The distance of east-west highway is 1.030 x 10^6m. How much is the value of mantissa in this number?
(a) 1.030
(b) 0.030
(d) 6
(c) 10​

The pilot should leave the shore at an angle of 14.04° in order to go directly to the child.

To find the angle, we can use the Pythagorean Theorem and trigonometry.

First, we need to find the distance between the child and the boat dock. Using the Pythagorean Theorem, we get:

d = √((200 m)^2 + (1500 m)^2) = 1520.69 m

Next, we need to find the time it will take for the child to reach the boat dock. Since the child is being carried by the current at a speed of 2.0 m/s, we can use the formula:

t = d/v = 1520.69 m / 2.0 m/s = 760.34 s

Now, we need to find the distance the boat will travel in this time. Since the boat has a speed of 8.0 m/s with respect to the water, we can use the formula:

d = v*t = 8.0 m/s * 760.34 s = 6082.72 m

Finally, we can use trigonometry to find the angle at which the pilot should leave the shore. Using the formula:

tan(θ) = opposite/adjacent = 200 m / 6082.72 m = 0.0329

θ = tan^-1(0.0329) = 14.04°

Therefore, the pilot should leave the shore at an angle of 14.04° in order to go directly to the child.

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(I) The focal length of the lens is 180.0 cm.

(II) The distance for which the image would be the same size as the object is 90.0 cm.

We can use the lens equation and the magnification formula to solve for the focal length of the lens and the distance for which the image would be the same size as the object.

The lens equation is:

1/f = 1/do + 1/di

where;

We are given that the distance between the image and the object is 90.0 cm and the magnification is 2. The magnification formula is:

m = - di/do

where;

Substituting these values into the magnification formula, we get:

2 = - di/do

di = -2do

Substituting these values into the lens equation, we get:

1/f = 1/do - 1/2do

Simplifying this equation, we get:

1/f = 1/2do

f = 2do

Substituting the given value of di, we get:

f = 2(-90.0 cm) = -180.0 cm

Since the focal length must be positive for a converging lens, we take the absolute value and get:

f = 180.0 cm

(ii) Distance for which the image would be the same size as the object:

When the image is the same size as the object, the magnification is 1. We can use the magnification formula to solve for the distance di:

m = - di/do

1 = - di/do

di = -do

Substituting the given value of do, we get:

di = -90.0 cm

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The distance between the boat and the camera is 0.0525 m. The boat must move closer by a distance of  35 mm to fill the width of the film.

you have a camera with a 35.0-mm-focal-length lens and 36.0-mm-wide film. you wish to take a picture of a 12.0-m-long sailboat but find that the image of the boat fills only the width of the film.

(a) Given data:

The focal length of the lens, f = 35.0 mm

Width of the film, w = 36.0 mm

Length of the sailboat, L = 12.0 m

Length of the image of the sailboat on film = Width of the film

Let the distance between the boat and the camera be x.

The object distance, u = ?

Image distance, v = f

Let M be the magnification of the image, then we have:

 M = v/u

Width of the image, W' = w

Since the image of the boat fills only half of the width of the film,

W' = w/2.

Now, using the relation,

1/f = 1/u + 1/v

Putting f = 35mm, v = 35mm, and solving for u, we get

u = 52.5 mm = 0.0525 m

Therefore, the distance between the boat and the camera is 0.0525 m.

(b)Width of the image, W = w = 36 mm

Width of the sailboat, L = 12 m

Let the new distance between the boat and the camera be y.

The new image distance, v' = f

We know that the magnification of the image is given by M = v'/u'.

And the width of the image of the boat is given by:

W = (M)(L)

Using the relation,

1/f = 1/u' + 1/v', we can find the object distance, u'.

Substituting the given values, we get

u' = 0.0175 m = 17.5 mm

Therefore, the boat must move closer by a distance of (52.5 - 17.5) mm = 35 mm to fill the width of the film.

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The minimum power density measurable is 0.1 times the dark current density divided by the elementary charge and the external quantum efficiency of the detector.

The minimum power density measurable can be calculated using the following formula:

P = Jsc / (q * η)

where P is the power density, Jsc is the short-circuit current density, q is the elementary charge, and η is the external quantum efficiency of the detector.

Assuming that the photogenerated current is 10% of the dark current, the total current density (J) can be expressed as:

J = Jdark + Jph

where Jdark is the dark current density and Jph is the photogenerated current density.

Since Jph is 10% of Jdark, we can express Jph as:

Jph = 0.1 * Jdark

The short-circuit current density (Jsc) can be approximated as Jph because the detector is operated under short-circuit conditions.

Thus, Jsc ≈ Jph = 0.1 * Jdark

Substituting this into the formula for P, we get:

P = Jsc / (q * η) ≈ Jph / (q * η) = 0.1 * Jdark / (q * η)

Therefore, the minimum power density measurable is 0.1 times the dark current density divided by the elementary charge and the external quantum efficiency of the detector.

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(a) To find the truck's velocity relative to you, we need to subtract your velocity from the truck's velocity.

Since you are driving north and the truck is approaching you in the other lane, the velocities are in opposite directions. Therefore, we can simply subtract the two speeds: Truck's velocity relative to you = Truck's velocity - Your velocity = 104 km/h - 88 km/h = 16 km/h north (since the truck is approaching you). Therefore, the truck's velocity relative to you is 16 km/h north. (b) To find your velocity relative to the truck, we need to subtract the truck's velocity from your velocity. Since you are driving north and the truck is approaching you from the opposite direction, the velocities are again in opposite directions. Therefore, we can simply subtract the two speeds: Your velocity relative to the truck = Your velocity - Truck's velocity = 88 km/h - 104 km/h = -16 km/h north (since you are moving away from the truck). Therefore, your velocity relative to the truck is 16 km/h south. (c) After you and the truck pass each other, the relative velocities change. The truck is now moving away from you, and you are moving away from the truck. Therefore, the velocity of the truck relative to you will decrease, while your velocity relative to the truck will increase. However, the magnitude of the relative velocities will remain the same, since the difference between the two velocities is still 16 km/h. The direction of the relative velocities will also change, with the truck's velocity relative to you now being 16 km/h south, and your velocity relative to the truck now being 16 km/h north.

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The forces acting on a skateboard include gravity, friction, air resistance, and applied force, and they all affect the motion of the skateboard in different ways.

When a skateboard is in motion, several forces act on it that affect its movement. The main forces acting on a skateboard are:

Gravity: This is a force that pulls the skateboard downwards towards the ground. The weight of the skateboard and the rider are also affected by gravity.

Friction: This is a force that opposes the motion of the skateboard. Friction is caused by the skateboard's wheels rolling on the ground or surface, and it can slow down the skateboard's speed.

Air resistance: This is a force that opposes the motion of the skateboard as it moves through the air. The faster the skateboard moves, the more air resistance it experiences.

Applied force: This is the force applied by the rider to propel the skateboard forward or to perform tricks. The amount of applied force determines how fast or slow the skateboard moves.

The combination of these forces affects the motion of the skateboard. If the applied force is greater than the forces of friction, air resistance, and gravity, the skateboard will accelerate. If the forces of friction, air resistance, and gravity are greater than the applied force, the skateboard will slow down or come to a stop.

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The electrostatic force each charge exerts on the other is  2.81N  .[tex]q_{1} =[/tex] [tex]3[/tex] × [tex]10^{-6} C[/tex][tex]\\q_2}[/tex]  [tex]= - 1.5[/tex] × [tex]10^{-6} C[/tex]   They are separated by a distance [tex]r = 0.12m[/tex]

Electrostatic force is [tex]F =[/tex] [tex]\frac{kq_{1} q_{2} }{r^{2} }[/tex]

[tex]F = 2.81N[/tex]

The term "scalar" refers to a category of quantities that may be fully defined by a single magnitude. Quantities called vectors can be fully represented by their direction as well as their magnitude.

How do you define a vector's magnitude?

A vector's length is considered to be its magnitude. The letter "a" stands for the dynamic array magnitude. For further information about a vector's magnitude, go to its introduction. This article deduces formulas for vectors magnitude in considerations of their coordinates in three and two dimensions, respectively.

Magnitude is simply "distance or amount," according to the definition given in physics. It illustrates the either the absolute or relative size or velocity direction for an item.

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Answer:

MCB (Miniature Circuit Breaker) is generally considered better than fuses for several reasons:

Cost-effective: MCBs are more cost-effective in the long run compared to fuses as they can be reset and reused after tripping, whereas fuses need to be replaced each time they blow.

Convenience: MCBs can be reset with the flip of a switch, while fuses need to be manually replaced, which can be a hassle.

Accuracy: MCBs are more accurate in protecting against overloading and short circuits. They can be set to trip at specific amperages, providing better protection against damage to electrical appliances and wiring.

Safety: MCBs are safer than fuses as they do not produce arc flashes during tripping, which can cause electrical fires or explosions.

Overall, MCBs are a more efficient and safer alternative to fuses for protecting electrical circuits.

Explanation:

Answer:

the correct answer is (c) -, -.

Explanation:

When an object is placed in front of a convex mirror at infinity, the image formed is a virtual image located at the focal point of the mirror.

According to the New Cartesian Sign Convention, the focal length of a convex mirror is negative. Therefore, the sign of the focal length in this case is "-".

Since the image is formed behind the mirror and is a virtual image, its distance from the mirror is negative. Therefore, the sign of the image distance in this case is also "-".

Therefore, the correct answer is (c) -, -.

The mass of the bullet is approximately 0.21 kg (to 2 significant figures).

We can use the principle of conservation of momentum to solve this problem. The total amount of momentum prior to and following the impact are equal.

Let the mass of the bullet be "m" kg.

The momentum of the bullet before the collision is:

p1 = m × 50 m/s = 50m

The momentum of the bullet and block after the collision is:

p2 = (m + 0.5 kg) × 15 m/s = 15(m + 0.5)

According to the principle of conservation of momentum:

p1 = p2

50m = 15(m + 0.5)

50m = 15m + 7.5

35m = 7.5

m = 7.5/35 = 0.2143 kg (rounded to 2 significant figures)

Therefore, the mass of the bullet is approximately 0.21 kg (to 2 significant figures).

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The color of an object has to do with the light wavelength. A white t-shirt looks yellow because it absorbs all the colors while black white all the wavelengths of light. A ripe mango is yellow because reflect all the reflects wavelengths.

The light wavelength has an impact on an object's color of which the object reflects. Because it absorbs all hues, a white t-shirt seems yellow, but a black one emits all light wavelengths. A ripe mango appears yellow because, when white light strikes it, its atomic impurities bounce all the wavelengths of light at our sight more than the other colors. ​

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The voltage dropped across R₃ is 52.68 volts.

To determine the voltage dropped across R₃, we can use the voltage divider formula:

V₃ = ( R₃ / ( R₁ + R₂ + R₃ + R₄ ) ) x ET

Substituting the given values, it can be calculated as:

V₃ = (36 Ω / ( 15 Ω + 18 Ω + 36 Ω + 13 Ω ) ) x 120 V

V₃ = ( 36 Ω / 82 Ω ) x 120 V

V₃ = 52.68 V

Therefore, the voltage dropped across R₃ is approximately 52.68 volts.

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1. The troposphere is the lowest layer of Earth's atmosphere. Most of the mass (about 75-80%) of the atmosphere is in the troposphere. Most types of clouds are found in the troposphere, and almost all weather occurs within this layer.

2. The stratosphere is very dry air and contains little water vapor. Because of this, few clouds are found in this layer and almost all clouds occur in the lower, more humid troposphere. Polar stratospheric clouds (PSCs) are the exception. PSCs appear in the lower stratosphere near the poles in winter.

3. The mesosphere lies between the thermosphere and the stratosphere. “Meso” means middle, and this is the highest layer of the atmosphere in which the gases are all mixed up rather than being layered by their mass. The mesosphere is 22 miles (35 kilometers) thick.

4. The thermosphere is a layer of Earth's atmosphere that is directly above the mesosphere and below the exosphere. It extends from about 90 km (56 miles) to between 500 and 1,000 km (311 to 621 miles) above our planet.

5. The ozone layer acts as a natural filter, absorbing most of the sun's burning ultraviolet ( UV ) rays. Stratospheric ozone depletion leads to an increase in UV -B that reach the earth's surface, where it can disrupt biological processes and damage a number of materials

Explanation: Hope This Helps!

The magnitude of the induced emf if magnetic field is reduced to zero in 0.12 seconds is 14.058 V.

The emf induced in a circular loop of wire is given by Faraday's law of electromagnetic induction:

emf = - N * (dΦ/dt)

where N is the number of turns in the loop,

Φ is the magnetic flux through the loop, and

(dΦ/dt) is the rate of change of magnetic flux through the loop.

The circular loop has a radius of 15 cm and consists of 53 turns of wire.

The magnetic field passing through the loop is perpendicular to its plane and has a magnitude of 0.45 T. The magnetic field is reduced to zero in 0.12 s.

The magnetic flux through the circular loop is given by:

Φ = B * A

where B is the magnetic field, and

A is the area of the circular loop.

Since the magnetic field is perpendicular to the plane of the loop, the area A is given by:

A = π * r²

where r is the radius of the circular loop.

Substituting the given values, we get:

A = π * (0.15 m)²= 0.0707 m²

Φ = B * A = (0.45 T) * (0.0707 m²) = 0.0319 Wb.

The rate of change of magnetic flux through the loop is given by:

(dΦ/dt) = -ΔΦ/Δt

where ΔΦ is the change in magnetic flux and

Δt is the time taken for the change.

Substituting the given values, we get:

(dΦ/dt) = -Φ/Δt = -(0.0319 Wb) / (0.12 s) = -0.266 Wb/s

Finally, substituting the values of N and (dΦ/dt) in Faraday's law, we get:

emf = - N * (dΦ/dt) = -(53) * (-0.266 V) = 14.058 V

Therefore, the magnitude of the induced emf in the circular loop is 14.058 V.

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Answer:

uhhh

Explanation:

Gino should tell his partner that she should have only done four quarter turns and should have finished the dance after 32 counts. This would have resulted in her facing only two walls at the end.

Quarter is a term used to describe the amount of rotation an object has undergone when it does a full turn. A quarter-turn is the equivalent of 90 degrees. Quarter are commonly used to measure the rotation of a wheel, a lever, a bolt, or any other object that can be rotated. It is also used in architecture to describe the amount of rotation of a staircase or other structure. In engineering, quarter-turns are used to measure the amount of torque that is being applied to a mechanism or device. Quarter are also used to measure the amount of rotation when adjusting the settings of a device.

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Answer:

50m/s.

Explanation:

Answer:

[tex]V_f=50m/s[/tex]

Explanation:

Use the the formula for final velocity (kinematic equation):

[tex]V_f=V_i+at[/tex]

Final velocity = Initial velocity + acceleration times time

Given values:

Rest - indicates 0 initial velocity

Acceleration - 10 m/s^2

Time - 5 seconds

Insert the given values:

[tex]V_f=0+5\times10[/tex]

[tex]V_f=50m/s[/tex]

To calculate time, divide the distance by speed.  hence, the time is 7.5 minutes.

Divide the distance by the speed to find the passing duration. Multiply the speed by the duration to get the distance. These formulae can be represented more simply as s=d/t, where s stands for speed, d for distance, and t for time.
Given: speed= 50 mph.
distance= 10km
1 mile = 1.6 km
1km = 1/1.6 miles= 0.625 milesmiles
We know that, time= distance/speedspeed
Time = 0.625*10/50= 0.125 hours.
1 hour = 60 minutes

hence, 0.125 hours= 60*0.125 = 7.5 minutes.

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Answer:

KE = 1/2 M V^2

V = (2 KE / M)^1/2 = (2 * 40 / .1)^1/2 = 28.3 m/s

At a pace of 8 km/hr, it would take the roller skater 3.125 hours to complete 25 km.

To determine how long it would take a roller skater to travel 25 km if their speed is 8 km/hr, we can use the formula:

time = distance / speed

In this case, the distance is 25 km and the speed is 8 km/hr. Substituting these values into the formula, we get:

time = 25 km / 8 km/hr

Simplifying, we get:

time = 3.125 hours

Therefore, it would take the roller skater 3.125 hours to travel 25 km at a speed of 8 km/hr. It's important to note that this is assuming a constant speed throughout the journey, and not taking into account any breaks or rest stops the skater may need to take.

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The fourth option best represents the proper graph, which resembles a branch of a hyperbola.

From Coulomb's Law.

F = k · (q₁ · q₂) / r²

where

Q1 and Q2 are the charges of the two particles, F is the electrostatic force separating them, k is the Coulomb's constant, and r is the separation between them.

The force becomes weaker as the two particles move apart.

Moreover, as r approaches zero, F approaches 1 / 0 =. F is endlessly powerful when there's no separation between the two particles. As a result, the force axis will never be touched by the force about distance graph.

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Answer:

a force

Explanation:

The speed of the wave remains unchanged.

The speed of a wave is determined by the product of its wavelength and frequency. This can be represented by the formula: speed = wavelength x frequency

If the frequency of a wave is doubled while the wavelength remains the same, then the product of the wavelength and frequency will remain constant. This means that the speed of the wave will also remain constant.

To understand this, let's consider an example:

Suppose the original frequency of a wave is f, and its wavelength is λ. Then, the speed of the wave is given by:

speed = λ x f

Now, if the frequency is doubled, the new frequency becomes 2f. The wavelength remains the same at λ. So, the new speed of the wave can be calculated as: new speed = λ x 2f = 2(λ x f) = 2 x speed

As we can see, the new speed is twice the original speed, which contradicts our assumption that the speed remains constant. Therefore, the speed of the wave remains unchanged when the frequency is doubled while the wavelength remains the same.

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Answer: To compare the force of electrostatic repulsion and gravitational attraction between two alpha particles, we need to calculate both forces using the given information:

Electrostatic force of repulsion between two alpha particles:

Coulomb's law states that the force of electrostatic repulsion between two charged particles is given by:

F = (kq1q2)/r^2

where k is Coulomb's constant, q1 and q2 are the charges of the two particles, and r is the distance between them.

Here, q1 = q2 = 3.2×10^-19 C (charge of alpha particle), and r = distance between two alpha particles.

We know that two alpha particles are doubly ionized helium nuclei, so they each contain two protons. Thus, their separation distance will be approximately equal to the sum of their radii, which is about 4 fm (4 × 10^-15 m).

Plugging in these values, we get:

F_elec = (8.85×10^-12 * (3.2×10^-19)^2) / (4×10^-15)^2

F_elec = 8.17×10^-28 N

Gravitational force of attraction between two alpha particles:

The force of gravitational attraction between two particles is given by:

F = G*(m1*m2)/r^2

where G is the gravitational constant, m1 and m2 are the masses of the two particles, and r is the distance between them.

Here, m1 = m2 = 6.68×10^-27 kg (mass of alpha particle), and r = distance between two alpha particles.

Plugging in these values, we get:

F_grav = (6.67×10^-11 * (6.68×10^-27)^2) / (4×10^-15)^2

F_grav = 3.71×10^-57 N

Therefore, we can see that the force of electrostatic repulsion between two alpha particles is much greater than the force of gravitational attraction between them.

F_elec/F_grav = (8.17×10^-28 N) / (3.71×10^-57 N) ≈ 2.2×10^29

Explanation:

Answer: The electrostatic force of repulsion between the two alpha particles is about 5.5 × 10^48 times stronger than the gravitational force of attraction between them.

Explanation:

The force of electrostatic repulsion between two alpha particles can be calculated using Coulomb's law:

F_e = k * (q1 * q2) / r^2

where k is the Coulomb constant (1/4πε0), q1 and q2 are the charges of the particles (in coulombs), and r is the distance between them (in meters).

For alpha particles, the charge is 3.2 × 10^-19 C and the distance between them can be assumed to be the sum of their radii, which is approximately 2 x 10^-15 m. Using the given value of the Coulomb constant and the values for charge and distance, we can calculate the force of electrostatic repulsion:

F_e = (9 × 10^9 Nm²/C²) * [(3.2 × 10^-19 C)^2 / (2 × 10^-15 m)^2]

= 1.15 × 10^-28 N

The force of gravitational attraction between two alpha particles can be calculated using Newton's law of gravitation:

F_g = G * (m1 * m2) / r^2

where G is the gravitational constant (6.67×10^-11 Nm²/kg²), m1 and m2 are the masses of the particles (in kg), and r is the distance between them (in meters).

For alpha particles, the mass is 6.68 × 10^-27 kg, and the distance between them can be assumed to be the sum of their radii, which is approximately 2 x 10^-15 m. Using the given value of the gravitational constant and the values for mass and distance, we can calculate the force of gravitational attraction:

F_g = (6.67 × 10^-11 Nm²/kg²) * [(6.68 × 10^-27 kg)^2 / (2 × 10^-15 m)^2]

= 2.1 × 10^-77 N

Comparing the two forces, we see that the electrostatic force of repulsion is much stronger than the gravitational force of attraction:

F_e / F_g = (1.15 × 10^-28 N) / (2.1 × 10^-77 N) = 5.5 × 10^48

Therefore, the electrostatic force of repulsion between the two alpha particles is about 5.5 × 10^48 times stronger than the gravitational force of attraction between them.

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The magnitude of the total displacement of Olivia's trip is 1.9 miles. The direction of the total displacement of her trip is southeast.

To find the magnitude of the total displacement, we need to use the Pythagorean theorem:
a^2 + b^2 = c^2

In this case, a = 0.9 miles (the distance she travels south), and b = 0.8 + 1.4 = 2.2 miles (the distance she travels east).

Plugging in these values, we get:
0.9^2 + 2.2^2 = c^2
0.81 + 4.84 = c^2
5.65 = c^2
c = √5.65
c = 2.376

So the magnitude of the total displacement is approximately 2.376 miles.

To find the direction of the total displacement, we need to use the tangent function:
tan θ = opposite/adjacent

In this case, opposite = 2.2 miles (the distance she travels east), and adjacent = 0.9 miles (the distance she travels south).

Plugging in these values, we get:

tan θ = 2.2/0.9
θ = tan^-1(2.2/0.9)
θ = 67.38 degrees

So the direction of the total displacement is approximately 67.38 degrees from south, or southeast.

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The following are the answers to the following questions:

There will not be a Doppler shift.

There will still be a Doppler shift observed by the radar gun.

Doppler shift is the change in frequency of a wave caused by relative motion between the source and the observer. This shift in frequency can be used to calculate the relative motion of the source and observer and is employed in a variety of applications such as radar, astronomy, and medical imaging.

1: The guns are adjusted for forward or backward motion towards or away from the gun. What happens if an automobile moves perpendicular to the gun's line of sight?

The radar gun will not detect any Doppler shift.

2: If the car is approaching or departing from a different angle, there is .

A Doppler change will still be detected, but it will be less than the maximum Doppler shift that would occur if the car moved directly towards or away from the gun.

3: As a result, if the radar gun indicates that a car is driving at a specific speed, the car is moving at the speed radar report.

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The current moving through a wire that has 0. 60 C of electrons passing through a point in 1. 5 min is  0.4 Amps (A).

The current is 0.4 Amps (A),

It is calculated using the formula

I = Q/t,

  = 0.60/1.5

  = 0.4 Amps

where Q is the charge (Coulombs) of electrons and t is the time (seconds) taken for them to pass through the point.

The positive sign for current corresponds to the direction a positive charge would move. In metal wires, current is carried by negatively charged electrons, so the positive current arrow points in the opposite direction the electrons move.

In fact, an electric current is defined as the net movement of charge(s). This includes slow drift of electrons inside a wire, or a stream of electrons, protons, or muons traveling near light speed through the vacuum inside a particle accelerator.

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The time taken by Super Mario  is 2.91 seconds to reach the mystery block.

To find the time it takes for Super Mario to reach the mystery block, we can use the kinematic equation:

∆y = v0t + (1/2)at2

Where ∆y is the change in a vertical position, v0 is the initial vertical velocity, a is the acceleration due to gravity, and t is the time.

Since the mystery block is 8.2 m above Mario's head, ∆y = 8.2 m. The initial vertical velocity, v0, is 0 m/s since Mario is starting from rest. The acceleration due to gravity in the Mushroom Kingdom, gmk, is 1/5th of that on Earth, so

a = (1/5)g = (1/5)(9.8 m/s2)

a = 1.96 m/s2.

Plugging these values into the equation, we get:

8.2 m = (0 m/s)t + (1/2)(1.96 m/s2)t2

Simplifying and rearranging, we get:

t2 = (8.2 m)/(0.98 m/s2)

Taking the square root of both sides, we get:

t = √((8.2 m)/(0.98 m/s2)) = 2.91 s

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The correct answer for a) the rocket was 19,671.65 meters (or 19.67 km) from the ground when the boosters turned off, b) rocket's upward acceleration while its boosters were firing was 11.7 m/s^2, c) the boosters were firing for 58.04 seconds, d)  the rocket is in the air for a total of 199.38 seconds

a) To find the distance from the ground when the boosters turned off, we can use the formula for displacement in terms of initial velocity, acceleration, and time:
d = v_i*t + 0.5*a*t^2

We know the initial velocity (v_i) is 0 m/s, the acceleration (a) is 11.7 m/s^2, and the final velocity (v_f) is 679 m/s.

We can use the formula v_f = v_i + a*t to find the time (t) that the boosters were firing:

679 m/s = 0 m/s + 11.7 m/s^2 * t

t = 58.04 s

Now we can plug this time into the displacement formula:

d = 0 m/s * 58.04 s + 0.5 * 11.7 m/s^2 * (58.04 s)^2

d = 19,671.65 m

So the rocket was 19,671.65 meters (or 19.67 km) from the ground when the boosters turned off.

b) The prototype rocket's upward acceleration while its boosters were firing was 11.7 m/s^2, as given in the question.

c) As calculated in part a), the boosters were firing for 58.04 seconds.

d) To find the total time the rocket is in the air, we need to find the time it takes to reach its maximum height and the time it takes to fall back to the ground.

We can use the formula d = v_i*t + 0.5*a*t^2 again, with the initial velocity (v_i) being 679 m/s, the acceleration (a) being -9.8 m/s^2 (due to gravity), and the displacement (d) being the difference between the maximum height and the height when the boosters turned off:

42.3 km - 19.67 km = 22.63 km = 22,630 m

22,630 m = 679 m/s * t + 0.5 * -9.8 m/s^2 * t^2

Solving this quadratic equation gives us two solutions for t, but only one is positive:

t = 70.67 s

So the time it takes to reach the maximum height is 70.67 seconds.

The time it takes to fall back to the ground is the same, since the initial and final velocities are the same but in opposite directions. Therefore, the total time the rocket is in the air is:

t_total = 58.04 s + 70.67 s + 70.67 s = 199.38 s

So the rocket is in the air for a total of 199.38 seconds before striking the ground.

To know more about acceleration, refer here:

https://brainly.com/question/28164989#

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