The total volume of the piece of wood is 1.33[tex]cm^3[/tex].
To calculate the total volume of the piece of wood, we can use the principle of displacement.
1. First, we need to find the difference in volume between the two water levels. The initial volume is 17.7 [tex]cm^3[/tex], and the final volume is 18.5 cm^3. The difference is 18.5 [tex]cm^3[/tex] - 17.7 [tex]cm^3[/tex] = 0.8 [tex]cm^3[/tex].
2. Now, we need to find the volume of water displaced by the piece of wood. Since the relative density of the wood is 0.60, it means that the wood is 0.60 times denser than water.
3. The volume of water displaced by the wood is equal to the difference in volume divided by the relative density of the wood. So, the volume of water displaced is 0.8 cm^3 / 0.60 = 1.33 [tex]cm^3[/tex].
4. Finally, the total volume of the piece of wood is equal to the volume of water displaced. Therefore, the total volume of the piece of wood is 1.33 [tex]cm^3[/tex].
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A place-kicker must kick a football from a point 36.0 m (about 40 yards) from the goal. Half the crowd hopes the ball will clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 23.2 m/s at an angle of 52.0° to the horizontal. Answer parts a-b.
(a) The ball falls short of clearing the crossbar by 3.05 m (negative value indicates falling short).
(b) The ball approaches the crossbar while falling since it doesn't reach a height greater than the crossbar's height during its trajectory.
To solve this problem, we'll analyze the vertical motion of the ball.
(a) To find how much the ball clears or falls short of clearing the crossbar vertically, we need to calculate the maximum height reached by the ball.
The initial velocity (V0) of the ball is 23.2 m/s, and the launch angle (θ) is 52.0° above the horizontal.
The vertical component of velocity (Vy) at the highest point of the trajectory is zero since the ball momentarily stops before falling back down.
To find the time taken to reach the highest point, we can use the equation:
Vy = V0 * sin(θ)
0 = 23.2 m/s * sin(52.0°)
Solving for sin(52.0°), we find:
sin(52.0°) ≈ 0.7880
Dividing both sides by 23.2 m/s, we get:
0.7880 = sin(52.0°)
Taking the inverse sine, we find:
52.0° ≈ arcsin(0.7880)
Using a calculator, we find:
52.0° ≈ 56.43°
Now we can calculate the time (t) it takes to reach the highest point using the equation:
t = (2 * Vy) / g
Since Vy = 0, we have:
t = 0
This means that the ball reaches its maximum height instantaneously and starts falling immediately. Therefore, the ball does not clear the crossbar.
To find how much the ball falls short of clearing the crossbar vertically, we can calculate the height of the ball at a horizontal distance of 36.0 m.
Using the equation for vertical displacement, we have:
Δy = V0y * t + (1/2) * g * [tex]t^2[/tex]
Plugging in the known values:
Δy = 0 * t + (1/2) * (-9.8 [tex]m/s^2[/tex]) * ([tex]t^2[/tex])
Since t = 0, the equation simplifies to:
Δy = 0
Therefore, the ball falls short of clearing the crossbar by 3.05 m vertically.
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Which statement best describes the refraction of light as it moves from air to glass?
A. Light bends due to the difference in the speed of light in air and glass.
B. Although the light bends, its speed remains the same as before.
C. Although the light changes speed, it continues in the same direction as before.
D. Light undergoes diffraction due to the difference in the speed of light in air and glass.
Explain the function of power supply, readout, peripheral, microcomputer, transducer and processor
The function of the power supply is to provide electrical energy to the device or system that needs it. The power supply converts the incoming voltage from the power source into a form that is usable by the device, such as DC voltage.
The readout is a device or component that displays data or information to the user. The readout could be a simple LED display or a complex graphical display.
A peripheral is a device or component that connects to a computer or other electronic device to provide additional functionality. Examples of peripherals include printers, scanners, and external hard drives.
A microcomputer is a type of computer that is designed to fit on a single microchip. Microcomputers are found in a wide range of devices, including smart phones, tablets, and embedded systems.
A transducer is a device that converts one form of energy to another. In electronics, transducers are commonly used to convert electrical energy into mechanical energy, or vice versa.
The processor is the central component of a computer or electronic device. The processor is responsible for executing instructions and controlling the other components of the system. The performance and capabilities of a device are largely determined by the speed and power of the processor.
.An electron of charge 1.6 x 10-19is situated in a uniform electric filed strength of 120 vm-1 Calculate the force acting on it
The force acting on the electron is 1.92 x 10^-17 N.
The problem states that an electron of charge 1.6 x 10^-19 is located in a uniform electric field of 120 Vm^-1, and it asks us to determine the force acting on it.
We can use Coulomb's law, which states that the force between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them. If the charges are of opposite signs, the force is attractive, while if the charges are of the same sign, the force is repulsive.
The formula for Coulomb's law is F = kq1q2/r^2, where F is the force between the charges, k is Coulomb's constant, q1 and q2 are the magnitudes of the charges, and r is the distance between them.
Since the electron has a charge of 1.6 x 10^-19 C, and the electric field strength is 120 Vm^-1, we can use the equation F = qE to find the force acting on it.
F = qE = (1.6 x 10^-19 C)(120 Vm^-1) = 1.92 x 10^-17 N.
Therefore, the force acting on the electron is 1.92 x 10^-17 N.
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With a force of 200 N a body is lifted 20 meters in 20 seconds. Calculate the weight of this body. Use the formula for distance as a function of acceleration with initial velocity equal to zero.
Answer:
The weight of the body is 3,924 N.
Explanation:
To solve this problem, we can use the formula for distance as a function of acceleration with initial velocity equal to zero:
distance = (1/2) x acceleration x time^2
We know that the distance the body is lifted is 20 meters, the time taken is 20 seconds, and the force applied is 200 N. We can use this information to calculate the weight of the body.
First, we need to calculate the acceleration:
distance = (1/2) x acceleration x time^2
20 = (1/2) x acceleration x (20)^2
acceleration = 0.5 m/s^2
Now that we know the acceleration, we can use the formula for weight:
force = mass x acceleration
We can rearrange this formula to solve for mass:
mass = force / acceleration
mass = 200 N / 0.5 m/s^2
mass = 400 kg
Finally, we can calculate the weight of the body using the formula:
weight = mass x gravity
Assuming a standard acceleration due to gravity of 9.81 m/s^2, we can calculate the weight:
weight = 400 kg x 9.81 m/s^2
weight = 3,924 N
Therefore, the weight of the body is 3,924 N.
Look at this graphic organizer of requirements to apply to become an astronaut.
Requirements for Astronauts
What does the graphic organizer most suggest about the job of an astronaut?
It is technical and potentially tedious.
It is detailed and potentially exhausting.
It is confidential and potentially exciting.
○ It is complex, demanding, and involves flight.
Save and Exit
Next
The graphic organizer suggests that the job of an astronaut is complex, demanding, and involves flight.
This conclusion can be drawn by examining the nature of the requirements listed in the graphic organizer. Firstly, the requirements listed in the organizer are numerous and encompass various aspects. They include educational qualifications, such as having a bachelor's degree in a relevant field, as well as specific experience, like piloting an aircraft.
These requirements highlight the complexity of the job and indicate that astronauts need to possess a diverse set of skills and knowledge. Additionally, the requirements for physical fitness and health demonstrate the demanding nature of the job.
Astronauts are expected to undergo rigorous physical training to ensure they can handle the physical stresses associated with space travel and the conditions they will encounter in space. This indicates that the job can be physically exhausting and requires individuals to be in excellent health.
Lastly, the inclusion of flight-related requirements, such as the need to pass a long-duration spaceflight physical and participate in aircraft flights, implies that the job of an astronaut involves actual flight experiences. This indicates that astronauts are directly involved in piloting spacecraft and are expected to have practical experience in flying.
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D 4.8
This is a harder question based on the Law of Conservation of Momentum. Take the time to work
your way through it. Start with a diagram.
A 400 kg bomb sitting at rest on a table explodes into three pieces. A 150 kg piece moves off to the
east with a velocity of 150 m s². A 100 kg piece moves off with a velocity of 200 m s at a direction of
south 60° west. What is the velocity of the third piece?
It is possible
The velocity of the third piece is v₃ = -12500 kg·m/s / m₃
How do we calculate?The law of conservation of momentum states that the total momentum before the explosion is equal to the total momentum after the explosion.
velocity of the third piece = v₃.
The total initial momentum before the explosion = 0
The total final momentum after the explosion= 0
Initial momentum = 0 kg·m/s (since the bomb is at rest)
Final momentum = m₁v₁ + m₂v₂ + m₃v₃
m₁ = mass of the first piece = 150 kg
v₁ = velocity of the first piece = 150 m/s (to the east)
m₂ = mass of the second piece = 100 kg
v₂ = velocity of the second piece = 200 m/s (south 60° west)
m₃ = mass of the third piece = unknown
v₃ = velocity of the third piece = unknown
0 = (150 kg)(150 m/s) + (100 kg)(200 m/s)(cos(60°)) + (m₃)(v₃)
final momentum = 0 and hence v₃ is found as :
0 = 22500 kg·m/s - 10000 kg·m/s + (m₃)(v₃)
-12500 kg·m/s = (m₃)(v₃)
v₃ = -12500 kg·m/s / m₃
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When white light reflects off of a green surface, which of the following occurs?
1. All wavelengths of light are absorbed.
2. Only the green wavelengths of light are absorbed.
3. Only the green wavelengths of light are reflected.
4. All wavelengths of light are reflected.
When white light reflects off of a green surface, only the green wavelengths of light are reflected (option d).
1. White light is a combination of all visible wavelengths of light, including red, orange, yellow, green, blue, indigo, and violet.
2. When white light hits a green surface, the surface absorbs some wavelengths of light and reflects others.
3. The color we perceive as "green" is the result of the green wavelengths of light being reflected by the surface.
4. In this case, the green surface absorbs all the wavelengths of light except for the green wavelengths, which are reflected back.
5. As a result, our eyes detect the reflected green light and interpret it as the color green.
6. This phenomenon occurs because the green surface selectively absorbs and reflects different wavelengths of light based on its molecular structure and the interactions between light and matter.
7. The absorption and reflection of specific wavelengths of light give objects their perceived color.
8. Therefore, when white light reflects off of a green surface, only the green wavelengths of light are reflected, while the other wavelengths are absorbed by the surface.
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The obliquity of the rotation of Uranus is over 90 degrees. Compared to the plane of the solar system, it rotates on its "side", unlike any other planet. It is surmised that this angle of rotation was caused by:
Two blocks, M1 and M2, are connected by a massless string that passes over a massless pulley as shown in the figure. M2, which has a mass of 19.0 kg,
rests on a long ramp of angle theta=25.0∘.
Ignore friction, and let up the ramp define the positive direction.
If the actual mass of M1 is 5.00 kg and the system is allowed to move, what is the acceleration of the two blocks?
What distance does block M2 move in 2.00 s?
The acceleration of the two blocks is[tex]2.14 m/s^{2[/tex]} and the distance does block M2 move in 2.00 s is 4.27 m.
Now we need to find the acceleration of the two blocks and the distance does block M2 move in 2.00 s.
We know that: mass of M1, m1 = 5.00 kg mass of M2, m2 = 19.0 kgθ = 25.0°Taking upward direction as positive for block M1 and downwards as positive for block M2.
Therefore, we can write the following equation of motion for the two blocks:
For M2: m2g - T = m2a ...(1)
For M1: T - m1g = m1a ...(2)
We can see from the figure that M2 is on an inclined plane making an angle θ with the horizontal.
We can resolve the weight of M2 into two components:
Perpendicular to the plane = m2gcosθParallel to the plane = m2gsinθ
The component parallel to the plane will tend to make the block move downwards.
Therefore, the effective weight will be:
mg = m2gsinθ ...(3)
From equation (1) we can write:
T = m2g - m2a ...(4)
Substituting equation (4) in equation (2), we get:
m2g - m2a - m1g = m1a ...(5)
On solving equation (5), we get the acceleration as:
a = g(m2sinθ - m1) / (m1 + m2)
On substituting the given values, we get:
[tex]a = 2.14 m/s^{2}[/tex]
The distance moved by M2 in 2 seconds can be found out using the formula:[tex]s = ut + \frac{1}{2} at^{2}[/tex]
Here, initial velocity, u = 0m/s Time, t = 2s Acceleration, [tex]a = 2.14 m/s^{2}[/tex]
On substituting these values, we get the distance travelled by M2 as: s = 4.27 m
Therefore, the acceleration of the two blocks is [tex]2.14 m/s^{2}[/tex]. And the distance does block M2 move in 2.00 s is 4.27 m.
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RHETORICAL ANALYSIS: How does Robinson use language in effective and engaging ways to develop his argument to his younger self-and, in the process, to young readers in the present? In your response, consider such techniques as metaphor, repetition, and sentence structure.
In "The Argonauts," Robinson effectively utilizes language techniques such as metaphor, repetition, and sentence structure to develop his argument to his younger self and engage young readers in the present. Through these techniques, Robinson creates a powerful and relatable narrative that resonates with his audience.
Robinson employs metaphors to convey complex ideas in a compelling and accessible manner. For instance, he compares his struggle with identity and gender to the mythical journey of the Argonauts, making it relatable and captivating for young readers. This metaphorical language enables readers to grasp the profound emotions and challenges he faced during his own personal journey.
Repetition is another technique Robinson employs to reinforce key ideas and create a rhythmic and memorable reading experience. By repeating certain phrases or concepts, he emphasizes their significance and invites readers to reflect on them. This repetition serves to engage young readers by encouraging them to contemplate their own experiences and perspectives.
Furthermore, Robinson carefully structures his sentences to create a sense of rhythm and flow, enhancing the overall readability and impact of his argument. Short, concise sentences create moments of clarity and emphasis, while longer, more descriptive sentences evoke a contemplative and introspective tone. This varied sentence structure adds depth and nuance to his narrative, captivating young readers and keeping them engaged throughout.
In conclusion, through the effective use of metaphor, repetition, and sentence structure, Robinson engages and captivates young readers, inviting them to reflect on their own identities and experiences. His language choices not only develop his argument to his younger self but also establish a connection with present-day young readers, making his work both impactful and relatable.
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deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W. Refer to appropriate Newton’s laws of motion.
The expression for the contribution of P to the pressure exerted on W is P = mV/(c^2t), derived using Newton's laws of motion and the definition of pressure.
In order to deduce an expression, in terms of m, c, and V, for the contribution of P to the pressure exerted on W, we can use the appropriate Newton’s laws of motion. Specifically, we can use the equation F = ma, where F represents force, m represents mass, and a represents acceleration.We know that pressure (P) is defined as force per unit area, or P = F/A. Rearranging this equation, we can solve for force: F = PA.Substituting this into the equation F = ma, we get PA = ma. Rearranging this equation, we can solve for pressure in terms of mass and acceleration: P = ma/A. Finally, we know that acceleration can be expressed in terms of velocity (V) and time (t): a = V/t.Substituting this into our equation for pressure, we get P = mV/(At). Since c represents the speed of sound, we can express A as [tex]A = c^2[/tex]. Therefore, our final expression for the contribution of P to the pressure exerted on W is:[tex]P = mV/(c^{2t})[/tex]In summary, we used the equation F = ma, the definition of pressure (P = F/A), and the relationship between acceleration (a), velocity (V), time (t), and the speed of sound (c) to deduce an expression for the contribution of P to the pressure exerted on W in terms of m, c, and V.For more questions on pressure
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A ball is thrown vertically upward with a speed of 15.0 m/s. Find a - How high does it rise? in meters, find b - How long does it take to reach its highest point? in seconds, find c - How long does the ball take to hit the ground after it reaches its highest point? in seconds, find d - What is its velocity when it returns to the level from which it started? in m/s.
Given that the initial velocity at which the ball is thrown vertically upward is 15m/s. Let us also assume that the value of acceleration due to gravity (g) = 9.8m/s² and in this case, the value will be -9.8m/s² as the ball is moving against gravity.
a) To calculate how high the ball rises, we can use the kinematic equation:
v² = u² + 2gs......(i)
where v ⇒ final velocity
u ⇒ initial velocity
g ⇒ acceleration and,
s ⇒ displacement (the height)
The final velocity will be 0 when the ball reaches its maximum height.
Substituting the values in equation (i), we get
0² = 15² + (2*-9.8*s)
0 = 225 - 19.6s
Thus, s = 225/19.6 = 11.48 m.
Therefore, the ball rises approximately 11.48 meters.
b) To find the time taken to reach the highest point, we can use the kinematic equation,
v = u + gt......(ii)
where t = time
Substituting the values in equation (ii)
0 = 15 - 9.8*t
t = -15/ -9.8 = 1.53 seconds
Thus, the time taken to reach the highest point = 1.53 seconds.
c) To find the time taken for the ball to hit the ground after it reaches its highest point, we can use the equation,
s = ut +1/2gt².....(iii)
As the ball is moving downwards, the initial velocity, u will be 0m/s.
Thus, substituting the values in equation (iii), we get
11.48 = 0*t + 1/2*9.8*t²
11.48 = 4.9t²
t² = 2.34
Therefore t = 1.53 seconds
Thus, the time taken for the ball to hit the ground is 1.53 seconds.
d) To find the velocity at which the ball returns to the level from which it started, we can use the equation
v = u+ gt.....(iv)
v = 0 + 9.8*1.53
Thus, v = 14.99 ≅ 15 m/s
Therefore, the velocity when it returns to the level from which it started is 15m/s.
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A rock with a mass of 0.2 kg with a velocity of 5 m/s strikes a stationary 1 kg wooden ball. After the
collision the rock flies back with a velocity of -2 m/s. What is the velocity of the wooden ball after the
collision?
A. -0.4 m/s
B. -1 m/s
C. 0.4 m/s
D. 1.4 m/s
Answer:
D. 1.4 m/s
Explanation:
forward direction is +
back direction is -
Momentum = P = mass x velocity = mv
let v = velocity of ball after collision
Law of Conservation of Momentum: total momentum before the collision must equal the total momentum after the collision
(0.20 kg)(5 m/s) + (1 kg)(0 m/s) = (0.2 kg)(-2 m/s) + (1 kg)v
1 kg·m/s + 0 = -0.4 kg·m/s + (1 kg)v
1 kg·m/s + 0.4 kg·m/s = (1 kg)v rearrange the equation and solve for v
(1 kg)v = 1.4 kg·m/s
v = (1.4 kg·m/s) / (1 kg) = 1.4 m/s
One strategy in a snowball fight is to throw a snowball at a high angle over level ground. Then, while your opponent is watching that snowball, you throw a second one at a low angle timed to arrive before or at the same time as the first one. Assume both snowballs are thrown with a speed of 26.5 m/s. The first one is thrown at an angle of 58.0° with respect to the horizontal. Find a - At what angle should the second snowball be thrown to arrive at the same point as the first?, find b - How many seconds later should the second snowball be thrown after the first in order for both to arrive at the same time?
The second snowball should be thrown at an angle of approximately 48.196° with respect to the horizontal to arrive at the same point as the first snowball.
the second snowball should be thrown 4.582 seconds later in order for both to arrive at the same time.
To find the angle at which the second snowball should be thrown, we can use the fact that the horizontal displacement of both snowballs must be the same.
Let's first find the horizontal and vertical components of the velocity for the first snowball. The initial speed is 26.5 m/s, and the angle is 58.0° with respect to the horizontal.
The horizontal component of the velocity for the first snowball is given by:
V1x = V1 * cos(angle1)
= 26.5 m/s * cos(58.0°)
= 26.5 m/s * 0.530
= 14.045 m/s
Now, let's find the vertical component of the velocity for the first snowball:
V1y = V1 * sin(angle1)
= 26.5 m/s * sin(58.0°)
= 26.5 m/s * 0.848
= 22.472 m/s
Since the vertical acceleration is the same for both snowballs (gravity), the time it takes for both to arrive at the same point is the same. Therefore, we can use the time of flight of the first snowball to calculate the vertical displacement for the second snowball.
The time of flight for the first snowball can be calculated using the vertical component of velocity and the acceleration due to gravity:
t = (2 * V1y) / g
= (2 * 22.472 m/s) / 9.8 m/s²
≈ 4.582 s
Now, let's find the vertical displacement for the second snowball:
Δy = V1y * t - (0.5 * g * t²)
= 22.472 m/s * 4.582 s - (0.5 * 9.8 m/s² * (4.582 s)²)
≈ 103.049 m
To find the angle at which the second snowball should be thrown, we can use the horizontal displacement and the vertical displacement:
tan(angle2) = Δy / Δx
= 103.049 m / (2 * 14.045 m/s * t)
= 103.049 m / (2 * 14.045 m/s * 4.582 s)
≈ 1.085
Now, we can find the angle2 by taking the arctan of both sides:
angle2 ≈ arctan(1.085)
angle2 ≈ 48.196°
Therefore,
To find how many seconds later the second snowball should be thrown, we can simply use the time of flight of the first snowball, which is approximately 4.582 seconds.
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Question 1 of 10
What is the slope of the line plotted below?
B. 2
5
10
C. 1
O A. 0.5
о
9
OD. -0.5
5
Particles q₁ = -66.3 μC, q2 = +108 μC, and
q3 = -43.2 μC are in a line. Particles q₁ and q2 are
separated by 0.550 m and particles q2 and q3 are
separated by 0.550 m. What is the net force on
particle q₂?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
7. Name the type of mirror used:-
(i) as a reflector in search light (iii) by the dentist
(ii) as side view mirror in vehicles. (iv) as a shaving mirror
Answer:
1. Concave mirror
2. Convex mirror
3. Concave mirror
4. Concave mirror
Explanation:
Concave mirror is placed near on an object it displays a virtual image
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Based on the information, we can infer that the temperature on the west and east coasts of the United States is higher than in the central part at latitude 35° North.
What do we see in the image?In the image you can see the map of the United States and two latitudinal lines of 35° and 45° North. Additionally we see the different temperatures that exist in various cities or locations in the United States.
Based on this information, we can infer that the temperatures on the east and west coasts are higher than the temperatures recorded in the central part. For example, at 35° latitude, the coasts register temperatures of more than 60°F while the central zone registers lower temperatures between 36 and 59°F.
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A tennis player moves in a straight-line path as shown in the figure below. Find her average velocity in the following time intervals. Find (a) - 0 to 1.0 s, find (b) 0 to 4s, find (c) - 1.0 s to 5.0 s, find (d) - 0 to 5.0 s.
(a) The average velocity of the tennis player at 0 to 1 s is 4 m/s.
(b) The average velocity of the tennis player at 0 to 4 s is -0.5 m/s.
(c) The average velocity of the tennis player at 1 to 5 s is 1 m/s.
(d) The average velocity of the tennis player at 0 to 5 s is 0.8 m/s.
What is the average velocity of the tennis player?The average velocity of the tennis player at the given time, is calculated by applying the formula for average velocity as follows;
average velocity = total displacement / total time
(a) The average velocity at 0 to 1 s;
average vel. = (4 m - 0 m ) / (1 s ) = 4 m/s
(b) The average velocity at 0 to 4 s;
average vel. = (-2 - 0 )m / 4 s = -0.5 m/s
(c) The average velocity at 1 to 5 s;
average vel. = (4 - 0 )m / (5 - 1) s = 1 m/s
(d) The average velocity at 0 to 5 s;
average vel. = (4 - 0 )m / (5 - 0) s = 0.8 m/s
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