A typical laboratory centrifuge rotates at 4000 rpm. Testtubes have to be placed into a centrifuge very carefully because ofthe very large accelerations.
Part A) What is the acceleration at the end of a test tubethat is 10 cm from the axis of rotation?
Part B) For comparison, what is the magnitude of theacceleration a test tube would experience if dropped from a heightof 1.0 m and stopped in a 1.0-ms-long encounter with a hardfloor?
Answers
Answer:
A) a_c = 1.75 10⁴ m / s², B) a = 4.43 10³ m / s²
Explanation:
Part A) The relation of the test tube is centripetal
a_c = v² / r
the angular and linear variables are related
v = w r
we substitute
a_c = w² r
let's reduce the magnitudes to the SI system
w = 4000 rpm (2pi rad / 1 rev) (1 min / 60s) = 418.88 rad / s
r = 1 cm (1 m / 100 cm) = 0.10 m
let's calculate
a_c = 418.88² 0.1
a_c = 1.75 10⁴ m / s²
part B) for this part let's use kinematics relations, let's start looking for the velocity just when we hit the floor
as part of rest the initial velocity is zero and on the floor the height is zero
v² = v₀² - 2g (y- y₀)
v² = 0 - 2 9.8 (0 + 1)
v =√19.6
v = -4.427 m / s
now let's look for the applied steel to stop the test tube
v_f = v + a t
0 = v + at
a = -v / t
a = 4.427 / 0.001
a = 4.43 10³ m / s²
Related Questions
The escape velocity for planet Earth is 11 km/s, and the escape velocity for Moon is 2.4 km/s. We'd like to know how much more energy it takes to escape the gravitational pull of the Earth, than to escape the pull of the Moon.
a. 20,000
b. 21
C.1
D. 5
Answers
More energy to escape gravitational pull of the Earth than to escape the pull of Moon is calculated as : D) 5
What is escape velocity?Escape velocity is the minimum speed needed for free, non-propelled object to escape from gravitational influence of primary body.
Energy required to escape the gravitational pull of a planet or a moon is:
\(\mathrm{E = (1/2)mv^2}\)
E is the energy required, m is mass of the object being launched, and v is escape velocity.
E(Earth) = (1/2)(1 kg)\(\mathrm{(11,000 m/s)^2}\) = \(\mathrm{6.05 \times 10^7 J}\)
E(Moon) = (1/2)(1 kg)\(\mathrm{(2,400 m/s)^2}\) = \(\mathrm{2.88 \times 10^6 \ J}\)
E(Earth) - E(Moon) = \(\mathrm{6.05 \times 10^7 J}\) - (\(\mathrm{2.88 \times 10^6 \ J}\)) = \(\mathrm{5.76 \times 10^7 \ J}\)
Answer D) is correct.
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Olive and her friend Wellington are playing down by Captain Don's docks when they find an old chain. The old chain has only three links. By measuring with an old fish scale which is a permanent feature of the dock area, they determine that the total mass of the chain is 3.12 kg (the scale reads in newtons, but Olive knows how to calculate the mass of the chain from its weight). While playing with the scale and the chain (the chain is hanging vertically from the end of the scale, and Olive is holding on to the top of the scale with both hands, either moving the entire system upwards or downwards), Olive notices that if she is accelerating the chain either upwards or downwards, the scale no longer accurately reads the weight of the chain. When the Scale Reading is Larger than the Weight of the Chain At one point in their experiments with the chain and the scale, Wellington observes that the scale reads 47.00 N . Part A When the scale reads 47.00 N , what is the tension in the chain at the point where the lowest two links connect
Answers
Answer:
T₁ = 15.66 N
Explanation:
From the given information:
Using the free body system in the chain.
T - mg = ma
47 - 3.12 × 9.8 = 3.12a
47 - 30.576 = 3.12a
16.424 = 3.12a
a = 16.424/3.12
a = 5.26 m/s²
Now, by the free body diagram of the lowest link; the tension (T₁ ) in the chain is:
T₁ - (3.12/3) × 9.8 = (3.12/3) × 5.26
T₁ - 1.04 × 9.8 = 1.04 × 5.26
T₁ - 10.192 = 5.4704
T₁ = 5.4704 + 10.192
T₁ = 15.6624
T₁ = 15.66 N
How might a quota on foreign cars affect US economy?
Answers
Answer:
The numerical limits imposed on imported goods through quotas ultimately leads to higher prices paid by consumers. Essentially, the import quota prevents or limits domestic consumers from buying imported goods. The import quota reduces the supply of imports.
Explanation:
WILL GIVE BRAINLIEST!!! NO LINKS PLZ!!!!
A 225 g hockey puck is sliding on ice in an arena towards the end boards that are 15.7 m away. The puck is travelling 12.0 m/s when it slides into some rough ice (coefficient of kinetic friction= 0.550).
Determine:
a) the acceleration of the puck on the rough ice.
b) the distance from the end boards the puck is when it comes to a stop.
Please show work.
Answers
Answer:
Explanation:
a) the acceleration of the puck on the rough ice.
a = μg = 0.550(9.81) = 5.3955 = 5.40 m/s²
(comes from μ = F/N = ma/mg = a/g)
b) the distance from the end boards the puck is when it comes to a stop.
v² = u² + 2as
0² = 12.0² + 2(-5.40)s
s = 13.3 ft
so distance from the boards is
15.7 - 13.3 = 2.4 m
by the way...that's some VERY rough ice...more like sand.
A particle moves with a velocity v in a circle of radius r, then its angular velocity is equal
to………….. and acts along the…………..
Answers
Answer:
Given that,
Speed = v
Radius = r
We have to ascertain the precise speed
Utilizing equation of speed
Where, v = velocity/speed
r = radius
= precise speed/ angular velocity
Angular Velocity :
The precise speed is characterized by the speed of turn.
The precise speed is straightforwardly corresponding to the direct speed and contrarily relative to the range of the molecule.
Subsequently, The precise speed is v/r
You push yourself on a skateboard with a force of 30 N east and accelerate at 0.5 m/s² east. Find the mass of theskateboard if your mass is 58 kg.
Answers
Answer:
2Kg
first you find the total mass of both the person und skateboard
f = m*a
30 = m* 0.5
m= 30/ 0.5
m= 60 kg
then you subtract the mass of the person from the total mass
m1 - m2
60 - 58
= 2 kg
Taking into account the Newton's second law, the mass of the skateboard is 2 kg.
Newton's Second Law establishes the relationship between the forces acting on an object and the acceleration it experiences. This law says that the acceleration that an object experiences when subjected to a net force (force applied to the body) is proportional to that force.
Mathematically, Newton's second law is expressed as:
F= m×a
where:
F = Force [N] m = Mass [kg] a = Acceleration [m/s²]In this case, you know:
F= 30 N m= ? a= 0.5 m/s²Replacing:
30 N= m× 0.5 m/s²
Solving:
m= 30 N÷0.5 m/s²
m= 60 kg
The mass of the sistem is the sum of the mass of the skateboard and the mass of the mass of your body:
Mass= mass of the skateboard + mass of your body
In this case:
60 kg= mass of the skateboard + 58 kg
Solving:
mass of the skateboard= 60 kg - 58 kg
mass of the skateboard= 2 kg
Finally, the mass of the skateboard is 2 kg.
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brainly.com/question/23845187?referrer=searchResults brainly.com/question/13959891?referrer=searchResultsYou need to construct a 400 pF capacitor for a science project. You plan to cut two L x L metal squares and place spacers between them. The thinnest spacers you have are 0.20 mm thick. What is the proper value of L?Express your answer in centimeters.
Answers
Consider that the formula for the capacitance of a square parallel plate capacitor is:
\(C=\epsilon_o\frac{A}{d}=\epsilon_o\frac{L^2}{d}\)where A=L^2 is the area of each plate, d is the separation between plates and
ε0 is the dielectric permitivity of vacuum ans is equal to 8.82*10^-12 F/m.
If you solve the previous expression for L and replace the given values for d and C, you obtain:
\(\begin{gathered} L=\sqrt[]{\frac{dC}{\epsilon_o}} \\ d=0.20mm=0.20\cdot10^{-3}m=2.0\cdot10^{-4}m \\ C=400pF=400\cdot10^{-12}F=4.00\cdot10^{-10}F \\ L=\sqrt[]{\frac{(2.0\cdot10^{-4}m)(4.00\cdot10^{-10}F)}{8.85\cdot10^{-10}\frac{F}{m}}} \\ L\approx0.0095m=0.95cm \end{gathered}\)Hence, the proper value of L to construct the required capacitor is approximately 0.95cm
measurement conversions [metric to metric] 568 cm = m
Answers
Answer:
5.68 meters
Explanation:
hope this helps!
Answer:
5.68
Explanation:
to convert cm to m you move the decimal point 2x to the left
1. How long will it take a car to accelerate from 15.2 m/s to 23.5 m/s if the car
has an average acceleration of 3.2 m/s2 ?
Solving for:
Formula:
Substitute known values:
Number answer:
Unit answer
Answers
Answer:
I have no clue I'm really really really really really but like really sorry this is so hard
19273)2
Woman
9272
What is the period of a wave with a speed of 10 m/s and frequency of 23 Hz?
Answers
The frequency calculator will let you find a wave's frequency given the wavelength and its velocity or period in no ... velocity equals 320 m/s and its wavelength is 8 m. Find its frequency. f = 320 m/s / 8 m. f = 40/s. f = 40 Hz.
What is the speed over the ground mosquito flying 2 m/s relative to the ar caught in a 2 m/s right angle crosswind
Answers
The speed over the ground is 0 m/s.
The speed of an object or body is the magnitude of the change of its position over time or the magnitude of the change of its position per unit of time; it is thus a scalar quantity.
As it is given that the mosquito is flying at 2 m/s relatives to the air caught in a 2 m/s right angle crosswind.
As we know that speed over the ground is the difference between the flight speed and resistance speed.
So, the general equation for the speed over the ground is :
v = Flight Speed of the mosquito - Resistance Speed of the crosswind
This implies, \(v= 2 {~}m / s - 2{~} m / s\)
\(v = 0 {~}m / s\)
Hence, the speed over the ground is 0 m/s.
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The net force on a box F as a function of the vertical position y is shown below.
What is the work done on the box from y=0 to 6.0m?
Answers
The work done on the box from y=0 to 6.0m is 120 J.
To calculate the work done on the box from y=0 to 6.0m, we need to find the area under the force vs. position graph over that interval.
First, we can find the work done from 0 m to 2 m. Since the force is constant at 40 N over this interval, the work done is simply:
W = F * d = 40 N * 2 m = 80 J
From 2 m to 6 m, the force is constant at -20 N, so the work done is:
W = F * d = (-20) N * 4 m = -80 J
Note that the negative sign indicates that the work is done by the box on the force (since the force is in the opposite direction of the displacement).
Therefore, the total work done on the box from y=0 to 6.0m is:
W_total = 80 J - 80 J = 0 J
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The following table lists the speed of sound in various materials. Use this table to answer the question.
Substance Speed (m/s)
Glass 5,200
Aluminum 5,100
Iron 4,500
Copper 3,500
Salt water 1,530
Fresh water 1,500
Mercury 1,400
Hydrogen at 0°C 1,284
Ethyl Alcohol 1,125
Helium at 0°C 965
Air at 100°C 387
Air at 0°C 331
Oxygen at 0°C 316
Sound will travel fastest in air at _____.
-5°C
0°C
10°C
15°C
Answers
Sound will travel fastest in air at 15°C.
Speed of sound in air
The speed of sound in air, given in the range of 100 degrees Celsius and 0 degree Celsius include;
Air at 100°C 387 m/s
Air at 0°C 331 m/s
From the date above, the speed of sound in air increases with increases in temperature. Thus, Sound will travel fastest in air at 15°C.
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A straight rod of length =22.80 cm carries a uniform charge density =1.80×10^−6 C/m. The rod is located along the ‑axis from 1=0.00 to 2=. The Coulomb force constant is =8.99×10^9 N·m2/C2.
A vertical line of positive charge of length capital L. A labeled point P is located above the line of charge along a line coaxial with the line of charge. The distance from the base of the line of charge to point P is y subscript zero, with y sub zero greater than L.
Find the expression for the electric field along the y‑axis Ey at a point P. What is the magnitude of the electric field at y0=60.00 cm?
Help is greatly appreciated!!
Answers
L
=
22.80
c
m
The value of the linear charge density on the rod is
λ
=
2.85
×
10
−
6
m
−
1
From the higher end of the rod, the distance of the test point is
y
∘
=
65
c
m
According to Coulomb's Law,
The value of the electric field due to the small segment of the rod is,
d
E
=
k
d
q
r
2
d
r
d
E
=
k
λ
L
r
2
d
r
dE =
k
d
q
r
2
dr dE =
k
λ
L
r
2
dr
Where k is the electrostatic constant.
The distance of the lower end from the origin is
y
1
=
0
y
1
=
0
The distance o the higher end of the rod from the origin is
y
2
=
L
y
2
=
L
Thus, the distance of the test point from the lower end is,
y
′
=
y
2
+
y
∘
y
′
=
y
2
+
y
∘
The net electric field at the test charge due to the linear charge density of the rod is,
E
=
∫
y
′
y
∘
k
λ
L
r
2
d
r
E
=
k
λ
L
∫
y
′
y
∘
1
r
2
d
r
E
=
k
λ
L
[
−
1
r
]
y
′
y
∘
E
=
k
λ
L
[
−
1
y
′
+
1
y
∘
]
E =
∫
y
∘
y
′
k
λ
L
r
2
d
r
E =kλL
∫
y
∘
y
′
1
r
2
d
r
E =kλL
[
−
1
r
]
y
∘
y
′
E =kλL
[
−
1
y
′
+
1
y
∘
]
Substituting the known values,
E
=
8.99
×
10
9
×
2.85
×
10
−
6
×
22.8
×
10
−
2
[
−
1
(
22.80
+
65
)
×
10
−
2
+
1
65
×
10
−
2
]
E
=
2.33
×
10
3
N
−
1
E =8.99×
10
9
×2.85×
10
−
6
×22.8×
10
−
2
[
−
1
(
22.80
+
65
)
×
10
−
2
+
1
65
×
10
−
2
]
E =2.33×
10
3
N
−
1
Thus, the electric field at the test point is
2.33
×
10
3
N
−
1
2.33
×
10
3
N
−
1
When did modern chemistry start
Answers
Question 2-5: You know from Activity 1-1 that an object will slow down if the force applied to the object (and the acceleration) are in the opposite direction to the velocity. Once you release the IOLab, no force is applied to the force sensor. What force is causing the IOLab to slow down
Answers
The force that is causing the IOLab to slow down is frictional force between the IOLab and the surface it is moving on.
As the force applied to an object increases, the acceleration of the object increases as well. This applied force is used to overcome the frictional force between the object and the surface it is moving on.
When the force applied to an object reduces, the effect of frictional force increases. This frictional force is the force that reduces the motion of an object.
Thus, we can conclude that the force that is causing the IOLab to slow down is frictional force between the IOLab and the surface it is moving on.
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Answer:I don’t if this helps The force that is causing the IOLab to slow down is frictional force between the IOLab and the surface it is moving on.As the force applied to an object increases, the acceleration of the object increases as well. This applied force is used to overcome the frictional forcebetween the object and the surface it is moving on.When the force applied to an object reduces, the effect of frictional forceincreases. This frictional force is the force that reduces the motion of an object.Thus, we can conclude that the force that is causing the IOLab to slow down is frictional force between the IOLab and the surface it is moving on.
Explanation:
A ball rolls of buildings that is 100m high calculate the time that it takes for ball to hit the ground
Answers
Answer:
2as=v2-u2
2000=v2
V=44
V=u+at
44/10=t
T=4.4seconds
A car speeds up from 18.54 m/s to
29.52 m/s in 13.84 s.
The acceleration of the car is:
Answers
Answer:
.7934\(m/s^{2}\)
Explanation:
Acceleration = change in velocity / change in time
A = 10.98\(m/s\) / 13.84\(s\)
A = .7934\(m/s^{2}\)
Answer:0.8 m/s^2
Explanation:
initial velocity(u)=18.54m/s
Final velocity(v)=29.52m/s
Time(t)=13.84 sec
Acceleration =(v-u)/t
acceleration =(29.52-18.54)/13.84
Acceleration =10.98/13.34
Acceleration=0.8 m/s^2
A cart slows with an acceleration of 2 m/s2 over a period of 6 s. If its initial speed was 12 m/s how far will it travel before it stops?
Answers
Answer:
36 m
Explanation:
a = -2 m/s² negative a because it's slowing down
t = 6 s
v₀ = 12 m/s
v = 0
Δx = v₀t + 1/2at²
Δx = (12 m/s)(6 s) + 1/2(-2 m/s²)(6 s)²
Δx = 72 m + (-36 m) = 36 m
the area under acceleration time garph represents?
Answers
Answer:
Change in Velocity because
\(at = v\)
Explanation:
Remeber area is length times Width. In this case, the area under a accleraton vs time graph is Accleration Times Time. Which is velocity
how to calculate the magnitude and direction of vector A-vectorB
A-B
Answers
Answer:
Let vector C(x,y) be A-B.
Explanation:
Now the direction is given by: tan-1(y/x)
Use the correct angle on the basis in which quadrant the vector C lies.
Magnitude = sqrt(x^2 + y^2)
A 3.0-kg sphere was moving at 25 m/s when it hit another object. This caused all KE to be converted into thermal energy. What was the increase in thermal energy?
I have been stuck on this problem for two hours. please help
Answers
Answer:
ΔT=0.81
o
C
Explanation:
As we know by energy conservation
All its kinetic energy will convert into thermal energy to raise its temperature
1
2
�
�
2
=
�
�
Δ
�
2
1
mv
2
=msΔT
now divide both sides by mass of the object
1
2
�
2
=
�
Δ
�
2
1
v
2
=sΔT
so change in temperature is given as
Δ
�
=
�
2
2
�
ΔT=
2s
v
2
Δ
�
=
2
5
2
2
×
387
ΔT=
2×387
25
2
Δ
�
=
0.8
1
�
�
ΔT=0.81
o
C
A 3.0-kg sphere was moving at 25 m/s when it hit another object. This caused all KE to be converted into thermal energy. What was the increase in thermal energy?
The initial kinetic energy of the sphere is given by:
KE = 0.5 * m * v^2
where m is the mass of the sphere and v is its velocity. Substituting the given values, we get:
KE = 0.5 * 3.0 kg * (25 m/s)^2
= 937.5 J
This is the initial kinetic energy of the sphere before it hit the other object. Since all of this energy is converted into thermal energy, the increase in thermal energy is simply equal to the initial kinetic energy, which is:
ΔE = KE = 937.5 J
Therefore, the increase in thermal energy is 937.5 Joules.
Suppose the U.S. national debt is about $15 trillion. If payments were made at the rate of $1,500 per second, how many years would it take to pay off the debt, assuming no interest were charged? Note: Before doing these calculations, try to guess at the answers. You may be very surprised. yr (b) A dollar bill is about 15.5 cm long. How many dollar bills attached end to end would it take to reach the Moon? The Earth-Moon distance is 3.84 108 m. dollar bills
Answers
Answer:
This question has already been answered.
Explanation:
https://brainly.com/question/13542582
If an airplane is accelerating down a runway at 2.5 m/s^2, how much is it’s velocity changing each second?
Answers
If an airplane is is accelerating down a runway at 2.5 m/s². It's velocity changes 2.5 m/s in each second.
What is acceleration?Acceleration is the rate at which speed and direction of velocity vary over time. A point or object going straight ahead is accelerated when it accelerates or decelerates.
Even if the speed is constant, motion on a circle accelerates because the direction is always shifting. Both effects contribute to the acceleration for all other motions.
Hence, the velocity of the airplane changes 2.5 m/s in each second when it is accelerating down a runway at 2.5 m/s².
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In the following description, QUOTE the portion that addresses weather in the area.
Portland, Oregon is located along the coast of the northwestern United States. Several famous volcanoes, including Mount St. Helens and Mount Ranier, are located nearby. Portland normally has rainy, but mild winters. Yesterday, it rained 0.75 inches in Portland.
Answers
Answer:
Portland normally had rainy, but mild winters. Yesterday it rained 0.75 inches in Portland.
Explanation:
This talks about the weather
True or False. In a parallel circuit, the voltage is the same anywhere in the circuit.
Answers
Answer:
true
Explanation:
in a parallel circuit, the voltage is the same
find the power of a lift that transfers 450 J of energy in 15 seconds.
Answers
Answer: P=30W
Explanation:
formula is p=w/t
p = power
w = work
t = elapsed time
input variables, solve then simplify.
The power of a lift that transfers 450 J of energy in 15 seconds is 30 watts.
Power is defined as the rate of doing work, i.e. the amount of work done per unit time.
Mathematically, it can be represented as follows:
Power = Work done / time taken
Therefore, the power of a lift that transfers 450 J of energy in 15 seconds can be calculated as follows:
Power = Work done / time taken= 450 J / 15 s= 30 W
Therefore, the power of the lift is 30 watts.
To explain further, we know that power is measured in watts (W), and it is the rate at which work is done or energy is transferred.
Here, we are given that the lift transfers 450 J of energy in 15 seconds.
We can find the power of the lift by dividing the amount of work done by the time taken to do it. By substituting the given values, we get the power of the lift as 30 W.
In simple terms, this means that the lift can transfer energy at a rate of 30 joules per second. This can also be interpreted as the lift can do 30 joules of work in one second.
Hence, we can conclude that the power of the lift is 30 watts.
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Un telescopio indica que la posición de un satélite de comunicaciones está determinada por el punto P (7x105m, 5x105m) . ¿A qué distancia se encuentra el satélite del telescopio? En que dirección viajará una señal que sale del telescopio hacia el satélite?
Answers
a) La Distancia entre el telescopio y el satélite es aproximadamente \(8.602\times 10^{5}\) metros.
b) La señal sale del telescopio y se desplaza \(7\times 10^{5}\) metros en Dirección + x y \(5\times 10^{5}\) metros en Dirección +y.
a) Si asumimos la Posición del telescopio como el Origen de un Marco de Referencia, entonces la Distancia entre el telescopio y el satélite (\(d\)), en metros, mediante la siguiente expresión:
\(d = \sqrt{(S(x,y) - O(x,y))\,\bullet\,(S(x,y) - O(x,y))}\) (1)
Donde:
\(S(x,y)\) - Coordenadas del satélite, en metros.\(O(x,y)\) - Coordenadas del telescopio, en metros.Si sabemos que \(O(x,y) = (0, 0)\) y \(S(x,y) = (7\times 10^{5}, 5\times 10^{5})\), entonces la Distancia entre el telescopio y el satélite es:
\(d = \sqrt{(7\times 10^{5})^{2}+(5\times 10^{5})^{2}}\)
\(d \approx 8.602\times 10^{5}\,m\)
Nota - (1) es una forma vectorial del Teorema de Pitágoras.
La Distancia entre el telescopio y el satélite es aproximadamente \(8.602\times 10^{5}\) metros.
b) La señal sale del telescopio y se desplaza \(7\times 10^{5}\) metros en Dirección + x y \(5\times 10^{5}\) metros en Dirección +y.
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A uniform magnetic field is in the positive z direction. A positively charged particle is moving in the positive x direction through the field. The net force on the particle can be made zero by applying an electric field in what direction
Answers
Answer:
We apply an electric field in the negative y direction
Explanation:
Since A uniform magnetic field is in the positive z direction and A positively charged particle is moving in the positive x direction through the field, the magnetic force acting on the positively charged particle is in the positive y direction according to Fleming's right-hand rule.
For the net force on the particle to be zero, we apply an electric field in the negative y direction to create an electric force on the positively charged particle, so as to cancel out the magnetic force.
A pilot wants to fly due north a distance of 125 km. The wind is blowing out of the west at a constant 35 km/h. If the plane can travel at 175km/h, how long will the trip take?
Answers
The time taken for the trip is determined as 0.73 hrs.
What is the time taken for the trip?The time taken for the trip is calculated by applying the following formula as follows;
time taken for the trip = ( total displacement ) / (average velocity)
To determine the total displacement of the pilot we will use Pythagoras theorem as follows;
(175 t)² = (35t)² + 125²
where;
t is the time of motion30625t² = 1225t² + 15625
Simplify the equation further and solve for t;
30625t² - 1225t² = 15625
29400t² = 15625
t² = 15625/29,400
t² = 0.53
t = √0.53
t = 0.73 hrs
Learn more about time of motion here: https://brainly.com/question/24739297
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