Which of the following would most likely result in a flash flood?

An ice storm

Melting snow on a plain

Excess precipitation in one location

A tornado with speeds of 110 km/h

The shortest time in which a car could accelerate from 0 mph to 60 mph is [tex]13.7s[/tex].

Further explanation:

The opposite force acting on the body is known as frictional force. It always acts in the opposite direction of motion of body.

Concept used:

The force applied to a body to keep it at rest is known as the static friction force. It always acts opposite to the direction of motion of body. It is defined as the product of coefficient of friction and the normal force acting on the body.

The expression for the normal reaction of the body is given as.

[tex]N = mg[/tex]  

The expression for the net force is given as.

[tex]{F_{net}} = ma[/tex]                                 …… (1)

The expression for the static friction is given as.

[tex]{F_s} = {\mu _s}N[/tex]

The expression for the balanced forces is given as.

[tex]{F_{net}} = {F_s} - {F_r}[/tex]

Substitute[tex]{\mu _s}N[/tex] for [tex]{F_s}[/tex] and  for[tex]{F_r}[/tex] in the above expression.

[tex]\begin{aligned}{F_{net}}&={\mu _s}N-{\mu _r}N\\&= \left( {{\mu _s} - {\mu _r}} \right)N \\ \end{aligned}[/tex]

Substitute [tex]mg[/tex] for [tex]N[/tex] in above expression.

[tex]{F_{net}}=\left({{\mu _s}-{\mu _r}}\right)\left( {mg}\right)[/tex]       …… (2)

Compare equation (1) and (2) we get.

[tex]a=g\left({{\mu _s}-{\mu _r}}\right)[/tex]                              …… (3)

Here, [tex]a[/tex] is the acceleration of the body, g is the acceleration due to gravity, [tex]{\mu _s}[/tex] is the coefficient of static friction and  is the coefficient of reactive force.

The expression for the first equation of motion is given as.

[tex]v = u + at[/tex]      

Rearrange the above expression for time is given as.

[tex]\fbox{\begin\\t = \dfrac{{\left( {v - u} \right)}}{a}\end{minispace}}[/tex]                              …… (4)

Here, [tex]v[/tex] is the final velocity, [tex]u[/tex] is the initial velocity and [tex]t[/tex] is the time.

Substitute [tex]1[/tex] for [tex]{\mu _s}[/tex], [tex]0.8[/tex] for[tex]{\mu _r}[/tex] and [tex]9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex] in equation (3).

[tex]\begin{aligned}a&=\left( {9.8\,{\text{m/}}{{\text{s}}^{\text{2}}}}\right)\left( {1 - 0.8}\right)\\&=1.96\,{\text{m/}}{{\text{s}}^{\text{2}}} \\ \end{aligned}[/tex]

Substitute [tex]0\,{\text{mph}}[/tex] for [tex]u[/tex], [tex]60\,{\text{mph}}[/tex] for [tex]v[/tex] and [tex]1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]a[/tex] in equation (4).

[tex]\begin{aligned}t&=\frac{{\left( {60\,{\text{mph}}-0\,{\text{mph}}}\right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&=\frac{{60\,{\text{mph}}\left( {\frac{{0.447\,{\text{m/s}}}}{{1\,{\text{mph}}}}} \right)}}{{1.96\,{\text{m/}}{{\text{s}}^{\text{2}}}}}\\&= 13.7\,{\text{s}} \\ \end{aligned}[/tex]

Thus, the time required to accelerate a car is [tex]\fbox{\begin\\13.7\, {\text{s}}\end{minispace}}[/tex].

Learn more:

1.  Conservation of momentum https://brainly.com/question/9484203.

2.  Motion under friction https://brainly.com/question/7031524.

3. Net force on the body https://brainly.com/question/4033012.

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Force, friction, Acceleration, acceleration due to gravity, normal, weight, mass, motion, sliding, sled, hill, inclined, plane, coefficient of friction, angle of inclination, 13.68 s, 13.69 s.  

Final Answer:

So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.

Explanation:

To solve this problem, we need to understand how the force of friction allows a car to accelerate and then apply the equations of motion to determine the shortest time it would take for the car to accelerate from 0 to 60 miles per hour (mph).

First, we need to define our variables and constants:
- The static friction coefficient between rubber and concrete, μs, which is 1.00. This is the factor that will limit our acceleration since it represents the maximum frictional force that can be exerted before sliding begins.
- The acceleration due to gravity, g, which is 9.81 m/s^2.
- The initial speed of the car, 0 mph, which needs to be converted to meters per second (m/s).
- The final speed of the car, 60 mph, which also needs to be converted to meters per second.

The force of static friction (f_friction) responsible for the acceleration can be calculated by multiplying the static friction coefficient (μs) with the normal force (N). The normal force, in this case, is equal to the weight of the car, which is the mass (m) times the acceleration due to gravity (g). Since we are looking for the highest possible acceleration, we can assume that the force of static friction is at its maximum value.

f_friction = μs * m * g

However, we don't need to know the mass of the car because it will cancel out in the equation of motion. The maximum possible acceleration (a_max) happens when the force of static friction is at its highest, and it can be calculated using:

a_max = f_friction / m = μs * g

Now we have the maximum possible acceleration the tires can provide on concrete, which is:

a_max = 1.00 * 9.81 m/s^2
a_max = 9.81 m/s^2

Next, let's convert the final velocity from mph to m/s:

v_final(mph) = 60 mph
v_final(m/s) = v_final(mph) * 0.44704 (conversion factor from mph to m/s)
v_final(m/s) = 60 * 0.44704
v_final(m/s) ≈ 26.8224 m/s

With the initial velocity (v_initial) being 0 m/s (since we start from rest), we can use one of the kinematic equations to find the time (t):

v_final = v_initial + a_max * t
26.8224 m/s = 0 m/s + 9.81 m/s^2 * t

Solving for t, we get:

t = v_final / a_max
t ≈ 26.8224 m/s / 9.81 m/s^2

t ≈ 2.734 seconds

So, under ideal conditions with the assumption of maximum static friction being available throughout the acceleration, the shortest time in which a typical car could accelerate from 0 to 60 mph on rubber-on-concrete would be approximately 2.734 seconds.

Answer:

9800 W

Explanation:

The elevation difference is 55.3 mm when a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, displacing water downward and kerosene upward.

When a pressure of 98.0 Pa (gage) is applied to the right tube of the manometer, the water in the right tube is displaced downward by the same distance that the kerosene in the left tube is displaced upward, maintaining equilibrium.

To find the elevation difference, we can use the hydrostatic pressure formula:

P = ρgh

Where:

P = pressure difference (98.0 Pa in this case)

ρ = density of the fluid (for water, ρ_water ≈ 1000 kg/m³, for kerosene, ρ_kerosene ≈ 820 kg/m³)

g = acceleration due to gravity (approximately 9.81 m/s²)

h = elevation difference

Since the water and kerosene columns have opposite elevation changes, we can write:

98.0 Pa = (ρ_water * g * h) - (ρ_kerosene * g * h)

Now, solve for h:

98.0 Pa = (1000 kg/m³ * 9.81 m/s² * h) - (820 kg/m³ * 9.81 m/s² * h)

98.0 Pa = (9810 h) - (8034 h)

98.0 Pa = 1776 h

h = 98.0 Pa / 1776 = 0.0553 meters = 55.3 mm

So, when a pressure of 98.0 Pa (gage) is applied to the right tube, the elevation difference between the two free surfaces in the manometer is 55.3 mm.

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The correct answer is a) droughts. Hurricanes are typically associated with heavy rain and high winds, which can lead to flooding.

Hurricanes are powerful storms that cause a variety of hazards, including inland flooding, storm surges, and tornadoes.

Inland flooding can occur when the heavy rainfall associated with a hurricane overflows rivers and streams. Storm surges are the rise in sea level that happens when a hurricane's strong winds push water onshore. This can cause significant flooding and damage in coastal areas. Tornadoes can be produced by hurricanes when thunderstorms present within the hurricane system become severe.

However, droughts are not typically a direct result of hurricanes. In fact, the heavy rainfall associated with hurricanes often alleviates drought conditions.

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The scale read, in grams, as the hamster slides down along the wedge-shaped block is 931.5 g.

Given that:

Mass of the hamster, m = 190 g

Mass of the block, M = 820 g

The angle of the wedge block = 40°

Since the block does not have friction, the only force acting on the hamster is the normal force.

The normal component of the force is:

[tex]F_n=mgcos(\theta)[/tex]

This normal force has a downward component equal to:

[tex]F_d=F_ncos(\theta)[/tex]

[tex]F_d=mgcos^2(\theta)[/tex]

The net force acting on the hamster is:

F = Mg + mgcos²(θ)

  = (0.82 × 9.8) + (0.19 × 9.8 × cos²(40°))

  = 9.128 N

In grams, this can be found as:

[tex]\text{F}=\frac{9.128}{9.8} \times1000[/tex]

  [tex]=931.5\text{ g}[/tex]

Hence, the force is 931.5 g.

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The complete question with the figure is given below.

The figure shows a 190 g hamster sitting on an 820 g wedge-shaped block. The block, in turn, rests on a spring scale. An extra-fine lubricating oil having μs=μk=0 is sprayed on the top surface of the block, causing the hamster to slide down. The friction between the block and the scale is large enough that the block does not slip on the scale. What does the scale read, in grams, as the hamster slides down?

The question explores a physics scenario where a hamster slides off a lubricated block due to negligible friction. The block doesn't move due to sufficient friction with the scale. Concepts of forces, friction, and mechanical equilibrium are integral to understanding this.

The subject in question revolves around a scenario in Physics involving a hamster on a wedge-shaped block sprayed with lubricating oil, creating a scenario with negligible friction. In this case, the block is on a spring scale and the friction between the block and scale is enough to prevent slipping. This situation explores concepts of mass, force, friction, and mechanical equilibrium.

When the block is sprayed with lubricant, the friction between the hamster and the block decreases to nearly zero (μs=μk=0). This causes the hamster to slide off. The block, however, doesn't move because the friction between the block and the scale is high enough.

This scenario can be analyzed in a few different ways. One way could be through the use of Newton's laws of motion, considering the forces applied to the hamster and block separately. Another way could be using principles of energy conservation, though this might be more complex due to the dynamic nature of the situation.

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In physics, the definition of work is simply the product of force exerted and the displacement. Therefore in this case, work done by the ant is:

Work = 0.00805 N * 9.80 cm * (1 m / 100 cm)

Work = 7.889 x 10^-4 J

The percent uncertainty of a 5 lb bag with an uncertainty of ±0.4 lb is calculated using the formula (SA / A) x 100%, resulting in a percent uncertainty of 8%.

To calculate the percent uncertainty for a 5 lb bag with an uncertainty of ±0.4 lb, you would use the following formula:

% uncertainty = (SA / A) × 100%

Where A is the expected value (5 lb in this case) and SA is the uncertainty in the value (0.4 lb). Plugging in the numbers:

% uncertainty = (0.4 lb / 5 lb) × 100% = 8%

This means the percent uncertainty of the bag's weight is 8%.

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The speed of sound within a substance depends on the bulk modulus and density of the material. An increase in bulk modulus, with density remaining constant, would increase the speed of sound within the material, given that the speed of sound is proportional to the square root of the bulk modulus.

The speed of sound within a substance is primarily determined by two physical properties of the material, namely the bulk modulus and density. The bulk modulus measures a substance’s resistance to uniform compression. The formula used to calculate the speed of sound within a substance is v = sqrt(B/p), where 'v' represents the speed of sound, 'B' signifies the bulk modulus of the material, and 'p' indicates the density of the material.

As such, if the bulk modulus of a material were to increase while the density remained constant, the speed of sound within that material would also increase because the speed of sound is directly proportional to the square root of the bulk modulus. For instance, since solids and liquids are generally more rigid (i.e., have higher bulk modulus values) than gases, the speed of sound tends to be greater in these media than in gases, assuming density remains constant.

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Answer;

A legislative body oversees city departments directly.

Explanation;

City commission government is a form of local government in the United States. In a city commission government, voters elect a small commission, typically of five to seven members, on a plurality-at-large voting basis.

The commissioners constitute the legislative body of the city and, as a group, are responsible for taxation, appropriations, ordinances, and other general functions.

Answer:

a legislative body oversees city departments directly

Explanation:

To find the time it takes for the box to fall off the truck, we use the equation of motion and the acceleration of the box relative to the truck. By setting the displacement equal to the distance from the rear of the truck and solving for time, we find that it takes approximately 1.34 seconds for the box to fall off.

To find the time it takes for the box to fall off the truck, we need to first find the acceleration of the box relative to the truck. Given that the truck is accelerating forward at 5.0 m/s² and the crate is accelerating forward at 2.94 m/s², we can subtract these accelerations to find the crate's acceleration relative to the truck, which is -2.06 m/s². Since the box falls off when it reaches a certain distance from the rear of the truck, we can use the equation of motion to find the time it takes to travel that distance.

The equation is:

d = ut + 1/2at²

Where:

Plugging in the values, we have:

1.81 m = 0 * t + 0.5 * (-2.06) * t²

Simplifying the equation:

1.81 = -1.03t²

Multiplying both sides by -1:

-1.81 = 1.03t²

Dividing both sides by 1.03:

t² = -1.81/1.03

When we take the square root of each side, we obtain:

t = ±√(-1.81/1.03)

The negative value of time implies an event before the start of motion, so we discard it. Therefore, the time it takes for the box to fall off the truck is approximately 1.34 seconds.

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The problem is about finding the reaction components at a ball-and-socket joint and the tension in wires supporting a sign. You can do this by breaking the forces and moments into their x,y,z components, setting up force balance and moment balance equations, and solving for the unknowns.

This problem pertains to static equilibrium, where both the sum of forces and the sum of torques are zero. To begin, let's establish a coordinate system where the ball-and-socket joint 'a' is considered the origin.

Assuming the tension in the wires is T and the angles of the wires to the x, y, z axes are known, you can break the tensions into their respective components using trigonometric principles. Let's consider Tbc and Tbd to be the tensions in the strings.

For the x-component we would sum up the forces in the x-direction and set it equal to zero. This can be represented as ∑Fx = Tbc,x + Tbd,x=0

The same goes for the y and z components with the z component taking into account the force due to gravity on the sign. Hence, ∑Fy = Tbc,y + Tbd,y=0 and ∑Fz = Tbc,z + Tbd,z - weight of the sign = 0. Here, the weight of the sign is 100kg*9.81m/s2

By balancing torques about each axis, you can find the tensions in the strings. The tension force should be the same in each wire if they make the same angles with the axes and are of equal length.

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This problem in Physics requires setting up equilibrium equations based on the given mass and calculating forces at the ball-and-socket joint and cable pressures.

The subject of this question falls under the branch of

Physics

referred to as statics, specifically the study of systems in equilibrium. Equilibrium implies that an object is neither accelerating nor rotating, meaning the sum of all forces and the sum of all torques acting on the object must both be zero. The 'x', 'y' and 'z' components referred to are likely part of a coordinate system used to describe the forces at work on the sign.

To determine these, one would need to set up equilibrium equations based on the known details about the system, which could include the known weight (provided by the mass of the sign and gravity), any relevant distances for calculating torques, and any other forces present such as tension in wires. The process of solving this system of equations would then yield the desired reaction components at the ball-and-socket joint and the tensions in the wires.

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Answer:

Electric charge, q = 323910 C

Explanation:

It is given that,

Net electric flux, [tex]\phi=3.66\times 10^{16}\ Nm^2/C[/tex]

We have to find the total electric charge on the planet. We can find it using Gauss's law. It is as follows :

[tex]\phi=\dfrac{q}{\epsilon_0}[/tex]

where

q is the net electric charge

[tex]\epsilon_0[/tex] is the electric permeability

So, net electric charge is given by :

[tex]q=\phi\times \epsilon_0[/tex]

[tex]q=3.66\times 10^{16}\ Nm^2/C\times 8.85\times 10^{-12}\ F/m[/tex]

[tex]q=323910\ C[/tex]

Hence, this is the required solution.

Answer:

Occulation

Explanation:

The sunlight is absorbed more on to the black color and no color is reflected back therefore, heat energy is released by the black color and make more hot.

When a light ray is incident on a transparent or translucent material, the light ray will return back after some fraction of it is absorbed on the material and the light ray travel in the straight direction from the material make it gleams.

White light absorbs all colors and reflects back all colors. Thus, the mixing of all reflected colors make the material white in color. A black light absorbs all colors and reflect no color.

The absorbed light energy in the black body is released as heat energy instead. This make the black colored material more hotter than other colors. Hence, a black shirt is most affected by sunlight.

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The coefficient of performance (COP) of a heat pump is a measure of its effectiveness and is defined as the ratio of the heat transfer into the warm space to the work input required by the heat pump.

The coefficient of performance (COP) of a heat pump is a measure of its effectiveness and is defined as the ratio of the heat transfer (Qh) into the warm space to the work input (W) required by the heat pump. Mathematically, COP = Qh/W.

In this case, the heat pump absorbs heat from the cold outdoors at 3°C and supplies heat to a house at 20°C. The rate of heat transfer is given as 30,000 kJ/h and the power consumed is 3 kW.

To find the COP, we need to convert the rate of heat transfer and power consumed to the same units. Since 1 W = 1 J/s and 1 kJ = 1000 J, we have Qh = 30,000 kJ/h x 1000 J/kJ / 3600 s/h = 8333.33 J/s and W = 3kW x 1000 J/W = 3000 J/s.

Substituting these values into the formula, we get COP = 8333.33 J/s / 3000 J/s = 2.78.

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To solve this problem, we use the formula:

I100 / I1 = [P / 4π(100m)^2] / [P / 4π(1m)^2]

I100 / I1 = 1 / 100^2

I100 / I1 = 10^-4

Therefore the change in intensity from 1m to 100m in decibels is:

B100 – B1 = 10 log(10^-4) dB = -40 dB

So the intensity at 100m is calculated as:

B100 = B1 – 40 dB = 140 dB – 40 dB = 100 dB

Answer:

100 dB

The intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex].

Further Explanation:

The vibration, produced in the medium, which travels as an wave of pressure or density through a medium is known as sound. The sound is a longitudinal wave and it requires a medium for its propagation.  

Given:

The distance between the place where firecracker is set off by a child and the family house is [tex]100\text{ m}[/tex].  

The intensity of the sound produced by the firecracker at the distance [tex]1\text{ m}[/tex] is[tex]140\text{ dB}[/tex].

Concept:

The intensity of the sound wave is defined as the power carried by the sound waves in the direction perpendicular to the direction of propagation per unit time.

The intensity of the sound wave is:

[tex]\fbox{\begin\\I=\dfrac{P}{4\pi r^2}\end{minispace}}[/tex]                                                        ...... (1)

Here, [tex]I[/tex] is the intesnity of the sound wave, [tex]P[/tex] is the power carried by the sound wave and [tex]r[/tex] is the distance between source and the listener.

The intensity of the sound in decible at the house is:

[tex]\fbox{\begin\\\beta _{100}=\beta _{1}+10\log \dfrac{I_{100}}{I_{1}}\end{minispace}}[/tex] ...... (2)

Here, [tex]\beta _{100}[/tex] is the intensity of the sound at the distance of [tex]100\text{ m}[/tex], [tex]\beta _{1}[/tex] is the intensity of sound produced by the firecracker at the distance of [tex]1\text{ m}[/tex], [tex]I_{100}[/tex] is the intensity of sound produced by firecracker at [tex]100\text{ m}[/tex] and [tex]I_{1}[/tex] is the intensity of sound produced by firecracker at [tex]1\text{ m}[/tex].

The ratio of the intensity of sound at distance [tex]100\text{ m}[/tex] and [tex]1\text{ m}[/tex] is:

[tex]\fbox{\begin\\\dfrac{{{I_{100}}}}{{{I_1}}}=\frac{{r_1^2}}{{r_2^2}}\end{minispace}}[/tex]                                                                                     …… (3)

Calculation:

Substitute the values in equation (3).

[tex]\begin{aligned}\frac{{{I_{100}}}}{{{I_1}}}&=\frac{{{{\left( 1 \right)}^2}}}{{{{\left( {100} \right)}^2}}}\\&={10^{ - 4}}\\\end{aligned}[/tex]

Substitute the values in equation (2).

[tex]\begin{aligned}{\beta _{100}}&=140\,{\text{dB}} - 40\,{\text{dB}} \\&=100\,{\text{dB}}\\\end{aligned}[/tex].

Thus, the intensity of the sound produced due to the firecracker at the distance of [tex]100\text{ m}[/tex] is [tex]\fbox{\begin\\100\text{ dB}\end{minispace}}[/tex]

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Answer Details:

Grade: College

Subject: Physics

Chapter: Waves and Oscillation

Keywords:

Intensity of sound, sound waves, vibration, firecracker, power carried by sound waves, 100db, 100 dB, 100dB, 100 m, bursting of firecrackers, sound inside the house, 140 db, 140 dB.

Answer:

Speed is slightly greater than 14 m/s.          

Explanation:

The ball will under go projectile motion. The horizontal velocity remains constant while there would be increase in vertical velocity due to acceleration due to gravity in downward direction.

Using first equation of motion, after 1 s, vertical velocity will be:

v = u +at

v=0+(9.8 m/s²)(1 s) = 9.8 m/s

Horizontal velocity = 10 m/s

Net velocity:

[tex]V= \sqrt{(9.8)^2+(10)^2}=\sqrt{196.04}=14.001 m/s[/tex]

Speed is slightly greater than 14 m/s.

The total length of the spring when a 17.20 kg mass is suspended from it is 34.85 cm, calculated using Hooke's Law and the original unstretched length of the spring.

To find the total length of the spring when a 17.20 kg mass is suspended from it, we use Hooke's Law, which is F = kx, where F is the force applied to the spring, k is the spring constant, and x is the extension from the spring's original unstretched length. First, we need to calculate the extension caused by the 8.40 kg mass: F = (8.40 kg)(9.8 m/s2) = 82.32 N. With k = 875 N/m, the extension (x) is F/k = 82.32 N / 875 N/m = 0.0941 m or 9.41 cm. This extension is in addition to the unstretched length of the spring (L0), given by the total length minus the extension for the 8.40 kg mass: L0 = 0.250 m - 0.0941 m = 0.1559 m.

Next, for the 17.20 kg mass, the force is F = (17.20 kg)(9.8 m/s2) = 168.56 N. The additional extension caused by this mass is x = F/k = 168.56 N / 875 N/m = 0.1926 m or 19.26 cm. Therefore, the total length of the spring with the 17.20 kg mass is L = L0 + x = 0.1559 m + 0.1926 m = 0.3485 m or 34.85 cm.

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Answer:

6.6°C

Explanation:

When both cars A and car B leave school at the same time, traveling in the same direction. car A travels at a constant speed of 75 km/h, while car B travels at a constant speed of 91 km/h, then car a would be 127.5 kilometers away from the school.

The total distance covered by any object per unit of time is known as speed. It depends only on the magnitude of the moving object. The unit of speed is a meter/second.

As given in the problem when both cars A and car B leave school at the same time, traveling in the same direction. car A travels at a constant speed of 75 km/h, while car B travels at a constant speed of 91 km/h,

The distance covered by car A after 1.7 hours = 75km/h × 1.7 hours

                                                                                  =127.5 kilometers

Thus, car A would be 127.5 kilometers away from the school after 1.7 hours.

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