance between the charges.

In this case, we have:

F = 60 N

q1 = \(-6.0 \times 10^{-6} C\)

r = 0.040 m

Plugging these values into the formula, we can solve for q2:

60 N = \(9.0\times 10^9 Nm^2/C^2 \times (-6.0 x 10^{-6}C)\times q2) / (0.040 m)^2\)

To simplify the equation, we can remove the absolute value since the charges are both negative. We also rearrange the equation to solve for q2:

q2 = \((60 N\times (0.040 m)^2) / (9.0\times 10^9 Nm^2/C^2\times 6.0\times 10^{-6} C)\)

Calculating the expression:

q2 =\((60 N\times 0.0016 m^2) / (5.4\times 10^3 C^2)\)

q2 ≈ 0.1111 C

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The frictional force is 11 N and the block’s acceleration is 0.14 m/s².

It is defined as the numerical value that indicates the amount of friction present between the surfaces of two bodies. The lower the coefficient of friction, the lower the friction between the surfaces, and the higher coefficient of friction the higher the friction force between them.

For part (a):

Taking x component:

F(x) = F cosθ

Taking y component:

F(y) = -F sinθ

Applying Newton’s second law to the y-axis:

F(N) = 15 sin40 + (3.5)(9.8)

F(N) = 44 N

coefficient U = 0.25

f(k) = 11 N

For part (b):

Applying Newton’s second law to the x-axis:

a = ((15) cos40 - 11)/3.5

a = 0.14 m/s²

Thus, the frictional force is 11 N and the block’s acceleration is 0.14 m/s².

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The density of paraffin at 100°C is approximately 419 kg/m³.

The density of a substance changes with temperature due to the expansion or contraction of its volume.

The change in density with temperature is given by the following formula:

ρ₂ = ρ₁/(1+β(T₂-T₁))

where;

Using this formula, we can calculate the density of paraffin at 100°C:

ρ₁ = 8.00 x 10² kg/m³ (at 0°C)

ρ₂ = ? (at 100°C)

β = 9.00 x 10^-4 K^-1

T₁ = 0°C

T₂ = 100°C

ρ₂ = ρ₁ / (1 + β(T₂ - T₁))

ρ₂ = 8.00 x 10² kg/m³ / (1 + 9.00 x 10^-4 K^-1 x (100°C - 0°C))

ρ₂ = 8.00 x 10² kg/m³ / (1 + 0.09)

ρ₂ = 4.19 x 10² kg/m³

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\(\huge\boxed{Answer:}\)

The current that would pass through your body if you touch the two terminals of a 9.0 V battery with your two hands depends on the resistance of your body, which varies. As a general guideline, the resistance of the human body is typically in the range of 1,000 to 100,000 ohms. Assuming a resistance of 10,000 ohms, the current passing through your body would be approximately 0.9 milliamperes (mA), which is generally considered safe for brief contact with the skin, but prolonged contact or higher currents can be dangerous and potentially lethal.

It is difficult to say precisely what current would pass through your body if you were to touch the terminals of a 9.0 V battery with your hands. There are a number of factors that would influence the actual current:

Your body's resistance - The resistance of your body depends on factors like skin moisture, calluses on your hands, and even how recently you have exercised. Dry hands with calluses have higher resistance, while sweaty hands have lower resistance.

Touching the terminals - The exact area of contact with the terminals and how firmly you grasp them will affect the current. Larger contact area and firmer grip leads to higher current.

Length of contact - The longer you touch the terminals, the higher the current will likely be as your skin resistance decreases over time.

Health of the battery - A new battery with full charge will provide higher current than one that is partially drained.

In general, for a healthy person with average resistance (around 1000 ohms), touching 9 V battery terminals briefly could result in currents on the order of milliamps (a few mA). However, due to the many variables at play, the current could potentially range from around 1 mA up to 10s of milliamps depending on factors like those listed above.

For reference, currents above 50-100 mA can start to cause involuntary muscle contraction and difficulty letting go. However, the 9 V battery's relatively low current capability makes severe injury unlikely in most cases. The main risks are minor shock and burns from the contacts.        

Answer: 0.5334

Explanation:

i got it right on accellus :p

Volume of value V = 0.5333 \(m^{3}\)  is below the water line .

Archimedes' principle states that a body immersed in a fluid is subjected to an upwards force equal to the weight of the displaced fluid.

F(b) = rho*g*V

F(b) = buoyant force

rho = fluid density

g = acceleration due to gravity

V = fluid volume

given

density of raft = 373 kg/\(m^{3}\)

volume of raft = 1.43 \(m^{3}\)

since, density  = mass / volume

mass of raft = density * volume

                    = 373 * 1.43 = 533.3 kg

using Archimedes principal

Force = density of liquid * volume of liquid displaced * acceleration due to                gravity

  mg        = 1000 * V * 9.8

  533.3 * 9.8 = 1000 * V * 9.8

  V = 0.5333 \(m^{3}\)

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The net force between Superman and Spiderman is 3 N, and it acts in the direction of Superman's force.

As per the question, the force exerted by :

Superman against Spiderman = 28 N

Spiderman against Superman = 25 N,

We can determine the net force and its direction by considering the following:

To find the net force, we need to subtract the forces exerted in opposite directions. Since Superman and Spiderman are pulling against each other, we have:

Net force = Force exerted by Superman - Force exerted by Spiderman

Net force = 28 N - 25 N

Net force = 3 N

The net force between Superman and Spiderman is 3 N.

To determine the direction of the net force, we need to consider the signs of the forces. Since Superman's force is greater than Spiderman's force, the net force will be in the direction of Superman's force.

Thus, the net force of 3 N is in the direction of Superman's force.

Therefore, the net force between Superman and Spiderman is 3 N, and it acts in the direction of Superman's force.

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Answer:

The cat ran a distance of 240 meters.

Explanation:

Given the following data;

Time, t = 30 secs.

Velocity, v = 8 m/s.

Distance =?

The formula for calculating distance is given by;

\( d = vt\)

Substituting into the equation, we have;

\( d = 8 * 30\)

d = 240m.

Answer:

Presión = 11111,11 N/m

Explanation:

Dados los siguientes datos;

Diámetro = 1,5 cm a metros = 1,5 / 100 = 0,015 m

Fuerza = 2N

Para encontrar la presión;

En primer lugar, determinaríamos el área del círculo usando la fórmula;

Área del círculo = πr²

Radio = diámetro/2 = 0.015/2 = 0.0075 m

Área de la jeringa = 3.142 * 0.0075²

Área de la jeringa = 3.142 * 0.00005625

Área de la jeringa = 0.00018 metros

Ahora, podemos encontrar la presión usando la fórmula;

Presión = fuerza/área

Presión = 2/0.00018

Presión = 11111,11 N/m

Answer:

B i believe

Explanation:

No, rays traveling parallel to the axis of a concave mirror do not pass through the center of curvature after they are reflected.

When parallel rays of light fall on a concave mirror, they are reflected and converge at a point called the focal point. The focal point is located on the principal axis, which is the line passing through the center of curvature and the midpoint of the mirror.

However, rays that pass through the center of curvature before reflection will reflect back upon themselves and pass through the center of curvature again after reflection. In other words, the rays that pass through the center of curvature are reflected back along their original path.

Rays that are not parallel to the principal axis will reflect and converge or diverge at different points depending on their angle of incidence and the position of the object relative to the mirror. The image formed by a concave mirror is a virtual or real image depending on the position of the object relative to the mirror and the distance of the image from the mirror.

In summary, parallel rays of light do not pass through the center of curvature of a concave mirror after reflection. Instead, they converge at a point called the focal point, which is located on the principal axis.

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No, rays traveling parallel to the axis of a concave mirror do not pass through the center of curvature of the mirror after they are reflected.

When a ray of light travels parallel to the axis of a concave mirror and strikes the mirror surface, it is reflected back towards the focal point of the mirror. This is known as the focal property of the concave mirror. The focal point lies on the principal axis, halfway between the vertex (center) of the mirror and the center of curvature.

However, the center of curvature is the point on the axis that is equidistant from every point on the surface of the mirror. Therefore, rays parallel to the axis will not necessarily pass through the center of curvature after they are reflected. In fact, rays passing through the center of curvature will be reflected back onto themselves, creating an image at the same location as the object (a 1:1 magnification).

So, while the focal point and center of curvature are related properties of a concave mirror, they serve different functions in determining the path of light rays as they reflect off the mirror surface.

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Answer: 10 blocks north

Explanation:

Answer:

4 blocks south

Explanation:

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Answer:

F = 4.48N

Explanation:

In order to calculate the net gravitational force on the rocket, you take into account the formula for the gravitational force between two objects, which is given by:

\(F=G\frac{m_1m_2}{r^2}\)         (1)

G: Cavendish's constant = 6.674*10^-11 m^3kg^-1s^-2

r: distance between the objects

You have a rocket at the middle of the distance between Earth and Moon, then, you have opposite forces on the rocket.

If you assume the origin of a system of coordinates at the rocket position, with the Moon to the left and the Earth to the right, you have:

\(F=G\frac{M_em}{r_1^2}-G\frac{M_mm}{r_2^2}\)       (2)

Me: mass of the Earth = 5.98*10^24 kg

Mm: mass of the Moon = 7.35*10^22 kg

m: mass of the rocket = 1200kg

r1: distance from the rocket to the Earth = 3.0*10^8m

r: distance between rocket and Moon = 3.84*10^8m - 3.0*10^8m = 8.4*10^7m

You replace the values of the parameters in the equation (2):

\(F=Gm[\frac{M_e}{r_1^2}-\frac{M_m}{r_2^2}]\\\\F=(6.674*10^{-11}m^3kg^{-1}s^{-2})(1200kg)[\frac{5.98*10^{24}kg}{(3.0*10^8m)^2}-\frac{7.35*10^{22}kg}{(8.4*10^7m)^2}]\\\\F=4.48N\)

The net force exerted over the rocket is 4.48N

The net gravitational force on the rocket from the earth and moon is 4.48N.

The gravitational force between two objects is:

F = Gm₁m₂/r²

Where m₁, m₂, are masses, r is the distance between them and G is gravitational constant = 6.67 * 10⁻¹¹ Nm²/kg²

Let us assume that the rocket is at the origin with the Moon to the left and the Earth to the right. Hence:

\(F=G[\frac{m_em}{r_1^2} -\frac{m_mm}{r_2^2} ]\)

Where me is the mass of earth =  5.98 × 10^24 kg,

r₁ is the distance from earth to rocket = 3.0 × 10^8 m,

m is the mass of rocket = 1200 kg,

mm is the mass of moon = 7.35 × 10^22 kg,

r₂ is the distance from moon to rocket = 3.84 × 10^8 m - 3.0 × 10^8 m = 0.84 × 10^8 m

Hence:

\(F=6.67*10^{-11}[\frac{1200*5.98*10^{24}}{3*10^8}-\frac{1200*3*10^{22}}{0.84*10^8} ]\\\\F=4.48\ N\)

The net gravitational force on the rocket from the earth and moon is 4.48N.

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(A) 300 J of labour are needed to stop the hoop. The hoop has a kinetic energy of \((1/2)mv2 = 300 J.\)  (B) The hoop will roll up the slope for 2.55 m before coming to a stop and rolling back down.

(A) The hoop has a kinetic energy of\((1/2)mv2 = 300 J\). The needed labour is also 300 J since an equivalent amount of work must be done to stop the hoop.

(b) When the hoop ascends the slope, its initial kinetic energy is transformed into potential energy. All of the kinetic energy has been transformed into potential energy at the highest point, making the potential energy there also 300 J. m represents the hoop's mass, and h is the height attained, therefore mgh = 300 J. We arrive at h = 3.06 m after solving for h. The distance climbed up the incline is then calculated using the formula \(d = h/sin(),\)  where is the incline's angle. With the values substituted, we obtain d = 2.55 m.

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Explanation:

For 787 dreamliner airplane,

Mass, \(m_1=2450\ kg\)

Height, \(h_1=11500\ m\)

Potential energy of 787 dreamliner airplane is given by :

\(E_1=m_1gh_1\\\\E_1=2450\times 9.8\times 11500\\\\E_1=2.76\times 10^8\ J\)

For a boeing 747,

Mass, \(m_2=3150\ kg\)

Height, \(h_2=9000 \ m\)

:

Potential energy of a boeing 747 airplane is given by

\(E_2=m_2gh_2\\\\E_2=3150\times 9.8\times 9000\\\\E_2=2.77\times 10^8\ J\)

So, the potential energy of a boeing 747 airplane is more than the 787 dreamliner airplane.

At first, we have to calculate the velocity of the car after 2.0 minutes and after that, we can calculate the velocity at the 90.0 m mark of the journey.It is given that a car travels at uniform acceleration from rest, So u = 0 (initial velocity).

We know that,v = u + atAlso, s = ut + (1/2)at^2where,v = final velocityt = time taken to reach that velocitya = accelerationand s = distancet = 2.0 minutes = 2.0 × 60 seconds = 120 secondsLet's calculate the velocity of the car after 2.0 minutes, v1:v1 = u + atv1 = 0 + a × tv1 = a × t --- equation (1)Again, when the car hits the 10.0 m mark, it begins to accelerate uniformly at 2.0 m/s^2.

Let's find the time taken to cover 10.0 m, using equation of motion,v^2 = u^2 + 2aswhere u = 0 (initial velocity), a = 2.0 m/s^2, s = 10.0 mv^2 = 2asv^2 = 2 × 2.0 × 10.0v^2 = 40.0v = √40.0v = 2√10.0 m/sLet's calculate the time taken, t1:t1 = (v - u) / at1 = v / at1 = (2√10.0) / 2.0t1 = √10.0 sSo, the total time taken to cover 10.0 m, t = t1 + 120 t = √10.0 + 120The distance covered after 10.0 m, s = 90.0 - 10.0 = 80.0 mLet's calculate the velocity of the car at 90.0 m mark using equation (1):v1 = a × tv1 = a(t - √10.0).

Now, using the first equation of motion,v2 = v1 + at2v2 = a(t - √10.0) + at2v2 = a(t - √10.0 + t2)On integrating the equation of motion, v = u + atWe get,s = ut + (1/2)at^2Now,v = u + at --- equation (2)By substituting the value of t in equation (2), we get the value of v.Using equation (2), v2 = a(t - √10.0 + t2)Let's find t2:v = u + atwhere, v = 0 (final velocity), u = v1 = a(t - √10.0) and a = 2.0 m/s^2v = a(t - √10.0)0 = 2.0(t - √10.0)t - √10.0 = 0t = √10.0Therefore, t2 = t - √10.0t2 = √10.0 + 120 - √10.0t2 = 120 seconds.

Finally, the velocity of the car at the 90.0 m mark of the journey,v2 = a(t - √10.0 + t2)v2 = 2.0 (120 - √10.0 + 120)v2 = 480 - 2√100v2 = 480 - 20v2 = 460 m/s.

Given,a car travels at a uniform acceleration from rest. After 2.0 minutes the car has hit the 10.0 m mark on it's journey and at that point begins to uniformly accelerate at 2.0 m/s2. We have to find out the velocity of the car at the 90.0 m mark of the journey. We can find it by calculating the velocity of the car after 2.0 minutes and then after that we will calculate the velocity at the 90.0 m mark of the journey. At first, we have to calculate the velocity of the car after 2.0 minutes and after that, we can calculate the velocity at the 90.0 m mark of the journey. It is given that a car travels at uniform acceleration from rest, So u = 0 (initial velocity) We know that,v = u + atAlso, s = ut + (1/2)at2where,v = final velocityt = time taken to reach that velocitya = accelerationand s = distance t = 2.0 minutes = 2.0 × 60 seconds = 120 seconds.

Let's calculate the velocity of the car after 2.0 minutes, v1:v1 = u + atv1 = 0 + a × tv1 = a × t --- equation (1)Again, when the car hits the 10.0 m mark, it begins to accelerate uniformly at 2.0 m/s2. Let's find the time taken to cover 10.0 m, using the equation of motion,v2 = u2 + 2aswhere u = 0 (initial velocity), a = 2.0 m/s2, s = 10.0 mv2 = 2asv2 = 2 × 2.0 × 10.0v2 = 40.0v2 = √40.0v2 = 2√10.0 m/s.

Let's calculate the time taken, t1:t1 = (v - u) / at1 = v / at1 = (2√10.0) / 2.0t1 = √10.0 sSo, the total time taken to cover 10.0 m, t = t1 + 120t = √10.0 + 120The distance covered after 10.0 m, s = 90.0 - 10.0 = 80.0 mLet's calculate the velocity of the car at 90.0 m mark using equation (1):v1 = a × tv1 = a(t - √10.0)Now, using the first equation of motion,v2 = v1 + at2v2 = a(t - √10.0) + at2v2 = a(t - √10.0 + t2)On integrating the equation of motion, v = u + atWe get,s = ut + (1/2)at2Now,v = u + at --- equation (2).

By substituting the value of t in equation (2), we get the value of v.Using equation (2), v2 = a(t - √10.0 + t2)v2 = 2.0 (120 - √10.0 + 120)v2 = 480 - 2√100v2 = 480 - 20v2 = 460 m/sThus, the velocity of the car at the 90.0 m mark of the journey is 460 m/s.

The velocity of the car after 2.0 minutes is a × t = 2.0 × 2.0 = 4.0 m/s.The velocity of the car at 90.0 m mark of the journey is 460 m/s.

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