Answer:
Explanation:
The total distance travelled by ball before first bounce
= 160 + 160 = 320 ft
Distance travelled between first bounce and second bounce
= 80 + 80 = 160 ft
Distance travelled between second bounce and third bounce
= 40 + 40 = 80 ft
Distance travelled between third bounce and fourth bounce
= 20 + 20 = 40 ft .
Distance travelled in fourth bounce = 10 ft
Total distance travelled = 320 + 160 + 80 + 40 + 10
= 610 ft .
A .2g sphere is suspended by a thread in an electreic field of 5000 N/C that is directed straight up. The tension in the string is 2.3 x 10^-3 N. Determine the charge on the shpere.
Answer:
6.8*10^-8C
Explanation:
The formula for calculating the Electric field on the sphere is expressed as;
E = Sum of force on the sphere/charge q
q is the charge on the sphere
E is the electric field
Given
E = 5000N/C
T = 2.3 x 10^-3 N.
q = ?
W = 0.0002 * 9.8
W = 0.00196N
W = 1.96*10^-3N
From the formula;
q = T-W/E
W is the weight of the sphere
T is the tension in the string
q = (2.3*10^-3 - 1.96 x 10^-3)/5000
q = 0.00034/5000
q = 6.8*10^-8C
Hence the charge on the sphere is 6.8*10^-8C
A 5 cm diameter orfice discharges fluid from a tank with a head of 5 meters. The discharge rate, Q, is measured at 0.015 m^3/s. The actual velocity at the vena contracta is 9 m/s. The coefficient of discharge is nearest to:________
Answer:
[tex]\mathbf{ C_d = 0.86}[/tex]
Explanation:
Given that:
diameter of the orifice D = 5 cm = 5 × 10⁻² m
discharge rate Q = 0015 m³/s
Actual velocity V = 9 m/s
By using the formula
[tex]Q = C_d \times A \times \sqrt{2gh}[/tex]
where;
[tex]v = \sqrt{2gh}[/tex] = 9 m/s
[tex]Q = C_d \times A \times v}[/tex]
where;
[tex]A = \dfrac{\pi D^2}{4}[/tex]
[tex]A = \dfrac{\pi (5 \times 10^{-2} ) ^2}{4}[/tex]
[tex]A = 0.00196 \ m^2[/tex]
Again,
[tex]Q = C_d \times A \times v}[/tex]
[tex]0.015 = C_d \times 0.00196 \times 9[/tex]
[tex]C_d = \dfrac{0.015 }{ 0.00196 \times 9}[/tex]
[tex]\mathbf{ C_d = 0.86}[/tex]
Hence, the coefficient of discharge [tex]\mathbf{ C_d = 0.86}[/tex]
An assessment tool that measures the amount of stress in a person’s life over a one-year period resulting from major life events is called
the Social Readjustment Rating Scale.
the Stress Scale.
the Annual Stress Scale.
the Social Scale.
Answer:
The correct answer is - the Social Readjustment Rating Scale.
Explanation:
The social readjustment rating scale is developed by Richard Rahe and Thomas Holmes to measure the stress caused by this important and major life event. Stress helps in developing a psychological approach for the particular person.
It is an assessment tool to calculate the impact of the major life events in the time period of one year with the help of units of 0 to 100. 100 is extreme or highest stress caused by the event such as the death of the spouse.
give two examples of uses of the expansion and contraction of materials
if we hold a very hot glass tumbler under cold water, it cracks. This is because the outer surface of the glass comes in direct contact with cold water and contracts more as compared to the inner surface. We observed that water expanded on heating.
Railway tracks consist of two parallel metal rails joined together. Small gaps, called expansion gaps, are deliberately left between the rails as there is an expansion of the rails in hot weather. Water expands on heating.
Assuming air is an incompressible fluid, enter an expression for an estimate of the density of air, in terms of the defined quantities and the acceleration due to gravity, g.
Answer:
Density of Air = ([tex]P_{1}[/tex] - [tex]P_{2}[/tex] )/(g x h)
Density of Air = 1.27 Kg/[tex]m^{3}[/tex]
Explanation:
Note: This question is not complete, and lacks its first part in which it contains important data to solve for the density of air. But, I have found the similar question and its data. So. I will be solving the question for the sack of understanding and concept.
Missing part: A weather balloon has an absolute-pressure sensor attached. On the ground the sensor reads [tex]P_{1}[/tex] = 1.01x [tex]10^{5}[/tex] Pa. At a height of h = 950 m, the sensor reads [tex]P_{2}[/tex]=8.92x[tex]10^{4}[/tex] Pa.
Solution:
Let
[tex]P_{1}[/tex] be the pressure of the balloon at ground.
[tex]P_{2}[/tex] be the pressure of the balloon at height h = 950 m
g = acceleration due to gravity,
In order to derive the expression, we need to find the pressure difference:
Pressure difference = ΔP
ΔP = [tex]P_{1}[/tex] - [tex]P_{2}[/tex]
As we know that,
Pressure difference = density x acceleration due to gravity x height.
So,
ΔP = [tex]P_{1}[/tex] - [tex]P_{2}[/tex] = (Density of Air) x (g) x (h)
We need expression for the density of air, so,
Density of Air = ΔP / (g x h)
Hence, the expression is:
Density of Air = ([tex]P_{1}[/tex] - [tex]P_{2}[/tex] )/(g x h)
Now, we can calculate the density of air as well, by putting the values given above in the data.
[tex]P_{1}[/tex] = 1.01 x [tex]10^{5}[/tex]
[tex]P_{2}[/tex] = 8.92 x [tex]10^{4}[/tex]
g = 9.8 m/s
h = 950 m
So,
Density of Air = ((1.01 x [tex]10^{5}[/tex]) - (8.92 x [tex]10^{4}[/tex]) )/ (9.8 x 950)
Density of Air = 1.27 Kg/[tex]m^{3}[/tex]
An expression for an estimate of the density of air is; ρ = ( P₁ - P₂)/gh
According to pascal's principle;
ΔP = ρgh
Where;
ΔP is change in pressure = P₁ - P₂
g is acceleration due to gravity
h is height of pressure change
Now, we want to find an expression for an estimate of the density of air. This means we want to make density the subject of the formula. Thus;
ΔP = ρgh
⇒ divide both sides by gh to get;
ρ = ΔP/(gh)
Recall that ΔP = P₁ - P₂
Thus; ρ = ( P₁ - P₂)/gh
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in the periodic table, the properties repeat in what direction?
Answer:
Left to right and top to bottom
Explanation:
On the periodic table, the properties repeat from left to right and from top to bottom.
Periodic properties have a pattern from the top to the bottom or down a group or family.
Also, across the period from left to right, they also show a repeating pattern.
Certain properties increase from left to right and decreases from top to bottom. E.g. electronegativity. Also, some properties decreases from left to right and increases from top to bottom e.g. atomic radius.Precisely 1.00 s after the speeder passes, the police officer steps on the accelerator; if the police car's acceleration is 2.70 m/s2 , how much time passes after the police car is passed by a speeder and before the police car overtakes the speeder (assumed moving at constant speed)
Answer:
t= 16.75 s
Explanation:
We will solve this exercise using the kinematic expressions
corridor that goes at constant speed, suppose that its speed is v₁ = 20 m/s, it does not appear in the statement, we start counting the time when it passes the policeman.
x₁ = v₁ t
The policeman starts from rest, so his initial velocity is zero and he has an acceleration a = 2.70 m /s², to use the same time counter we take into account that the policeman left at = 1.00 s after passing the corridor
x₂ = v₀ (t-t₀) + ½ a (t-t₀)²
x₂ = ½ a (t-1)²
at the point where the two meet, the position must be the same
x₁ = x₂
v₁ t = ½ a (t-1)²
(t-1)² = [tex]\frac{2 v_1 t}{a}[/tex]
t² - 2t + 1 - \frac{2 v_1 t}{a} +1 = 0
t² - 2(1 + [tex]\frac{v_1}{a}[/tex]) t +1
let's we solve the second degree equation
t² - 2 ( 1 + [tex]\frac{20}{2.7}[/tex]) t + 1=0
t² - 16.81 t +1=0
t = [ 16.81 ± [tex]\sqrt{ 16.81^2 - 4 )}[/tex] ] /2
t = [16.81 ± 16.695]/2
t₁= 16.75 s
t2= 0.06 s
Time t₂ is less than the reaction time of humans, so the correct answer is the first time
t= 16.75 s
Place the left charge at the 2 cm position and the right one at the 4 cm position. Vary the left and right charge to the values provided below and record the resulting forces.
Left Charge Righ Charge Resulting force(N)
1μC 4μC
4μC 1μC
2μC 2μC
1μC 2μC
1μC 8μC
2μC 8μC
Answer:
Explanation:
Force between two charges can be expressed as follows
F = k q₁ q₂ / d²
q₁ and q₂ are two charges , d is distance between them , k is a constant whose value is 9 x 10⁹
distance between charges is fixed which is 4 -2 = 2 cm = 2 x 10⁻² m
force between 1μC and 4μC
= 9 x 10⁹ x 1 x 4 x 10⁻¹² / ( 2 x 10⁻² )²
= 9 x 10 = 90 N
force between 4μC and 1μC
= 9 x 10⁹ x 4 x 1 x 10⁻¹² / ( 2 x 10⁻² )²
= 9 x 10 = 90 N
force between 2μC and 2μC
= 9 x 10⁹ x 2 x 2 x 10⁻¹² / ( 2 x 10⁻² )²
= 9 x 10 = 90 N
force between 1μC and 2μC
= 9 x 10⁹ x 1 x 2 x 10⁻¹² / ( 2 x 10⁻² )²
= 4.5 x 10 = 45 N
force between 1μC and 8μC
= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²
= 18 x 10 = 180 N
force between 2μC and 8μC
= 9 x 10⁹ x 1 x 8 x 10⁻¹² / ( 2 x 10⁻² )²
= 36 x 10 = 360 N
Left Charge Right Charge Resulting force(N)
1μC 4μC 90 N
4μC 1μC 90 N
2μC 2μC 90 N
1μC 2μC 45 N
1μC 8μC 180 N
2μC 8μC 360 N
A girl walks 20.0 m east then 70.0 meters north. What is the girl’s displacement (mag. And direction)? What is the girl’s distance?
Explanation:
Given that,
A girl walks 20.0 m east then 70.0 meters north.
Displacement is the shortest path covered by an object. Let it is d. It is calculated as :
[tex]d=\sqrt{20^2+70^2} \\\\=72.80\ m[/tex]
For direction,
[tex]\theta=\tan^{-1}(\dfrac{70}{20})\\\\\theta=74.05^{\circ}[/tex]
Girl's distance = 20 m + 70 m
= 90 m
Hence, this is the required solution.
Two pieces of steel wire with identical cross sections have lengths of L and 2L. The wires are each fixed at both ends and stretched so that the tension in the longer wire is four times greater than in the shorter wire. If the fundamental frequency in the shorter wire is 60 Hz, what is the frequency of the second harmonic in the longer wire?
Answer:
Explanation:
Expression for fundamental frequency of tone produced in a wire under tension of T and length L is given as follows
[tex]f=\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
m is mass per unit length .
We shall apply this formula for given wires .
For shorter wire
[tex]60 =\frac{1}{2L} \times \sqrt{\frac{T}{ m} }[/tex]
For longer wire for second harmonic
length of wire is 2L , tension is 4T ,
[tex]f =\frac{2}{4L} \times \sqrt{\frac{4T}{ m} }[/tex]
[tex]f =\frac{2\times 2}{4L} \times \sqrt{\frac{T}{ m} }[/tex]
f = 2 x 60 = 120 Hz .
Ashudent drops an object from rest above a force plate that records information about the force exerted on the object as a function of time during the time interval in which the objects in contact with the force plate Which of the following measurements should the student take in addition to the measurements from the force platit. to determine the change in momentum of the object from immediately before the collision to immediately after the collision?
A The mass of the object
B The final speed of the object MOH 5000
C The distance fallen by the object
D The student has enough information to make the determination
Answer:
A The mass of the object
Explanation:
i hope it's helpful
You have now seen examples during lecture on how to calculate the electric field for a line of charge and a ring of charge - both uniformly distributed. This activity will ask you to solve for the electric field, on-axis, of a uniformly-charged disk sitting in the yz plane. Below is a picture of the situation of interest. Note: Treat it as a totally flat disk and ignore its thickness in the x direction. Also, let x be the distance between the center of the disc and point P.
Answer:
1/4πε₀[Hx/ (√x² + b²)^3/2]i.
Explanation:
So, without mincing words let's dive straight into the solution to the question above. There is need to determine the electric field on-axis of a uniformly-charged disk sitting.
The electric field in the x-component, dεₓ = 1/4πε₀[H/ x² + b²] cos .
Thus, the total electric field in the x-component, εₓ = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2.
Therefore, the electric field = 1/4πε₀ [ xdH/ (x^2 + b^2)^3/2i.
Where x=0
A 0.1ohm resistor has power rating 5w. is the resistor safe when conducting a current of 10A.
Answer:
1.5 Amp is rated for 5 W so it would not be possible
Starting from the front door of your ranch house, you walk 50.0 m due east to your windmill, and then you turn around and slowly walk 30.0 m west to a bench where you sit and watch the sunrise. It takes you 27.0 s to walk from your house to the windmill and then 47.0 s to walk from the windmill to the bench. For the entire trip from the front door to the bench, what are your :
a. average velocity
b. average speed
Answer:
Explanation:
Total displacement for entire trip = final position - initial position
= 50 m - 30 m = 20 m
Total time = 27 + 47 = 74 s
Average velocity = Total displacement / total time
= 20 / 74 = .27 m /s
Total distance covered in entire trip = 50 + 30 = 80 m
Total time = 74 s
Average speed = Total distance covered / total time
= 80 / 74 = 1.08 m /s .
A spring with k = 58 N/m hangs vertically next to a ruler. The end of the spring is next to the 17-cm mark on the ruler.
If a 2.5-kg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?
Express your answer to two significant figures and include the appropriate units.
Answer:
Explanation:
The force occurring on a spring F = mg
If x should be the expansion in the spring.
Then F = mg = -kx
where;
[tex]x = \dfrac{mg}{k}[/tex]
[tex]x = \dfrac{2.5 \ kg*9.81 \ m/s^2 }{58 \ N/m}[/tex]
x = 0.423 m
x = 42.3 cm
The final reading of the spring when the end of the spring lines up with the ruler mark is = 42.3 cm + 17 cm
= 59.3 cm
≅ 60 cm
7Which of the following terms describes how glaciers move?
A Quickly
B Gradually
C Aggressively
D Rapidly
Answer:
D is the answer I think (0 w 0 )
Explanation:
The glaciers move gradually. Hence, option (C) is correct.
What is glacier?A glacier is a long-lasting mass of heavy ice that is perpetually moving. When the ablation of snow is greater than the accumulation over a long period of time, frequently centuries, a glacier forms.
As it slowly flows and deforms under forces brought on by its weight, it gains distinctive features like crevasses and seracs. Cirques, moraines, and fjords are the result of the erosion of rock and debris from its substrate as it travels.
The considerably thinner sea ice and lake ice that form on the surface of bodies of water are not the same as glaciers, which form only on land and may flow into water bodies.
The huge ice sheets, commonly referred to as "continental glaciers," in the polar areas of the planet contain 99 percent of the planet's glacial ice.
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A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in the magnetic field, the iron filings arranged themselves, creating lines of force. How do the energy and the lines of force change when a stronger bar magnet is used?
The answer is "The energy increases, and the lines of force are denser"
Answer:
The energy increase, and the lines of force are denser
Explanation:
The rest of the answers
1.) The field energy will increase.
3.) It points toward the field of earths magnetic poles.
4.) l, ll, and lll only
5.) ll, lV, l, lll
Based on the information given, the energy and the lines of force change as the energy increases, and the lines of force are dense.
Energy
In physics, it should be noted that energy is the quantitative property which must be transferred to a body or physical system to perform work on the body.
From the information, the bar magnet was placed underneath a sheet of paper where a pile of iron filings sits and in the presence of the energy stored in the magnetic field, the iron filings arranged themselves, creating lines of force. Therefore, when a stronger bar magnet is used, the energy increases and the lines of force are denser.
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A student is drinking a cup of hot chocolate. This method of energy transfer is
Answer:
conduction I believe if not its convection
Answer: Conduction transfers energy from the spoon to the hot chocolate.
Explanation: Heated water molecules and steam rise in the beaker, carrying heat by convection.
A small rock is thrown vertically upward with a speed of 17.0m/s from the edge of the roof of a 26.0m tall building. The rock doesn't hit the building on its way back down and lands in the street below. Air resistance can be neglected.
Part A
What is the speed of the rock just before it hits the street?
Express your answer with the appropriate units.
Part B
How much time elapses from when the rock is thrown until it hits the street?
Express your answer with the appropriate units.
Answer:
A) v = 28.3 m/s
B) t = 4.64 s
Explanation:
A)
Assuming no other forces acting on the rock, since the accelerarion due to gravity close to the surface to the Earth can be taken as constant, we can use one of the kinematic equations in order to get first the maximum height (over the roof level) that the ball reaches:[tex]v_{f}^{2} - v_{o}^{2} = 2* g* \Delta h (1)[/tex]
Taking into account that at this point, the speed of the rock is just zero, this means vf=0 in (1), so replacing by the givens and solving for Δh, we get:[tex]\Delta h = \frac{-v_{o} ^{2}}{2*g} = \frac{-(17.0m/s)^{2} }{2*(-9.8m/s2)} = 14.8 m (2)[/tex]
So, we can use now the same equation, taking into account that the initial speed is zero (when it starts falling from the maximum height) and that the total vertical displacement is the distance between the roof level and the ground (26.0 m) plus the maximum height that we have just found in (2) , 14.8m:Δh = 26.0 m + 14. 8 m = 40.8 m (3)Replacing now in (1), we can solve for vf, as follows:[tex]v_{f} =\sqrt{2*g*\Delta h} = \sqrt{2*9.8m/s2*40.8m} = 28.3 m/s (4)[/tex]
B)
In order to find the total elapsed from when the rock is thrown until it hits the street, we can divide this time in two parts:1) Time elapsed from the the rock is thrown, until it reaches to its maximum height, when vf =02) Time elapsed from this point until it hits the street, with vo=0.For the first part, we can simply use the definition of acceleration (g in this case), making vf =0, as follows:[tex]v_{f} = v_{o} + a*\Delta t = v_{o} - g*\Delta t = 0 (5)[/tex]
Replacing by the givens in (5) and solving for Δt, we get:[tex]\Delta t = \frac{v_{o}}{g} = \frac{17.0m/s}{9.8m/s2} = 1.74 s (6)[/tex]
For the second part, since we know the total vertical displacement from (3), and that vo = 0 since it starts to fall, we can use the kinematic equation for displacement, as follows:[tex]\Delta h = \frac{1}{2} * g * t^{2} (7)[/tex]
Replacing by the givens and solving for t in (7), we get:[tex]t_{fall} =\sqrt{\frac{2*\Delta h}{g}} = \sqrt{\frac{2*40.8m}{9.8m/s2} } = 2.9 s (8)[/tex]
So, total time is just the sum of (6) and (8):t = 2.9 s + 1.74 s = 4.64 s1. 9mA electric current is flowing through a conducting wire. Then the number of electron
passing through it in 3mimute is?
A)2x10^18
B) 1X10^18
C) 2x10^19
D) 1.01x10^19
Answer:
use the formula negative -eE/A×t
Energy stored in the nucleus of an atom which releases energy through
fission or fusion *
please help thank you
Answer:
[tex]\theta \approx 59.036^{\circ}[/tex], [tex]T_{2} \approx 23.324\,N[/tex]
Explanation:
First we build the Free Body Diagram (please see first image for further details) associated with the mass, we notice that system consist of a three forces that form a right triangle (please see second image for further details): (i) The weight of the mass, (ii) two tensions.
The requested tension and angle can be found by the following trigonometrical and geometrical expressions:
[tex]\theta = \tan^{-1} \frac{W}{T_{2}}[/tex] (1)
[tex]T_{1} = \sqrt{W^{2}+T_{2}^{2}}[/tex] (2)
Where:
[tex]W[/tex] - Weight of the mass, measured in newtons.
[tex]T_{1}[/tex], [tex]T_{2}[/tex] - Tensions from the mass, measured in newtons.
If we know that [tex]W = 20\,N[/tex] and [tex]T_{2} = 12\,N[/tex], then the requested values are, respectively:
[tex]\theta = \tan^{-1} \frac{20\,N}{12\,N}[/tex]
[tex]\theta \approx 59.036^{\circ}[/tex]
[tex]T_{2} = \sqrt{(20\,N)^{2}+(12\,N)^{2}}[/tex]
[tex]T_{2} \approx 23.324\,N[/tex]
Kieran caught 18 more Pokémon on Saturday than
Noah did. Kieran caught 53 Pokémon. Write and solve
an equation to find how many Pokémon, p, Noah
caught on Saturday.
Answer:
Noah caught 35 Noah on Satuday.
Explanation:
Given that,
Kieran caught 53 Pokemon
Kieran caught 18 more Pokémon on Saturday than Noah did.
Let Noah caught x Pokémon on Saturday.
ATQ,
18 + x = 53
Thie is the equation that can be used to find the value of x.
Subtract 18 from both sides.
18 + x -18 = 53 - 18
x = 35
Hence, Noah caught 35 Noah on Satuday.
When you use an array to implement the ADT list, retrieving an entry at a given position is slow.
a) true
b) false
When you use an array to implement the ADT list, adding an entry at a the end of the list is fast.
a) true
b) false
When you use a vector to implement the ADT list, retrieving an entry at a given position is slow.
a) true
b) false
When you use a vector to implement the ADT list, adding an entry at a the end of the list is fast.
a) true
b) false
Answer:
1)FALSE
2)TRUE
3)FALSE
4) FALSE
Explanation:
1) Retrieving an entry at a given position is not slow ( it is done using index number ) hence the answer is FALSE
2) Adding an entry at the end of the list is Fast this is because in using array implementation a new entry is made immediately after the last position and it is done very fast as well. hence the answer is TRUE
3) Retrieving an entry at a given position ( from an ADT list ) that is been implemented using Vector is very fast this is simply because Indexing in vector is done at a fixed time hence the answer is FALSE
4) Adding an entry at the end of the list is slow using vector hence answer is FALSE
Describe an experiment to show that the maximum attractive property is at the poles of
a magnet?
Answer:
Take a bar magnet and place a steel pin at some distance. ... Now, bring the steel pin near the pole of the bar magnet. We notice that pin sticks to the magnet. This experiment shows that maximum magnetic force acts at the poles of the magnet.
Explanation:
hope that is help
write your thoughs about this why do you think there were improvents or decline in your physical fitness assessment result PA HELP PO
cal ulate a moment of force o. 50 meter distance and 10 newton force
Answer:
500N/M
Explanation:
given that
force=10N
distance=50M
moment=force*distance
=10×50=500j
Which of the following is a mood disorder?
phobic disorder
bipolar disorder
schizophrenia
conversion disorder
Answer:
bipolar disorder
Explanation:
bipolar disorder, is previously known as manic depression, is a mental disorder characterized by periods of depression and periods of abnormally elevated mood that last from days to weeks.
A tangent line drawn on a velocity-time graph has a rise of 19 m/s and a run of 4.0 m/s. How large is the acceleration? What type of acceleration Is this?
Answer:
Acceleration = 4.8 m/s²
Explanation:
Given:
Change in velocity = 19 m/s
Change in time = 4 s
Find:
Acceleration
Computation:
Acceleration = Change in velocity / Change in time
Acceleration = 19/4
Acceleration = 4.8 m/s²
Positive acceleration
A car is traveling at 120 km/h (75 mph). When applied the braking system can stop the car with a deceleration rate of 9.0 m/s2. The typical reaction time for an alert driver is 0.8 s versus 3 s for a sleepy driver. Assuming a typical car length of 5 m, calculate the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver. Group of answer choices
Answer:
the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15
Explanation:
Given that;
speed of car V = 120 km/h = 33.3333 m/s
Reaction time of an alert driver = 0.8 sec
Reaction time of an alert driver = 3 sec
extra time taken by sleepy driver over an alert driver = 3 - 0.8 = 2.2 sec
now, extra distance that car will travel in case of sleepy driver will be'
S_d = V × 2.2 sec
S_d = 33.3333 m/s × 2.2 sec
S_d = 73.3333 m
hence, number of car of additional car length n will be;
n = S_n / car length
n = 73.3333 m / 5m
n = 14.666 ≈ 15
Therefore, the number of additional car lengths approximately it takes the sleepy driver to stop compared to the alert driver is 15