Part of understanding the physical effects on Mars, we must understand
first that our laws of Physics on Earth must apply in the same manner that it
is on Mars. Discuss the Three Laws of Motion as set forth by Isaac Newton.
Following this, write out the mathematical description of these laws. Provide
three examples to con rm your results, and include free body diagrams
Answer:
so easy add the subtract then multiplay the add
Explanation:
1: Give one word answer.
The bouncing back of the light rays after hitting a smooth surface _____
Answer:
RThe answer is Reflection....
Using a simply pulley/rope system, a crewman on an Arctic expedition is trying to lower a 6.17-kg crate to the bottom of a steep ravine of height 23.8 meters. The 55.6-kg crewman is walking along holding the rope, being careful to lower the crate at a constant speed of 1.50 m/s. Unfortunately, when the crate reaches a point 13.2 meters above the ground, the crewman steps on a slick patch of ice and slips. The crate immediately accelerates toward the ground, dragging the hapless crewman across the ice and toward the edge of the cliff.
If we assume the ice is perfectly slick (that is, no friction between the crewman and the ice once he slips and falls down), at what speed will the crate hit the ground? Assume also that the rope is long enough to allow the crate to hit the ground before the crewman slides over the side of the cliff. At what speed will the crewman hit the bottom of the ravine?
Answer:
a. Vc = 5.06 m/s
b. Vp = 22.18 m/s
Explanation:
The acceleration of the pulley-mass system is as follows:
a = [tex]\frac{mg}{m + M}[/tex]
Solving for acceleration, we get:
a = [tex]\frac{6.17 *9.8}{6.17 + 55.6}[/tex]
a = 0.97
So, for the part a:
Calculate the velocity of the crewman by using the following equation:
Vc = [tex]\sqrt{Vi^{2} + 2ay}[/tex]
Substituting the values into the equation, we get:
Vc = [tex]\sqrt{1.50^{2} + 2*0.97*13.2}[/tex]
Vc = 5.06 m/s
Now, for part b:
Calculate the final velocity of the pulley by using the following expression:
Vp = [tex]\sqrt{Vi^{2} +2gy }[/tex]
Just plugging in the values.
Vp = [tex]\sqrt{5.06^{2} +2*9.8*23.8 }[/tex]
Vp = 22.18 m/s
A crate rests on a flatbed truck which is initially traveling at 13.6 m/s on a level road. The driver applies the brakes and the truck is brought to a halt in a distance of 38.1 m. If the deceleration of the truck is constant, what is the minimum coefficient of friction between the crate and the truck that is required to keep the crate from sliding
Answer:
0.248
Explanation:
Initial speed u = 13.6
Final speed v = 0
Distance s = 38.1
We have umg = ma
We make u subject of the formula
u = a/g
V² = u² + 2as
a = v²-u²/2s
We substitute the values into the above
a = 0-(13.6)²/2*38.1
a = 184.96/76.2
a = 2.427m/sec
Remember that
u = a/g
u = 2.427/9.8
= 0.2476
This is approximately 0.248
This is the minimum coefficient of friction required to keep the crate from sliding.
Visualizing yourself crossing the finish line and how'd you'd feel is
a method of blocking unwanted feelings
a way to cope with stress
utilizing positive values
O a method of influence on others
Answer:
I believe you put how you think you'd feel it's that simple
Answer:
utilizing positive values
Explanation:
Please help I’m almost done with exam
Which phrase is the best description of what a telescope does?
O A. Causes objects to grow larger
B. Transports equipment to space
C. Converts solar energy to electricity
D. Detects electromagnethwaves
Answer:
D
Explanation:
Telescopes detect electromagnetic waves from space and it travels back to the telescope lens in order for you to see.
Answer:
D
Explanation:
Ap ex science exam lol
difine precision and accuracy
Which statement is true of a glass lens that diverges light in air?
A.
It is thick near the center and thin at the edges.
B.
It is thin near the center and thick at the edges.
C.
It is uniformly thick.
D. It is uniformly thin.

Answer: it is thin near the center and thick at the edges
Explanation: took the test on Plato :)
There is a parallel plate capacitor. Both plates are 4x2 cm and are 10 cm apart. The top plate has surface charge density of 10C/m^2 and the bottom plate has a surface charge density of -10C/m^2. Find the total charge on each plate. Find the electric field at the point exactly midway between the plates. Find the electric potential between the two plates. If an electron was in the middle the two plates, find the force on it.
Answer:
1) The total charge of the top plate is 0.008 C
b) The total charge of the bottom plate is -0.008 C
2) The electric field at the point exactly midway between the plates is 0
3) The electric field between plates is approximately 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates is approximately 1.807 × 10⁻⁷ N
Explanation:
The given parameters of the parallel plate capacitor are;
The dimensions of the plates = 4 × 2 cm
The distance between the plates = 10 cm
The surface charge density of the top plate, σ₁ = 10 C/m²
The surface charge density of the bottom plate, σ₂ = -10 C/m²
The surface area, A = 0.04 m × 0.02 m = 0.0008 m²
1) The total charge of the top plate, Q = σ₁ × A = 0.0008 m² × 10 C/m² = 0.008 C
b) The total charge of the bottom plate, Q = σ₂ × A = 0.0008 m² × -10 C/m² = -0.008 C
2) The electrical field at the point exactly midway between the plates is given as follows;
[tex]V_{tot} = V_{q1} + V_{q2}[/tex]
[tex]V_q = \dfrac{k \cdot q}{r}[/tex]
Therefore, we have;
The distance to the midpoint between the two plates = 10 cm/2 = 5 cm = 0.05 m
[tex]V_{tot} = \dfrac{k \cdot q}{0.05} + \dfrac{k \cdot (-q)}{0.05} = \dfrac{k \cdot q}{0.05} - \dfrac{k \cdot q}{0.05} = 0[/tex]
The electric field at the point exactly midway between the plates, [tex]V_{tot}[/tex] = 0
3) The electric field, 'E', between plates is given as follows;
[tex]E =\dfrac{\sigma }{\epsilon_0 } = \dfrac{10 \ C/m^2}{8.854 \times 10^{-12} \ C^2/(N\cdot m^2)} \approx 1.1294 \times 10^{12}\ N/C[/tex]
E ≈ 1.1294 × 10¹² N/C
The electric field between plates, E ≈ 1.1294 × 10¹² N/C
4) The force on an electron in the middle of the two plates
The charge on an electron, e = -1.6 × 10⁻¹⁹ C
The force on an electron in the middle of the two plates, [tex]F_e[/tex] = E × e
∴ [tex]F_e[/tex] = 1.1294 × 10¹² N/C × -1.6 × 10⁻¹⁹ C ≈ 1.807 × 10⁻⁷ N
The force on an electron in the middle of the two plates, [tex]F_e[/tex] ≈ 1.807 × 10⁻⁷ N
Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance DA beyond the starting line at t=0. The starting line is at x=0. Car A travels at a constant speed vA. Car B starts at the starting line but has a better engine than Car A, and thus Car B travels at a constant speed vB, which is greater than vA?
The question is incomplete. Here is the complete question.
Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance [tex]D_{A}[/tex] beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed [tex]v_{A}[/tex]. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed [tex]v_{B}[/tex], which is greater than [tex]v_{A}[/tex].
Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.
Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.
Answer: Part A: [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Part B: [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Explanation: First, let's write an equation of motion for each car.
Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:
[tex]x=x_{0}+vt[/tex]
where
[tex]x_{0}[/tex] is initial position
v is velocity
t is time
Car A started the race at a distance. So at t = 0, initial position is [tex]D_{A}[/tex].
The equation will be:
[tex]x_{A}=D_{A}+v_{A}t[/tex]
Car B started at the starting line. So, its equation is
[tex]x_{B}=v_{B}t[/tex]
Part A: When they meet, both car are at "the same position":
[tex]D_{A}+v_{A}t=v_{B}t[/tex]
[tex]v_{B}t-v_{A}t=D_{A}[/tex]
[tex]t(v_{B}-v_{A})=D_{A}[/tex]
[tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex]
Car B meet with Car A after [tex]t=\frac{D_{A}}{v_{B}-v_{A}}[/tex] units of time.
Part B: With the meeting time, we can determine the position they will be:
[tex]x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )[/tex]
[tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex]
Since Car B started at the starting line, the distance Car B will be when it passes Car A is [tex]x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}[/tex] units of distance.
The distance traveled by the car A and car B should be equal to the as they meet at the same position.
The time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
How to calculate the distance traveled by body?The distance is the product of the speed of the body and the time taken to travel the distance.
Given information-
Car A has a head start and is a distance DA beyond the starting line at,
[tex]t=0[/tex]
Car A travels at a constant speed [tex]v_A[/tex].
Car B travels at a constant speed [tex]v_B[/tex].
The distance is the product of the speed of the body and the time taken to travel the distance.
The position equation from the motion for car A can be given as,
[tex]x_A=v_At+D_A[/tex]
The position equation from the motion for car B can be given as,
[tex]x_B=v_Bt[/tex]
The distance traveled by the car A and car B should be equal to the as they meet at the same position. Thus,
[tex]x_A=x_B[/tex]
Put the values,
[tex]v_At+D_A=v_Bt\\v_At-v_Bt=-D_A\\t(v_B-v_A)=D_A\\t=\dfrac{D_A}{v_B-v_A}[/tex]
Hence the time car B will catch the car A after is,
[tex]\dfrac{D_A}{v_B-v_A}[/tex]
Learn more about the speed of the object here;
https://brainly.com/question/4931057
Even though Alice visits the wishing well frequently and always tosses in a coin for good luck, none of her wishes have come true. As a result, she decides to change her strategy and make a more emphatic statement by throwing the coin downward into the well. If the water is 5.43 m below the point of release and she hears the splash 0.85 seconds later, determine the initial speed at which she threw the coin. (Take the speed of sound to be 343 m/s.)
Answer:
Explanation:
Total time taken = 0.85 s .
Time taken by sound to travel 5.43 m + time taken by coin to fall by 5.43 m = .85
5.43 / 343 + time taken by coin to fall by 5.43 m = .85
time taken by coin to fall by 5.43 m = .85 - 5.43/343 = .834 s
Let the initial velocity of throw of coin = u
displacement of coin s = 5.43 m
time take to fall t = .834 s
s = ut + 1/2 gt²
5.43 = u x .834 + .5 x 9.8 x .834²
5.43 = u x .834 + 3.41
u x .834 = 2.02
u = 2.42 m /s .
Four-wheel drive trucks do not stop better on icy
roads than a car. Is what law of motion (Newton's laws)
What are two things that happen to the sugars that are made by the plant during photosynthesis?
I
Answer:
The sugars produced by photosynthesis can be stored, transported throughout the tree, and converted into energy which is used to power all cellular processes. Respiration occurs when glucose (sugar produced during photosynthesis) combines with oxygen to produce useable cellular energy.
Explanation:
I think this is correct lol.
Plzz help me with this
Answer: d
Explanation:
Which of the following hydrocarbons are SATURATED hydrocarbons?
I. alkanes II. alkenes III. alkynes IV. cycloalkanes
A. I and IV
B. II and III
C. I and III
D. II and IV
Answer:
i think c
Explanation:
Plzz answer this question correctly
Answer:
by reducing friction.....
1) In order to get work done, what must be present?
a) Energy
b) Oxygen
please help
Answer:
In order to accomplish work on an object there must be a force exerted on the object and it must move in the direction of the force
Explanation:
option a is right
A flat screen tv uses 120 watts. How much energy is used up if it is left on for 15 min?
A.) 4j
B.) 15j
C.) 0.67j
D.) 108,000j
Answer:
d
Explanation:
While talking to a friend, a construction worker momentarily set her cell phone down on one end of an iron rail of length 7.50 m. At that moment, a second worker dropped a wrench so that it hit the other end of the rail. The person on the phone detected two pulses of sound, one that traveled through the air and a longitudinal wave that traveled through the rail. (Assume the speed of sound in iron is 5,950 m/s and the speed of sound in air is 343 m/s).
A) Which pulse reaches the cell phone first?
B) Find the separation in time (in s) between the arrivals of the two pulses.
Answer:
A)
The impulse that reaches the cell phone first is the Longitudinal wave
B) 0.0206 seconds
Explanation:
length of Iron rail = 7.5 m
speed of sound in Iron = 5950 m/s
speed of sound in Air = 343 m/s
A) Determine which pulse reaches the cell phone first
The impulse that reaches the cell phone first is the Longitudinal wave
Time for longitudinal pulse to be detected = 7.5 / 5950 = 0.00126 s
Time for pulse through air to be detected = 7.5 / 343 = 0.02186 s
B) separation in time between the arrivals of the two pulses
ΔT = 0.02186 - 0.00126 = 0.0206 seconds
A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is?
Answer:
h over 2 dg
Explanation:
brainliest!!!!!!!
A football quarterback runs 15.0 m straight down the playing field in 2.30 s. He is then hit and pushed 3.00 m straight backward in 1.74 s. He breaks the tackle and runs straight forward another 28.0 m in 5.20 s.
Calculate his average velocity (in m/s) for each of the three intervals. (Assume the quarterback's initial direction is positive. Indicate the direction with the sign of your answer.)
Answer:
6.52 m/s
1.72 m/s
5.38 m/s
Explanation:
this question requires us to find the average velocity.
1. velocity in straight down direction:
velocity = distance/time
= 15.0/2.30
= 6.52 m/s
2. velocity in straight backward direction:
velocity = distance/time
= 3.00 /1.74
= 1.72m/s
3. velocity in straight forward direction
velocity = distance/time
= 28.0/5.20
= 5.38 m/s
these are the his velocities for each if the intervals.
thank you!
10. A change in
indicates the acceleration of an object
O A the time of travel
OB the distance from a given point
O c displacement
OD velocity
Answer:
d velocity will be the one according to me
An object initially traveling in a straight line with
a speed of 5.0 meters per second is accelerated
at 2.0 meters per second squared for 4.0 seconds.
The total distance traveled by the object in the
4.0 seconds is
Answer:
We conclude that the total distance traveled by the object in the 4 seconds is 36 m.
Explanation:
Given
Initial velocity u = 5.0 m/sAcceleration a = 2.0 m/s²Time t = 4 sTo determine
The total distance traveled by the object in the 4.0 seconds is
Important Tip:
We can determine the total distance traveled by the object in the 4.0 seconds by using the equation of motion such as
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = distanceu = initial velocitya = accelerationt = timesubstituting u = 5.0, a = 2, and t = 4 in the formula
[tex]s=ut+\frac{1}{2}at^2[/tex]
[tex]s=\left(5\right)\left(4\right)+\frac{1}{2}\left(2\right)\left(4\right)^2[/tex]
[tex]s=20+16[/tex]
[tex]s=36[/tex] m
Therefore, we conclude that the total distance traveled by the object in the 4 seconds is 36 m.
The total distance traveled by the object is 36 meters.
Given the following data:
Initial velocity = 5.0 m/s Acceleration = 2.0 [tex]m/s^2[/tex] Time = 4.0 seconds.To find the total distance traveled by the object, we we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2}at^2[/tex]
Where:
S is the total distance traveled.
u is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting the values into the formula, we have;
[tex]S = 5(4) + \frac{1}{2}(2)(4^2)\\\\S = 20 + 1(16)[/tex]
Total distance, S = 36 meters.
Therefore, the total distance traveled by the object is 36 meters.
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an object of 5kg is attached to a rope of length 4m is Rotating horizontally at 8m/s horizontally 20m above the ground if the rope is suddenly cut what is the horizontal distance travelled by the object? Please guys help
Answer:
16 meters
Explanation:
When the rope is suddenly cut the object moving tangent at the circle. In that moment the gravity act in the object making it falls.
First we need to find how much time de object take to reach at the ground.
VERTICALLY EQUATION:[tex]h(t)=h-v*t-\frac{g}{2} t^{2} \\[/tex]g=acceleration of gravity=10m/s²
v= vertical velocity =0m/s
h=vertical altitude =20m
We will find t such that h(t)=0
[tex]0=20-5t^{2} \\\\5t^{2} =20\\\\t^{2} =4\\\\t=2s[/tex]
HORIZONTALLY EQUATION:*horizontally we do not have acceleration[tex]D(t)=v*t[/tex]v=horizontal velocity
D(t=2) is the horizontal distancetravelled by the object:
[tex]D(2)=8*2\\\\D(2)=16m[/tex]
If a car's speed triples, how does the momentum and kinetic energy of the
car change? Answer in form (momentum change, kinetic energy change)
Answer: When the car speed triples, momentum also triples but Kinetic energy increases 9 times or by 9 fold.
Explanation:
The momentum of a car (an object) is
p= mv
where
m is =the mass of the object( in this case car)
v is its= velocity
While the kinetic energy is is given by the formulae
K=1/2mv²
To determine how momentum and kinetic energy of the car changes when the speed of the object triples, We have that the new velocity,
v¹= 3v
So that the momentum change becomes
p¹=mv¹=m (3v)= 3mv
mv=p
therefore p¹= 3p
we can see that the momentum also triples.
And the kinetic energy change becomes
K¹=1/2m(v¹)²= 1/2m (3v)²
= 1/2m9v²= 1/2 x m x 9 x v²=9 x1/2mv²
1/2mv²=K
K¹= Kinetic energy = 9k
but Kinetic energy increases 9 times
A plastic cup weighing 100 g floats on water so that 1/4 of the volume of the cup is immersed in water. How much volume of oil can be poured into the glass to keep it still sinking? The density of the oil is 900 kg / m
Answer:
Any floating object displaces a volume of water equal in weight to the object's MASS. ... If you place water and an ice cube in a cup so that the cup is entirely full to the ... If you take a one pound bottle of water and freeze it, it will still weigh one ... Fresh, liquid water has a density of 1 gram per cubic centimeter (1g = 1cm^3, ...
how does the strength of the forces that hold the basic particles of a substance together relate to the temperature at which the substance changes state
The kinetic energy keeps the molecules apart and moving around, and is a function of the temperature of the substance. ... Increasing the pressure on a substance forces the molecules closer together, which increases the strength of intermolecular forces.
A block of mass m1 = 19.5 kg slides along a horizontal surface (with friction, μk = 0.35) a distance d = 2.6 m before striking a second block of mass m2 = 8.25 kg. The first block has an initial velocity of v = 6.5 m/s.
(a) Assuming that block one stops after it collides with block two, what is block two's velocity after impact in m/s?
(b) How far does block two travel, d2 in meters, before coming to rest after the collision?
Answer:
19.5 m/s
87.8 m
Explanation:
The acceleration of block one is:
∑F = ma
-m₁gμ = m₁a
a = -gμ
a = -(9.8 m/s²) (0.22)
a = -2.16 m/s²
The velocity of block one just before the collision is:
v² = v₀² + 2aΔx
v² = (8.25 m/s)² + 2 (-2.16 m/s²) (2.3 m)
v = 7.63 m/s
Momentum is conserved, so the velocity of block two just after the collision is:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂
m₁u₁ = m₂v₂
(18.5 kg) (7.63 m/s) = (7.25 kg) v
v = 19.5 m/s
The acceleration of block two is also -2.16 m/s², so the distance is:
v² = v₀² + 2aΔx
(0 m/s)² = (19.5 m/s)² + 2 (-2.16 m/s²) Δx
Δx = 87.8 m
Explanation:
By using conservation of linear momentum and also by equating work done to kinetic energy, [tex]V_{2}[/tex] = 15.36 m/s and [tex]d_{2}[/tex] = 4.32 meters
Parameters given are :
[tex]m_{1}[/tex] = 19.5 kg
friction, μk = 0.35
distance d = 2.6 m
mass [tex]m_{2}[/tex] = 8.25 kg.
initial velocity of [tex]U_{1}[/tex] = 6.5 m/s.
a.) Since we assumed that the block one stops after it collides with block two, the final velocity for block one will be zero. That is, [tex]V_{1}[/tex] = 0 so its final momentum = 0
Let us also assume that block two was initially at rest. Therefore, it initial velocity and its momentum will be equal to zero.
The formula to use will be :
[tex]m_{1}U_{1} = m_{2}V_{2}[/tex]
Substitute all the parameters into the formula above
19.5 x 6.5 = 8.25[tex]V_{2}[/tex]
Make [tex]V_{2}[/tex] the subject of formula
[tex]V_{2}[/tex] = 126.75/8.25
[tex]V_{2}[/tex] = 15.36 m/s
b.) Let us first calculate the work done in by block one.
The K.E = [tex]1/2mU^{2}[/tex]
substitute its mass and velocity into the formula
K.E = 1/2 x 19.5 x [tex]6.5^{2}[/tex]
K.E = 411.94 Joule
The work done = Kinetic energy
But the resultant Force F = force f - friction
where Frictional force = 0.35 x 19.5 x 9.8
Frictional force = 66.89N
Work done will be the product of resultant Force F and the distance travelled
(F - 66.89) x 2.6 = 411.94
F - 66.89 = 411.94/2.6
F - 66.89 = 158.44
F = 225.3 N
The second block will experience the same force which is equal to 225.3N
Find the kinetic energy of the second block.
K.E = [tex]1/2mV^{2}[/tex]
K.E = 0.5 x 8.25 x 15.36^2
K.E = 973.2
Using The work done = Kinetic energy
225.3[tex]d_{2}[/tex] = 973.2
[tex]d_{2}[/tex] = 973.2/225.3
[tex]d_{2}[/tex] = 4.32 meters
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[RM.03H]Which of these is the most likely impact of extensive mining of uranium to produce energy?
land becomes unfit for food production
rainfall decreases because of harmful gases
greenhouse gases are absorbed by the mineral
radiations are better absorbed by the atmosphere
Answer:
land becomes unfit for food production
Blythe and Geoff compete in a 1.00-km race. Blythe's strategy is to run the first 600 m of the race at a constant speed of 4.10 m/s, and then accelerate to her maximum speed of 7.30 m/s, which takes her 1.00 min, and then finish the race at that speed. Geoff decides to accelerate to his maximum speed of 8.30 m/s at the start of the race and to maintain that speed throughout the rest of the race. It takes Geoff 3.00 min to reach his maximum speed. Assume all accelerations are constant.
Required:
a. Calculate the time of Blythe's run.
b. Calculate the time of Geoff s run.
Answer:
Explanation:
1.00 km = 1000 m .
Blythe's run : -------
Time taken to run 600 m speed
= distance / speed
T₁= 600 / 4.10 = 146.34 s
Next time T₂ = 1 min = 60 s
acceleration of Blythe
a = (7.30 - 4.10) / 60 = .053 m /s²
displacement during acceleration
= ut + 1/2 at²
= 4.10 x 60 + .5 x .053 x 60²
= 246 + 95.4
= 341.4 m
Rest of the distance to be covered = 1000 - ( 600 + 341.4 )
= 58.9 m
Time taken to cover this distance
T₃= 58.9 / 7.3 = 8.06 s
Total time = T₁ + T₂ + T₃ = 214.4 s
Geoff s run : ---------
initial acceleration during first 3 min
= (8.3 - 0 ) / (3 x 60 )
= .046 m /s²
displacement
s = ut + 1/2 a t²
= 0 + .5 x .046 x ( 3 x 60 )²
= 745.2 m
Rest of the distance of race
= 1000 - 745.2 = 254.8 m
This distance is covered at speed of 8.3 m/s
time taken to cover this distance
T₂ = 254.8 / 8.3
= 30.7 s
Total time taken to complete the race
= 180 + 30.7
= 210.7 s .