A orbiting piece of debris with a mass of 140 kg is orbiting the Earth at a height of 51,000 km above the surface.
The Earth's mass and radius are 5.98×10^24 kg and 6,385 km. Resolution At a distance of 51000 km from the surface of the Earth, the gravitational field strength of the debris is obtained using the formula;$$g=\frac{GM}{r^2}$$
Where; G is the universal gravitational constant M is the mass of the Earth r is the radius of orbit of the debris
We know that G=6.6743×10^-11 m^3 kg^-1 s^-2,
M=5.98×10^24 kg,
and r= (6385 km + 51000 km)
= 57385 km
= 5.7385 ×10^7m.
Substitute these values into the formula,
$$g=\frac{6.6743×10^{-11} × 5.98×10^{24}}{(5.7385×10^7)^2}
= 0.22 N/kg$$
Thus, the gravitational field strength at this height is 0.22 N/kg.
Weight of the satellite: We can find the weight of the satellite using the formula;$$F=mg$$
Where; F is the weight of the satellite g is the gravitational field strength m is the mass of the satellite We know that m=140 kg and g=0.22 N/kg.
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A 18 Hz, 16.5 V source produces a 1.95 mA current when connected
to a capacitor.
a) What is the capacitance in μF?
The capacitance of the capacitor can be determined by using the formula relating current and capacitance in an AC circuit. By calculating the capacitance, we can find the value in microfarads (μF). Hence, the capacitance of the capacitor is approximately 0.002 microfarads (μF).
In an AC circuit, the relationship between current (I), capacitance (C), voltage (V), and frequency (f) is given by the equation:
I = 2πfCV
Rearranging the equation, we can solve for the capacitance (C):
C = I / (2πfV)
Given the following values:
Frequency (f) = 18 Hz
Voltage (V) = 16.5 V
Current (I) = 1.95 mA = 1.95 × 10^(-3) A
Substituting the values into the equation, we have:
C = (1.95 × 10^(-3) A) / (2π × 18 Hz × 16.5 V)
Simplifying the expression:
C ≈ 0.002 μF
Therefore, the capacitance of the capacitor is approximately 0.002 microfarads (μF).
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A 65.0 kg skater moving initially at 3.20 m/s on rough horizontal ice comes to rest uniformly in 3.05 s due to friction from the ice. What force does friction exert on the skater? Express your answer with the appropriate units
When a skater slides on a rough horizontal ice and comes to rest uniformly, the force of friction exerted by the ice can be found. To find the force of friction, we can use the following equation:f = maWhere f is the force, m is the mass, and a is the acceleration.
We can find the acceleration of the skater using the formula for uniformly accelerated motion:Δx = vit + 1/2at²where Δx is the distance covered, vi is the initial velocity, t is the time, and a is the acceleration.
Rearranging the formula, we get:
a = 2(Δx - vit)/t²where Δx is the distance covered, vi is the initial velocity, and t is the time.
Substituting the given values:
a = 2(0 - 3.20*3.05)/3.05² = -2.10 m/s² (negative because the skater is decelerating)
Now we can substitute the values of m and a into the equation for force:
f = ma = 65.0 kg x -2.10 m/s² = -136
The force of friction exerted by the ice on the skater is 136 N.
The negative sign indicates that the force is opposite to the direction of motion.
The units of force are Newtons (N). Answer:
The force of friction exerted on the skater is -136 N.
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You throw a ball straight up with an initial velocity of 16.3 m/s. It passes a tree branch on the way up at a height of 7.01 m. How much additional time (in s) will pass before the ball passes the tree branch on the way back down?
The additional time it will take for the ball to pass the tree branch on the way back down after passing it on the way up is given by the time taken to reach the maximum height subtracted from the total time taken.
Given:
Coefficient of static friction (μs) = 0.50
Angle of inclination (θ) = 45°
To find the mass of the block (M), we can equate the maximum static friction force (fstatic max) to the component of the gravitational force acting down the slope.
The maximum static friction force is given by:
fstatic max = μsN
where N is the normal force.
The normal force can be calculated as:
N = Mg cos θ
where M is the mass of the block and g is the acceleration due to gravity.
The component of the gravitational force down the slope is given by:
Mg sin θ
Setting fstatic max equal to Mg sin θ, we have:
μsN = Mg sin θ
μs(Mg cos θ) = Mg sin θ
μs cos θ = sin θ
μs = sin θ / cos θ
Now, substituting the given values:
0.50 = sin 45° / cos 45°
Using the trigonometric identity sin θ / cos θ = tan θ, we have:
0.50 = tan 45°
Taking the inverse tangent (arctan) of both sides, we find:
45° = arctan(0.50)
Using a calculator, we can determine that the angle is approximately 26.565°.
Therefore, the correct mass of the block is approximately 0.391 kg.
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Shaggy throws a ball straight up while Velma times it. She sees that the ball goes by the top of a flagpole after 0.50 s and returns to the level of the top of the pole after a total elapsed time of 4.1 s. Calculate the speed of the ball at the moment it passed the top of the flagpole on the way up. Neglect air resistance. Show work on back. A) 6.4 m/s B) 16 m/s C) 18 m/s D) 29 m/s E) 33 m/s
The speed of the ball, when it passes the top of the flagpole on the way up, is approximately 29 m/s.
From the given data, the time for the ball to reach the top of the flagpole can be calculated using the formula v = u + at. Since the initial velocity (u) is 0, the formula can be simplified as v = gt, where g is the acceleration due to gravity (9.8 m/s²) and t is the time taken for the ball to reach the top.
Hence, t = v/g = 0.50 s.
Subsequently, the time taken for the ball to return to the level of the top of the pole is (4.1 - 0.50) s = 3.6 s.
Using the formula v = u + at, the final velocity (v) of the ball can be calculated. Since the final velocity is 0 when the ball returns to the top, the formula can be simplified as u = -at = -g(3.6) = -35.28 m/s.
Therefore, the speed of the ball, when it passes the top of the flagpole on the way up, is approximately 29 m/s, which is the absolute value of the initial velocity. Hence, the correct option is D) 29 m/s.
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With the aid of a string, a gyroscope is accelerated from rest to 29 rad/s in 0.39 s. ω = 29 rad/s t = 0.39 s (a) What is its angular acceleration, in radians per square seconds? (b) How many revolut
(a) The angular acceleration is approximately 74.36 rad/s².
(b) The gyroscope goes through approximately 4.61 revolutions in the process.
(a). The angular acceleration of a gyroscope can be determined using the formula:
angular acceleration (α) = change in angular velocity (Δω) / time (t)
In this case, the change in angular velocity (Δω) is given as 29 rad/s (final angular velocity) minus 0 rad/s (initial angular velocity), which simplifies to 29 rad/s. The time (t) is given as 0.39 s.
Therefore, the angular acceleration (α) can be calculated as:
α = Δω / t
α = 29 rad/s / 0.39 s
Calculating this, we find that the angular acceleration is approximately 74.36 rad/s².
(b). To determine the number of revolutions the gyroscope goes through in the process, we need to convert the final angular velocity from radians per second to revolutions per second.
1 revolution is equal to 2π radians. The number of revolutions (n) can be calculated using the formula:
n = final angular velocity (ω) / (2π)
In this case, the final angular velocity (ω) is given as 29 rad/s.
Substituting the values into the formula, we get:
n = 29 rad/s / (2π)
Calculating this, we find that the gyroscope goes through approximately 4.61 revolutions in the process.
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Complete question is,
With the aid of a string, a gyroscope is accelerated from rest to 29 rad/s in 0.39 s. ω = 29 rad/s, t = 0.39 s
(a) What is its angular acceleration, in radians per square seconds?
(b) How many revolutions does it go through in the process?
car driving at 60.0 km/h slams the brakes, imparting an acceleration of −10.0 km/h
2
. How far does the car travel from the moment it hit the brakes, in meters? (Note: 1 km=1,000 m.)
The car travels a distance of 49.8 from the moment it hit the brakes.
In order to find the distance traveled by the car, we can use the equation of motion:
v^2 = u^2 + 2as
where,
v is the final velocity (0 m/s, since the car comes to a stop)
u is the initial velocity (60.0 km/h)
a is the acceleration (-10.0 km/h^2)
s is the distance traveled (what we need to find)
Motion is represented in terms of displacement, distance, velocity, acceleration, speed, and time. Equations of motion are equations that describe the behavior of a physical system in terms of its motion as a function of time. They are mathematical formulae that describes the position, velocity, or acceleration of a body relative to a given frame of reference. The following are the three equations of motion:
First Equation of Motion : v = u + at.
Second Equation of Motion : s = u t + (1/2) at^2.
Third Equation of Motion : v^2 = u^2 + 2as.
In the given question, first, we need to convert the velocities from km/h to m/s.
Since 1 km = 1,000 m and 1 hour = 3,600 seconds:
Therefore, 60.0 km/h = 60.0 * (5/18) = 50/3 = 16.7 m/s
and, -10.0 km/h = (-10) * (5/18) = -25/9 = -2.8 m/s^2
Now, we can substitute the values into the equation of motion:
0^2 = (16.7)^2 + 2 * (-2.8) * s
Simplifying the equation, we have:
(278.89 - 5.6) s = 0
Rearranging the equation, we can solve for \(s\):
=> 5.6s = 278.89
=> s = 278.89 / 5.6
=> s = 49.8 m
Therefore, the car travels a distance of 49.8 meters from the moment it hits the brakes.
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6. Near the end of the life of our Sun, it will become a white dwarf star. This means it will be size of the Moon, but have the same mass as it does now. What will be the gravitational field strength on the surface of the Sun at this time? 7. When humans last went to the Moon on an Apollo space mission, the command module orbited the Moon while the lunar module landed on the Moon's surface. The command module orbited 100 km above the surface of the Moon. How long did it take the command module to complete one orbit of the Moon?
6. The gravitational field strength on the surface of the Sun at the end of its life as a white dwarf star would be much greater than it is now.
Despite the fact that the star will be only the size of the Moon, it will still have the same mass as it does now. As a result, the gravitational field strength on the surface of the Sun will be quite high.7.
The time it takes for an object to complete one orbit of the Moon is determined by the distance between the Moon and the object. The distance between the Moon and the command module is given as 100 km. Because the lunar module has landed on the Moon's surface, we can assume that the mass of the Moon is negligible and does not have an impact on the time taken to orbit the Moon.
The command module's orbit, on the other hand, is determined by the gravitational force of the Moon, which is determined by the Moon's mass. The formula for the time taken to complete one orbit of a planet or satellite is as follows:T = 2π √(a3/GM)Where T is the period of orbit, a is the semi-major axis of the orbit, G is the universal gravitational constant, and M is the mass of the object being orbited.
Using the given data, we can calculate the period of the command module's orbit of the Moon as follows:a = r + hWhere r is the radius of the Moon (1737 km) and h is the height of the orbit [tex](100 km).a = 1837 kmM = 7.35 × 1022 kgT = 2π √(a3/GM)T = 121[/tex] minutes (approximately)Therefore, the command module took about 121 minutes to complete one orbit of the Moon.
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A 4.0 mili grams particle with electric charge of +110.0μC orbits a charged wire, making 12000 revolutions per second. The radius of the orbit is 1.5 cm and length of the wire is 90.0 cm. What is the wire's linear charge density?
The linear charge density of the wire is approximately 3.26 × 10^-6 C/m. The force that maintains the particle's circular motion is the electrostatic force between the particle and the charged wire. This force is given by Coulomb's law.
The force that maintains the particle's circular motion is the electrostatic force between the particle and the charged wire. This force is given by Coulomb's law:
F = k * (q1 * q2) / r^2
where k is the Coulomb constant (9 × 10^9 Nm^2/C^2), q1 and q2 are the charges of the two particles, and r is the distance between them.
In this case, the wire is negatively charged (since the orbiting particle has a positive charge), so the electrostatic force acts as a centripetal force that keeps the particle in its circular orbit:
F = m * a
where m is the mass of the particle and a is the centripetal acceleration, given by:
a = v^2 / r
where v is the orbital velocity of the particle.
To find the linear charge density of the wire, we need to use the fact that the sum of the charges along the wire creates an electric field that interacts with the orbiting particle. The electric field at the position of the particle is given by:
E = k * λ / r
where λ is the linear charge density of the wire. This electric field provides the centripetal force needed to maintain the particle's circular motion:
F = q * E
where q is the charge of the particle.
Equating these two expressions for F and solving for λ, we get:
λ = q * r * v^2 / (k * 2πr^2)
Plugging in the values given in the problem, we get:
v = 2πr * f = 2π * 1.5 cm * 12000 s^-1 = 113.1 m/s (where f is the frequency of the particle's orbit)
q = 110.0 μC = 110.0 × 10^-6 C
m = 4.0 mg = 4.0 × 10^-6 kg
r = 1.5 cm = 0.015 m
Substituting these values, we get:
λ = (110.0 × 10^-6 C) * (0.015 m) * (113.1 m/s)^2 / (9 × 10^9 Nm^2/C^2 * 2π * (0.015 m)^2)
λ ≈ 3.26 × 10^-6 C/m
Therefore, the linear charge density of the wire is approximately 3.26 × 10^-6 C/m.
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A person walks 60.0 m east and then 11.0 m west. Find both the distance he has traveled and his displacement.
Question 2
An object moves 20 m east in 30 s and then returns to its starting point taking an additional 50 s. If west is chosen as the positive direction, what is the average velocity of the object?
Question 3
You are driving home on a weekend from school at 55 mi/h for 110 miles. It then starts to snow and you slow to 25 mi/h. You arrive home after driving 4 hours and 35 minutes. How far is your hometown from school?
Question 1:The distance traveled is 71.0 m, and the displacement is 49.0 m east.
Question 2:The average velocity of the object is 0 m/s.
Question 3:The distance from school to your hometown is 327.415 miles.
Question 1:
The distance traveled is the sum of the magnitudes of the individual displacements: 60.0 m + 11.0 m = 71.0 m.
The displacement is the vector difference between the initial and final positions: 60.0 m east - 11.0 m west = 49.0 m east.
Question 2:
The total distance traveled is 20 m + 20 m = 40 m (as the object returns to its starting point).
The average velocity is the total displacement divided by the total time taken: 0 m / 80 s = 0 m/s.
Question 3:
The total driving time is 4 hours and 35 minutes, which is equal to 4.583 hours.
The distance traveled at 55 mi/h is 55 mi/h * 4.583 h = 252.415 mi.
The remaining distance is covered at 25 mi/h for the remaining time, which is 25 mi/h * (4.583 h - 110 mi / 55 mi/h) = 75 mi.
Therefore, the total distance from school to your hometown is 252.415 mi + 75 mi = 327.415 mi.
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Funner A is initally 1.0 m west of a flacpole and is running with a constant velocty of 4.0minh due east. Funner 8 is lintially 2.0mi east of the fapole and is running with a constant velooty of 7.0 m/h di west. Mara far are the runners freen the flagpole when ther meet?
When the runners meet, Funner B is approximately 2012.73 meters west of the flagpole. The distance between the runners and the flagpole when they meet, is their individual distances from the flagpole at any given time.
Let's denote the distance of Funner A from the flagpole as DA and the distance of Funner B from the flagpole as DB. We want to find the value of DB when the two runners meet.
Initially, Funner A is 1.0 m west of the flagpole, so DA = -1.0 m. Funner B is 2.0 mi east of the flagpole, so DB = 2.0 mi.
Funner A is running with a constant velocity of 4.0 m/h due east, and Funner B is running with a constant velocity of 7.0 m/h due west.
Let's assume that they meet after t hours. The distance covered by Funner A during this time is given by DA = 4.0 m/h × t, and the distance covered by Funner B is given by DB = 2.0 mi - 7.0 m/h × t.
The time at which they meet, we set DA equal to DB:
4.0 m/h × t = 2.0 mi - 7.0 m/h × t
For t, we can convert the units of the distances to a common unit, such as meters:
4.0 m/h × t = 2.0 mi × (1609.34 m/mi) - 7.0 m/h × t
Simplifying the equation gives:
4.0 m/h × t + 7.0 m/h × t = 2.0 mi × (1609.34 m/mi)
11.0 m/h × t = 2.0 mi × (1609.34 m/mi)
t = 2.0 mi × (1609.34 m/mi) / 11.0 m/h
Calculating the value of t gives:
t ≈ 291.39 hours
Now that we know the time at which they meet, we can substitute this value back into either DA or DB to find the distance between the runners and the flagpole at that time.
Using DB = 2.0 mi - 7.0 m/h × t:
DB ≈ 2.0 mi - 7.0 m/h × (291.39 hours)
DB ≈ -2012.73 m
Therefore, when the runners meet, Funner B is approximately 2012.73 meters west of the flagpole.
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Two 10-cm-diameter charged rings face each other, 22.0 cm apart. Both rings are charged to + 20.0 nC . What is the electric field strength
Part A) at the midpoint between the two rings?
Part B) at the center of the left ring?
Please help with both parts
Given that the charged rings face each other, 22.0 cm apart. Both rings are charged to +20.0 nC. Let us calculate the electric field strength.
Part AAt the midpoint between the two rings, the electric field strength is given by;E = kQ/d2Where
k = 9 x 109 Nm2/C2 is the Coulomb constant,
Q = 20 nC
= 20 x 10-9 C is the charge on the rings, and
d = 22 cm
= 0.22 m is the separation distance between the rings.So,
E = (9 x 109) x (20 x 10-9) / (0.11)2
E = 82000 N/CPart BAt the center of the left ring, the electric field strength is given by;
E = kQ/d2Where
k = 9 x 109 Nm2/C2 is the Coulomb constant,
Q = 20 nC
= 20 x 10-9 C is the charge on the rings, and
d = 5 cm
= 0.05 m is the separation distance between the point and the center of the left ring.So,
E = (9 x 109) x (20 x 10-9) / (0.05)2
E = 1.3 x 1012 N/C Therefore, the electric field strength at the midpoint between the two rings is 82000 N/C and the electric field strength at the center of the left ring is 1.3 x 1012 N/C.
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Question 3
a) What happens when a positively-charged rod is touched to a
negatively-charged rod in the absence of the charged balloon? What
are the charges on the rods after
touching?
When a positively-charged rod is touched to a negatively-charged rod in the absence of the charged balloon, electrons from the negatively-charged rod move to the positively-charged rod.
The positively-charged rod gains electrons and becomes less positive, while the negatively-charged rod loses electrons and becomes less negative, after touching, the positively-charged rod will have a slightly less positive charge, while the negatively-charged rod will have a slightly less negative charge.
This process is called the transfer of charge.The transfer of charge is the process by which electric charges are moved from one place to another. It can occur in a number of ways, including conduction, induction, and polarization.
In the case of touching a positively-charged rod to a negatively-charged rod, the transfer of charge occurs by conduction.
This is because the electrons flow directly from one rod to the other through physical contact. Overall, the transfer of charge plays a crucial role in many aspects of physics and engineering, from the operation of batteries and generators to the behavior of lightning and other electrical phenomena in the atmosphere.
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To start a car engine, the car battery moves 3.8×1021 electrons through the starter motor- How many coulombs of charge are moved?
To start a car engine, the car battery moves 3.8×1021 electrons through the starter motor-the number of coulombs of charge that are moved is approximately 6.08 × 102 C.
The electric charge of a coulomb is equivalent to the electric charge transferred by a current of one ampere flowing for one second.
Therefore, the number of coulombs of charge that are moved when a car battery moves 3.8 × 1021 electrons through the starter motor is a unit conversion problem.
The formula for electric charge is as follows: q = n × e, where q is the total charge in coulombs, n is the total number of electrons, e is the electric charge of a single electron.
Using the formula above, we can calculate the total charge that is moved as follows:
q = (3.8 × 1021) × (1.6 × 10-19)= 6.08 × 102 charge.
Therefore, the number of coulombs of charge that are moved is approximately 6.08 × 102 C.
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The number of coulombs of charge moved is 6.08.
In order to understand the calculation of the number of coulombs of charge moved, let's break down the steps involved.
First, it is known that the charge of each electron is 1.6 × 10^-19 C. This value represents the fundamental unit of charge carried by an electron.
According to Coulomb's law, the total charge (Q) is equal to the product of the number of electrons (n) moving and the charge of a single electron (e). Mathematically, it can be represented as Q = n x e.
In this specific scenario, the problem states that 3.8×10²¹ electrons are moving through the starter motor of a car battery. By substituting the given values into the equation, we can calculate the total charge moved.
Q = (3.8 × 10²¹) × (1.6 × 10^-19)
To simplify the calculation, we can use the properties of scientific notation. When multiplying numbers in scientific notation, we add the exponents and multiply the coefficients:
Q = 3.8 × 1.6 × 10²¹ × 10^-19
Multiplying the coefficients gives us:
Q = 6.08 × 10²¹ × 10^-19
Now, we can simplify the expression by adding the exponents of 10:
Q = 6.08 × 10²¹-19
Finally, we have our result in scientific notation. The number 6.08 represents the coefficient, and 10²¹-19 represents the exponent. This can be expressed in decimal notation as:
Q = 6.08 C
Therefore, the number of coulombs of charge moved is 6.08 C.
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1. An object is held by 4 wires inside a box which is located on a frictionless horizontal surface. The arrangements of the wires and their tensions are given on the diagram. What is the acceleration
The object will experience zero acceleration because the net force acting on it is zero. This is because the tensions in the wires balance out each other and the force of gravity is also balanced by the normal force of the surface.
Therefore, the object will remain at rest or continue moving with a constant velocity.According to Newton's second law of motion, the net force acting on an object is directly proportional to its acceleration and inversely proportional to its mass. That is,Fnet=maWhere Fnet is the net force acting on the object,
m is the mass of the object, and a is the acceleration of the object.In this case, the object is held by four wires whose tensions are given in the diagram. The tensions in the wires are the forces that act on the object in the upward direction. These tensions balance out the force of gravity that acts on the object in the downward direction. The force of gravity and the normal force of the surface on which the box rests are equal in magnitude and opposite in direction. Since the tensions balance out the force of gravity, the net force acting on the object is zero. Therefore, the acceleration of the object is zero.
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A planet with a radius of 6.00x 10^7 m has a gravitational field
of magnitude 55.7 m/s^2 at the surface.
What is the escape speed from the planet?
the escape speed from the planet is approximately 40,284 m/s.
Escape speed is the minimum speed at which a moving object must travel to escape the gravitational pull of a massive body. It can be calculated using the formula:v = √(2GM/r
Where v is the escape speed, G is the universal gravitational constant, M is the mass of the planet, and r is the radius of the planet.
To determine the value of v, we must first calculate the mass of the planet.Mass = (g × r²)/G, where g is the gravitational field strength and r is the radius of the planet. Therefore, the mass of the planet is:
M =[tex](55.7 m/s² × (6.00 × 10^7 m)²)/6.67430 × 10^-11 N(m/kg)²M = 8.43 × 10^25 kg[/tex]
Now that we have the mass of the planet, we can calculate the escape speed:v = [tex]√(2GM/r)v = √(2 × 6.67430 × 10^-11 N(m/kg)² × 8.43 × 10^25 kg / 6.00 × 10^7 m)v = 40,284.34 m/s[/tex]
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A speed trap is set up with two pressure activated strips placed across a highway,118 meters apart.A car is speeding along at 34.4 meters per second,while the speed limit is only 18.6 meters per
second.At the instant the car activates the first strip,the driver begins slowing down.What minimum deceleration(magnitude only) is needed so that the driver's average speed is within the limit by the time the car crosses the second strip?
The minimum deceleration required to ensure that the driver's average speed is within the limit by the time the car crosses the second strip is 1.53 m/s². Hence, the minimum required deceleration (magnitude only) is 1.53 m/s².
As per the given problem, we have a car traveling at a speed of 34.4 meters per second on a highway with a speed limit of 18.6 meters per second. Two pressure-activated strips are placed across the highway, separated by a distance of 118 meters. We need to determine the minimum deceleration required to ensure that the driver's average speed is within the limit by the time the car crosses the second strip.
Let's consider the time taken for the car to travel the distance between the two strips as 't'. The car's velocity when it crosses the first strip is 34.4 m/s, while its velocity when it crosses the second strip is 18.6 m/s.
We can express the time 't' using the given velocities and deceleration 'a' as:
[tex]t = \frac{118}{34.4 - at} + \frac{118}{18.6}[/tex]
Multiplying both sides by (34.4 - at)(18.6), we obtain:
[tex]$$t(34.4 - at)(18.6) = 118(18.6) + t(34.4 - at)(118)$$[/tex]
Simplifying further:
[tex]$$t(34.4 \times 18.6 - 34.4at + 118 \times 18.6 + 118 \times 34.4at) = 118 \times 53.0$$[/tex]
This equation can be rewritten as:
[tex]$$6330.24t - 4087.2at = 6247.4$$[/tex]
Dividing both sides by 3.3, we have:
[tex]$$1921.53t - 1236.48at = 1892.85$$[/tex]
Therefore, the minimum deceleration required to ensure that the driver's average speed is within the limit by the time the car crosses the second strip is 1.53 m/s². Hence, the minimum required deceleration (magnitude only) is 1.53 m/s².
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The electron drift speed in a copper wire of diameter 1.8 mm is 3.6×10−4ms−1 The number of free electrons per unit volume for copper is 8.5×1028m−3 Estimate the current in the wire.
The current in the wire is 0.00274 A.
Given, the electron drift speed in a copper wire of diameter 1.8 mm is 3.6 × 10⁻⁴ms⁻¹
The number of free electrons per unit volume for copper is 8.5 × 10²⁸m⁻³
To estimate the current in the wire we use the relation, `I = nAvq`.
Where,I is the currentn is the number of free electrons per unit volumeV is the volume of the conductorq is the charge on a single electrona is the cross-sectional area of the conductor.
The volume of the conductor is given byV = πr²LWhere,r is the radius of the wire andL is the length of the wire.
The cross-sectional area of the conductor is given bya = πr².
Substituting the values, we getV = π(0.9 × 10⁻³m)²(1m)
V = 2.54 × 10⁻⁶m³a = π(0.9 × 10⁻³m)²
a = 2.54 × 10⁻⁶m²q = 1.6 × 10⁻¹⁹C
Using the above formula, I = nAvq
I= 8.5 × 10²⁸m⁻³ × (2.54 × 10⁻⁶m³) × (3.6 × 10⁻⁴ms⁻¹) × (1.6 × 10⁻¹⁹C)
I = 0.00274 A
Therefore, the current in the wire is 0.00274 A.
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a closed system is observed to have 39 objects in it. initially, the total energy of the system is 6507 j. after 5.3 h have passed, how much total energy will the system contain?
In a closed system with 39 objects, the total energy initially is 6507 J. After 5.3 hours, the total energy of the system will remain the same in accordance with the law of conservation of energy.
According to the law of conservation of energy, the total energy in a closed system remains constant unless there is an external energy transfer. In this case, the closed system consists of 39 objects, and initially, it has a total energy of 6507 J. The law of conservation of energy states that the total energy of the system will remain constant over time. Therefore, after 5.3 hours have passed, the total energy of the system will still be 6507 J.
Since no information is given about any energy transfers or external influences on the system, we can conclude that the system is isolated and no energy is added or removed from it. Thus, the initial energy of 6507 J will be retained after 5.3 hours. It is important to note that this conclusion assumes an ideal closed system with no energy exchanges with the surroundings. In practical situations, factors such as energy dissipation, friction, and other external influences may cause some energy loss or gain in the system.
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"
A soccer ball is kicked with an initial horizontal velocity of \( 20 \mathrm{~m} / \mathrm{s} \) and an initial vertical velocity of \( 10 \mathrm{~m} / \mathrm{s} \). a) What is the speed of the ball"
The speed of the soccer ball is 22.36 m/s (rounded to two decimal places).
To determine the speed of a soccer ball that is kicked with an initial horizontal velocity of 20 m/s and an initial vertical velocity of 10 m/s, we can use the Pythagorean theorem.
The Pythagorean theorem states that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, as shown below:
[tex]$$\mathrm{hypotenuse^2 = opposite^2 + adjacent^2}$$[/tex]
In this case, the horizontal velocity is the adjacent side and the vertical velocity is the opposite side. Therefore, the hypotenuse is the speed of the ball. We can use the following formula to find the speed:
[tex]$$\mathrm{speed = \sqrt{v_{x}^2 + v_{y}^2}}$$[/tex]
where \(v_{x}\) is the horizontal velocity and \(v_{y}\) is the vertical velocity.
Substituting the given values, we have:
[tex]$$\mathrm{speed = \sqrt{(20\ m/s)^2 + (10\ m/s)^2}}$$[/tex]
[tex]$$\mathrm{speed = \sqrt{400\ m^2/s^2 + 100\ m^2/s^2}}$$[/tex]
[tex]$$\mathrm{speed = \sqrt{500\ m^2/s^2}}$$[/tex]
[tex]$$\mathrm{speed = 22.36\ m/s}$$[/tex]
Therefore, the speed of the soccer ball is 22.36 m/s (rounded to two decimal places).
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Julie drives 53.0mi to Grandmother's house. On the way to Grandmother's, Julie drives half the distance at 36.0mph and half the distance at 64.0mph. What is Julie's average speed on the way to Grandmother's house?
Julie's average speed to her grandmother's house requires calculating total time from the distances traveled at different speeds.
To find Julie's average speed on the way to Grandmother's house, we can use the formula for average speed, which is the total distance traveled divided by the total time taken.
Given:
Distance traveled at 36.0 mph = Distance traveled at 64.0 mph = Half the total distance
Total distance traveled = 53.0 mi
Let's calculate the time taken for each segment:
Time taken for the first half of the distance at 36.0 mph:
Distance = (1/2) * 53.0 mi = 26.5 mi
Time = Distance / Speed = 26.5 mi / 36.0 mph
Time taken for the second half of the distance at 64.0 mph:
Distance = (1/2) * 53.0 mi = 26.5 mi
Time = Distance / Speed = 26.5 mi / 64.0 mph
Now, let's calculate the total time taken:
Total time = Time for the first half + Time for the second half
Finally, we can calculate the average speed:
Average speed = Total distance / Total time
By substituting the appropriate values and performing the calculations, we can find Julie's average speed on the way to Grandmother's house.
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Lap runner ( 10pts.) A runner is initially jogging due east on a 500 m track, at a speed of 2.2
s
m
. After reaching the end of the track they run back ( 500 m due west), at a speed of 5
s
m
. A. (2 pts.) Find the runner's displacement and distance covered during their run. 2 B. (2 pts.) Find the runner's average velocity during their run. C. ( 3pts.) Find the runner's average speed during their run. D. (3pts.) Find the average acceleration of the runner during their run.
A runner was initially jogging due east on a 500 m track, at a speed of 2.2m/s. After reaching the end of the track, they run back (500 m due west), at a speed of 5m/s. Therefore, we need to find the runner's displacement and distance covered during their run.
A runner was initially jogging due east on a 500 m track, at a speed of 2.2m/s. After reaching the end of the track, they run back (500 m due west), at a speed of 5m/s. Therefore, we need to find the runner's displacement and distance covered during their run.
Displacement
The distance from where the runner started and where they ended is known as the displacement. Here, the displacement is 0, as the runner ends where they started from.
Distance
The distance covered is the actual distance covered during the run. The distance covered by the runner is 500m + 500m = 1000m = 1km.
Average Velocity
The average velocity of the runner is the total displacement divided by the total time taken. Here, the total time taken is 500m/2.2m/s + 500m/5m/s = 227s. Therefore, the average velocity is 0/227s = 0 m/s.
Average Speed
The average speed of the runner is the total distance covered divided by the total time taken. Here, the total time taken is 500m/2.2m/s + 500m/5m/s = 227s. Therefore, the average speed is 1km/227s = 0.0044 km/s.
Average Acceleration
The average acceleration of the runner is the total change in velocity divided by the total time taken. Here, the total time taken is 500m/2.2m/s + 500m/5m/s = 227s. Therefore, the average acceleration is (5m/s - 2.2m/s)/227s = 0.008m/s².
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In an effort to impress its friends, an acrobatic cockroach runs and jumps
off the top of a flight of stairs. Each step is 20 cm tall and 25 cm wide. The cockroach
jumps from the very edge of the top step with an initial velocity of 1.5 m/s at 20◦ above the
horizontal. On which step does the cockroach land?
The cockroach jumps off the top step of the stairs. With an initial velocity of 1.5 m/s at 20° above the horizontal, it lands on the second step. The steps are 20 cm tall and 25 cm wide.
To determine on which step the cockroach lands, we can analyze the horizontal and vertical motion separately.
Let's start with the horizontal motion:
The horizontal velocity of the cockroach remains constant at 1.5 m/s throughout its jump. Since the cockroach jumps from the edge of the top step, there is no horizontal displacement. Therefore, the horizontal position of the cockroach remains the same as it jumps.
Next, let's analyze the vertical motion:
The cockroach jumps with an initial velocity of 1.5 m/s at an angle of 20° above the horizontal. We need to find the time it takes for the cockroach to reach the ground.
Using the vertical motion equation:
h = v₀y * t + (1/2) * g * t²,
where h is the vertical displacement, v₀y is the vertical component of the initial velocity, g is the acceleration due to gravity, and t is the time.
The initial vertical velocity can be calculated as:
v₀y = v₀ * sin(θ),
where v₀ is the initial velocity and θ is the launch angle.
Substituting the given values:
v₀y = 1.5 m/s * sin(20°),
v₀y ≈ 0.514 m/s.
The vertical displacement h can be determined by finding the time it takes for the cockroach to reach the ground. The equation becomes:
0 = 0.514 m/s * t - (1/2) * 9.8 m/s² * t².
Simplifying the equation:
4.9 t² - 0.514 t = 0.
Solving for t using the quadratic formula:
t = 0.514 s or t = 0 s (discarding the solution t = 0 s as it represents the initial time).
Therefore, the time it takes for the cockroach to reach the ground is approximately 0.514 seconds.
Now, let's calculate the distance covered horizontally during this time:
distance = v₀x * t,
where v₀x is the horizontal component of the initial velocity.
Substituting the given values:
distance = 1.5 m/s * cos(20°) * 0.514 s,
distance ≈ 0.730 meters.
Since each step is 25 cm wide, and the cockroach covers a horizontal distance of approximately 0.730 meters, it would land on the second step.
Therefore, the cockroach would land on the second step.
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A bicyclist makes a trip that consists of three parts, each in the same direction (due north) along a straight road. During the first part, she rides for 24.4 minutes at an average speed of 9.15 m/s. During the second part, she rides for 37.7 minutes at an average speed of 4.13 m/s. Finally, during the third part, she rides for 12.8 minutes at an average speed of 17.6 m/s. (a) How far has the bicyclist traveled during the entire trip? (b) What is the average speed of the bicyclist for the trip? (a) Number Units (b) Number Units
Answer:
(a) The bicyclist has traveled approximately 36255.46 meters during the entire trip.
(b) The average speed of the bicyclist for the trip is approximately 8.07 m/s.
Explanation:
To calculate the distance traveled during each part of the trip, we can use the formula:
Distance = Speed × Time
Given:
First part:
Time (t1) = 24.4 minutes = 24.4 × 60 seconds (convert minutes to seconds)
Speed (v1) = 9.15 m/s
Second part:
Time (t2) = 37.7 minutes = 37.7 × 60 seconds (convert minutes to seconds)
Speed (v2) = 4.13 m/s
Third part:
Time (t3) = 12.8 minutes = 12.8 × 60 seconds (convert minutes to seconds)
Speed (v3) = 17.6 m/s
(a) Calculating the distance for each part:
Distance1 = v1 × t1
Distance1 = 9.15 m/s × (24.4 × 60 s)
Distance1 = 9.15 m/s × 1464 s
Distance1 ≈ 13389.6 m (rounded to one decimal place)
Distance2 = v2 × t2
Distance2 = 4.13 m/s × (37.7 × 60 s)
Distance2 = 4.13 m/s × 2262 s
Distance2 ≈ 9349.06 m (rounded to two decimal places)
Distance3 = v3 × t3
Distance3 = 17.6 m/s × (12.8 × 60 s)
Distance3 = 17.6 m/s × 768 s
Distance3 ≈ 13516.8 m (rounded to one decimal place)
(b) To find the average speed of the entire trip, we can use the formula:
Average Speed = Total Distance / Total Time
Total Distance = Distance1 + Distance2 + Distance3
Total Time = t1 + t2 + t3
Substituting the calculated values:
Total Distance = 13389.6 m + 9349.06 m + 13516.8 m
Total Distance ≈ 36255.46 m (rounded to two decimal places)
Total Time = 24.4 × 60 s + 37.7 × 60 s + 12.8 × 60 s
Total Time = 1464 s + 2262 s + 768 s
Total Time = 4494 s
Average Speed = 36255.46 m / 4494 s
Average Speed ≈ 8.07 m/s (rounded to two decimal places)
(a) The bicyclist has traveled approximately 36255.46 meters during the entire trip.
(b) The average speed of the bicyclist for the trip is approximately 8.07 m/s.
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(a) What is the resistance of a 120Ω, a 2.50kΩ, and a 3.90kΩ resistor connected in series? Ω (b) What is the resistance if they are connected in parallel? x What is the expression for the effective resistance of two or more resistors connected in parallel? Ω
Resistance of a 120Ω, a 2.50kΩ, and a 3.90kΩ resistor connected in series is:We know that the resistance of resistors in series is simply the sum of their individual resistances:
Rtotal = R1 + R2 + R3R
total = 120Ω + 2.50kΩ + 3.90kΩ
= 6.52 kΩ
(b) The resistance if they are connected in parallel is given as follows:We know that the effective resistance of resistors connected in parallel is given by the formula: 1/Rtotal = 1/R1 + 1/R2 + 1/R3
Therefore, the total resistance (Rtotal) of the resistors connected in parallel can be given by:Rtotal = 1/ (1/R1 + 1/R2 + 1/R3)
When we substitute the values, we get
Rtotal = 1/ (1/120Ω + 1/2.50kΩ + 1/3.90kΩ) = 0.0826 kΩ or 82.6Ω
Expression for the effective resistance of two or more resistors connected in parallel:
The formula for the effective resistance of two or more resistors connected in parallel is given as:
1/Rtotal = 1/R1 + 1/R2 + 1/R3 + .......1/RnSo, Rtotal = 1/ (1/R1 + 1/R2 + 1/R3 + .....1/Rn)
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Inpolarized light passes through two ideal polarizers in turn with polarization axes at 52.4∘ to each other. What is the fraction of the incident light tensity that is transmitted? Numeric Response
The fraction of the incident light intensity that is transmitted when polarized light passes through two ideal polarizers in turn with polarization axes at 52.4∘ to each other is 0.5.
Polarized light is a type of light that has waves that oscillate in one direction. By passing unpolarized light through a polarizing filter, polarized light can be created. Polarizers are also known as polarizing filters or polarization filters. When polarized light passes through two ideal polarizers in turn with polarization axes at 52.4∘ to each other, the fraction of the incident light intensity that is transmitted is 0.5. This is because the maximum light intensity is equal to the cosine squared of the angle between the polarization axes of the polarizers.
The cosine of 52.4° is equal to 0.612, and the square of this value is 0.3756. Thus, the maximum transmitted intensity is 37.56% of the original intensity. When this transmitted light passes through the second polarizer, half of it is blocked because the polarization axis of the second polarizer is perpendicular to that of the first. Therefore, the fraction of the incident light intensity that is transmitted is 0.5.
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Light from a laser passes through a grating with 500 slits per millimeter. The deflection angle for light in the fourth order is 58.0 degrees. What is the wavelength of light? please answer in word prg
The wavelength of light passing through the grating is approximately 5.38 × 10^(-7) meters or 538 nm. We can use the grating equation: nλ = d * sinθ.
To calculate the wavelength of light passing through a grating, we can use the grating equation:
nλ = d * sinθ
where:
n is the order of the interference,
λ is the wavelength of the light,
d is the spacing between the slits of the grating,
θ is the deflection angle.
Given:
Order of interference (n) = 4
Spacing between slits (d) = 500 slits/mm = 500 × 10^3 slits/m = 5 × 10^5 slits/m
Deflection angle (θ) = 58.0 degrees = 58.0 × π/180 radians
Rearranging the equation, we have:
λ = (d * sinθ) / n
Substituting the known values:
λ = (5 × 10^5 slits/m * sin(58.0 × π/180 radians)) / 4
Using a calculator, λ ≈ 5.38 × 10^(-7) meters
Therefore, the wavelength of light passing through the grating is approximately 5.38 × 10^(-7) meters or 538 nm.
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the gestalt principle of proximity refers to the idea that
The Gestalt Principle of Proximity refers to the concept that elements that are near to each other appear to be grouped together.
The Gestalt Principle:
This principle asserts that individuals tend to perceive objects in a group, group items that are similar in size, shape, or colour, and make associations based on location, time, or appearance.
The human brain seems to group items together that are near to one another, according to the Gestalt Principle of Proximity. This makes it simpler to recognize and remember patterns and meanings within a larger picture.
This principle works to sort out the many visual stimuli that our eyes detect in our surroundings by grouping similar things together into larger chunks. This allows us to perceive these items more rapidly and accurately.
Therefore, the idea that objects that are close to one another appear to be grouped together is known as the Gestalt Principle of Proximity.
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In the circuit R 1=8Ω,R 2=4Ω and R 3=2Ω and the ideal batteries have emfs ε 1=12 V, and ε 2 =6 V. (a) Using the indicated directions for the currents, write the junction equation for this circuit: (b)Using the indicated directions for the currents, write two loop equations for this circuit. (c)Find the currents in the branches (d) At what rate is energy being dissipated as heat in resistor R
2?
The rate at which energy is being dissipated in resistor [tex]$R_2$[/tex] is [tex]$16 W$[/tex].
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In the circuit:
(a) junction equation is I₁ + I₂ = I₃(b) Loop 1: -12 + 8I₁ - 4I₂ = 0 Loop 2: 6 + 4I₂ - 2I₃ = 0(c) the currents are I₁ = 3 A, I₂ = 2 A, I₃ = 5 A(d) 16 watts of energy is being dissipated as heat in resistor R₂.How to calculate a circuit?(a) The junction equation for the circuit is:
I₁ + I₂ = I₃
This equation states that the sum of the currents flowing into a junction must equal the sum of the currents flowing out of the junction.
(b) The two loop equations for the circuit are:
Loop 1:
-12 + 8I₁ - 4I₂ = 0
Loop 2:
6 + 4I₂ - 2I₃ = 0
These equations are obtained by applying Kirchhoff's loop rule to two different loops in the circuit.
(c) The currents in the branches can be found by solving the junction and loop equations. Solving the junction equation for I₃:
I₃ = I₁ + I₂
Substituting this into the loop equations:
Loop 1:
-12 + 8I₁ - 4(I₁ + I₂) = 0
=> 4I₁ - 4I₂ = 12
Loop 2:
6 + 4I₂ - 2(I₁ + I₂) = 0
=> 2I₂ - 2I₁ = -6
Solving these equations:
I₁ = 3 A
I₂ = 2 A
I₃ = 5 A
(d) The rate at which energy is being dissipated as heat in resistor R2 is:
P = I₂² R₂
= (2 A)² (4 Ω)
= 16 W
Therefore, 16 watts of energy is being dissipated as heat in resistor R₂.
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A battery operated toy car starts from rest (zero initial velocity) and accelerates in
a straight line at a constant rate of 3.5 ⁄ .
(a) What is the speed of the car after 20 seconds?
A. 3.5
B. 70
C. 1,050
(b) How far did the toy car travel during this time?
A. 35
B. 0.25
C. 700
(c) After one minute (60 seconds) of motion from its start, the toy car battery "goes dead" (it
lost all of its electrical power), and the toy truck comes to a stop 10 seconds later because of
friction in the wheels.
How fast (speed) was the toy truck moving at the moment the battery lost its power?
A. 210
B. 0.11
C. 3.8 × 105
(d) What is the magnitude of the acceleration (deceleration) of the toy truck to bring it to a stop
after the battery "goes dead?" Assume that this acceleration (deceleration) is constant.
A. 0.34 ⁄
B. 9.8 ⁄
C. 21 ⁄
(a) The speed of the object after 20 seconds is 70 m/s.
(b) The distance traveled by the object in 20 seconds is 700 meters.
(c) The speed of the toy truck after 60 seconds is 210 m/s.
(d) The deceleration of the toy truck is approximately 0.34 m/s².
Acceleration, a = 3.5 m/s²
Time, t = 20 s
(a) Speed, v:
Using the formula v = u + at, where u is the initial velocity (which is 0 in this case), we can calculate the speed as:
v = 0 + 3.5 × 20 = 70 m/s
(b) Distance, s:
Using the formula s = ut + 1/2 at² and considering the initial velocity u = 0, we can find the distance as:
s = 0 + 1/2 × 3.5 × 20² = 700 m
(c) Speed, v:
Using the formula v = u + at, where the initial speed u is 0 and the time t is 60 s, and the acceleration a is 3.5 m/s², we can determine the speed as:
v = 0 + 3.5 × 60 = 210 m/s
(d) Acceleration (deceleration), a:
Using the formula s = ut + 1/2 at², where s represents the distance travelled, and considering the initial speed u as 210 m/s, the time t as 70 s (60 s + 10 s), and solving for a, we find:
1/2 a(60)² = 210 × 60 + 1/2 a(10)²
Solving for a gives a = 0.34 m/s².
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An airplane is flying due north at a constant speed of 200
s
m
. A sudden storm causes them to turn due northwest at the same speed, over a time interval of 50 s. A. Graphically sketch
v
0
and
v
f
, and use them to sketch
a
ave
. B. Find the components of
a
ave
during the turn. Check with your sketch above to make sure you get the trig functions and the signs If each component correct. a
ave,x
= a
ave ,y
=
Answer:
aᵥₑₓ = 0 m/s²
aᵥₑᵧ = 0 m/s²
Explanation:
To answer this question, we need to break it down into two parts: graphically sketching the initial and final velocities and finding the components of the average acceleration during the turn.
A. Graphical Sketch:
Let's first sketch the initial velocity (v₀) and final velocity (vₙ) vectors.
- The initial velocity vector (v₀) is directed due north and has a magnitude of 200 m/s. We represent it as a straight arrow pointing upward.
- The final velocity vector (vₙ) is directed due northwest. Since it has the same magnitude of 200 m/s, we draw it as a line segment making a 45-degree angle with the horizontal. This line segment starts from the same initial point as v₀.
B. Components of Average Acceleration:
To find the components of the average acceleration (aᵥₑ) during the turn, we need to determine the change in velocity (∆v) and divide it by the time interval (∆t).
Since the speed remains constant at 200 m/s, the change in velocity (∆v) is zero because the magnitude of the velocity does not change.
∆v = 0 m/s
Therefore, both components of the average acceleration, aᵥₑₓ, and aᵥₑᵧ, are zero. This means that during the turn, there is no change in the speed or direction of the airplane.
By considering the given information and the graphical sketch, we can conclude:
aᵥₑₓ = 0 m/s²
aᵥₑᵧ = 0 m/s²
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