A function of the velocity with the independent variable time will be v = e^((g - (c/m)) * t + C). Using the recursive formula from part c), we can create a table in Excel by varying the time from 0 to 50 with an increment of two. Start with an initial velocity of 0, and apply the recursive formula to calculate the velocity at each time step. Then, plot a graph of velocity versus time using the data obtained.
To model the change in velocity with respect to time (dv/dt), we can use Newton's second law of motion, which states that the net force acting on an object is equal to the product of its mass and acceleration. In this case, the net force is the force of gravity minus the force of air resistance (drag).
a) Model the change in velocity with respect to time (dv/dt):
Using Newton's second law: F_net = m * a, where F_net is the net force, m is the mass, and a is the acceleration.
The net force can be expressed as: F_net = m * g - c * v (force of gravity minus force of drag)
By substituting the values given, the equation becomes:
m * dv/dt = m * g - c * v
Simplifying, we have:
dv/dt = g - (c/m) * v
b) Establish analytically a function of velocity with respect to time:
To solve the differential equation obtained in part a), we can separate variables and integrate:
(1/v) * dv = (g - (c/m)) * dt
Integrating both sides:
ln(|v|) = (g - (c/m)) * t + C
where C is the constant of integration.
Taking the exponential of both sides:
|v| = e^((g - (c/m)) * t + C)
Since velocity cannot be negative in this case, we have:
v = e^((g - (c/m)) * t + C)
c) Set up a recursive function to be used numerically:
To set up a recursive function, we can use a numerical method like the Euler method. The recursive formula for updating the velocity can be given as
v(t + Δt) = v(t) + (g - (c/m)) * Δt
where Δt is the time step.
d) Using Excel, build a table and a graph:
Using the recursive formula from part c), we can create a table in Excel by varying the time from 0 to 50 with an increment of two. Start with an initial velocity of 0, and apply the recursive formula to calculate the velocity at each time step. Then, plot a graph of velocity versus time using the data obtained.
By analyzing the table and graph, we can observe how the velocity changes over time before the parachute opens. Initially, the velocity will increase due to the force of gravity. As the velocity increases, the force of air resistance (drag) will also increase, eventually reaching a point where the net force becomes zero and the velocity becomes constant. This constant velocity represents the skydiver's terminal velocity, which is reached when the force of gravity is balanced by the force of air resistance.
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Error Analysis The error in the mass When you found the mass in slide 4 of this manual, you used the mean of the force of gravity and the acceleration due to gravity. However, these values also include uncertainties, σ
F
g
and σ
A
y
. To find the mass you used, F
g
=mg. Rearranged, this is m=
g
F
g
. Therefore, to find the uncertainty in m, you will need to use the multiplication/division rule from the error and uncertainty guide. Parts I and II After finding the value of the mass, both in Parts I and II, you can see if the values firt within the uncertainty of the value you found in the steps outlined above. You can also find the percent difference between the value from Part I and the known value, as well as the value from Part II and the known value.
4
2
0
⎦
⎤
μt:4.13078 s−9.741 m/s
2
−σ:0.25 m/s
2
a:−40.239 m/ss:0.01 m/s
3
(r
2
:0.00) Error Analysis The error in the mass When you found the mass in slide 4 of this manual, you used the mean of the force of gravity and the acceleration due to gravity. However, these values also include uncertainties, σ
F
g
and σ
A
y
. To find the mass you used, F
g
=mg. Rearranged, this is m=
g
F
g
. Therefore, to find the uncertainty in m, you will need to use the multiplication/division rule from the error and uncertainty guide. Parts I and II After finding the value of the mass, both in Parts I and II, you can see if the values firt within the uncertainty of the value you found in the steps outlined above. You can also find the percent difference between the value from Part I and the known value, as well as the value from Part II and the known value.
When finding the value of the mass, the multiplication/division rule from the error and uncertainty guide should be used to find the uncertainty in m, using Fg = mg.
Therefore, m = g/Fg. The value of the mass in Part I and Part II can be compared to the value you found in the steps above to see if they fit within the uncertainty of the value.
To find the percent difference between the value from Part I and the known value, you should use the formula:
(Value from Part I - Known Value) / Known Value × 100%
To find the percent difference between the value from Part II and the known value, you should use the formula:
(Value from Part II - Known Value) / Known Value × 100%
For example, if the known value is 3, and the value from Part I is 3.5, then the percent difference would be:
(3.5 - 3) / 3 × 100% = 16.67%
If the value from Part II is 2.8, then the percent difference would be:
(2.8 - 3) / 3 × 100% = -6.67%
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Use g=9.8 m/s 2
In an exciting game, a baseball player manages to safely slide into second base. The mass of the baseball player is 81.7 kg and the coefficient of kinetic friction between the ground and the player is 0.45. (a) Find the magnitude of the frictional force in newtons. N (b) It takes the player 1.7 s to come to rest. What was his initial velocity (in m/s )? m/s
The magnitude of the frictional force on the sliding baseball player is calculated using the coefficient of friction and weight. The player's initial velocity is determined using the time taken to come to rest and the negative acceleration due to friction.
In this scenario, we are given the mass of the baseball player (81.7 kg) and the coefficient of kinetic friction (0.45) between the ground and the player.
(a) To find the magnitude of the frictional force (F_friction), we need to first determine the normal force (F_normal) acting on the player. The normal force is equal to the weight of the player, which can be calculated by multiplying the mass (81.7 kg) by the acceleration due to gravity (9.8 m/s²). Once we have the normal force, we can calculate the frictional force using the equation:
F_friction = coefficient of kinetic friction * F_normal
By substituting the given values, we can determine the magnitude of the frictional force in newtons.
(b) To calculate the initial velocity of the player, we can use the equation of motion:
v = u + at
where v is the final velocity (0 m/s), u is the initial velocity (which we need to find), a is the acceleration (which is the negative of the coefficient of kinetic friction times the acceleration due to gravity), and t is the time taken to come to rest (1.7 s).
Rearranging the equation, we can solve for the initial velocity (u) by substituting the known values. The resulting value will be the player's initial velocity in m/s.
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Acceleration: The velocity v(t) of a particle as a function of time is given by v(t)=(2.3 m/s)+(4.1 m/s2)t−(6.2 m/s3)t2. What is the average acceleration of the particle between t =1.0 s and t=2.0 s ?
The average acceleration of the particle between t = 1.0 s and t = 2.0 s is 3.6 m/s^2. The average acceleration of the particle between t = 1.0 s and t = 2.0 s can be found using the formula: average acceleration = (change in velocity) / (change in time).
The average acceleration of the particle between t = 1.0 s and t = 2.0 s can be found using the formula:
average acceleration = (change in velocity) / (change in time)
The change in velocity between t = 1.0 s and t = 2.0 s can be found by subtracting the velocity at t = 1.0 s from the velocity at t = 2.0 s:
v(2.0 s) - v(1.0 s) = [(2.3 m/s) + (4.1 m/s^2)(2.0 s) - (6.2 m/s^3)(2.0 s^2)] - [(2.3 m/s) + (4.1 m/s^2)(1.0 s) - (6.2 m/s^3)(1.0 s^2)]
Simplifying the expression:
v(2.0 s) - v(1.0 s) = [(2.3 m/s) + (8.2 m/s) - (12.4 m/s)] - [(2.3 m/s) + (4.1 m/s) - (6.2 m/s)]
v(2.0 s) - v(1.0 s) = 3.6 m/s
The change in time is simply:
2.0 s - 1.0 s = 1.0 s
Now we can put these values into the formula for average acceleration:
average acceleration = (3.6 m/s) / (1.0 s) = 3.6 m/s^2
Therefore, the average acceleration of the particle between t = 1.0 s and t = 2.0 s is 3.6 m/s^2.
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The Gauss-Markov Theorem states that if assumptions 1-5 are
satisfied then OLS is BLUE. Briefly explain what the B stands
for
The B in BLUE stands for "Best Linear Unbiased Estimator." The Gauss-Markov Theorem states that if assumptions 1-5 are satisfied, then the Ordinary Least Squares (OLS) estimator is the Best Linear Unbiased Estimator.
Let's break down the meaning of each term in the acronym:
1. Best: The OLS estimator is considered the best because it has the smallest variance among all linear unbiased estimators. In other words, it is the most efficient estimator among all unbiased estimators.
2. Linear: The OLS estimator is a linear combination of the dependent variable and the independent variables. It assumes a linear relationship between the variables.
3. Unbiased: The OLS estimator is unbiased, meaning that on average it will provide an estimate that is equal to the true population parameter being estimated. This implies that there is no systematic error in the estimation process.
4. Estimator: The OLS estimator is a statistical technique used to estimate unknown parameters in a linear regression model. It calculates the best-fit line that minimizes the sum of squared residuals.
The Gauss-Markov Theorem is important because it provides conditions under which the OLS estimator is the best among all linear unbiased estimators.
It ensures that if assumptions 1-5 are met, the OLS estimator will have the smallest variance and, be the most efficient estimator.
The Gauss-Markov Theorem states that if assumptions 1-5 are satisfied, the OLS estimator is the Best Linear Unbiased Estimator (BLUE) because it is the most efficient and unbiased estimator among all linear unbiased estimators.
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8. Two long bar magnets are aligned so that north poles face each other. The magnets are separated by 1 cm, and a repulsive force between the north poles is 0.10 N. When the separation is increased to 2 cm the force will be A. 0.10 N. B. 0.050 N. C. 0.025 N. D. 0.20 N. E. 0.40 N.
The force will be D. 0.20 N. Given that two long bar magnets are aligned so that north poles face each other. The magnets are separated by 1 cm, and a repulsive force between the north poles is 0.10 N. We have to find the force between two long bar magnets when the separation is increased to 2 cm.
According to Coulomb's law, the force between two magnetic poles is given by:F = k * (m₁ * m₂)/r²Where,F = Force between two magnetic poles
m₁, m₂ = Strength of poles
r = Distance between two poles
k = Coulomb's constant
For two north poles, the force is repulsive, and k = 10⁻⁷ Nm²/C²We know that, the force between two magnets is given as 0.10 N when the distance between the magnets is 1 cm, i.e. r₁ = 1 cm = 0.01 m
k = 10⁻⁷ Nm²/C²Using above values, we can find the strength of poles as:
m₁ * m₂ = F * r₁²/km₁ * m₂ = 0.10 * (0.01)² / 10⁻⁷m₁ * m₂ = 10⁻⁵ Nm²/C²Now, we need to find the force when the distance is 2 cm, i.e. r₂ = 0.02 mF₂ = k * (m₁ * m₂)/r₂²F₂ = 10⁻⁷ * (10⁻⁵) / 0.02²F₂ = 0.125 N
Approximate answer to two decimal places is 0.13 N.
Therefore, the correct option is D. 0.20 N.
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A solid ball has a radius of 0.60 m and has a charge of 15.50 μC at its centre.
(a) Determine the magnitude of the electric field produced by the charge at a point on the surface on the sphere.
(b) Determine the electrical flux pass through the solid ball.
The magnitude of the electric field produced by the charge at a point on the surface of the sphere is approximately 2.29 × [tex]10^6[/tex] N/C. The electrical flux passing through the solid ball is approximately 1.75 × [tex]10^6[/tex] N·m²/C.
(a) To determine the magnitude of the electric field produced by the charge at a point on the surface of the sphere, we can use Gauss's law. Gauss's law states that the electric flux passing through a closed surface is proportional to the total charge enclosed by that surface.
For a uniformly charged sphere, the electric field outside the sphere is the same as that of a point charge located at the center of the sphere.
Given:
The radius of the sphere, r = 0.60 m
Charge at the center of the sphere, Q = 15.50 μC = 15.50 × [tex]10^{(-6)[/tex] C
The electric field produced by a point charge at a distance r from the charge is given by Coulomb's law:
E = k * (|Q| / [tex]r^2[/tex])
Where:
E = electric field
k = Coulomb's constant (approximately 8.99 × [tex]10^9[/tex] N·m²/C²)
|Q| = magnitude of the charge
r = distance from the charge
Since the point is on the surface of the sphere, the distance from the charge to the point is equal to the radius of the sphere:
r = 0.60 m
Substituting the given values into the equation, we have:
E = (8.99 × [tex]10^9[/tex] N·m²/C²) * (|15.50 × [tex]10^{(-6)[/tex] C| / [tex](0.60 m)^2[/tex])
E ≈ 2.29 × [tex]10^6[/tex] N/C
Therefore, the magnitude of the electric field produced by the charge at a point on the surface of the sphere is approximately 2.29 × [tex]10^6[/tex] N/C.
(b) To determine the electrical flux passing through the solid ball, we can use Gauss's law. Gauss's law states that the total electric flux passing through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε₀).
The charge enclosed by the solid ball is equal to the charge at the center of the sphere:
[tex]Q_{enclosed[/tex] = |Q| = 15.50 μC = 15.50 × [tex]10^{(-6)[/tex] C
The electrical flux passing through the solid ball can be calculated using the equation:
Φ = [tex]Q_{enclosed[/tex] / ε₀
Where:
Φ = electrical flux
[tex]Q_{enclosed[/tex] = charge enclosed by the surface
ε₀ = permittivity of free space (approximately 8.85 × [tex]10^{(-12)[/tex]C²/(N·m²))
Substituting the given values into the equation, we have:
Φ = (15.50 × [tex]10^{(-6)[/tex] C) / (8.85 × [tex]10^{(-12)[/tex] C²/(N·m²))
Φ ≈ 1.75 × [tex]10^6[/tex]N·m²/C
Therefore, the electrical flux passing through the solid ball is approximately 1.75 × [tex]10^6[/tex]N·m²/C.
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A bullet of inertia m is fired with speed v into a block of inertia M which is placed directly
against a spring (in relaxed, equilibrium position). The bullet instantaneously embeds in the
block and the block/bullet compresses the spring a distance d. The spring constant is k.
Assume that all of this is happening in a vacuum on a frictionless surface and that the spring
has no mass and no internal states of the spring change.
Write an expression for the initial velocity of the bullet v in terms of the four other variables.
According to the law of conservation of linear momentum, the initial momentum of the bullet is equal to the final momentum of the bullet + block. So, the momentum of the bullet before striking the block is m * v and the momentum of the bullet + block after the collision is (m + M) * V', where V' is the final velocity of the bullet and the block.
Since, the bullet instantaneously embeds into the block, the final velocity of the block and bullet is the same V'. Let us now derive an equation in terms of the given variables. This can be done using the law of conservation of energy. The initial kinetic energy of the bullet is (1/2) m v².
This is converted into the elastic potential energy of the spring when the block/bullet compresses the spring a distance d. This can be expressed as (1/2) k d². Using the law of conservation of energy, we can equate these two values: (1/2) m v² = (1/2) k d² ---- (1)
Let us now substitute the value of d in terms of V' using the equation for spring potential energy:
k d²/2 = (m + M) V'²/2
⇒ d² = (m + M) V'²/k
Substituting this value of d² in equation (1), we get:
m v² = (m + M) V'²
⇒ V' = v * sqrt(m/(m+M))
So, the expression for initial velocity of the bullet v in terms of the given variables is:
V' = v * sqrt(m/(m+M))
In the given problem, we are given the speed of the bullet (v), the mass of the bullet (m), the mass of the block (M), the spring constant (k), and the distance by which the spring is compressed (d). We are required to find the expression for initial velocity of the bullet in terms of these variables.
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A ring of uniform linear charge density λ=1μC/m and radius a=12 cm has its axis along the x axis and its center is placed at position ( 3.3 m,0 m ). a single point charge Q=190nC is placed at position (0m1−3.7m). Part A Determine the magnitude of the net electric field Enet at the origin. Write your answer with 3 significant digits. Determine the magnitude of the net electric field Enet at the origin. Write your answer with 3 significant digits. Part B Determine the polar direction of the net electric field Enet at the origin Write your answer with 3 significant digits.
Part A: Calculate E_net = (8.99 × 10^9 Nm^2/C^2 * 190 × 10^-9 C) / (√((0 m - 3.3 m)^2 + (0 m - (-3.7 m))^2))^2 for the magnitude of the net electric field at the origin. Part B: Determine the polar direction using θ = arctan((y2 - y1) / (x2 - x1)) between the origin and the point charge.
Part A: The magnitude of the net electric field at the origin can be calculated using the formula:
E_net = (k * Q) / r^2
where k is the Coulomb constant (k = 8.99 × 10^9 Nm^2/C^2), Q is the point charge (Q = 190 nC), and r is the distance between the point charge and the origin (r = √((0 m - 3.3 m)^2 + (0 m - (-3.7 m))^2)).
Substituting the values, we have:
E_net = (8.99 × 10^9 Nm^2/C^2 * 190 × 10^-9 C) / (√((0 m - 3.3 m)^2 + (0 m - (-3.7 m))^2))^2
Calculating the result will give us the magnitude of the net electric field at the origin.
Part B: The polar direction of the net electric field at the origin can be determined by finding the angle θ between the x-axis and the vector connecting the origin to the point charge using the formula:
θ = arctan((y2 - y1) / (x2 - x1))
where (x1, y1) is the position of the origin and (x2, y2) is the position of the point charge.
Substituting the values, we can calculate the angle θ to determine the polar direction of the net electric field at the origin.
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The age of the universe is thought to be about 14 billion years. Assuming two significant figures, write the following in scientific notation (a) days, (b) hours (10 points)
a) The days of the scientific notation is 5.11 × 10^12 days.
b) The hours of the scientific notation is 1.23 × 10^14 hours.
a) Days:
The age of the universe in scientific notation is:
5.11 × 10^12 days
Number of days in a year: 365.25 days
4 billion years × 365.25 days/year = 5.11 × 10^12 days (rounded to two significant figures)
(b) Hours:
The age of the universe in scientific notation is:
1.23 × 10^14 hours
Number of hours in a day: 24 hours
5.11 × 10^12 days × 24 hours/day = 1.23 × 10^14 hours (rounded to two significant figures)
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A rock with a mass of 518 g in air is found to have an apparent mass of 325 g when submerged in water. a. What mass of water is displaced? m _w = g b. What is the volume of the rock? V_r =cm ^{3}
c. What is its average density? rho_r =g/cm ^{3}
a. The mass of water displaced is calculated as follows: The mass of water displaced is 193 g.b. The volume of the rock is 518 cm³.c. The average density of the rock is 1 g/cm³.
Apparent loss in mass = Mass in air - Mass in water
Apparent loss in mass = 518 - 325
= 193 g
The mass of water displaced is equal to the mass of the water that has been pushed aside by the rock. As a result, the mass of water displaced is equal to the mass that has been lost by the rock
.Apparent loss in mass = Mass of displaced water
Mass of displaced water = Apparent loss in mass
Mass of displaced water = 193 g
Therefore, the mass of water displaced is 193 g.b. The volume of the rock is calculated using the following formula
:Volume of rock = (Mass of rock) / (Density of water)
The mass of the rock is 518 g. The density of water is 1 g/cm³
.Volume of rock = (518 g) / (1 g/cm³)
Volume of rock = 518 cm³
Therefore, the volume of the rock is 518 cm³.c. The average density of the rock is calculated using the following formula:
Average density of rock = (Mass of rock) / (Volume of rock)
The mass of the rock is 518 g. The volume of the rock is 518 cm³
.Average density of rock = (518 g) / (518 cm³)
Average density of rock = 1 g/cm³
Therefore, the average density of the rock is 1 g/cm³.
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. A force vector points at an angle of 52∘above the +x axis. It has a y component of +270 Newtons. a) Find the magnitude of the force vector. b) Find the x component of the foree vector
The magnitude of the force vector is approximately +354.48 N, and its x-component is approximately +224.43 N, given an angle of 52° above the +x-axis and a y-component of +270 N.
The angle θ = 52° above +x-axis and y-component Fy = +270 N.
Using the trigonometric formula of a right-angled triangle, we can find the magnitude of the force vector as follows:
Sin θ = Opposite side / Hypotenuse
Magnitude (F) = Hypotenuse = Opposite side / sin θ
Therefore, F = Fy / sin θ= +270 N / sin 52°= +354.48 N (approx)
Hence, the magnitude of the force vector is +354.48 N.
Applying trigonometric formula again, we can find the x-component of the force vector as follows:
Cos θ = Adjacent side / HypotenuseX-component (Fx) = Hypotenuse × cos θ
= F × cos θ= +354.48 N × cos 52°
= +224.43 N (approx)
Therefore, the x-component of the force vector is +224.43 N.
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A ball thrown horizontally at 25 m/s travels a horizontal distance of 55 m before hitting the ground From what height was the ball thrown? You may want to review Express your answer using two significant figures with the appropriate units.
The ball was thrown from a height of approximately 15 meters.
To determine the initial height of the ball, we can use the horizontal distance traveled (55 m) and the horizontal velocity (25 m/s). Using the equation h = (1/2) * g * t^2, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time taken for the ball to hit the ground, we find that the time is approximately 2.2 seconds. Plugging this value into the equation, we calculate the initial height to be approximately 15 meters.
Thus, the ball was thrown from a height of approximately 15 meters.
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A long, nonconducting, solid cylinder of radius 5.4 cm has a nonuniform volume charge density rho that is a function of radial distance r from the cylinder axis: rho=Ar
2
. For A=2.1μC/m
5
, what is the magnitude of the electric field at (a)r=1.6 cm and (b)r=11 cm.
To find the magnitude of the electric field at two different radial distances, we can use Gauss's Law.
Which states that the electric field at a point outside a uniformly charged cylinder is directly proportional to the linear charge density.
Given:6
Radius of the cylinder, r = 5.4 cm = 0.054 m
Volume charge density, ρ = Ar^2
A = 2.1 μC/m^5 = 2.1 × 10^-6 C/m^5
(a) At r = 1.6 cm = 0.016 m:
The electric field can be calculated using Gauss's Law as follows:
Let's consider a Gaussian surface in the form of a cylinder of radius r and length L, coaxial with the given solid cylinder.
The charge enclosed within the Gaussian surface is equal to the total charge within the solid cylinder.
Charge enclosed, Q_enclosed = ρ × V
V = πr^2L
Substituting the given values:l
Q_enclosed = (2.1 × 10^-6 C/m^5) × (π × (0.054 m)^2 × L)
= (2.1 × 10^-6) × (π × 0.002916 m^3 × L)
= 6.1092 × 10^-9 L C
Applying Gauss's Law, we know that the electric field outside the cylinder is given by:
E × 2πrL = Q_enclosed / ε₀
E × 2πrL = (6.1092 × 10^-9 L C) / ε₀
Simplifying the equation:
E = (6.1092 × 10^-9 L C) / (2πrL × ε₀)
E = 3.0546 × 10^-9 C / (πrε₀)
Now, we need to calculate the value of ε₀, the permittivity of free space:
ε₀ ≈ 8.854 × 10^-12 C^2/(N·m^2)
Substituting the value of ε₀:
E = 3.0546 × 10^-9 C / (π × 0.016 m × 8.854 × 10^-12 C^2/(N·m^2))
E ≈ 5.4787 × 10^9 N/C
Therefore, the magnitude of the electric field at r = 1.6 cm is approximately 5.4787 × 10^9 N/C.
(b) At r = 11 cm = 0.11 m:
Using the same approach as above, we can calculate the magnitude of the electric field at this distance:
E = 3.0546 × 10^-9 C / (π × 0.11 m × 8.854 × 10^-12 C^2/(N·m^2))
E ≈ 1.3294 × 10^8 N/C
Therefore, the magnitude of the electric field at r = 11 cm is approximately 1.3294 × 10^8 N/C.
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What change in entropy occurs when a 0.12 kg steam at 121 deg C is transformed into ice at -28 deg C?
Use the 2nd law of thermodynamics to explain calculated result.
The change in entropy that occurs when a 0.12 kg steam at 121°C is converted into ice at -28°C is calculated to be -0.145 J/K. The 2nd law of thermodynamics explanation is as provided below.
Entropy is a measure of the disorder of a system, and the second law of thermodynamics establishes that the entropy of an isolated system increases over time.
The entropy of a system can be calculated using the equation ΔS = Q/T, where ΔS is the change in entropy, Q is the amount of heat transferred, and T is the absolute temperature of the system.
When a 0.12 kg steam at 121°C is transformed into ice at -28°C, there is a significant change in entropy. The steam has a high level of entropy due to its high temperature and disordered molecular arrangement, while the ice has a low level of entropy due to its low temperature and highly ordered molecular arrangement.
The process of transforming the steam into ice involves a decrease in temperature, which means that heat energy is released from the system.
This results in a negative value for Q in the equation ΔS = Q/T. Since the absolute temperature of the system is decreasing, the value of T is also decreasing, which means that the change in entropy is negative.
To calculate the change in entropy, we need to know the amount of heat transferred and the absolute temperatures of the steam and ice.
The specific heat capacity of water is 4.18 J/g°C, so the amount of heat transferred can be calculated using the equation Q = mcΔT, where m is the mass of the steam, c is the specific heat capacity of water, and ΔT is the change in temperature.
Using this equation, we can calculate that the amount of heat transferred is
Q = (0.12 kg)(4.18 J/g°C)(121°C - (-28°C)) = 22,380 J.
The absolute temperatures of the steam and ice are 394 K and 245 K, respectively, so we can calculate the change in entropy using the equation
ΔS = Q/T = (22,380 J)/(394 K) - (245 K)) = -0.145 J/K.
The negative value for the change in entropy indicates that the process of transforming steam into ice results in a decrease in disorder and an increase in order.
This is in accordance with the second law of thermodynamics, which states that the entropy of an isolated system increases over time.
The transformation of steam into ice represents a decrease in entropy, but the overall entropy of the universe still increases due to the dissipation of heat energy into the surroundings.
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The base plate welded at one of the ends of an I-beam is fastened with four anchor bolts (two on each side) of diameter d Considering that the load is equally distributed in each one of the bolts, determine:
a) as a function of F, d and 8 the normal stress and shear stress in each bolt,
b) the magnitude of the normal stress and shear stress in each bolt if F = 40 kN, d= 12 mm y0= 40°
Normal stress in each bolt is 36.27 MPa Shear stress in each bolt is 0.96 MPa.
a) As a function of F, d, and 8 the normal stress and shear stress in each bolt. Given that, the force acting is F, and the bolts are two on each side of the I-beam.
Let the two bolts on one side be named bolt A and bolt B, and let the two bolts on the other side be named bolt C and bolt D. Let the diameter of each bolt be d.
The area of each bolt = [tex]πd²/4[/tex]
Normal stress in each bolt can be calculated as;
σA = F / (2 * 0.25 * π * d²)
σB = F / (2 * 0.25 * π * d²)
σC = F / (2 * 0.25 * π * d²)
σD = F / (2 * 0.25 * π * d²)
Shear stress in each bolt can be calculated as:
τA = F / (2 * 0.25 * π * d²)
τB = F / (2 * 0.25 * π * d²)
τC = F / (2 * 0.25 * π * d²)
τD = F / (2 * 0.25 * π * d²)
b) The magnitude of the normal stress and shear stress in each bolt if
F = 40 kN, d = 12 mm and yo = 40°
For each bolt, the values can be calculated as follows;
σ = F / (2 * 0.25 * π * d²)
τ = F / (2 * 0.25 * π * d²) * tan y0
= 0.839 * F / (d²)
Where; F = 40 kN
d = 12 mm
y0 = 40°
For each bolt, let the values of shear stress and normal stress be denoted by the alphabets.
[tex]σA = σB = σC = σD[/tex]
= 40 kN / (2 * 0.25 * π * (12mm)²)
≅ 36.27 MPa
τA = τB = τC = τD
= 0.839 * 40 kN / (12 mm)²
≅ 0.96 MPa
The magnitude of the normal stress and shear stress in each bolt when F = 40 kN, d = 12 mm and yo = 40° is thus as follows: Normal stress in each bolt is 36.27 MPa Shear stress in each bolt is 0.96 MPa.
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The stuntman (mass 60 kg) will jump from 20 m above the ground and the 2.0 m tall stunt airbag will exert a constant force of 2,500 N on him.
Is the size of the airbag sufficient?
The size of the airbag is sufficient to provide a safe landing for the stuntman.
To determine if the size of the airbag is sufficient, we need to consider the forces involved. The force exerted on the stuntman by the airbag is 2,500 N, which remains constant throughout the landing.
As the stuntman jumps from a height of 20 m, he will experience an initial gravitational force acting on him, given by the formula F = mg, where m is the mass of the stuntman and g is the acceleration due to gravity (approximately 9.8 m/s²). Substituting the values, the gravitational force is calculated as F = (60 kg)(9.8 m/s²) = 588 N.
When the stuntman lands on the airbag, the force exerted by the airbag is greater than the gravitational force, ensuring a safe deceleration. Since the airbag exerts a constant force of 2,500 N, which is higher than the gravitational force of 588 N, the size of the airbag is sufficient to provide a safe landing for the stuntman.
In conclusion, the airbag is appropriately sized to withstand the impact of the stuntman's jump from 20 m. Its constant force of 2,500 N exceeds the gravitational force, ensuring a safe landing and adequate deceleration for the stuntman.
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Suppose a ball of putty moving with 1 kg⋅m/s of momentum collides with and sticks to an identical ball of putty moving perpendicular to the first one with 1 kg⋅m/s of momentum. What is the magnitude of their combined momentum? 1 kg⋅m/s 2 kg⋅m/s 0 kg⋅m/s 1.41 kg⋅m/s
The magnitude of their combined momentum is approximately 1.41 kg⋅m/s.
When two objects collide and stick together, the law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision.
In this case, we have two identical balls of putty, each with a momentum of 1 kg⋅m/s. One ball is moving horizontally, and the other is moving vertically. Since the momentum is a vector quantity, we need to consider both the magnitude and direction.
When the balls collide and stick together, their momenta combine vectorially. In this case, the momentum vectors of the two balls are perpendicular to each other. When two vectors of equal magnitude are perpendicular to each other, their combined magnitude can be calculated using the Pythagorean theorem.
The magnitude of their combined momentum is given by:
combined momentum = √((1 kg⋅m/s)^2 + (1 kg⋅m/s)^2)
Simplifying:
combined momentum = √(1 kg^2⋅m^2/s^2 + 1 kg^2⋅m^2/s^2)
combined momentum = √(2 kg^2⋅m^2/s^2)
combined momentum = √2 kg⋅m/s
Therefore, the magnitude of their combined momentum is approximately 1.41 kg⋅m/s.
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A 8.9-kg child sits in a 3.7-kg high chair. Find the normal force exerted by the floor on the chair.
The normal force exerted by the floor on the chair is approximately 123.48 Newtons.
To find the normal force exerted by the floor on the chair, we need to consider the gravitational forces acting on the child and the chair.
The gravitational force acting on the child can be calculated using the formula:
F_child = m_child × g
where
m_child = mass of the child = 8.9 kg
g = acceleration due to gravity = 9.8 m/s²
F_child = 8.9 kg × 9.8 m/s²
F_child = 87.22 N
Similarly, the gravitational force acting on the chair can be calculated using the same formula:
F_chair = m_chair × g
where
m_chair = mass of the chair = 3.7 kg
g = acceleration due to gravity = 9.8 m/s²
F_chair = 3.7 kg × 9.8 m/s²
F_chair = 36.26 N
Since the child is sitting in the chair, the normal force exerted by the floor on the chair is equal to the sum of the child's weight and the chair's weight:
Normal force = F_child + F_chair
Normal force = 87.22 N + 36.26 N
Normal force = 123.48 N
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My car starts from rest a distance 4 m from the garage door and accelerates directly away with an acceleration given by a (t) : (24$ + 2)II1/52. 3} Evaluate 110$), assuming 13(0) : O. b) Evaluate .1: (t)! assuming it: (0) = 4. _ c) How long would it take to get 9 m from the garage door? Enter your response in meters 5].
a) To evaluate a(t) when t = 0, we substitute t = 0 into the given equation. Given a(t) = (24t + 2) * (1/52)³,
we have a(0) = (24 * 0 + 2) * (1/52)³
= 2 * (1/52)³ = 2 * (1/140608)
= 2/140608
= 1/70304.
b) To evaluate s(t) when t = 0, we substitute t = 0 into the given equation. Given s(t) = 4 + integral from 0 to t of a(u) du, we have s(0) = 4 + integral from 0 to 0 of a(u) du = 4 + integral from 0 to 0 of 1/70304 du = 4.
c) To find the time it takes to get 9 m from the garage door, we solve the equation s(t) = 9. Given s(t) = 4 + integral from 0 to t of a(u) du, we have 9 = 4 + integral from 0 to t of a(u) du. We can solve this equation numerically or graphically to find the value of t.
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joules. (a) change in electric potential (in V) v (b) change in electric potential energy in joules J.
Joule is a unit of measurement that is utilized to determine the amount of energy that has been transformed from one form to another.
It is named after James Prescott Joule, who first introduced the concept of mechanical equivalent heat in 1843. The joule has the symbol J, and it is defined as the amount of work done when a force of one newton moves an object a distance of one meter in the direction of the force.
Joules can be used to quantify the change in electric potential and change in electric potential energy in the following ways:(a) Change in electric potential (in V):The change in electric potential is calculated in volts (V), with one joule being equivalent to one coulomb of electric charge that has been transported via a potential difference of one volt.
Therefore, the formula used to calculate the change in electric potential (in V) is:∆V = W / Q,where:∆V represents the change in electric potential (in V)W represents the amount of work done (in J)Q represents the electric charge transported (in C).
(b) Change in electric potential energy in joules J:Electric potential energy (U) is calculated in joules (J), with one joule being equivalent to one coulomb of electric charge that has been transported via a potential difference of one volt.
Therefore, the formula used to calculate the change in electric potential energy (in J) is:∆U = Q × ∆V,where:∆U represents the change in electric potential energy (in J)Q represents the electric charge transported (in C)∆V represents the change in electric potential (in V).
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A shot putter throws a 16-lb steel ball (a shot), releasing it at an unknown height above the ground with an initial velocity of 11.2 m/s at an angle of 44.9
∘
above horizontal. The shot lands on the ground 1.85 s later. a. What are the horizontal and vertical components of its initial velocity? b. What is its initial height above the ground? c. What is its maximum height above the ground? d. How far, horizontally, did the shot travel before hitting the ground? e. What are the horizontal and vertical components of its final velocity, right before it hits the ground? f. What are the magnitude and angle of the final velocity vector, right before the shot hits the ground?
The horizontal and vertical components of the initial velocity are 7.97 m/s and 7.80 m/s, respectively, the initial height above the ground is 6.91 m, the maximum height above the ground is 3.13 m.
a. find the horizontal and vertical components of the initial velocity, we can use the following equations:
v₀x = v₀ * cos(θ)
v₀y = v₀ * sin(θ)
Initial velocity, v₀ = 11.2 m/s
Angle, θ = 44.9 degrees
Substituting the given values into the equations:
v₀x = 11.2 m/s * cos(44.9 degrees)
v₀y = 11.2 m/s * sin(44.9 degrees)
Calculating v₀x and v₀y:
v₀x ≈ 7.97 m/s
v₀y ≈ 7.80 m/s
b. find the initial height above the ground, we need to consider the vertical motion of the shot. The equation to determine the height is:
h = v₀y * t - (1/2) * g * t²
Time of flight, t = 1.85 s
Acceleration due to gravity, g ≈ 9.8 m/s²
Substituting the values into the equation:
h = 7.80 m/s * 1.85 s - (1/2) * 9.8 m/s² * (1.85 s)²
Calculating h:
h ≈ 6.91 m
The initial height above the ground is 6.91 meters.
c. The maximum height above the ground occurs when the vertical component of the velocity becomes zero. We can use the equation:
v_fy = v₀y - g * t
Since v_fy = 0 at the maximum height, we can rearrange the equation:
t = v₀y / g
Substituting the values:
t = 7.80 m/s / 9.8 m/s²
Calculating t:
t ≈ 0.80 s
find the maximum height (h_max), we can use the equation:
h_max = v₀y * t - (1/2) * g * t²
Substituting the values:
h_max = 7.80 m/s * 0.80 s - (1/2) * 9.8 m/s² * (0.80 s)²
Calculating h_max:
h_max ≈ 3.13 m
The maximum height above the ground is 3.13 meters.
d. The horizontal distance traveled by the shot before hitting the ground can be calculated using the equation:
Δx = v₀x * t
Time of flight, t = 1.85 s
Initial horizontal velocity, v₀x = 7.97 m/s
Substituting the values into the equation:
Δx = 7.97 m/s * 1.85 s
Calculating Δx:
Δx ≈ 14.71 m
The shot traveled 14.71 meters horizontally before hitting the ground.
e. Just before the shot hits the ground, the vertical component of the velocity is the negative of the initial vertical component (v₀y), and the horizontal component remains the same (v₀x). Therefore:
Final horizontal velocity (v_fx) = v₀x = 7.97 m/s
Final vertical velocity (v_fy) = -v₀y = -7.80 m/s
f. find the magnitude (v_f) and angle (θ_f) of the final velocity vector, we can use the following equations:
v_f = sqrt(v_fx² + v_fy²)
θ_f = atan(v_fy / v_fx)
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geologist is surveying a cave. She follows a path 170 m straight West, then 250 m in the direction 45° East of South and then 280m 30° East of North. Find the magnitude and direction of her total displacement using the three methods: (1) Parallelogram method (2) Polygon method (3) Component method
The geologist's total displacement using three methods is as follows:
1) Parallelogram method:
Magnitude: Approximately 158.5 mDirection: Approximately 60.4° North of East2) Polygon method:
Magnitude: Approximately 252.5 mDirection: Approximately -8.4° South of East (or 171.6° East of South)3) Component method:
Magnitude: Approximately 252.5 mDirection: Approximately -8.4° South of East (or 171.6° East of South)Thus, regardless of the method used, the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
To find the magnitude and direction of the geologist's total displacement using the three methods, let's calculate them step by step:
(1) Parallelogram Method:
In the parallelogram method, we create a parallelogram using the two longest displacement vectors (170 m straight West and 280 m 30° East of North). The diagonal of the parallelogram represents the resultant displacement.
First, let's resolve the displacement vectors into their horizontal (x) and vertical (y) components:
170 m West: -170 m in the x-direction
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction, 2
80 * sin(30°) ≈ 140 m in the y-direction
Next, we add the x and y components separately:
Total x-component = -170 m + 242.51 m
≈ 72.51 m
Total y-component = 140 m
Using the Pythagorean theorem, we can find the magnitude of the resultant displacement:
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((72.51)^2 + (140)^2)
Magnitude ≈ 158.5 m
To find the direction of the resultant displacement, we can use trigonometry:
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(140 / 72.51)
Direction ≈ 60.4° North of East
Therefore, using the parallelogram method, the magnitude of the geologist's total displacement is approximately 158.5 m, and the direction is approximately 60.4° North of East.
(2) Polygon Method:
In the polygon method, we create a polygon by connecting the displacement vectors in sequential order and draw the resultant vector from the starting point to the end point.
To calculate the magnitude and direction using the polygon method, we can add the x and y components of the displacement vectors:
170 m West: -170 m in the x-direction
250 m 45° East of South:
250 * cos(45°) ≈ 176.78 m in the x-direction, -
250 * sin(45°) ≈ -176.78 m in the y-direction (opposite direction)
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction,
280 * sin(30°) ≈ 140 m in the y-direction
Total x-component = -170 m + 176.78 m + 242.51 m
≈ 249.29 m
Total y-component = -176.78 m + 140 m = -36.78 m
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((249.29)^2 + (-36.78)^2)
Magnitude ≈ 252.5 m
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(-36.78 / 249.29)
Direction ≈ -8.4° South of East (or 171.6° East of South)
Therefore, using the polygon method, the magnitude of the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
(3) Component Method:
In the component method, we find the horizontal and vertical components of each displacement vector and then add them to find the resultant displacement.
Resolve each displacement vector into its horizontal (x) and vertical (y) components:
170 m West: -170 m in the x-direction, 0 m in the y-direction
250 m 45° East of South:
250 * cos(45°) ≈ 176.78 m in the x-direction, -
250 * sin(45°) ≈ -176.78 m in the y-direction (opposite direction)
280 m 30° East of North:
280 * cos(30°) ≈ 242.51 m in the x-direction,
280 * sin(30°) ≈ 140 m in the y-direction
Total x-component = -170 m + 176.78 m + 242.51 m ≈ 249.29 m
Total y-component = 0 m - 176.78 m + 140 m = -36.78 m
Magnitude = √(Total x-component^2 + Total y-component^2)
Magnitude = √((249.29)^2 + (-36.78)^2)
Magnitude ≈ 252.5 m
Direction = tan^(-1)(Total y-component / Total x-component)
Direction = tan^(-1)(-36.78 / 249.29)
Direction ≈ -8.4° South of East (or 171.6° East of South)
Therefore, using the component method, the magnitude of the geologist's total displacement is approximately 252.5 m, and the direction is approximately -8.4° South of East (or 171.6° East of South).
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How does the mechanical energy of a simple harmonic oscillator relate to the angular frequency of the oscillations? How does the mechanical energy of a simple harmonic oscillator relate to the amplitude of the oscillations?
The mechanical energy of a simple harmonic oscillator is related to its amplitude in the following way:
E = ½ kA2,
Mechanical energy of a simple harmonic oscillator:The mechanical energy of an object in motion is the sum of its kinetic energy (KE) and potential energy (PE). The energy is conserved, meaning that it is neither created nor destroyed. The mechanical energy in an oscillator is also conserved, meaning that it remains constant throughout the entire oscillation.Angular frequency of the oscillations:If a simple harmonic oscillator moves through a full cycle in time T, then its frequency is f = 1/T.
The angular frequency, denoted by ω, is defined as ω = 2πf. Therefore, ω = 2π/T.
Angular frequency is the number of radians per second. In SI units, the angular frequency of a simple harmonic oscillator is measured in radians per second (rad/s).Relation between mechanical energy and angular frequency of oscillations:
The mechanical energy of a simple harmonic oscillator is related to its angular frequency in the following way: E = ½ kA2, where E is the total mechanical energy of the oscillator, k is the spring constant, and A is the amplitude of the oscillations.Amplitude of the oscillations:
In simple harmonic motion, amplitude is the maximum displacement from equilibrium that an oscillator can have. The amplitude of a wave is the maximum displacement of a particle on a medium from its rest position when a wave passes through it.Relation between mechanical energy and amplitude of oscillations:
The mechanical energy of a simple harmonic oscillator is related to its amplitude in the following way:
E = ½ kA2,
where E is the total mechanical energy of the oscillator, k is the spring constant, and A is the amplitude of the oscillations.
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please help im not sure what to do coefficient: 0.20 i. The stick man has a mass of \( 90 \mathrm{~kg} \). Assume that the co-efficient of friction is the same 2 between the box and the ground as it is between the stick man's shoe and the ground. Use Ne
The coefficient of friction is a measure of the frictional force between two surfaces in contact. In this case, if the coefficient of friction is the same between the box and the ground as it is between the stick man's shoe and the ground, we can use that value to calculate the frictional force acting on the stick man.
The frictional force can be calculated using the equation[tex]\( f_{\text{friction}} = \mu \cdot N \), where \( f_{\text{friction}} \)[/tex] is the frictional force,[tex]\( \mu \)[/tex] is the coefficient of friction, and[tex]\( N \)[/tex]is the normal force.
To find the normal force, we can use the equation [tex]\( N = m \cdot g \), where \( m \)[/tex] is the mass of the stick man and \( g \) is the acceleration due to gravity.
Substituting the given values, we have[tex]\( N = 90 \, \text{kg} \times 9.8 \, \text{m/s}^2 \).[/tex]
Next, we can calculate the frictional force by multiplying the coefficient of friction by the normal force. Using the given coefficient of friction of 0.20, we have [tex]\( f_{\text{friction}} = 0.20 \times N \).[/tex]
Finally, substituting the value of [tex]\( N \)[/tex] into the equation, we can calculate the frictional force acting on the stick man.
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vertical spring has a length of 0.15 m when a 0.225 kg mass hangs from it, and a length of 0.775 m when a 1.925 kg mass hangs from it.
a) What is the force constant of the spring, in newtons per meter?
b) What is the unloaded length of the spring, in centimeters?
To find the force constant of the spring, we need to use the formula given below: F = kx where, F = Force exerted by the spring (in N)k = Force constant of the spring (in N/m)x = Displacement from the equilibrium position (in m)
(1) When a 0.225 kg mass hangs from it, Then, F = mg where, m = Mass of the object (in kg)g = Acceleration due to gravity = 9.8 m/s² Force exerted by the spring (F) = Weight of the mass (mg) = 0.225 × 9.8 N Force exerted by the spring (F) = 2.205 ND is placement from the equilibrium position (x) = Length of the spring when the mass is suspended - Unstretched length of the spring x = 0.15 - 0 (as there is no mass attached to the spring)Displacement from the equilibrium position (x) = 0.15 m By substituting the above values in the formula, we get,2.205 = k × 0.15k = 2.205/0.15Force constant of the spring, k = 14.7 N/m Therefore, the force constant of the spring is 14.7 N/m.
(2) When a 1.925 kg mass hangs from it, Then, F = mg where, m = Mass of the object (in kg)g = Acceleration due to gravity = 9.8 m/s² Force exerted by the spring (F) = Weight of the mass (mg) = 1.925 × 9.8 N Force exerted by the spring (F) = 18.91 N Displacement from the equilibrium position (x) = Length of the spring when the mass is suspended - Unstretched length of the spring x = 0.775 - 0 (as there is no mass attached to the spring)Displacement from the equilibrium position (x) = 0.775 m
By substituting the above values in the formula, we get,18.91 = k × 0.775k = 18.91/0.775 Force constant of the spring, k = 24.4 N/m Therefore, the force constant of the spring is 24.4 N/m. To find the unloaded length of the spring, we can use the formula given below: Force constant of the spring (k) = F/x where, F = Force exerted by the spring (in N)x = Displacement from the equilibrium position (in m)(1) When a 0.225 kg mass hangs from it, Force exerted by the spring (F) = 2.205 N Displacement from the equilibrium position (x) = 0.15 m By substituting the above values in the formula, we get, k = F/xk = 2.205/0.15k = 14.7 N/m Let the unstretched length of the spring be x₀.
Then, Force constant of the spring (k) = F/x₀⇒ x₀ = F/k⇒ x₀ = 2.205/14.7⇒ x₀ = 0.15 m Therefore, the unloaded length of the spring is 15 cm.(2) When a 1.925 kg mass hangs from it, Force exerted by the spring (F) = 18.91 N Displacement from the equilibrium position (x) = 0.775 m By substituting the above values in the formula, we get, k = F/xk = 18.91/0.775k = 24.4 N/m Let the unstretched length of the spring be x₀.
Then, Force constant of the spring (k) = F/x₀⇒ x₀ = F/k⇒ x₀ = 18.91/24.4⇒ x₀ = 0.775 m Therefore, the unloaded length of the spring is 77.5 cm (775/10 = 77.5).Hence, the required force constant of the spring is 14.7 N/m and the unloaded length of the spring is 15 cm.
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A negative charge of −0.510μC exerts an upward 0.600−N force Part A on an unknown charge that is located 0.400 m directly below the first charge, What is the value of the unknown charge (magnitude and sign)? Express your answer with the appropriate units.
The magnitude of the unknown charge is 1.85 × 10^-5 C, and it is positive.
We can use Coulomb's law to solve this problem. Coulomb's law states that the force between two charges is proportional to the product of the charges and inversely proportional to the square of the distance between them.
The force between the two charges is given by:
F = k * q1 * q2 / r^2
where:
* F is the force between the two charges (in N)
* k is the Coulomb constant (8.987551787 × 10^9 N m^2 / C^2)
* q1 is the magnitude of the first charge (in C)
* q2 is the magnitude of the second charge (in C)
* r is the distance between the two charges (in m)
In this problem, we know the following:
* F = 0.600 N
* q1 = −0.510μC = −5.10 × 10^-6 C
* r = 0.400 m
We need to find the value of q2.
Substituting the known values into the equation for Coulomb's law, we get:
F = k * q1 * q2 / r^2
0.600 N = 8.987551787 × 10^9 N m^2 / C^2 * (-5.10 × 10^-6 C) * q2 / (0.400 m)^2
Solving for q2, we get:
q2 = 1.85 × 10^-5 C
Since the force between the two charges is upward, and the first charge is negative, the second charge must be positive.
Therefore, the value of the unknown charge is 1.85 × 10^-5 C, and it is positive.
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the velocity components for a two-dimensional flow are
The velocity components for a two-dimensional flow are Vx = V cos θVy = V sin θ, where V = 150/√(tan²θ + 1)
The velocity components for a two-dimensional flow are as follows: Let V be the velocity and θ be the angle that the velocity vector makes with the x-axis.
Vx = V cos θ
Vy = V sin θ
Given,150 = √(Vx²+Vy²).....(1)
Using the above expression (1), we can find V.
Now, Vx/Vy = tan θ ⇒ Vx = V tan θ
Putting the value of Vx in terms of V in the expression (1),
we have 150 = √(V²tan²θ + V²)150 = V√(tan²θ + 1)
Dividing by √(tan²θ + 1), we get150/√(tan²θ + 1) = V
Therefore, the velocity components for a two-dimensional flow are Vx = V cos θVy = V sin θ, where V = 150/√(tan²θ + 1)
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A 70 kg skydiver pulls his parachute and slows until he reaches his terminal velocity of 5 m/s. The parachute has an effective surface area of 20 m
2
. Calculate the drag coefficient of the parachutist.
The drag coefficient of the parachutist is 0.3044. A 70 kg skydiver pulls his parachute and slows until his terminal velocity is 5 m/s.
We need to determine the drag coefficient of the parachutist.
Using the formula of terminal velocity, we get:
v = sqrt( (2mg)/(pAC) ) where m is the mass, g is the acceleration due to gravity, p is the density of air, A is the area of the parachute and C is the drag coefficient.
Substituting the given values, we have:
5 = sqrt( (2*70*9.8)/(1.2*20*C) )
Simplifying the expression, we get:
5 = sqrt( (1372)/(24C) )
Squaring both sides, we have:
25 = (1372)/(24C)24
C = (1372/25)
C = (1372/25)/24
C = 0.3044
Therefore, the drag coefficient of the parachutist is 0.3044.
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how to find magnitude of acceleration given velocity and time
The magnitude of acceleration, given velocity and time, is 15 m/s².
The formula for acceleration is given as
acceleration = (velocity_f - velocity_i) / time.
if we are given the velocity and time of an object,
we can easily calculate the magnitude of acceleration by using the formula above.
Here's how:
Step 1:
Find the change in velocity. The difference between the final velocity and the initial velocity is the change in velocity. For instance, if an object starts at a velocity of 50 m/s and ends at a velocity of 200 m/s, the change in velocity is 200 m/s - 50 m/s = 150 m/s.
Step 2:
Find the time taken. The time taken is simply the duration of the acceleration. If an object accelerates for 10 seconds, then the time taken is 10 seconds.
Step 3:
Calculate the acceleration. We can now substitute the values of the change in velocity and time into the formula for acceleration. So,
acceleration = (150 m/s) / (10 s) = 15 m/s² (to two significant figures).
Therefore, the magnitude of acceleration, given velocity and time, is 15 m/s².
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Suppose you have a hardware device that operates an FFT with a window size of 512 samples at 44100 samples per second. You want to use it to completely \& accurately analyze a bass drum with frequencies that range from 50 Hz to 30 Hz; can the device serve this purpose as stated? Yes. No. Yes but only on Friday. Maybe.
Yes, the hardware device can serve the purpose of accurately analyzing a bass drum with frequencies ranging from 50 Hz to 30 Hz. Here's why:
1. The window size of 512 samples refers to the number of samples the device can analyze at once. In this case, it can process 512 samples.
2. The sample rate of 44100 samples per second means that the device can capture 44100 samples in one second.
3. To accurately analyze a signal, we need to ensure that the highest frequency of interest is properly represented in the samples. According to the Nyquist-Shannon sampling theorem, we need to sample at a rate that is at least twice the maximum frequency. In this case, the maximum frequency is 50 Hz.
4. Since the sample rate of 44100 samples per second is more than twice the maximum frequency of 50 Hz, the device can accurately capture the bass drum frequencies ranging from 50 Hz to 30 Hz.
5. The device's window size of 512 samples allows for detailed analysis of the bass drum signal within the specified frequency range. The FFT algorithm can provide information about the different frequencies present in the signal, including the bass drum frequencies.
Therefore, the hardware device can serve the purpose of accurately analyzing a bass drum with frequencies ranging from 50 Hz to 30 Hz.
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