The time taken to reach the maximum height is approximately 3.009 seconds. The maximum height (relative to ground level) is approximately 52.063 meters.
To solve this problem, we can use the equations of motion for projectile motion. Let's break down each part of the problem.
Given:
Initial height (y0) = 7.80 m
Initial speed (v0) = 36.0 m/s
Launch angle (θ) = 55.0°
Acceleration due to gravity (g) = 9.8 m/s² (assuming no air resistance)
(a) To find the time taken to reach the maximum height, we need to consider the vertical motion only. At the maximum height, the vertical velocity becomes zero. We can use the following equation:
v = v0y - g * t
At maximum height, v = 0, and v0y is the vertical component of the initial velocity, which is given by:
v0y = v0 * sin(θ)
Setting v = 0, the equation becomes:
0 = v0 * sin(θ) - g * t
Solving for t:
t = v0 * sin(θ) / g
Substituting the given values:
t = (36.0 m/s) * sin(55.0°) / (9.8 m/s²)
Therefore, the time taken to reach the maximum height is approximately 3.009 seconds.
Calculate t to find the time taken to reach the maximum height.
(b) The maximum height (hmax) can be calculated using the equation:
hmax = y0 + v0y^2 / (2g)
Substituting the given values:
hmax = 7.80 m + (36.0 m/s * sin(55.0°))^2 / (2 * 9.8 m/s²)
the maximum height (relative to ground level) is approximately 52.063 meters.
Calculate hmax to find the maximum height.
(c) To find the total time of flight, we need to consider the vertical motion again. The total time of flight (T) is given by:
T = 2t
Substitute the previously calculated value of t to find the total time of flight.
(d) The speed of the projectile just before hitting the ground is equal to the initial speed, as there is no horizontal acceleration. Therefore, the speed (magnitude of velocity) is:
speed = v0
Substitute the given value to find the speed of the projectile before it hits the ground.
Please provide the values of θ, v0, and y0, and I'll calculate the results for you.
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Light from a helium-neon laser (λ=633 nm) passes Part A through a circular aperture and is observed on a screen 4.70 m behind the aperture. The width of the central What is the diameter (in mm ) of the hole? maximum is 2.20 cm. You may want to review
Two −2.7×10
−9
−C charged point-like objects are separated by 0.20 m. - Part A Determine the potential (assuming zero volts at infinity) at a point halfway between the objects. Express your answer with the appropriate units. - Part B Determine the potential (assuming zero volts at infinity) at a point 0.20 m to the side of one of the objects (and 0.40 m from the other) along a line joining them. Express your answer with the appropriate units.
The potential at a point halfway between two point-like objects is -5400 V (volts) while the potential at a point 0.20 m to the side of one of the objects (and 0.40 m from the other) along a line joining them is -13.5 kV (kilo volts).
A positive work done implies that the potential energy has increased, while negative work done implies that the potential energy has decreased.
The potential energy at a point p in the field of two point charges Q1 and Q2 separated by a distance r is given as follows;
Vp = k(Q1/r1 + Q2/r2) where k = 1 / 4πε0, ε0 is the permittivity of free space and r1 and r2 are the distances from p to Q1 and Q2 respectively.
The point halfway between the two charges is equidistant from each of them and at the mid-point between them.
Using the above formula, the potential energy is given by
Vp = k(Q1/r1 + Q2/r2)where Q1 = Q2 = -2.7 × [tex]10^-9[/tex] C, r1 = r2 = 0.10 m and k = 1 / 4πε0.
From the above equation,Vp = 8.99 × [tex]10^9[/tex] × (-2.7 × [tex]10^-9[/tex] / 0.1 + (-2.7 × [tex]10^-9[/tex]/ 0.1))= -5.4 × [tex]10^3[/tex] V
The potential at a point 0.20 m to the side of one of the objects (and 0.40 m from the other) along a line joining them can be calculated as follows:
Vp = k(Q1/r1 + Q2/r2) where Q1 = -2.7 × [tex]10^-9[/tex] C, Q2 = -2.7 × [tex]10^-9[/tex]C, r1 = 0.2 m and r2 = 0.4 m.
From the above equation,
Vp = 8.99 × 10^9 × (-2.7 × [tex]10^-9[/tex] / 0.2 - 2.7 × [tex]10^-9[/tex] / 0.4)= -1.35 × [tex]10^4[/tex] V.
Therefore, the potential at a point halfway between two point-like objects is -5400 V (volts) while the potential at a point 0.20 m to the side of one of the objects (and 0.40 m from the other) along a line joining them is -13.5 kV (kilo volts).
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A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. At what angle from the beam axis will the first order peak occur if the tube emits light with wavelength of 590.8 nm ? (in deg) Tries 0/12 At what angle will the second order peak occur? (in deg) Tries 0/12
A diffraction grating with 230 lines per mm is used in an experiment to study the visible spectrum of a gas discharge tube. The second-order peak will occur at an angle of around 43.36° from the beam axis.
To determine the angle at which the first-order peak will occur using a diffraction grating, we can use the formula for the angle of diffraction (Young's diffraction):
sin(θ) = m * λ / d
Where:
θ is the angle of diffraction,
m is the order of the peak (in this case, first order, m = 1),
λ is the wavelength of the light,
and d is the spacing between adjacent lines on the Young's diffraction.
Given:
m = 1
λ = 590.8 nm = 590.8 × [tex]10^{(-9)[/tex] m
d = 1 mm / 230 lines = (1 / 230) × [tex]10^{(-3)[/tex] m
Let's substitute these values into the formula to find the angle of the first-order peak:
sin(θ) = (1 * 590.8 × [tex]10^{(-9)[/tex]) / ((1 / 230) × [tex]10^{(-3)[/tex])
sin(θ) = 590.8 × 230
θ = sin^(-1)(590.8 × 230)
Using a calculator, we can find the value of θ to be approximately 21.85°.
Therefore, the first-order peak will occur at an angle of approximately 21.85 degrees from the beam axis.
To determine the angle at which the second-order peak will occur, we use the same formula, but with m = 2:
sin(θ) = (2 * 590.8 × [tex]10^{(-9)[/tex]) / ((1 / 230) × [tex]10^{(-3)[/tex])
sin(θ) = 2 * 590.8 × 230
θ = [tex]sin^{(-1)[/tex](2 * 590.8 × 230)
Using a calculator, we find the value of θ to be approximately 43.36°.
Therefore, the second-order peak will occur at an angle of approximately 43.36 degrees from the beam axis.
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A single conservative force F(x) acts on a 1.0 kg particle that moves along an x axis. The potential energy U(x) associated with F(x) is given by U(x)=−4xe−x/4J, where x is in meters. At x = 5.0 m the particle has a kinetic energy of 2.0 J. Determine (a) the least value of x the particle can reach and (b) the greatest value of x the particle can reach. (c) the maximum kinetic energy of the particle and (d) the value of x at which it occurs. (e) Determine an expression in newtons and meters for F(x) as a function of x. (f) For what (finite) value of x does F(x) = 0?
The least value of x the particle can reach is 8 m, the greatest value of x is 0 m, the maximum kinetic energy is 2 J, and it occurs at x = 8 m. The expression for F(x) as a function of x is [tex]4e^(-x/4) - xe^(-x/4)/2 N[/tex]. The force F(x) is equal to zero at x = 8 m.
(a) To find the least value of x the particle can reach, we need to determine the point where the potential energy is at its minimum. We can do this by finding the point where the derivative of the potential energy function is zero:
[tex]dU/dx = -4e^(-x/4) + xe^(-x/4)/2 = 0[/tex]
Simplifying this equation gives:
[tex]-4e^(-x/4) + xe^(-x/4)/2 = 0[/tex]
Multiplying both sides by [tex]2e^(x/4)[/tex] gives:
-8 + x = 0
Solving for x, we find:
x = 8
Therefore, the least value of x the particle can reach is 8 m.
(b) To find the greatest value of x the particle can reach, we need to determine the point where the potential energy is zero. We can set U(x) equal to zero and solve for x:
[tex]-4xe^(-x/4) = 0[/tex]
Since the exponential term can never be zero, the only solution is x = 0. Therefore, the greatest value of x the particle can reach is 0 m.
(c) The maximum kinetic energy of the particle occurs when the potential energy is at its minimum. From part (a), we found that the minimum potential energy occurs at x = 8 m. At this point, the potential energy is 0 J, so the entire energy is in the form of kinetic energy. Therefore, the maximum kinetic energy of the particle is 2 J.
(d) The value of x at which the maximum kinetic energy occurs is the same as the value of x at which the potential energy is at its minimum, which is x = 8 m.
(e) To determine an expression for F(x) as a function of x, we can calculate the force as the negative derivative of the potential energy:
F(x) = -dU/dx
Differentiating the potential energy function [tex]U(x) = -4xe^(-x/4)[/tex] with respect to x gives:
[tex]F(x) = -(-4e^(-x/4) + xe^(-x/4)/2)[/tex]
Simplifying this expression gives:
[tex]F(x) = 4e^(-x/4) - xe^(-x/4)/2[/tex]
Therefore, the expression for F(x) as a function of x is [tex]4e^(-x/4) - xe^(-x/4)/2 N[/tex].
(f) To find the value of x at which F(x) = 0, we can set the expression for F(x) equal to zero and solve for x:
[tex]4e^(-x/4) - xe^(-x/4)/2 = 0[/tex]
Multiplying both sides by[tex]2e^(x/4)[/tex] gives:
8 - x = 0
Solving for x, we find:
x = 8
Therefore, for x = 8 m, the force F(x) is equal to zero.
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A car travels along the perimeter of a vertical circle (radius = 250 m) at a constant speed of 23 m/s. The driver has a mass of 62 kg. Take g = 9.8 m/s2 What is the magnitude of the force exerted by the seat on the driver at the lowest point on this circular path? O a. 130 N O b. 1300 N O c. 610N O d. 740 N e. 480 N The lowest vibration frequency of guitar string of length 0.8 m is 300 Hz. (a) What is the wavelength of the waves for this vibration? Use a diagram to explain your reasoning. (b) What is the speed of waves on the string?
The magnitude of the force exerted by the seat on the driver at the lowest point on this circular path is **610 N**. At the lowest point on the circular path, the driver experiences both the force due to gravity and the centripetal force.
The force due to gravity is given by the formula F_gravity = m * g, where m is the mass of the driver (62 kg) and g is the acceleration due to gravity (9.8 m/s^2). The centripetal force is provided by the seat and is given by the formula F_centripetal = m * v^2 / r, where v is the velocity of the car (23 m/s) and r is the radius of the circular path (250 m).
The total force exerted by the seat on the driver is the vector sum of the force due to gravity and the centripetal force. By calculating the magnitudes of both forces and adding them together, we get a result of approximately 610 N.
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A 20cm-high object is placed 150 cm away from a concave mirror that has a focal length of
100 cm. The mirror has a diameter of 120 cm.
(a) Create the ray diagram using the simulation. Record the focal length, radius of curvature, and object distance and height. Then measure and record the image distance and height. Be sure to include signs. Is the image real or virtual, upright or inverted, reduced or enlarged?
(b) Now use the mirror equations to find the image distance, image height, and magnification. Show all equations and calculations. Do experiment and theory agree?
Creation of Ray Diagram and analysis: A concave mirror has the focal length of 100 cm.
The object distance and height are given to be 150 cm and 20 cm.
The diameter of the mirror is 120 cm.
Here, we need to calculate the image distance and height of the object along with its nature.
In order to calculate the image distance and height, first, we need to create a ray diagram.
The diagram is given below.
From the diagram, it can be observed that the image is formed in front of the mirror, which shows that the image is virtual.
The image is inverted, which means that the image is also inverted.
The height of the image is 6.67 cm and the distance of the image from the mirror is 50 cm.
The positive sign for the object distance shows that the object is in front of the mirror.
The negative sign for the image distance shows that the image is formed in front of the mirror.
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If a point has 30 J of energy from a 5 C charge, what must be
the electric potential at that point?
A. 3 V
B. 6 V
C. 30 V
D. 150 V
The electric potential at a point is calculated by dividing the electric potential energy by the charge at that point. In this case, the electric potential is 6 V for a point with 30 J of energy from a 5 C charge. The correct option is B.
The electric potential at a point can be calculated by dividing the electric potential energy by the charge at that point. The formula for electric potential is:
V = U / Q
where V is the electric potential, U is the electric potential energy, and Q is the charge.
U = 30 J (electric potential energy)
Q = 5 C (charge)
Substituting the values into the formula:
V = 30 J / 5 C
V = 6 V
Therefore, the electric potential at that point is 6 V.
The correct option is B. 6 V.
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Question 9 (1 point) 4) Listen ▶ A crest with an amplitude of 20 cm meets a trough with an amplitude of 30 cm. The resultant waveform will be: a trough with an amplitude of 10 cm. a trough with an amplitude of 50 cm. a crest with an amplitude of 50 cm. a crest with an amplitude of 10 cm.
When a crest with an amplitude of 20 cm meets a trough with an amplitude of 30 cm, the resultant waveform will be a trough with an amplitude of 10 cm.A waveform is a graphical representation of the sound wave. Waveform displays wave properties, such as amplitude, wavelength, phase shift, and frequency, over time.
The amplitude of a wave is the distance from the centre line to the highest point of the wave. The distance from the centre line to the lowest point of the wave is equal to the amplitude of the trough. Amplitude is usually measured in decibels (dB) or volts. The amplitude of a waveform determines how loud or soft the sound is.
The frequency of a wave is the number of times it oscillates per second, and it is measured in hertz (Hz). A wave's wavelength is the distance between two crests or troughs, measured in meters or feet.
The time it takes for a wave to complete one full cycle is referred to as the period of the wave, measured in seconds. The period of a wave is determined by its frequency.
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A crate with a mass of 82kg sits on a tilted rampand experiences friction so that it remains motionless. The ramp is tilted at an angle of 22 degrees relative to the ground. What is the magnitude of the normal force in newtons exerted on the crate from the ramp?
A crate with a mass of 82kg sits on a tilted ramp and experiences friction so that it remains motionless. The magnitude of the normal force in newtons exerted on the crate from the ramp is 327.89 N.
To determine the magnitude of the normal force exerted on the crate from the ramp, we need to consider the forces acting on the crate in the vertical direction.
The normal force (N) is the force exerted perpendicular to the ramp by the surface, counteracting the gravitational force pulling the crate downward.
The gravitational force acting on the crate can be calculated using the formula:
[tex]Force_{gravity[/tex] = mass * gravity
where the mass of the crate is 82 kg and the acceleration due to gravity is approximately 9.8 [tex]m/s^2[/tex]
[tex]Force_{gravity[/tex] = 82 kg * 9.8 [tex]m/s^2[/tex]
Next, we need to decompose the gravitational force into its components parallel and perpendicular to the ramp. The component perpendicular to the ramp is equal to the normal force (N), and the component parallel to the ramp is equal to the force due to gravity acting down the ramp.
The component of force due to gravity acting down the ramp is given by:
[tex]Force_{parallel[/tex] = [tex]Force_{gravity[/tex]* sin(theta)
where theta is the angle of the ramp, which is 22 degrees in this case.
[tex]Force_{parallel[/tex]l = 82 kg * 9.8 [tex]m/s^2[/tex] * sin(22 degrees)
Finally, since the crate remains motionless, the normal force (N) must balance the force parallel to the ramp. Therefore, the normal force can be calculated as:
N = [tex]Force_{parallel[/tex]
Substituting the values:
N = 82 kg * 9.8 [tex]m/s^2[/tex]* sin(22 degrees)
Calculating the value:
N ≈ 327.89 N
Therefore, the magnitude of the normal force exerted on the crate from the ramp is approximately 327.89 N.
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If the velocity versus time graph of an object is a horizontal line, parallel to the +x axis, the object is a) at rest b) moving with zero acceleration c) moving with decreasing speed d) moving with constant non-zero acceleration e) moving with increasing speed
The velocity versus time graph is a graphical representation of an object’s motion. In this case, if the velocity versus time graph of an object is a horizontal line parallel to the +x axis, it means that the object is not accelerating.
Hence, the correct answer is option B – moving with zero acceleration.If the velocity versus time graph is a horizontal line, it implies that the object's velocity is constant with time. A horizontal line indicates that the object's velocity is not changing with time; this means that the object is not accelerating.
Therefore, if an object has a horizontal line parallel to the +x axis in its velocity versus time graph, the object is moving with zero acceleration and a constant velocity, thus, option B is the correct answer.In conclusion, the velocity versus time graph of an object shows the motion of the object. A horizontal line indicates that the object's velocity is constant with time; hence, the object is not accelerating.
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A pendulum with a period of 2.00096 s in one location (g = 9.794 m/s2) is moved to a new location where the period is now 1.99597 s. Help on how to format answers: units What is the acceleration due to gravity at its new location? g=
The acceleration due to gravity at the new location is approximately 9.746 m/s².
The period of a simple pendulum is determined by the formula T = 2π√(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity.
In the first location, the pendulum has a period of 2.00096 s. Let's call the length of the pendulum in the first location L₁. Using the formula, we have:
2.00096 = 2π√(L₁/9.794)
Squaring both sides of the equation, we get:
4.00385 = 4π²(L₁/9.794)
Simplifying further, we find:
L₁/9.794 = 4.00385/(4π²)
L₁ = (4.00385/(4π²)) * 9.794
Now, let's move the pendulum to the new location, where the period is 1.99597 s. Let's call the length of the pendulum in the new location L₂. Using the formula again, we have:
1.99597 = 2π√(L₂/g₂)
where g₂ is the acceleration due to gravity at the new location.
Squaring both sides of the equation and substituting the expression for L₁, we get:
3.98391 = 4π²((4.00385/(4π²)) * 9.794)/g₂
Simplifying further, we find:
g₂ = (4π² * ((4.00385/(4π²)) * 9.794))/3.98391
Evaluating this expression, we find that g₂ is approximately 9.746 m/s².
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An object has a circular path with radius 8.00 cm. The angular velocity of the object is 150
rad/s. Determine (a) tangential velocity and (b) centripetal force.
Therefore, the tangential velocity of the object is 12 m/s and the centripetal force acting on the object is 22500 N
To determine the tangential velocity and centripetal force of an object moving in a circular path, we can use the following formulas:
(a) Tangential velocity (v):
v = r * ω
where r is the radius of the circular path and ω is the angular velocity.
(b) Centripetal force (F):
F = m * a = m * ([tex]v^2[/tex] / r)
where m is the mass of the object, v is the tangential velocity, and a is the centripetal acceleration.
Radius, r = 8.00 cm = 0.08 m
Angular velocity, ω = 150 rad/s
(a) Tangential velocity:
v = r * ω
v = 0.08 m * 150 rad/s
Calculate the value:
v = 12 m/s
(b) Centripetal force:
F = m * ([tex]v^2[/tex] / r)
F = m * (12 [tex]m/s)^2[/tex] / 0.08 m
Simplify the equation and substitute the appropriate values:
F = m * 1800 [tex]m^2/s^2[/tex] / 0.08 m
Calculate the value:
F = m * 22500 N.
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A cyclist races around a circular track at the constant speed of 20 m/s. The radius of the track is 50 m. The acceleration of the cyclist is
A. zero.
B. 8 m/s2, toward the center of the track.
C. 10 m/s2, downward.
D. 20 m/s2, in the direction of travel.
The correct option is A. zero.
Acceleration is a vector quantity that represents the rate of change of an object's velocity with respect to time. It is a physical quantity that measures how much the speed and/or direction of an object changes per unit time.Acceleration and velocity in circular motion A cyclist races around a circular track at a constant speed of 20 m/s. As the cyclist is moving in a circle, it has a velocity vector that is constantly changing in direction. As a result, the cyclist has an acceleration.The acceleration of an object in circular motion is always directed towards the center of the circle. Because the cyclist is moving in a circle, the direction of the cyclist's acceleration is towards the center of the circle.The magnitude of the acceleration of an object moving in a circle is given by the following equation:a = v2/r where
:a is acceleration is velocity is the radius of the circle For the given cyclist, the speed is given as 20 m/s and the radius of the circular track is 50 m. Using the equation, we geta = (20 m/s)2/50 m= 400/50= 8 m/s2Thus, the acceleration of the cyclist is 8 m/s2, directed towards the center of the circular track.
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[5] Inelastic collision preserves: a) Velocities Y N b) Masses Y N c) Momentum. Y N d) Kinetic energy. Y N [6] Energy of Simple Harmonic Motion consist of: a) Kinetic energy Y N b) Potential energy Y N c) Internal energy Y N d) Kinetic and potential energy Y N [7] Main characteristics of Simple Harmonic Motion are: a) Constant period b) Constant amplitude c) Independence between period and amplitude. d) Displacement is sine or cosine function. e) Velocity is linear function. f) Acceleration is quadratic function [8] Complete set of features of components of vectors contains: a) Magnitude, direction and orientation Y b) Angle and magnitude Y c) Starting point, orientation, direction and magnitude Y d) Magnitude and orientation Y yoooooo zoooooo N N N N Z Z N
1. Inelastic collision preserves: c) Momentum. [Yes] d) Kinetic energy. [No]
2. Energy of Simple Harmonic Motion consists of: d) Kinetic and potential energy. [Yes]
3. Main characteristics of Simple Harmonic Motion are: a) Constant period [Yes] b) Constant amplitude [Yes] d) Displacement is sine or cosine function. [Yes] e) Velocity is linear function. [No] f) Acceleration is quadratic function [No]
4. Complete set of features of components of vectors contains: a) Magnitude, direction and orientation [Yes] b) Angle and magnitude [No] c) Starting point, orientation, direction and magnitude [No] d) Magnitude and orientation [No]
1. In an inelastic collision, momentum is preserved. This means that the total momentum before and after the collision remains the same. However, kinetic energy is not necessarily conserved in an inelastic collision as some energy may be converted into other forms such as heat or deformation.
2. The energy of simple harmonic motion consists of both kinetic energy and potential energy. As the oscillating object moves back and forth, it alternates between kinetic energy (when it is in motion) and potential energy (when it is at its maximum displacement).
3. The main characteristics of simple harmonic motion are:
a) Constant period, which means that the time taken for one complete oscillation remains the same.
b) Constant amplitude, which indicates that the maximum displacement from the equilibrium position remains constant.
d) Displacement follows a sine or cosine function, showing a periodic pattern.
e) Velocity is not a linear function but rather varies with the position of the object.
f) Acceleration is not a quadratic function but rather varies with the position of the object.
4. The complete set of features of components of vectors includes magnitude, direction, and orientation. The magnitude represents the size or length of the vector, while the direction indicates the line along which the vector is pointing. The orientation specifies the sense or rotation of the vector in space.
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Using Wien's Law, what is the maximum wavelength of light that a
star with a temperature of 6500 Kelvin emits?
Wien's Law describes the relationship between the wavelength of light that a star emits and its temperature. This law states that the wavelength at which a star emits the most light is inversely proportional to its temperature. In other words, hotter stars emit shorter wavelengths of light than cooler stars.
Wien's Law can be represented as: λmax = b / T Where λmax is the wavelength of maximum emission, b is Wien's constant (2.898 x 10^-3 m K), and T is the temperature of the star in Kelvin (K).
Now, let's use the given temperature of 6500 K to determine the maximum wavelength of light that the star emits.
λmax = b / Tλmax = 2.898 x 10^-3 m K / 6500 Kλmax = 4.457 x 10^-7 meters.
Therefore, the maximum wavelength of light that a star with a temperature of 6500 Kelvin emits is 4.457 x 10^-7 meters.
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Two sinusoidal waves with identical wavelengths and amplitudes travel in opposite directions along a string with a speed of 5.7 cm/s. If the time interval between instants when the string is flat is 0.49 s, what is the wavelength of the waves? Number Units
The wavelength of the waves is 2.793 cm.
Two sinusoidal waves are traveling in opposite directions with identical wavelengths and amplitudes, as shown in the figure below. We can see that when the string is flat, the two waves are in phase.
Therefore, the distance between the two flat regions is half a wavelength. If we measure this distance and multiply it by 2, we can find the wavelength of the waves. [tex]\lambda=2x[/tex]
We can use the formula λ = vt, where λ is the wavelength, v is the speed, and t is the time interval between two flat regions. In this problem, we are given the speed v = 5.7 cm/s and the time interval t = 0.49 s. Therefore, the wavelength is: λ = vtλ = 5.7 cm/s × 0.49 sλ = 2.793 cm
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A 46.9 turns circular coil with radius 8.99 cm and resistance 0.482 2 is placed in a magnetic field directed perpendicular to the plane of the coil. The magnitude of the mag- netic field varies in time according to the ex- pression B=ayt+a₂t². where a 0.0658 T/s, a2 = 0.0779 T/s² are constants, time t is in seconds and field B is in Tesla. Find the magnitude of the induced emf in the coil at t = 8.79 s. Answer in units of V. Answer in units of V
The magnitude of the induced EMF in the coil at t = 8.79 s is 0.6632 V (to 4 significant figures).According to Faraday's Law of Electromagnetic Induction, a changing magnetic field induces an electromotive force (EMF) in a conductor or coil in that field.
The magnitude of the EMF induced in a coil can be determined using the formula E = -N (dΦ/dt), where E is the induced EMF, N is the number of turns in the coil, and dΦ/dt is the rate of change of the magnetic flux through the coil.
We can find the magnitude of the induced EMF in the given coil as follows:
Number of turns, N = 46.9, Radius of the coil, r = 8.99 cm = 0.0899 m, Resistance of the coil, R = 0.482 2 T and Magnetic field, B = ayt + a2t2 = 0.0658 t/s × 8.79 s + 0.0779 t/s2 × (8.79 s)2= 0.7128 .
TEMF induced in the coil, E = -N (dΦ/dt).
We know that magnetic flux, Φ = B.A, where A is the area of the coil.
For a circular coil, A = πr2. Hence, Φ = B.πr2.
Substituting the given values in the above equation, we haveΦ = (0.7128 T) × π(0.0899 m)2= 0.00017813 Wb.
Taking the derivative with respect to time t on both sides, we getdΦ/dt = d/dt (B.πr2) = πr2 × dB/dt.
Substituting the given values in the above equation, we have:dΦ/dt = π(0.0899 m)2 × (0.0658 t/s + 2 × 0.0779 t/s2 × 8.79 s)= 0.01416 V.
Using the above values in the equation for EMF induced in the coil, we get E = -N (dΦ/dt)=-46.9 × 0.01416 V=-0.6632 V.
Therefore, the magnitude of the induced EMF in the coil at t = 8.79 s is 0.6632 V (to 4 significant figures). Hence, the correct option is the following:0.6632 V.
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A well-thrown ball is caught in a well-padded mitt. If the deceleration of the ball is 1.80×10
4
m/s
2
, and 1.76 ms (1 ms=10
−3
s) elapses from the time the ball first touches the mitt until it stops, what was the initial velocity (in m/s) of the ball? (Enter the magnitude.) m/s
The initial velocity of the ball was 38.85 m/s.
Determine the initial velocity of the ball, we can use the formula that relates acceleration, time, and initial velocity:
This value is obtained by using the equation v = u + at, where v is the final velocity (0 m/s since the ball stops), u is the initial velocity (what we want to find), a is the deceleration of the ball (-2.10 × [tex]10^4 m/s^2[/tex]), and t is the time elapsed (1.85 ms or 1.85 × [tex]10^{-3[/tex]s).
By rearranging the equation and plugging in the given values, we can solve for u. The result indicates that the ball was initially moving at a speed of about 38.85 m/s before being caught.
v = u + at
v = final velocity (0 m/s, as the ball stops)
u = initial velocity (unknown)
a = acceleration (-[tex]2.10 * 10^4 m/s^2[/tex], negative because it opposes the initial velocity)
t = time taken (1.85 ms = 1.85 × [tex]10^{-3[/tex] s)
Plugging in the given values into the equation, we have:
0 = u + (-2.10 ×[tex]10^4 m/s^2[/tex]) × (1.85 × [tex]10^{-3[/tex] s)
Simplifying the equation, we can solve for u:
0 = u - (2.10 ×[tex]10^4 m/s^2[/tex]) × (1.85 × [tex]10^{-3[/tex] s)
Rearranging the equation:
u = (2.10 × [tex]10^4 m/s^2[/tex]) × (1.85 × [tex]10^{-3[/tex] s)
Calculating the expression:
u ≈ 38.85 m/s
The initial velocity of the ball was 38.85 m/s.
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Two uniform charged disks are parallel and share the same axis, both having a radius of 6 meters, and separated by a distance of 0.3 m. The E-field along the axis of a single thin uniform disk is given by 2πkσ[1−1/(1+(R/x)
2
)
1/2
] A) From the information above, determine the area charge density (σ) of each disk. B) Calculate the electric field halfway between these disks, along their shared axis. C) Calculate the electric field 15 cm above the top disk, Q
1
(along the central axis). D) Calculate the same for a point 15 cm below Q
2
. What did you notice about these fields? How did they compare to the field in between the plates (in part B)?
The formula for the electric field along the axis of a single disk may not be applicable in this case. Without further information, it is not possible to determine this value accurately.
A) To determine the area charge density (σ) of each disk, we can use the given formula for the electric field along the axis of a single disk. The formula is:
E = 2πkσ[1 - 1/(1 + (R/x)^2)^(1/2)]
At the center of the disk (x = R), the electric field is zero. We can substitute this value into the formula:
0 = 2πkσ[1 - 1/(1 + (R/R)^2)^(1/2)]
Simplifying this equation gives:
1 = 1/(1 + 1)^(1/2)
1 = 1/2^(1/2)
Squaring both sides:
1 = 1/2
This is not a valid result, which means our assumption that the electric field is zero at the center of the disk is incorrect. Therefore, the formula for the electric field along the axis of a single disk may not be applicable in this case.
B) Since the formula for the electric field along the axis of a single disk may not be valid for the given configuration, we need an alternative approach to calculate the electric field halfway between the disks. Without further information, it is not possible to determine this value accurately.
C) Similarly, without additional information or a different approach, it is not possible to calculate the electric field 15 cm above the top disk (Q1) along the central axis.
D) Likewise, without further information or a different method, it is not possible to calculate the electric field 15 cm below Q2 along the central axis.
In summary, based on the information provided, we cannot accurately determine the electric field values between the disks or at specific points above or below the disks using the given formula. Additional details or alternative approaches are required to calculate these values.
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A parallel-plate capacitor with area of 200 mm2 and plate separation of 5 mm can store 8.70 x 10-11 C of charge when 5.5 volts is applied to the plates. Calculate the capacitance and the dielectric constant for this capacitor. dielectric constant is 39.5 dielectric constant is 2.77 V capacitance is 5 E-12 F dielectric constant is 1000 capacitance is 100 E-11 F dielectric constant is 181 F dielectric constant is 895 capacitance is 44 E-12 F dielectric constant is 44.7 capacitance is 1.58 E-11 F capacitance is 4400 E-11 F dielectric constant is 9395 V/m capacitance is 400 E-10 F capacitance is 17 E-12 F None of the answers is correct. dielectric constant is 90 V/m
The capacitance of the parallel plate capacitor with a dielectric constant of 90 V/m is 4.4 x 10⁻¹³ F.area of parallel plate capacitor is 200mm² = 2x10⁻⁴ m², separation of the plates is 5mm = 5x10⁻³ m, and 5.5 volts is applied to the plates.
The charge stored in a capacitor can be calculated using the formula,Q = CV where Q is the charge, C is the capacitance, and V is the voltage.
Substitute the given values,Q = 8.70 x 10⁻¹¹ C, V = 5.5 V, and C = ?C = Q/VC = 8.70 x 10⁻¹¹ C / 5.5 Vc = 1.58 x 10⁻¹² F.
The capacitance of the parallel plate capacitor is 1.58 x 10⁻¹² F.
The capacitance of a parallel plate capacitor with air as the dielectric medium is given by the formula, C = ε₀A/dwhere C is the capacitance, ε₀ is the permittivity of free space, A is the area of the plates, and d is the separation between the plates.
The permittivity of free space, ε₀ = 8.85 x 10⁻¹² F/m².
Substitute the given values,C = ε₀A/dC = 8.85 x 10⁻¹² F/m² x 2x10⁻⁴ m² / 5x10⁻³ mC = 3.54 x 10⁻¹² F.
The capacitance of the parallel plate capacitor with a dielectric constant of 39.5 is given by the formula,C = kε₀A/d where k is the relative permittivity or the dielectric constant.
Substitute the given values,C = kε₀A/dC = 39.5 x 8.85 x 10⁻¹² F/m² x 2x10⁻⁴ m² / 5x10⁻³ mC = 44.07 x 10⁻¹² FC = 4.4 x 10⁻¹¹ F.
The capacitance of the parallel plate capacitor with a dielectric constant of 90 V/m is given by the formula,C = εrε₀A/d where εr is the relative permittivity or the dielectric constant in terms of the electric flux density.
Substitute the given values,C = εrε₀A/dC = (90 V/m / 8.85 x 10⁻¹² F/m²) x 2x10⁻⁴ m² / 5x10⁻³ mC = 0.44 x 10⁻¹² FC = 4.4 x 10⁻¹³ F.
Therefore, the capacitance of the parallel plate capacitor with a dielectric constant of 90 V/m is 4.4 x 10⁻¹³ F.
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Find the density of dry air at 17.31’Hg and -1 degree F.
The temperature and pressure are the two main factors affecting the density of air. The density of air is proportional to its pressure and inversely proportional to its temperature.
In order to calculate the density of air, we'll need to use the following formula:Density of air = (pressure * molecular weight)/(temperature * R)where R is the gas constant.The molecular weight of dry air is 28.97 gm/mole. At 17.31” Hg pressure and -1°F temperature, the density of dry air can be calculated using the formula as follows:Density = (pressure * molecular weight) / (temperature * R)Let’s find the value of R first:R = 1545.348/PsiK, where PsiK = 14.696 psi and K = 459.67°F.Substituting the values:R = 1545.348 / (14.696 + 459.67)R = 53.3528 lb/ft3 °FRounding the value of R to two decimal places we get, R = 53.35 lb/ft3°FNow let’s substitute the given values of pressure and temperature into the formula to find the density of dry air: Density = (pressure * molecular weight) / (temperature * R)Density = (17.31 * 28.97) / (-1 + 459.67) * 53.35Density = 0.07438 lb/ft3The density of dry air at 17.31” Hg and -1°F is 0.07438 lb/ft3.
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which component is responsible for converting digital audio into sound
The component responsible for converting digital audio into sound is a speaker or a transducer.
The speaker receives an electrical signal containing digital audio data and converts it into sound waves that can be heard by the human ear.
The digital audio signal is typically in the form of binary code, which represents the audio waveform in a series of discrete samples. The speaker uses this digital information to vibrate a diaphragm or a membrane, creating pressure variations in the air that result in sound waves.
These sound waves then travel through the air and reach our ears, where they are perceived as audible sound.
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A book rests on the surface of a table. Consider the following four forces that arise in this situation:
1. the off the ground pulling the book
2. the force of the table pushing the book
3. the force of the book pushing the table
4. the force of pulling the earth
Which two forces form an "action-reaction" pair that obeys Newton's third law?
d. 2 and 4
a. 1 and 2
c. 1 and 4
b. 1 and 3
e. 3 and 4
Newton's Third Law of motion states that for every action, there is an equal and opposite reaction. When analyzing the forces acting on a book resting on a table, we can identify the action-reaction pairs that follow this law. Given the forces:
1. The force of gravity pulling the book downwards
2. The force of the table pushing the book upwards
3. The force of the book pushing the table downwards
4. The force of the Earth pulling the book towards it
We need to determine which two forces form an action-reaction pair. The force of gravity (force 1) is an action force, and the force of the table pushing the book upwards (force 2) is the reaction force. These forces are equal in magnitude and opposite in direction, satisfying Newton's third law.
Therefore, the action-reaction pair that obeys Newton's third law is forces 1 and 2.
Answer: Option (a) 1 and 2.
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The radius of the earth's very nearly circular orbit around the sun is 1.5×10
11
m. Find the magnitude of the earth's velocity, angular velocity, and centripetal acceleration as it travels around the sun. (Exercise 4.33) (v=3.0×10
4
m/s,ω=2.0×10
−7
rad/s,a
r
=6.0×10
−3
m/s
2
)
Given data:
Radius of the earth's orbit = r = 1.5 x 10^11 m
Linear velocity of earth = v = 3.0 x 10^4 m/s
Angular velocity of earth = ω = 2.0 x 10^-7 rad/s
Centripetal acceleration = ar = 6.0 x 10^-3 m/s^2
To find:
Magnitude of velocity of the earth Magnitude of angular velocity of the earth Magnitude of the centripetal acceleration of the earth The magnitude of velocity of the earth The magnitude of velocity of the earth is given as
:v = rω
Where,
v = magnitude of velocity of earth
r = radius of the earth's orbit around the sun
ω = angular velocity of the earth
Substituting the given values,
v = rω= (1.5 x 10^11 m) (2.0 x 10^-7 rad/s)
v = 30 m/s
Therefore, the magnitude of the centripetal acceleration of the earth is 6.0 x 10^-3 m/s^2.
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does the path length difference δℓδℓ increase or decrease as you move from one bright fringe of a two-slit experiment to the next bright fringe farther out?
In a two-slit experiment, the path length difference δℓ between light waves passing through the two slits is crucial to the interference pattern.
The answer to the question is that the path length difference δℓ increases as you move from one bright fringe to the next bright fringe farther out.In an ideal two-slit experiment, light is diffracted as it passes through a small aperture, and the resulting wave fronts diffract again as they pass through a pair of parallel slits. The waves from each slit interfere, producing a pattern of bright and dark fringes on a screen that is located a distance D from the slits. The distance between the slits is d, and the angle between a line from the center of the screen to a bright fringe and a line from the center of the screen to the center of the interference pattern is θ.In such an experiment, the path length difference δℓ between light waves passing through the two slits is a factor in the interference pattern. The path length difference δℓ is given by δℓ = d sin θ.As the angle θ increases, the distance between bright fringes increases, which means that the path length difference δℓ increases. This is because the distance between the slits d remains constant, while the angle θ increases. Therefore, the path length difference δℓ increases as you move from one bright fringe to the next bright fringe farther out.In conclusion, the path length difference δℓ increases as you move from one bright fringe to the next bright fringe farther out in a two-slit experiment.
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μ, (intensive), that G = a) Show using the definitions of Gibbs free energy G (extensive), and Chemical potential μN where N is the number of particles. Discuss why do not have such a relation for Helmholtz free energy F(extensive) per particle with any intensive thermodynamic quantity. b) Obtain the Gibbs-Durhem relation c) Draw schematically the PV diagram for a van der Wall's gas, showing the Maxwell's construction. d) What is the implications on this diagram from the results of part (b) above?
a) The Gibbs free energy G is an extensive thermodynamic quantity that depends on the number of particles N, whereas the chemical potential μ is an intensive thermodynamic quantity that describes the change in Gibbs free energy with respect to the number of particles N.
Therefore, the relation between G and μ is G = μN.
On the other hand, the Helmholtz free energy F is also an extensive thermodynamic quantity, but it does not have a direct relation with any intensive thermodynamic quantity per particle. This is because the Helmholtz free energy is primarily concerned with the internal energy and entropy of a system, whereas the chemical potential μ is related to the change in Gibbs free energy due to changes in the number of particles.
b) The Gibbs-Duhem relation is given by:
dG = -SdT + VdP + μdN,
where G is the Gibbs free energy, S is the entropy, T is the temperature, V is the volume, P is the pressure, μ is the chemical potential, and N is the number of particles. The Gibbs-Duhem relation describes the relationship between the different thermodynamic variables in a system.
c) The PV diagram for a van der Waals gas typically exhibits non-ideal behavior due to intermolecular forces. It shows a region of non-linear behavior where the gas transitions between the gas and liquid phases. The Maxwell's construction is a technique used to construct an idealized curve in the PV diagram that separates the two-phase regions.
d) The results from part (b) imply that the chemical potential μ plays a crucial role in understanding the phase transitions and equilibrium conditions of the system. The presence of the Maxwell's construction in the PV diagram indicates the coexistence of two phases during the phase transition, and it ensures that the area enclosed by the curve represents the work done during the transition.
The implications of the Gibbs-Duhem relation and the presence of the Maxwell's construction highlight the importance of considering non-ideal behavior and phase transitions in thermodynamic systems.
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An ideal diatomic gas undergoes an adiabatic compression during which time its volume changes from what is the final pressure How are the pressure and volume related for an ideal gas subjected to an adiabatic compression? Check units for consistency. atm
Using the equation for adiabatic compression of an ideal gas:
[tex]\(P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma\)[/tex]
Substitute the given values for initial pressure[tex](\(P_1\)), initial volume (\(V_1\)), final volume (\(V_2\)), and γ.[/tex]
Step 1: Identify the given values
- Initial pressure (P1)
- Initial volume (V1)
- Final volume (V2)
- Heat capacity ratio (γ) for the gas (for an ideal diatomic gas, γ = 7/5 or 1.4)
Step 2: Plug in the given values into the adiabatic compression equation
[tex]\[P_2 = P_1 \left(\frac{V_1}{V_2}\right)^\gamma\][/tex]
Step 3: Calculate the final pressure
- Substitute the given values into the equation
[tex]\[P_2 = P_1 \left(\frac{V_1}{V_2}\right)^{\gamma}\][/tex]
- Calculate [tex]\(\left(\frac{V_1}{V_2}\right)^{\gamma}\)\[\left(\frac{V_1}{V_2}\right)^{\gamma} = \left(\frac{V_1}{V_2}\right)^{1.4}\][/tex]
- Calculate the final pressure [tex]\(P_2\) by multiplying \(P_1\) with \(\left(\frac{V_1}{V_2}\right)^{\gamma}\)[/tex]
Step 4: Express the final pressure with the appropriate units (atm)
Remember to ensure that the units for volume and pressure are consistent throughout the calculations.
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most cranial nerves carry both sensory and motor innervation. a. true b. false
The statement "most cranial nerves carry both sensory and motor innervation" is true
As most of the cranial nerves carry both sensory and motor innervation.
Sensory fibers carry the sensations of sight, sound, and smell from various parts of the body to the brain, while motor fibers stimulate or control the muscles of the body and glands. The cranial nerves are a set of 12 nerves that arise from the brainstem and control the various functions of the head, neck, and internal organs.
The nerves are numbered I through XII, and each nerve is responsible for a particular function or group of functions in the body. They are responsible for sensory and motor innervation for various parts of the head and neck, as well as some visceral organs in the body.
The motor and sensory functions of cranial nerves are intermingled, so that most of the nerves carry both sensory and motor fibers.
For example, the trigeminal nerve is responsible for both facial sensation and the control of the muscles of the face, while the glossopharyngeal nerve is responsible for both taste sensation and the control of the muscles of the tongue. In conclusion, the statement "most cranial nerves carry both sensory and motor innervation" is true.
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Consider a spherical conducting shell with inner radius 5 cm, outer radius 10 cm with a charge of 50 nC, concentric with a solid insulating sphere with radius 2 cm and charge of −10nC. Calculate the electric field 8 cm away from the center in N/C.
The electric field contribution from the solid insulating sphere at a point 8 cm away from the center is [tex]-1.405 * 10^6 N/C.[/tex]
To calculate the electric field at a point 8 cm away from the center, we need to consider the contributions from both the conducting shell and the solid insulating sphere.
Electric field contribution from the conducting shell:
Since the point is outside the conducting shell, the electric field inside a conductor is zero. Therefore, the conducting shell does not contribute to the electric field at this point.
Electric field contribution from the solid insulating sphere:
To calculate the electric field from a charged solid sphere at a point outside the sphere, we can use the formula:
E = k * (Q / r²)
where:
E is the electric field,
k is Coulomb's constant ([tex]8.99 * 10^9 N m^2/C^2[/tex]),
Q is the charge of the sphere, and
r is the distance from the center of the sphere.
In this case, the charge of the solid insulating sphere is -10 nC and the distance from the center to the point is 8 cm.
[tex]E_{sphere} = (8.99 * 10^9 N m^2/C^2) * (-10 * 10^{-9} C) / (0.08 m)^2[/tex]
[tex]E_{sphere} = (8.99 *10^9 N m^2/C^2) * (-10 * 10^{-9} C) / (0.08^2 m^2)[/tex]
[tex]E_{sphere} = (-8.99 * 10^9 N m^2/C^2) * (10 * 10^{-9} C) / (0.0064 m^2)[/tex]
[tex]E_{sphere} = -1.405 * 10^6 N/C[/tex]
Therefore, the electric field contribution from the solid insulating sphere at a point 8 cm away from the center is [tex]-1.405 * 10^6 N/C[/tex]. Note that the negative sign indicates the direction of the electric field vector.
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A fan is rotating at a constant 362rev/min. What is the magnitude of the acceleration of a point on one of its blades 10 cm from the axis of rotation? a=m/s^2 A particle travels in a circle of radius 14.9 m at a constant speed of 20 m/s. What is the magnitude of the acceleration? a_c=m/s^2
The magnitude of the acceleration of a point on the blade of the fan, 10 cm from the axis of rotation, is 381.6 m/s². The magnitude of the acceleration of a particle traveling in a circle of radius 14.9 m at a constant speed of 20 m/s is 28.4 m/s².
For both scenarios, we can use the formula for centripetal acceleration:
a_c = (v²) / r
where a_c is the centripetal acceleration, v is the linear velocity, and r is the radius of the circular path.
(a) For the fan rotating at 362 rev/min, we need to convert the angular velocity to linear velocity and convert the radius to meters. Given:
Angular velocity (ω) = 362 rev/min
Radius (r) = 10 cm = 0.1 m
First, we convert the angular velocity to radians per second:
ω = 362 rev/min * (2π rad/rev) * (1 min/60 s) ≈ 38.01 rad/s
Next, we calculate the linear velocity using the formula:
v = ω * r
Substituting the values, we get:
v = 38.01 rad/s * 0.1 m ≈ 3.801 m/s
Now we can calculate the centripetal acceleration using the formula:
a_c = (v²) / r
Substituting the values, we find:
a_c = (3.801 m/s)² / 0.1 m ≈ 381.6 m/s²
Therefore, the magnitude of the acceleration of a point on the fan blade is approximately 381.6 m/s².
(b) For the particle traveling in a circle of radius 14.9 m at a constant speed of 20 m/s, we can directly use the formula for centripetal acceleration:
a_c = (v²) / r
Linear velocity (v) = 20 m/s
Radius (r) = 14.9 m
Substituting the values into the formula, we find:
a_c = (20 m/s)² / 14.9 m ≈ 28.4 m/s²
Therefore, the magnitude of the acceleration of the particle is approximately 28.4 m/s².
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