On a post flight inspection of your aircraft after an aborted takeoff due to an elevator malfunction, you find that the elevator control cable has broken. According to NTSB 830, you
a) must immediately notify the nearest NTSB office
b) should notify the NTSB within 10 days
c) must file a NASA report immediately

Kinetic friction has a coefficient of 0.52 while static friction has a coefficient of 0.67.

A force called friction prevents motion from occurring when two surfaces come into contact. Based on the characteristics of the two surfaces, their roughness, and the pressure used to press them together, it moves the objects in the opposite direction from the motion or planned motion.

a. The formula friction = force of friction / normal force is used to calculate the coefficient of static friction.

The frictional force needed to move the chair is 165 N, and the normal force equals the chair's weight (25 kg * 9.8 m/s2 = 245 N).

The static friction coefficient is therefore equal to 165 N / 245 N = 0.67.

b. The formula friction = force of friction / normal force is used to determine the coefficient of kinetic friction.

The frictional force needed to maintain the chair's motion at a constant speed is 127 N, whereas the normal force remains 245 N.

Therefore, the kinetic friction coefficient is equal to 0.52 for 127 N/245 N.

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Answer:

Their wavelength

Explanation:

Light is a form of energy that travels as waves. Their length — or wavelength — determines many of light's properties. For instance, wavelength accounts for light's colour and how it interacts with matter. The range of wavelengths, from super short to very long, is known as the light spectrum.. Please mark this answer the brainliest. I need it. thanks.

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The speed of the package just before hitting the ground is 46.942 m/s.

We can solve this problem using the kinematic equations of motion. Let's assume that the positive direction is upwards and take the acceleration due to gravity to be -9.8 m/s².

(a) To find how long the package takes to reach the ground, we can use the equation:

y = y0 + v0t + (1/2)at²

Plugging in the values, we get:

0 = 59 m + (11 m/s)t + (1/2)(-9.8 m/s^2)t²

Simplifying and solving for t, we get:

t = 4.79 seconds

Therefore, the package takes approximately 4.79 seconds to reach the ground.

(b) To find the speed at which the package hits the ground, we can use the equation:

v = v0 + at

Plugging in the values, we get:

v = 11 m/s + (-9.8 m/s²)(4.79 s)

= -46.942 m/s (the negative sign indicates that the package is moving downwards)

The speed of the package just before hitting the ground is 46.942 m/s.

The pace at which distance changes over time is referred to as speed. It has a dimension of time and distance. As a result, the fundamental unit of length as well as the basic unit if duration are combined to form the High silica content of speed. As a result, the Si derived unit for speed is the meter per second.

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Time taken by the ball to reach the ground is  1.0 second.

Acceleration is rate of change of velocity with time. Due to having both direction and magnitude, it is a vector quantity. Si unit of acceleration is meter/second² (m/s²).

The initial height of the  balls = 5.0 meter

Initial time = 0 second

The initial speed of the ball = 01 m/s

Acceleration due to gravity = 10.0m/s²

Hence, Time taken by the ball to reach the ground is = √(2 × 5.0/10.0) second

= 1.0 second.

Therefor it take 1.0 second to ball

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Friction is the force that resists the motion of one object against another. It is a force that acts in the opposite direction of the motion. Friction occurs when two objects rub against each other, resulting in a force that resists their relative motion. This force can be caused by a variety of factors, including the surface roughness of the objects, the material they are made of, the weight of the objects, and the environment they are in.

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The horizontal distance travelled by the golf is 7.75 m.

The time taken for the golf to fall from the given height is calculated by applying the following equation.

t = √ ( 2h / g )

where;

t = √ ( 2 x 46 / 9.8 )

t = 3.1 seconds

The horizontal distance travelled by the golf is calculated as follows;

d = vt

where;

d = 2.5 m/s x 3.1 s

d = 7.75 m

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The complete question is below:

a golf ball is suit from a ship which is at a height 46 m above water surface the horizontal distance covered by Golf is​ if the horizontal velocity is 2.5 m/s.

The tension at the top of the horizontal circle is T = m (v²/r - g ).

The tension at the bottom of the horizontal circle is m (v²/r + g ).

The tension at the bottom and top of he rope is calculated by applying the following formula as shown below;

The tension at the top of the horizontal circle is calculated as;

T = ma - mg

T = mv²/r - mg

T = m (v²/r - g )

where;

The tension at the bottom of the horizontal circle is calculated as;

T = ma + mg

T = mv²/r + mg

T = m (v²/r + g )

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The complete question is below:

Find the expression for the tension at the bottom and top of the circle

Answer:

3.92 m/s²

Explanation:

It's hard to read your sketch, but it looks like m2 = 30 kg?  If so, here's the answer:

m1 = 20 kg

m2 = 30 kg

a = (m2·g) - (m1·g·sin∅) / (m1 + m2)   you can see how this equation is derived from Newton's 2nd Law in any physics textbook or website

a = ((30 kg)(9.8 m/s²) - (20 kg)(9.8 m/s²)(sin30)) / (20 kg + 30 kg)

a = (294 kg·m/s² - 98 kg·m/s²) / (50 kg) = 3.92 m/s²

Answer:

Approximately [tex]7.61 \times 10^{3}\; {\rm m\cdot s^{-1}}[/tex].

Explanation:

Add the radius of the Earth to the altitude of the satellite to find the radius of the orbit:

[tex]\begin{aligned}r &= (6370 + 150)\; {\rm km} \\ &= 6880\; {\rm km} \\ &= 6.88\times 10^{6}\; {\rm m}\end{aligned}[/tex].

Look up the gravitational constant:

[tex]G \approx 6.6743\times 10^{-11}\; {\rm m^{3}\, kg^{-1}\, s^{-2}[/tex].

Let [tex]m[/tex] denote the mass of the satellite. At an orbital velocity of [tex]v[/tex], the (centripetal) net force on the satellite would be:

[tex]\displaystyle (\text{net force}) = \frac{m\, v^{2}}{r}[/tex].

Let [tex]M[/tex] denote the mass of planet Earth. At a distance of [tex]r[/tex] from the center of the Earth, the gravitational attraction on the satellite would be:

[tex]\displaystyle (\text{gravitational attraction}) = \frac{G\, M\, m}{r^{2}}[/tex].

When the satellite is at the correct orbital velocity, the net force on the satellite would be equal to the gravitational attraction from the Earth. In other words:

[tex]\displaystyle (\text{net force}) = (\text{gravitational attraction})[/tex].

[tex]\displaystyle \frac{m\, v^{2}}{r} = \frac{G\, M\, m}{r^{2}}[/tex].

Rearrange the equation and solve for orbital velocity [tex]v[/tex]:

[tex]\begin{aligned}v^{2} &= \frac{G\, M}{r}\end{aligned}[/tex].

[tex]\begin{aligned}v &= \sqrt{\frac{G\, M}{r}} \\ &\approx \sqrt{\frac{(6.6743 \times 10^{-11})\, (5.97 \times 10^{24})}{6.88 \times 10^{6}}}\; {\rm m\cdot s^{-1}} \\ &\approx 7.61 \times 10^{3}\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].

Answer:

You will observe the transition from the first energy level to the fourth energy level. This is because the energy of the photons is equal to the amount needed to move the electron from the first energy level to the fourth energy level. Any energy higher than 51.0 electron-volts would not be observed because the electron would not be able to transition to a higher energy level.

It is given that, the heat gained is 4500 btu per hour. The temperature difference here is  30 F and the specific heat of air is  0.24 btu/lb°F. Then the cubic feet of air per minute is 138.8 CFM.

The sensible heat transfer in a system can be calculated using the equation below:

q = CFM × 1.08 ×ΔT

q = CFM x 0.075 lb/ft3 x 60 min/hour x 0.24 btu/lb°F x ∆T

where, 0.24 btu/lb°F is the specific heat of the dry air.

Given that q = 4500 btu/hour.

temperature difference = 93 F - 63 F.

Then 4500 btu/hr = CFM   × 1.08 × 30 F

CFM of air = 4500  btu/hr /(1.08 × 30 F ) = 138.8 CFM.

There for the number of cubic feat of air per minute is 138.8.

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Answer:

Explanation:

(a) The time necessary for the car to reach the truck can be calculated using the equation for average velocity:

v_avg = v0 + at

where v0 is the initial velocity of the car (which is 0 m/s at the instant the light turns green), a is the acceleration of the car (2.00 m/s^2), and t is the time elapsed.

Since the truck is traveling with a constant velocity of 15.0 m/s, we can equate the average velocity of the car to the velocity of the truck:

v0 + at = 15.0 m/s

Solving for t, we get:

t = (15.0 m/s - v0) / a = 15.0 m/s / 2.00 m/s^2 = 7.50 s

(b) The distance beyond the traffic light that the car will pass the truck can be calculated using the equation for displacement:

d = v0t + 1/2at^2

Plugging in the values we found above, we get:

d = (0 m/s)(7.50 s) + 1/2(2.00 m/s^2)(7.50 s)^2 = 56.25 m

(c) The speed of the car when it passes the truck can be calculated using the equation for velocity:

v = v0 + at = 0 m/s + (2.00 m/s^2)(7.50 s) = 15.0 m/s

Small air bubbles will form as the solubility of its dissolved air decreases when a glass of cold water is warmed to room temperature. Option B. is correct.

It's not as easy as it seems to pour water into a glass. Gases have been dissolved in them.

Furthermore, before we warm a glass of ice-cold water to room temperature, we must understand what we are doing. In essence, we are raising the water's temperature.

We now need to think about how rising temperatures will affect how soluble gases in the water.

By raising the temperature, we are supplying thermal energy in the form of heat to the gaseous molecules. The mobility and propensity to escape from the solution both increase as the thermal energy of the dissolved gaseous molecules rises. Because of this, the solubility of gases decreases as temperature rises.

As a result, when water is heated, gaseous molecules that have been dissolved begin to rise up from the water.

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1.) The potential energy of an object can be calculated using the equation: PE = mgh

Where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

So, the potential energy of a 60 kg ball falling from a height of 20 m would be:

PE = 60 kg * 10 m/s² * 20 m = 12000 J (joules)

2.) The average power can be calculated using the equation: P = W / t

Where W is the work done and t is the time taken.

In this case, the work done can be calculated as W = mgh, where m is the mass of the girl (30 N), g is the acceleration due to gravity (10 m/s²), and h is the height of the stairs (5 m).

So, W = 30 N * 10 m/s² * 5 m = 1500 J

And the average power can be calculated as:

P = 1500 J / 6 s = 250 W (watts)

Answer:

Answer is in attached photo.

Explanation:

The solution is in the attached photo, do take note to solve these questions, multiple formulas are needed:

Q1) Potential Energy = Mass x Gravitational Accceleration x Height

= m x g x h (SI Unit: J)

Q2) Work Done = Force x Distance

= F x d (SI Unit: J)

Average Power = [tex]\frac{Total \ Work \ Done}{Total \ Time}[/tex] (SI Unit: W)

The required three examples of a workshop are given as 1) Woodworking Workshop, 2) Photography Workshop, and 3) a Writing Workshop.

Workshops are the places where production operations take place.

Woodworking Workshop: A workshop that teaches the skills and techniques involved in woodworking such as using tools, working with different types of wood, and constructing wooden objects such as furniture, toys, and crafts.

Photography Workshop: A workshop that focuses on teaching the fundamentals of photography such as composition, lighting, and editing techniques. Participants learn how to use a camera and develop their own style through practical exercises and group discussions.

Writing Workshop: A workshop that provides a supportive environment for writers to develop their craft and receive feedback from peers and a writing instructor. Topics covered in a writing workshop may include character development, plot structure, dialogue, and revision techniques.

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The speed of the charged particles or proton or electron will be 2.75 × 10⁵ m/s.

The kinetic energy of an object is the energy which it possesses due to its motion. Kinetic energy is defined as the work needed to accelerate an object of a given mass from the state of rest to its stated velocity.

At the center of the square, all the kinetic energy of the proton is converted to potential energy of the proton. Therefore, far away the kinetic energy of proton must be equal to the potential energy of the person at the center.

Speed of the electron far away from the square, v = 2.75 × 10⁵ m/s

Given, that the charge, q = 15 × 10⁻⁹C

Each side of square, l = 0.02m

Distance of center of square from any charge r = √2l/2 = r= √2 × 0.02/ 2

r = 0.014m

q = 15nC

q = 15nC

I = 0.02m

q = 15nC

q = 15nC

Electric potential at the center of the square is given by:

Vc = 4 × (kq/r)

Electric potential at point far away is V∞ = 0

Potential Energy difference between the two point is

ΔV = Ve - V∞

ΔV = 4×(kq/r)

For the proton to pass through the center of square the proton must have kinetic energy equal to the potential energy at the center of the square.

The kinetic energy of the proton K = mv²

The potential energy of the proton U = eΔV

For the proton to pass through

K(at ∞) = U(Center)

1/2 mv² = 4kqe/ r

v = [tex]\sqrt{(8kq)/(mr)}[/tex]

v = √(8 × 9 × 10⁹ × 15 × 10⁻⁹ × 1.6 × 10⁻¹⁹)/ (1.67 × 10⁻²⁷ × 0.014) m/s

v = 2.75 × 10⁵ m/s

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Answer:

The speedometer reading of the car in km/h is 100.1 km/h. This can be calculated by using the formula K = 0.5mv2, where K is the kinetic energy, m is the mass, and v is the velocity. Rearranging this equation to solve for v yields v = √(2K/m). In this case, m = 66 kg and K = 1.1 x 104 J, so v = √(2 x 1.1 x 104 J/ 66 kg) = 100.1 km/h.

Answer:

Explanation:

To determine the displacement of the jogger using a graphical method, we can create a vector diagram to represent the individual components of the jogger's motion. We can then find the resulting displacement by adding the vectors end to end.

Assuming that "50cm = 2km" is the scale, we can convert the distances to represent them on the diagram:

2.0 km due east = 100 cm

10 km at 45° north of east = 70.71 cm (using Pythagorean theorem)

0.5 km due north = 25 cm

Next, we can plot the vectors on a coordinate grid, with east as the x-axis and north as the y-axis.

The jogger's first move, 2 km due east, can be represented by a vector starting from the origin and extending 100 cm to the right.

The second move, 10 km at 45° north of east, can be represented by a vector starting from the end of the first vector and extending 70.71 cm up and to the right.

The third move, 0.5 km due north, can be represented by a vector starting from the end of the second vector and extending 25 cm straight up.

Finally, to find the displacement, we add the vectors end to end. The displacement is the vector that starts at the origin and extends to the endpoint of the final vector. The magnitude of the displacement can be found using the Pythagorean theorem.

Using this method, we can determine the displacement of the jogger using a graphical method and the given scale.

The correct order of steps involved in performing a primary assessment for a child is check the patient for responsiveness, summon more advanced medical personnel, check for breathing and a pulse, and quickly scan for severe bleeding. The correct order is 2, 1, 4, and 3.

The first step in performing a primary assessment is to check the patient for responsiveness, if the child is responsive or not. This involves calling out their name, tapping their shoulder, or gently shaking them to see if they respond.

The next step is to summon more advanced medical personnel. If the child is not responsive or if you suspect a serious injury, it's important to call for more advanced medical personnel such as paramedics or emergency medical technicians (EMTs) as soon as possible.

Next, check for breathing and a pulse by checking if the child's airway, breathing, and pulse to make sure they're getting enough oxygen. If the child isn't breathing, begin cardiopulmonary resuscitation (CPR) immediately.

Then, quickly scan for severe bleeding. Look for any severe bleeding, such as from a major wound or a deep cut. If the child is bleeding severely, apply pressure to the wound to stop the bleeding.

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Your question seems incomplete, but I suppose the question was:

"After you size- up the scene, the following are steps involved with performing a primary assessment for a child. Place them in the order in which they should occur.

1. summon more advanced medical personnel

2. check the patient for responsiveness

3. quickly scan for severe bleeding

4. check for breathing and a pulse."

The distances from the axis of the rod (a long, straight metal rod has a radius of 4.60 cm and a charge per unit length of 27.4 nc/m.), where distances are measured perpendicular to the rod's axis

(a) 3.20 cm = 15.39 x 10³ N/c

(b) 2.0 cm = 2.43 x 10³ N/c

(c) 200 cm = 2.4 x 10² N/c

Given a long, infinite length and a constant charge per unit, the electric field is given by:

E = λ /2πε₀r

Where,

λ = wavelength (m)

ε₀ = 8.8542 x 10⁻¹²

r = radius (m)

Hence,

The distance are measured perpendicular to the rod's axis:

a. 3.20 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.0320)

= 15.39 x 10³ N/c

b. 20 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (0.2)

= 2.43 x 10³ N/c

c. 200 cm

E = (27.4 x 10⁻⁹) / 2π (8.8542 x 10⁻¹²) (2)

= 2.4 x 10² N/c

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The oscillation frequency of the two-block system is 1.6 Hz.

When a system is in equilibrium, neither its internal energy state nor its state of motion tend to change over time. A simple mechanical body is said to be in equilibrium if it neither experiences linear acceleration nor angular acceleration; unless it is disturbed by an external force, it will remain in that state indefinitely. If all of the forces acting on a single particle are vector summated to zero, equilibrium results. In addition to the states described above for the particle, a rigid body is said to be in equilibrium if the vector sum of all torques acting on it equals zero, keeping its state of rotational motion constant.

[tex]$$The oscillation frequency of the two-block is system is then found from Equation (3):$$\begin{aligned}f & =\frac{1}{2 \pi} \sqrt{\frac{k}{2 m}} \end{aligned}[/tex]

[tex]\begin{aligned}& =\frac{1}{2 \pi} \sqrt{\frac{196 \mathrm{~s}^{-2}}{2}} \\& =1.6 \mathrm{~Hz}\end{aligned}$$\\\\\text{where $2 m$ is the mass of the two blocks}[/tex]

Thus, the oscillation frequency of the two-block system is 1.6 Hz.

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Answer:

the ocean liner

Explanation:

KE = 1/2mv²

bullet:  KE = 1/2(0.0011 kg)((389 m/s)² = 83.2 J

ship:  KE = 1/2(7.8x10⁷ kg)(11 m/s)² = 4.7x10⁹ J

The optimal course of action is to allow the laptop to warm up to room temperature before turning it on.

The room atmosphere and the outside temperature of the laptop are very different in terms of temperature.

When the user was at work and the laptop was discovered on the deck, it was delivered to his house. He is unable to recall how long it has been really chilly outside. The user opens his laptop and checks to see if everything is working correctly. Instead, turn on the machine at room temperature and let it preheat up.

When a computer is still in a frigid environment, it must warm up to room temperature for roughly 6 to 24 hours before being turned on. Otherwise, the heat the components produce might cause the laptop to become moist with water. The system's components might then be harmed by that water.

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Answer:

2.6 A

Explanation:

Firstly here we need to find out the net resistance. 30Ω and 40Ω are grouped in parallel and to these 10Ω and 15Ω are in series. So we can find the net resistance as ,

[tex]\implies R_{net}=\dfrac{R_1R_2}{R_1+R_2} \\[/tex]

[tex]\implies R_{net}=\dfrac{30\times 40}{30+70}\Omega \\[/tex]

[tex]\implies R_{net}=\dfrac{120}{7}\Omega \\[/tex]

Again,

[tex]\implies R_{net}= R_1 + R_2+R_3 \\[/tex]

[tex]\implies R_{net}=\dfrac{120}{7}+10+15\Omega \\[/tex]

[tex]\implies R_{net}= \dfrac{120+105+70}{7}\Omega\\[/tex]

[tex]\implies R_{net}= \dfrac{295}{7}\Omega \approx 42\Omega \\[/tex]

Now use Ohm's law as ,

[tex]\implies V = iR \\[/tex]

[tex]\implies 110V = i\times 42\Omega\\[/tex]

[tex]\implies i =\dfrac{110}{42} A\\[/tex]

[tex]\implies \underline{\underline{ i = 2.6\ A }} \\[/tex]

and we are done!

Flared tubing is commonly used in hydraulic systems. Carbon Steel fittings are typically used in hydraulic equipment when attaching tubing to a hose.

Liquid fluid power is used by hydraulic machines to do operations. Heavy-duty construction vehicles are a typical illustration. Hydraulic fluid is pumped to numerous hydraulic motors and hydraulic cylinders located all around the machine in this sort of machine and is pressured in accordance with the resistance present. Control valves direct or automate the distribution of the fluid through hoses, tubes, or pipes.

Pascal's law, which asserts that any pressure applied to a fluid inside a closed system would transmit that pressure equally everywhere and in all directions, is the foundation of hydraulic systems, just like pneumatic systems. An incompressible liquid, as opposed to a compressible gas, serves as the fluid in a hydraulic system.

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The force acting on the electron is dependent upon the distance seen between positive charge and indeed the electron, the charges on the electron, and indeed the charge of something like the positive charge.

Why is there a positive charge?

A material that has more protons than electrons is said to have a positive charge. We are aware that electrons seem to be positively charged and protons become positively charged. Consequently, positively charged objects have more protons than electrons.

The positive charge is absent:

The positive charge on the protons and neutrons (red) results from an imbalance between protons and electrons, more particularly, when there are more protons than electrons. Protons can be added to an atom or other substance with both a neutral charge to produce a positive charge.

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The electric field points away from positive charges and toward negative charges. A distribution of charges creates an electric field that can be found by taking the vector sum of the fields created by individual point charges. Note that if a charge is placed in an electric field created by q', q will not significantly affect the electric field if it is small compared to q'.

Imagine an isolated positive point charge with a charge Q (many times larger than the charge on a single electron).

There is a single electron at a distance from the point charge. On which of the following quantities does the force on the electron depend?

Check all that apply.

A the distance between the positive charge and the electron

B the charge on the electron

C the mass of the electron

D the charge of the positive charge

E the mass of the positive charge

F the radius of the positive charge

G the radius of the electron

I got A, B, and D for this part of the problem and was correct. However,

For the same situation as in Part A, on which of the following quantities does the electric field at the electron's position depend?

Check all that apply.

A the distance between the positive charge and the electron

B the charge on the electron

C the mass of the electron

D the charge of the positive charge

E the mass of the positive charge

F the radius of the positive charge

G the radius of the electron

The intensity of pressure at section 2 is approximately 345.64 kN/m^2.

Pressure is described as the force applied perpendicular to the surface of an object per unit area over which that force is distributed.

To determine the intensity of pressure at section 2, we need to use the equation of continuity and the Bernoulli equation:

we will substitute the values from the continuity equation into the Bernoulli equation which will give us :

P_1 + 1/2 * p * (Q_1 / A_1)^2 + p * g * h_1 = P_2 + 1/2 * p * (Q_2 / A_2)^2 + p * g * h_2

Substitute all the given values into the equation

Solving for P_2 = 420,000 N/m^2 + 47.06 KN/m^2 - 71.42 KN/m^2 = 345.64 KN/m^2

In conclusion,  the intensity of pressure at section 2 is approximately 345.64 kN/m^2.

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Answer:

The sensors are approximately [tex]2.83\; {\rm m}[/tex] apart.

Sensor [tex]\texttt{a}[/tex] is approximately [tex]12.2\; {\rm m}[/tex] from the top of the building.

Height of the building is approximately [tex]135\; {\rm m}[/tex].

(Assumptions: air resistance on the ball is negligible; [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)

Explanation:

Make use of the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex], where [tex]x[/tex] represents displacement, [tex]v[/tex] and [tex]u[/tex] are the final and initial velocity, and [tex]a[/tex] is the acceleration.

In other words, as the velocity of the object changes from [tex]u\![/tex] to [tex]v\![/tex] at a rate of [tex]a[/tex], position of the object would have changed by [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex].

When the ball is at sensor [tex]\texttt{a}[/tex], velocity of the ball was [tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].

Shortly after, when the ball is at sensor [tex]\texttt{b}[/tex], velocity of the ball was [tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex].

Under the assumptions, [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex]. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between sensor [tex]\texttt{a}[/tex] ([tex]u = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]) and [tex]\texttt{b}[/tex] ([tex]v = (-17.2)\; {\rm m\cdot s^{-2}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-17.2)^{2} - (-15.5)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-2.83)\; {\rm m}\end{aligned}[/tex].

Hence, the distance between sensor [tex]\texttt{a}[/tex] and [tex]\texttt{b}[/tex] is approximately [tex]2.83\; {\rm m}[/tex].

Since the ball was released from rest, the initial velocity of the ball would be [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex].

Let [tex]v[/tex] denote the velocity of the ball at sensor [tex]\texttt{a}[/tex]: [tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex].

Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the roof ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and sensor [tex]\texttt{a}[/tex] ([tex]v = (-15.5)\; {\rm m\cdot s^{-1}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-15.5)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-12.2)\; {\rm m}\end{aligned}[/tex].

In other words, the distance between sensor [tex]\texttt{a}[/tex] and the top of the roof would be approximately [tex]12.2\; {\rm m}[/tex].

Another SUVAT equation, [tex]v = u + a\, t[/tex], gives the velocity of an object after accelerating for a duration of [tex]t[/tex] (at a rate of [tex]a[/tex], starting from an initial velocity of [tex]u[/tex].)

Make use of this SUVAT equation to find the velocity of the ball right before hitting the ground. Specifically, velocity of the ball was [tex]u = (-17.2)\; {\rm m\cdot s^{-1}}[/tex] at sensor [tex]\texttt{b}[/tex]. After accelerating at [tex]a = (-g) = (-9.81)\; {\rm m\cdot s^{-2}}[/tex] for [tex]t = 3.5\; {\rm s}[/tex], velocity of the ball would be:

[tex]\begin{aligned}v &= u + a\, t \\ &= (-17.2)\; {\rm m\cdot s^{-1}} + (-9.81)\, (3.5)\; {\rm m\cdot s^{-1}} \\ &= (-51.535)\; {\rm m\cdot s^{-1}} \end{aligned}[/tex].

Again, the velocity of the ball was [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] at the top of the building. Apply the SUVAT equation [tex]x = (v^{2} - u^{2}) / (2\, a)[/tex] to find the change in the position of the ball between the top of the building ([tex]u = 0\; {\rm m\cdot s^{-1}}[/tex]) and right before hitting the ground ([tex]v = (-51.535)\; {\rm m\cdot s^{-1}}[/tex]):

[tex]\begin{aligned}x &= \frac{(-51.535)^{2} - (0)^{2}}{2\, (-9.81)} \; {\rm m} \approx (-135)\; {\rm m}\end{aligned}[/tex].