The answer is c. 456 °C. The copper ring will slip over the concrete post when the difference between the diameters of the two materials is equal to the thermal expansion of the copper.
The thermal expansion coefficient of copper is 17.3 * 10^-6 m/m*°C. So, the copper ring will expand by 0.0346 cm when heated by 1°C.
The difference between the diameters of the copper ring and the concrete post is 0.05 cm. So, the copper ring will slip over the concrete post when it is heated to 0.05 / 0.0346 = 14.4°C.
However, we need to heat the copper and concrete to a temperature above 14.4°C, because the concrete will also expand when heated. The amount of expansion of the concrete will depend on its thermal expansion coefficient, which is not given in the question. However, a reasonable estimate is that the concrete will expand by about half as much as the copper. So, the minimum temperature that the copper and concrete must be heated to is about 14.4 + 7.2 = 45.6°C.
So the answer is (c).
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The concept of resonance explains .. A. the cooking of food by microwaves B. the reception of radio waves by antennae
C. the collapse of the Tacoma Narrows Bridge
D. all of these
The correct answer is D: all of these. The concept of resonance explains various phenomena, including the cooking of food by microwaves, the reception of radio waves by antennae, and the collapse of the Tacoma Narrows Bridge.
Resonance occurs when an object or system vibrates at its natural frequency in response to an external force or stimulus. In the case of microwaves, the concept of resonance is utilized to cook food efficiently.
Microwaves generate electromagnetic waves that match the resonant frequency of water molecules, causing them to vibrate and generate heat. Similarly, radio waves are received by antennae through resonance.
The antennae are designed to resonate at specific frequencies, allowing them to capture the radio signals and convert them into electrical signals for transmission. In the case of the Tacoma Narrows Bridge, resonance played a detrimental role.
The bridge's structural design and the wind conditions caused the bridge to vibrate at its natural frequency, resulting in destructive oscillations and ultimately leading to its collapse. Therefore, resonance explains these phenomena, making option D, "all of these," the correct answer.
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Vectors À and B both have
positive y- components, and make angles of a = 35° and
B= 10° with the positive and negative x-axis, respectively.
Vector C points along the negative y axis with a magnitude of 19. If
the vector sum À + B+ C= 0, what are the magnitudes of À
and B?
Two vectors À and B both have positive y- components, and make angles of a = 35° and B= 10° with the positive and negative x-axis, respectively. Vector C points along the negative y axis with a magnitude of 19.
If the vector sum À + B+ C= 0, we have to find the magnitudes of À and :Let's solve the problem by drawing the diagram. The direction of vectors A and B are shown below:As we know that the vector sum of A, B, and C is zero. It means that the direction of the vectors A, B and C is such that A and B lie on the x-y plane and C is along the negative y-axis. Now let's find out the vector sum
À + B+ CÀ + B+ C = 0mÀ cos(35°) i + m À sin(35°) j + m B cos(10°) i + m B sin(10°)j + (-19j) = 0
Since the vector sum is equal to zero, it means the magnitude of the vector sum should be zero and also the x and y component of the vector sum should be zero. Hence we can write,
cos(35°) m À + cos(10°) m B = 0---------(1)sin(35°)m À + sin(10°) m B - 19 = 0 ------(2)
Solving equation (1) and (2) will give us the value of
m À and m B. m À = -7.64mB = 20.04The magnitude of À will be |A| = m À = 7.64
The magnitude of B will be |B| = m B = 20.04The magnitude of the vectors
À and B are 7.64 and 20.04 respectively.
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please answer both im reviewing for a final :) Question 23 of 37 ) A car travels in the positive x-direction in the reference frame S at an ordinary speed. The reference frame s' moves at a speed of 0.80c, along the x-axis. The rest length of the car is 3.10 m. Calculate the length of the car according to observers in the S' frame 00 L 1100 Question 22 of 37 > Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54 x 107 light-years from Earth. If the lifetime of a human is taken to be 70.0 years, a spaceship would need to achieve some minimum speed min to deliver a living human being to this galaxy. How close to the speed of light would this minimum speed be? Express your answer as the difference between Umin and the speed of light c. - Umin m/s
The length of the car, as observed in the S' frame, is shorter due to relativistic effects.
The minimum speed required to travel to the Andromeda Galaxy is very close to the speed of light.
According to the theory of relativity, when an object moves relative to an observer, its length appears shorter in the direction of motion. This phenomenon is known as length contraction.
In this case, the car is moving in the positive x-direction in the S frame, while the S' frame is moving at a speed of 0.80 times the speed of light (0.80c) along the x-axis.
The rest length of the car is given as 3.10 m in the S frame. To calculate the length of the car in the S' frame, we can use the formula for length contraction:
Length_s' = Length_s / γ
where γ is the Lorentz factor, given by γ = 1 / √(1 - v^2/c^2), with v being the velocity of the S' frame relative to the S frame. Plugging in the values, we can calculate the length of the car as observed in the S' frame.
The Andromeda Galaxy is located at a distance of 2.54 x 10^7 light-years from Earth. Since the lifetime of a human is taken to be 70.0 years, a spaceship would need to travel this immense distance within that timeframe to deliver a living human being.
To determine the minimum speed required, we can divide the distance by the time:
Minimum speed = Distance / Time = (2.54 x 10^7 light-years) / (70.0 years)
However, it's important to convert this distance and time into a common unit to perform the calculation accurately. Since the speed of light is approximately 3 x 10^8 meters per second, we can convert the distance to meters by multiplying it by the number of meters in a light-year (9.461 x 10^15 m).
Similarly, we convert the time to seconds by multiplying it by the number of seconds in a year (3.156 x 10^7 s). Substituting the values, we can calculate the minimum speed required.
The resulting speed will be very close to the speed of light (c), and the difference between the minimum speed (Umin) and the speed of light (c) will be negligible.
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Consider a sealed cylindrical container released from a height h and rolling without friction on an inclined plane. If water is added to the container, would the velocity of the cylinder when it reaches the end of the incline be faster than of the empty one? Consider that water slides without friction inside the container and in both cases the cylinder is released from the
same height.
Select one:
O True
O False
False. If water is added to the container, the velocity of the cylinder when it reaches the end of the incline will not be faster than of the empty one
The velocity of the cylinder when it reaches the end of the incline would not be affected by the addition of water to the container. The key factor determining the velocity of the cylinder is the height from which it is released and the incline angle of the plane. The mass of the water inside the container does not affect the acceleration or velocity of the cylinder because it is assumed to slide without friction.
The cylinder's velocity is determined by the conservation of mechanical energy, which depends solely on the initial height and the angle of the incline. Therefore, the addition of water would not make the cylinder reach the end of the incline faster or slower compared to when it is empty.
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A small coherent asteroid with a density of 2,300 kgm3 is hit with a large metal projectile from a spacecraft experiment and breaks up completely. The sizes of the pieces follow the approximate power-law size distribution:
N(R) = C(R/1.00 m)-3 dR
where C = 100 m-1 and 1.00 mm ≤ R ≤ 1.00 m.
Determine the total mass of the asteroid by integrating over the size distribution. Hint: Determine dm the total mass of number of particles N(R) of radius exactly R (actually within the radius interval [R, R + dR])
(Answer in kg and in scientific notation using 2 sig. figs.)
To determine the total mass of the asteroid, we need to integrate the size distribution function over the range of sizes.
The size
distribution function
is given by N(R) = C(R/1.00 m)^(-3)dR, where C = 100 m^(-1) and 1.00 mm ≤ R ≤ 1.00 m.
By integrating this function, we can calculate the total mass of the asteroid.
Given:
Density
of the asteroid (ρ) = 2,300 kg/m^3
Size distribution function: N(R) = C(R/1.00 m)^(-3)dR
C = 100 m^(-1)
Integrate the size distribution function to find the total
mass
:
The total mass (m) is given by:
m = ∫ N(R) * ρ * dV
Since the volume
element
dV is related to the radius R as dV = 4/3 * π * R^3, we can substitute it into the equation:
m = ∫ N(R) * ρ * (4/3 * π * R^3) * dR
Substitute the given values and simplify the equation:
m = ∫ (100 m^(-1)) * (R/1.00 m)^(-3) * (2,300 kg/m^3) * (4/3 * π * R^3) * dR
Integrate the equation over the
range
of sizes:
m = ∫ (100 * 2,300 * 4/3 * π) * (R/1.00)^(-3+3) * R^3 * dR
m = (100 * 2,300 * 4/3 * π) * ∫ R^3 * dR
Evaluate the integral:
m = (100 * 2,300 * 4/3 * π) * [1/4 * R^4] evaluated from R = 1.00 mm to R = 1.00 m
Calculate the total mass:
m = (100 * 2,300 * 4/3 * π) * [1/4 * (1.00 m)^4 - 1/4 * (1.00 mm)^4]
Answer:
The total mass of the asteroid is approximately 6.09 × 10^9 kg (to 2 significant figures).
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Question 2 (MCQ QUESTION: answer in ULWAZI) Consider the normalised eigenstates for a particle in a 1 dimensional box as shown: Eigenstates v The probability of finding a particle in any of the three energy states is: Possible answers (order may change in ULWAZI Greatest on the left of the box Greatest on the right of the box Greatest in the centre of the box The same everywhere inside the box Zero nowhere in the box [3 Marks] [3].
The probability of finding a particle in any of the three energy states is the same everywhere inside the box.
The probability of finding a particle in any of the three energy states is the same everywhere inside the box. Consider the normalised eigenstates for a particle in a 1-dimensional box as shown: Eigenstates. The normalised eigenstates for a particle in a 1-dimensional box are as follows:Here, A is the normalization constant.\
To find the probability of finding a particle in any of the three energy states, we need to find the probability density function (PDF), ψ²(x).Probability density function (PDF), ψ²(x) is given as follows:Here, ψ(x) is the wave function, which is the normalised eigenstate for a particle in a 1-dimensional box.
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An RLC series circuit has a 2.80Ω resistor, a 200μH inductor, and a 78.0μF capacitor. (a) Find the circuit's impedance (in Ω ) at 120 Hz. Ω (b) Find the circuit's impedance (in Ω ) at 5.00kHz. Ω (c) If the voltage source has Vrms=5.60 V, what is Irms (in A) at each frequency? Irms,120 Hz=Irms,5.00kHz=AA (d) What is the resonant frequency (in kHz ) of the circuit? kHz (e) What is Irms (in A) at resonance? A
(a) The impedance of an RLC series circuit is given by the formula Z = √(R^2 + (Xl - Xc)^2), where R is the resistance, Xl is the inductive reactance, and Xc is the capacitive reactance.
At 120 Hz, the inductive reactance (Xl) can be calculated using the formula Xl = 2πfL, where f is the frequency and L is the inductance.
Similarly, the capacitive reactance (Xc) can be calculated using the formula Xc = 1 / (2πfC), where C is the capacitance. Plugging in the given values, we can calculate the impedance.
(b) Using the same formula as in part (a), we can calculate the impedance at 5.00 kHz by substituting the given frequency and the values of R, L, and C.
(c) To find the current (Irms) at each frequency, we can use Ohm's law, which states that I = V / Z, where V is the voltage and Z is the impedance. Given the voltage (Vrms), we can calculate the current using the impedance values obtained in parts (a) and (b).
(d) The resonant frequency of an RLC series circuit is given by the formula fr = 1 / (2π√(LC)). By substituting the given values of L and C, we can find the resonant frequency in kHz.
(e) At resonance, the current (Irms) is determined by the resistance only since the reactances cancel each other out. Therefore, the current at resonance is equal to Vrms divided by the resistance (R).
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Verify that a 2.3 x 1017kg mass of water at normal density would make a cube 61 km on a side. (This mass at nuclear density would make a cube 1.0 m on a side. Submit a file with a maximum size of 1 MB
To verify the given information, let's calculate the volume of water represented by a mass of 2.3 x 10^17 kg at normal density and check if it would form a cube with a side length of 61 km.
Density of water at normal conditions is approximately 1000 kg/m³.
Volume = Mass / Density
Volume = (2.3 x 10^17 kg) / (1000 kg/m³)
Volume = 2.3 x 10^14 m³
Side length = ∛Volume
Side length = ∛(2.3 x 10^14 m³)
Side length ≈ 611.6 km
Therefore, the calculated side length is approximately 611.6 km, which is close to the given value of 61 km. It seems there might be an error in the given information, as the side length would be much larger than stated.
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9. [0/10 Points] DETAILS PREVIOUS ANSWERS MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Using a diffraction grating with 4500 lines/cm, the third order of a wavelength appears at 10º. Determine the wavelength and then determine at what angle the first order will appear. λ=12.73 nm 8₁=
The wavelength of the light is determined to be 12.73 nm (nanometers). The angle at which the first order will appear is approximately 21.08°.
Diffraction grating with 4500 lines/cm
Third order of a wavelength appears at 10ºWe have to determine the wavelength and then determine at what angle the first order will appear.
1: Calculating the Wavelength
Formula to calculate the wavelength is given by:dsinθ = nλHere, d = 1/N, where N is the number of lines per unit length, i.e., d = 1/4500 = 0.000222 m.
θ = 10º (given)
n = 3 (third order)
λ = ?d × sin θ = nλ0.000222 × sin 10° = 3λ
λ = 0.00000001273 m = 12.73 nm
2: Calculating the Angle for the First OrderWe know that the angle of diffraction for the first order is given by:dsinθ = λ
Here, d = 1/N, where N is the number of lines per unit length, i.e., d = 1/4500 = 0.000222 m.
λ = 12.73 nm = 12.73 × 10^−9 m
θ = ?
d × sin θ = λsin
θ = λ/dθ = sin−1(λ/d)
θ = sin−1(12.73 × 10^−9 / 0.000222)
θ = 21.08° (approx)
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Part A You have a special lightbulb with a very delicate wire filament. The wire will break if the current in it ever exceeds 1.70 A , even for an instant. What is the largest root-mean-square current you can run through this bulb? Pal AΣφ PE ? Irms A Submit Request Answer
The largest root-mean-square current that can be run through this bulb is approximately 1.70 A.
To determine the largest root-mean-square (rms) current that can be run through the lightbulb without breaking the filament, we need to consider the relationship between rms current and peak current.
The rms current (Irms) is related to the peak current (Ipeak) through the following equation:
Irms = Ipeak / √2
Given that the wire filament will break if the current exceeds 1.70 A, we can set up the following equation:
1.70 A = Ipeak / √2
To solve for Ipeak, we can multiply both sides of the equation by √2:
Ipeak = 1.70 A * √2
Ipeak ≈ 2.404 A
Therefore, the largest rms current (Irms) that can be run through the bulb without breaking the filament is:
Irms = Ipeak / √2 ≈ 2.404 A / √2 ≈ 1.70 A
So the largest root-mean-square current that can be run through this bulb is approximately 1.70 A.
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How do you specify the z component of an electrons total angular
momentum in units of h/2pi?
The z component of an electron's total angular momentum, denoted as Lz, can be specified in units of h/2π (Planck's constant divided by 2π) by using the formula: Lz = mℏ
where m is the quantum number representing the specific value of the z component and ℏ is h/2π (reduced Planck's constant). The quantum number m can take on integer or half-integer values (-ℓ, -ℓ+1, ..., ℓ-1, ℓ), where ℓ is the orbital angular momentum quantum number.
Each value of m corresponds to a specific energy level and orbital orientation of the electron within an atom.
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A ball of mass 100g is dropped from a hight of 12.0 m. What is the ball's linear momentum when it strikes the ground? Input the answer in kgm/s using 3 significant fugures
The linear momentum of the ball is 1.534 kg m/s.
The mass of the ball is 100 g, and the height from which it is dropped is 12.0 m. We have to calculate the linear momentum of the ball when it strikes the ground. To find the velocity of the ball, we have used the third equation of motion which relates the final velocity, initial velocity, acceleration, and displacement of an object.
Let's substitute the given values in the equation, we get:
v² = u² + 2asv² = 0 + 2 × 9.8 × 12.0v² = 235.2v = √235.2v ≈ 15.34 m/s
Now we can find the linear momentum of the ball by using the formula p = mv. We get:
p = 0.1 × 15.34p = 1.534 kg m/s
Therefore, the ball's linear momentum when it strikes the ground is 1.534 kg m/s.
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The separation between the two slits is 0.020μm,1μm=10−6 m, and the first-order minimum (equivalent to dark fringe formed by light) is formed at an angle of 8.63∘ relative to the incident electron beam. Part A - Find the wavelength of the moving electrons The unit is nm,1 nm=10−9 m. Keep 2 digits after the decimal point. Part B - Find the momentum of each moving electron. Use scientific notations, format 1.234⋆10n.
In the given double-slit experiment with electrons, the separation between the two slits is 0.020 μm.
The first-order minimum (dark fringe) is observed at an angle of 8.63 degrees relative to the incident electron beam. The task is to determine the wavelength of the moving electrons (Part A) and the momentum of each moving electron (Part B).
Part A: To find the wavelength of the moving electrons, we can use the formula for the wavelength of a particle diffracted by a double slit, given by λ = (d * sinθ) / n, where λ is the wavelength, d is the separation between the slits, θ is the angle of the first-order minimum, and n is the order of the minimum (which is 1 in this case). By substituting the given values, we can calculate the wavelength of the moving electrons.
Part B: The momentum of each moving electron can be determined using the de Broglie wavelength equation, which states that the momentum of a particle is equal to h / λ, where h is Planck's constant. By substituting the calculated wavelength from Part A into the equation, we can find the momentum of each moving electron in scientific notation format.
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Question 1 (1 point)
A force, F, is applied to an object with a displacement, Δd. When does the equation W = FΔd equal the work done by the force on the object?
Question 1 options:
always
when the force is in the same direction as the displacement
when the force is perpendicular to the displacement
when the force is at an angle of 450 to the displacement
Question 2 (1 point)
At a construction site, a constant force lifts a stack of wooden boards, which has a mass of 500 kg, to a height of 10 m in 15 s. The stack rises at a steady pace. How much power is needed to move the stack to this height?
Question 2 options:
1.9 x 102 W
3.3 x 102 W
3.3 x 103 W
1.6 x 104 W
Question 3 (1 point)
Saved
A mover pushes a sofa across the floor of a van. The mover applies 500 N of horizontal force to the sofa and pushes it 1.5 m. The work done on the sofa by the mover is
Question 3 options:
285 J
396 J
570 J
750J
Question 4 (1 point)
A cart at the farmer's market is loaded with potatoes and pulled at constant speed up a ramp to the top of a hill. If the mass of the loaded cart is 5.0 kg and the top of the hill has a height of 0.55 m, then what is the potential energy of the loaded cart at the top of the hill?
Question 4 options:
27 J
0.13 J
25 J
130 J
Question 6 (1 point)
Suppose that a spacecraft of mass 6.9 x 104 kg at rest in space fires its rockets to achieve a speed of 5.2 x 103 m/s. How much work has the fuel done on the spacecraft?
Question 6 options:
2.2 x 106 J
1.8 x 109 J
3.6 x 109 J
9.3 x 1011 J
Question 7 (1 point)
A 60 kg woman jogs up a hill in 25 s. Calculate the power the woman exerts if the hill is 30 m high.
Question 7 options:
706W
750W
650W
380W
Question 8 (1 point)
A shopper pushes a loaded grocery cart with a force of 15 N. The force makes an angle of 300 above the horizontal. Determine the work done on the cart by the shopper as he pushes the cart 14.2 m.
Question 8 options:
166J
213J
185J
225J
Question 9 (1 point)
A car of mass 1.5 x 105 kg is initially travelling at a speed of 25 m/s. The driver then accelerates to a speed of 40m/s over a distance of 0.20 km. Calculate the work done on the car.
Question 9 options:
3.8x105 J
7.3x107 J
7.3x105 J
7.3x103 J
Question 10 (1 point)
A 86g golf ball on a tee is struck by a golf club. The golf ball reaches a maximum height where its gravitational potential energy has increased by 255 J from the tee. Determine the ball's maximum height above the tee.
303m
34m
0.3m
30m
Answer:
1.) The equation W = FΔd equal the work done by the force on the object when the force is in the same direction as the displacement.
2.) The equation W = FΔd equal the work done by the force on the object when the force is in the same direction as the displacement.
3.) The work done on the sofa by the mover is 285 J.
4.) The potential energy of the loaded cart at the top of the hill is 27 J.
6.) The amount of work done by the fuel on the spacecraft is 3.6 x 109 J
7.) The power the woman exerts when jogging up the hill is 706 W.
8.) The work done on the cart by the shopper is 166 J.
9.) The work done on the car is 7.3 x 107 J.
10.) The ball's maximum height above the tee is 30 m.
Explanation:
1.) The equation W = FΔd equal the work done by the force on the object when the force is in the same direction as the displacement.
2.) The equation W = FΔd equal the work done by the force on the object when the force is in the same direction as the displacement.
Power = Work / Time
Power = (Mass * Acceleration * Height) / Time
Power = (500 kg * 9.8 m/s^2 * 10 m) / 15 s
Power = 3.3 x 103 W
3.) The work done on the sofa by the mover is 285 J.
Work = Force * Distance
Work = 500 N * 1.5 m
Work = 285 J
4.)The potential energy of the loaded cart at the top of the hill is 27 J.
Potential Energy = Mass * Gravitational Constant * Height
Potential Energy = 5.0 kg * 9.8 m/s^2 * 0.55 m
Potential Energy = 27 J
6.) The amount of work done by the fuel on the spacecraft is 3.6 x 109 J
Work = Kinetic Energy
Work = (1/2) * Mass * Velocity^2
Work = (1/2) * 6.9 x 10^4 kg * (5.2 x 10^3 m/s)^2
Work = 3.6 x 10^9 J
7.) The power the woman exerts when jogging up the hill is 706 W.
Power = Work / Time
Power = (Mass * Gravitational Potential Energy) / Time
Power = (60 kg * 9.8 m/s^2 * 30 m) / 25 s
Power = 706 W
8.) The work done on the cart by the shopper is 166 J.
Work = Force * Distance * Cos(theta)
Work = 15 N * 14.2 m * Cos(30)
Work = 166 J
9.) The work done on the car is 7.3 x 107 J.
Work = Force * Distance
Work = (Mass * Acceleration) * Distance
Work = (1.5 x 10^5 kg * (40 m/s - 25 m/s)) * 0.20 km
Work = 7.3 x 10^7 J
10.) The ball's maximum height above the tee is 30 m.
Potential Energy = Mass * Gravitational Constant * Height
255 J = 0.086 kg * 9.8 m/s^2 * Height
Height = 30 m
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We know now that kWh (or GJ) is a unit of energy and kW is a unit of power, and energy = power x time. But, what is the difference between energy and power? or how would you define each? (hint: think units, how is a watt represented in joules?). Please provide some examples to illustrate the difference; could be from any system (lights, motors, etc).
Energy and power are related concepts in physics, but they represent different aspects of a system. Energy refers to the capacity to do work or the ability to produce a change.
It is a scalar quantity and is measured in units such as joules (J) or kilowatt-hours (kWh). Energy can exist in various forms, such as kinetic energy (associated with motion), potential energy (associated with position or state), thermal energy (associated with heat), and so on.
Power, on the other hand, is the rate at which energy is transferred, converted, or used. It is the amount of energy consumed or produced per unit time. Power is a scalar quantity measured in units such as watts (W) or kilowatts (kW).
It represents how quickly work is done or energy is used. Mathematically, power is defined as the ratio of energy to time, so it can be expressed as P = E/t.
To illustrate the difference between energy and power, let's consider the example of a light bulb. The energy consumed by the light bulb is measured in kilowatt-hours (kWh) and represents the total amount of electrical energy used over a period of time.
The power rating of the light bulb is measured in watts (W) and indicates the rate at which electrical energy is converted into light and heat. So, if a light bulb has a power rating of 60 watts and is switched on for 5 hours, it will consume 300 watt-hours (0.3 kWh) of energy.
Similarly, in the case of an electric motor, the energy consumed would be measured in kilowatt-hours (kWh), representing the total amount of electrical energy used to perform work.
The power of the motor, measured in kilowatts (kW), would indicate how quickly the motor can convert electrical energy into mechanical work. The higher the power rating, the more work the motor can do in a given amount of time.
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A charge of +54 µC is placed on the x-axis at x = 0. A second charge of -38 µC is placed on the x-axis at x = 50 cm. What is the magnitude of the electrostatic force on a third charge of 4.0 µC placed on the x-axis at x = 15 cm? Give your answer in whole numbers.
The magnitude of the electrostatic force on a third charge placed at a specific location can be calculated using Coulomb's law.
In this case, a charge of +54 µC is located at x = 0, a charge of -38 µC is located at x = 50 cm, and a third charge of 4.0 µC is located at x = 15 cm on the x-axis. By applying Coulomb's law, the magnitude of the electrostatic force can be determined.
Coulomb's law states that the magnitude of the electrostatic force between two charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. Mathematically, it can be expressed as F = k * |q1 * q2| / r^2, where F is the electrostatic force, q1, and q2 are the charges, r is the distance between the charges, and k is the electrostatic constant.
In this case, we have a charge of +54 µC at x = 0 and a charge of -38 µC at x = 50 cm. The third charge of 4.0 µC is located at x = 15 cm. To calculate the magnitude of the electrostatic force on the third charge, we need to determine the distance between the third charge and each of the other charges.
The distance between the third charge and the +54 µC charge is 15 cm (since they are both on the x-axis at the respective positions). Similarly, the distance between the third charge and the -38 µC charge is 35 cm (50 cm - 15 cm). Now, we can apply Coulomb's law to calculate the electrostatic force between the third charge and each of the other charges.
Using the equation F = k * |q1 * q2| / r^2, where k is the electrostatic constant (approximately 9 x 10^9 Nm^2/C^2), q1 is the charge of the third charge (4.0 µC), q2 is the charge of the other charge, and r is the distance between the charges, we can calculate the magnitude of the electrostatic force on the third charge.
Substituting the values, we have F1 = (9 x 10^9 Nm^2/C^2) * |(4.0 µC) * (54 µC)| / (0.15 m)^2, where F1 represents the force between the third charge and the +54 µC charge. Similarly, we have F2 = (9 x 10^9 Nm^2/C^2) * |(4.0 µC) * (-38 µC)| / (0.35 m)^2, where F2 represents the force between the third charge and the -38 µC charge.
Finally, we can calculate the magnitude of the electrostatic force on the third charge by summing up the forces from each charge: F_total = F1 + F2.
Performing the calculations will provide the numerical value of the magnitude of the electrostatic force on the third charge in whole numbers.
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Question 5: Consider a long tube (Dube - 10 mm) with air flow (Vlowe=0.1 m/s). Aerosol particles (diameter Dp = 2 µm and settling velocity 0.1 mm/s (a) Verify what kind of airflow (laminar or turbulent) in the tube? (b) Verify what kind of particle motion (laminar or turbulent) while settling in the tube? (c) What is the minimum length of the tube need for all particles not to pass out the tube?
(a) Reynolds number is less than 2300, hence the airflow is laminar.
(b) Reynolds number is less than 1, the settling of the particles in the tube is laminar.
(c) The minimum length of the tube needed for all particles not to pass out the tube is 0.69 mm.
(a) Flow of air is laminar. To verify this:
Reynolds number (Re) = Vd/v (where V = velocity of fluid, d = diameter of the tube, v = kinematic viscosity of the fluid)
Re = (0.1 × 2 × 10^-6) / (1.5 × 10^-5)
= 1.33
Since Reynolds number is less than 2300, hence the airflow is laminar.
(b) The particle motion in the tube is laminar since the flow is laminar. Settling particles are affected by the gravitational force, which is a body force, and the viscous drag force, which is a surface force.
When the particle's Reynolds number is less than 1, it is said to be in the Stokes' settling regime, and the drag force is proportional to the settling velocity.
Dp = 2 µm
settling velocity = 0.1 mm/s.
The Reynolds number of the particles can be calculated as follows:
Rep = (ρpDpVp)/μ
= (1.2 kg/m³)(2 × 10⁻⁶ m)(0.1 mm/s)/(1.8 × 10⁻⁵ Pa·s)
≈ 0.13
Since the Reynolds number is less than 1, the settling of the particles in the tube is laminar.
(c) The particle will not pass out of the tube if it reaches the bottom of the tube without any further settling. Therefore, the settling time of the particle should be equal to the time required for the particle to reach the bottom of the tube.
Settling time, t = L / v
The particle settles at 0.1 mm/s, hence the time taken to settle through the length L is L/0.1 mm/s
Therefore, the minimum length L of the tube required is:
L = settling time × settling velocity
= t × v
= 6.9 × 10^-5 × 0.1 mm/s
= 0.69 mm
Total length of the tube should be more than 0.69 mm so that all the particles settle down before exiting the tube. So, the minimum length of the tube needed for all particles not to pass out the tube is 0.69 mm.
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A proton (q=+e, m-u), a deuteron (q=+e, m-2u), and an alpha particle (q m-4u) all having the same kinetic energy enter a region of uniform magnetic field of them are moving perpendicular to the magnetic field, what is the ratio of: a) the radius ra of the deuteron path to the radius rp of the proton path and b) the radius ra of the alpha particle path to rp?
a)The ratio of the radius of the deuteron path to the radius of the proton path is 2:1. b) the ratio of the radius of the alpha particle path to the radius of the proton path is also 2:1. The radius of the circular path followed by a charged particle in a uniform magnetic field can be determined using the equation: r = (m * v) / (q * B).
where: r is the radius of the path, m is the mass of the particle,v is the velocity of the particle, q is the charge of the particle, B is the magnetic field strength.In this case, we have three particles: a proton, a deuteron, and an alpha particle. The kinetic energy of each particle is the same, but their masses and charges differ. Let's denote the radius of the deuteron path as rd, the radius of the proton path as rp, and the radius of the alpha particle path as ra.
a) Ratio of the radius of the deuteron path to the radius of the proton path (rd/rp): To find this ratio, we need to compare the mass and charge values for the deuteron and proton:
- Deuteron (D): q = +e, m = 2u
- Proton (P): q = +e, m = u
Using the equation for the radius of the path, we can calculate the ratio:
(rd/rp) = ((m_D * v) / (q_D * B)) / ((m_P * v) / (q_P * B))
(rd/rp) = (2u * v) / (u * v)
(rd/rp) = 2/1
(rd/rp) = 2
Therefore, the ratio of the radius of the deuteron path to the radius of the proton path is 2:1.
b) Ratio of the radius of the alpha particle path to the radius of the proton path (ra/rp):
To find this ratio, we compare the mass and charge values for the alpha particle and proton:
- Alpha particle (α): q = +2e, m = 4u
- Proton (P): q = +e, m = u
Using the equation for the radius of the path, we can calculate the ratio:
(ra/rp) = ((m_α * v) / (q_α * B)) / ((m_P * v) / (q_P * B))
(ra/rp) = (4u * v) / (u * 2v)
(ra/rp) = 4/2
(ra/rp) = 2
Therefore, the ratio of the radius of the alpha particle path to the radius of the proton path is also 2:1.
In conclusion:
a) The ratio of the radius of the deuteron path to the radius of the proton path is 2:1.
b) The ratio of the radius of the alpha particle path to the radius of the proton path is also 2:1.
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If a have a 30 g of paraffin wax, how much will expand in 10 seconds and how long will it take to fully expand?
note: the expansion rate of paraffin wax is 15% percent, activation temperature 17C , melting point 50 C
The paraffin wax will expand by approximately 4.5 grams in 10 seconds, and it will take approximately 1 hour and 40 minutes to fully expand.
Paraffin wax expands when heated due to the phase change from solid to liquid. Given that the activation temperature of the paraffin wax is 17°C and its melting point is 50°C, we can calculate the expansion rate.
Calculate the amount of expansion in 10 seconds.
The expansion rate of paraffin wax is 15%. So, if we have 30 grams of paraffin wax, the expansion in 10 seconds can be calculated as follows:
Expansion in 10 seconds = 15% of 30 grams = (15/100) * 30 grams = 4.5 grams.
Calculate the time required for full expansion.
To determine the time required for the paraffin wax to fully expand, we need to consider the rate at which it expands. Since we know the expansion rate and the amount of wax, we can calculate the time as follows:
Total expansion = 15% of 30 grams = (15/100) * 30 grams = 4.5 grams.
To fully expand from its solid state to liquid, the paraffin wax needs to go through the entire phase change process, which takes time. Unfortunately, the provided information does not specify the specific rate of expansion or the time required for the paraffin wax to reach its melting point.
In general, the time required for full expansion depends on various factors such as the amount of wax, the rate of heating, the surroundings, and the thermal conductivity. Therefore, without additional information, it is not possible to determine the exact time required for the paraffin wax to fully expand.
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REMARKS The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems. Enough uranium deposits are known so as to provide the world's current energy requirements for a few hundred years. Breeder reactor technology can greatly extend those reserves. QUESTION Estimate the average mass of 235
U needed to provide power for the average American family for one year. kg PRACTICE IT Use the worked example above to help you solve this problem. (a) Calculate the total energy released if 1.02 kg of 235
U undergoes fission, taking the disintegration energy per event to be Q=208MeV. MeV (b) How many kilograms of 235
U would be needed to satisfy the world's annual energy consumption (about 4.0×10 20
J )? kg EXERCISE HINTS: GETTING STARTED I I'M STUCK! How long can 1.02 kg of uranium-235 keep a 75 watt lightbulb burning if all its released energy is converted to electrical energy? t= years
The average mass of 235U needed to provide power for the average American family for one year is 1.15 x 10^-6 kg.
The amount of joules used in one year by the average American family is around 3.75 x 10^7 J. The energy that would be released if 1.02 kg of 235U undergoes fission is 3.24 x 10^13 J. Therefore, to produce the amount of energy needed for the average American family, 3.75 x 10^7 J ÷ 3.24 x 10^13 J/kg = 1.15 x 10^-6 kg of 235U is needed.
So, the average mass of 235U needed to provide power for the average American family for one year is 1.15 x 10^-6 kg. The calculation implicitly assumes perfect conversion to usable power, which is never the case in real systems. Enough uranium deposits are known so as to provide the world's current energy requirements for a few hundred years. Breeder reactor technology can greatly extend those reserves.
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A man holds a 2kg watermelon above his head 1.8m above the ground. He holds the watermelon steady so it is not moving. How much work is done by the man as he is holding the watermelon?
The man does approximately 35.28 Joules of work while holding the watermelon steady above his head.
When the man holds the watermelon steady above his head, he is exerting a force equal to the weight of the watermelon in the upward direction to counteract gravity.
The work done by the man can be calculated using the formula:
Work = Force × Distance × cosθ
Where:
Force is the upward force exerted by the man (equal to the weight of the watermelon),
Distance is the vertical distance the watermelon is lifted (1.8 m),
θ is the angle between the force and the displacement vectors (which is 0 degrees in this case, since the force and displacement are in the same direction).
Mass of the watermelon (m) = 2 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Distance (d) = 1.8 m
Weight of the watermelon (Force) = mass × gravity
Force = 2 kg × 9.8 m/s^2
Force = 19.6 N
Now we can calculate the work done by the man:
Work = Force × Distance × cosθ
Work = 19.6 N × 1.8 m × cos(0°)
Work = 19.6 N × 1.8 m × 1
Work = 35.28 Joules
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Two objects, A and B, start from rest. Object A starts with acceleration 1.6 m/s^2 and 4.0 seconds later after A, object B starts in the same direction with acceleration 3.4 m/s^2. How long will it take for object B to reach object A from the moment when A started to accelerate?
A car moving with over-speed limit constant speed 31.8 m/s passes a police car at rest. The police car immediately takes off in pursuit, accelerating with 9.6 m/s^2. How far from initial point police car will reach the speeder?
It will take approximately 2.747 seconds for Object B to reach Object A from the moment when Object A started to accelerate.
To find the time it takes for Object B to reach Object A, we need to consider the time it takes for Object A to reach its final velocity. Given that Object A starts from rest and has an acceleration of 1.6 m/s^2, it will take 4.0 seconds for Object A to reach its final velocity. During this time, Object A will have traveled a distance of (1/2) * (1.6 m/s^2) * (4.0 s)^2 = 12.8 meters.After the 4.0-second mark, Object B starts accelerating with an acceleration of 3.4 m/s^2. To determine the time it takes for Object B to reach Object A, we can use the equation of motion:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since Object B starts from rest, the equation simplifies to:
distance = (1/2) * acceleration * time^2
Substituting the known values, we have:
12.8 meters = (1/2) * 3.4 m/s^2 * time^2
Solving for time, we find:
time^2 = (12.8 meters) / (1/2 * 3.4 m/s^2) = 7.529 seconds^2
Taking the square root of both sides, we get: time ≈ 2.747 seconds
Therefore, it will take approximately 2.747 seconds for Object B to reach Object A from the moment when Object A started to accelerate.
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The distance between two planets A and B is 8 light years. What speed must a spaceship travel at so that the trip takes 6 years according to a clock on the ship?
The spaceship must travel at approximately 0.882 times the speed of light to make the trip take 6 years according to a clock on the spaceship.
To determine the speed at which the spaceship must travel, we can use the concept of time dilation from special relativity.
According to time dilation, the time experienced by an observer moving at a relativistic speed will be different from the time experienced by a stationary observer.
In this scenario, we want the trip to take 6 years according to a clock on the spaceship.
Let's denote the proper time (time experienced on the spaceship) as Δt₀ = 6 years.
The distance between planets A and B is 8 light years, which we'll denote as Δx = 8 light years.
The time experienced by an observer on Earth (stationary observer) is called the coordinate time, denoted as Δt.
Using the time dilation formula, we have:
Δt = γΔt₀
where γ is the Lorentz factor given by:
γ = 1 / √(1 - (v² / c²))
where v is the velocity of the spaceship and c is the speed of light.
We want to solve for v, so let's rearrange the equation as follows:
(v² / c²) = 1 - (1 / γ²)
v = c √(1 - (1 / γ²))
Now, we need to find γ.
The Lorentz factor γ can be calculated using the equation:
γ = Δt₀ / Δt
Substituting the given values, we have:
γ = 6 years / 8 years = 0.75
Now we can substitute γ into the equation for v:
v = c √(1 - (1 / γ²))
v = c √(1 - (1 / 0.75²))
v = c √(1 - (1 / 0.5625))
v = c √(1 - 1.7778)
v = c √(-0.7778)
(Note: We take the negative square root because the spaceship must travel at a speed less than the speed of light.)
v = c √(0.7778)
v ≈ 0.882 c
Therefore, the spaceship must travel at approximately 0.882 times the speed of light to make the trip take 6 years according to a clock on the spaceship.
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Electroncoration Part A Wandectron is accelerated from rest through a potential difference of 9.9 kV, what is the magnitude (absolute value) of the change in potential energi
When an electron is accelerated from rest through a potential difference of 9.9 kV, its resulting speed is approximately 5.9 x 10⁷ m/s.
The resulting speed of an electron accelerated through a potential difference can be calculated using the formula [tex]v = \sqrt{(2qV/m)}[/tex], where v is the speed, q is the charge of the electron, V is the potential difference, and m is the mass of the electron.
In this case, the charge of the electron (q) is [tex]1.60 \times 10^{-19} C[/tex], and the potential difference (V) is 9.9 kV, which can be converted to volts by multiplying by 1000. The mass of the electron (m) is [tex]9.11 \times 10^{-31} kg[/tex].
Plugging these values into the formula, we get [tex]v = \sqrt{(\frac {2 \times 1.60 \times 10^{-19} C \times 9900 V}{9.11 \times 10^{-31} kg}}[/tex]. Evaluating this expression gives us v ≈ 5.9 x 10⁷ m/s.
Therefore, the resulting speed of the electron accelerated through a potential difference of 9.9 kV is approximately 5.9 x 10⁷ m/s.
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The complete question is:
If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed? [tex](e = 1.60 \times 10{-19} C, k= 8.99 \times 10^9 N \cdot m^2/C^2, m_{el} = 9.11 \times 10^{-31} kg)[/tex]
A. 5.9 x 10⁷ m/s B. 2.9 x 10⁷ m/s C. 4.9 x 10⁷ m/s D. 3.9 x 10⁷ m/s
A stationary object in a flow of speed 37 m/s produces a drag force of 15 N. The flow speed is then changed to 25 m/s. What will be the drag force if the Drag Coefficient and density are assumed constant? Give your answer in Newtons to 2 decimal places.
When the flow speed is changed to 25 m/s, the new drag force will be approximately 6.70 N. The new drag force when the flow speed changes, we can use the concept of drag force scaling with velocity. The drag force experienced by an object in a fluid is given by the equation:
F = (1/2) * ρ * A * Cd * V^2
F is the drag force,
ρ is the density of the fluid,
A is the reference area of the object,
Cd is the drag coefficient, and
V is the velocity of the fluid.
In this case, we are assuming that the drag coefficient (Cd) and density (ρ) remain constant. Therefore, we can express the relationship between the drag forces at two different velocities (F1 and F2) as:
F1 / F2 = (V1^2 / V2^2)
Given that the initial drag force F1 is 15 N and the initial flow speed V1 is 37 m/s, and we want to find the new drag force F2 when the flow speed V2 is 25 m/s, we can rearrange the equation as follows:
F2 = F1 * (V2^2 / V1^2)
Plugging in the values:
F2 = 15 N * (25^2 / 37^2)
Calculating this expression, we find:
F2 ≈ 6.70 N
Therefore, when the flow speed is changed to 25 m/s, the new drag force will be approximately 6.70 N
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A snowmobile is originally at the point with position vector 31.1 m at 95.0° counterclockwise from the x axis, moving with velocity 4.73 m/s at 40.0°. It moves with constant acceleration 1.93 m/s2 at 200°. After 5.00 s have elapsed, find the following. (Express your answers in vector form.)
(a) its velocity vector
v= m/s
(b) its position vector
r= m
Given that, A snowmobile is originally at the point with position vector 31.1 m at 95.0° counterclockwise from the x axis, moving with velocity 4.73 m/s at 40.0°. It moves with constant acceleration 1.93 m/s2 at 200°.
Let's calculate velocity vector of the snowmobile after 5 seconds. Initial velocity of the snowmobile, u = 4.73 m/s at an angle of 40° with the horizontal. Time taken to reach the final velocity, t = 5 seconds. Acceleration, a = 1.93 m/s² at an angle of 200° with the horizontal. Using the second equation of motion, v = u + at. Here, v, u, and a are vectors. Let v⃗ be the velocity vector ,v⃗ = u⃗ + at⃗, v⃗ = 4.73(cos40°i^ + sin40°j^) + (1.93(cos200°i^ + sin200°j^))(i^,j^ are unit vectors in x and y directions respectively).By substituting the values, we get v⃗ = (4.73cos40° + 1.93cos200°)i^ + (4.73sin40° + 1.93sin200°)j^. So, the velocity vector is v⃗ = (3.27i^ + 5.37j^) m/s.
Now, let's calculate the position vector of the snowmobile after 5 seconds. Initial position vector of the snowmobile, r⃗ = 31.1(cos95°i^ + sin95°j^)(i^,j^ are unit vectors in x and y directions respectively)The final position vector, s⃗, can be calculated using the following equation. s⃗ = r⃗ + ut⃗ + 1/2 a t²t = 5.00 seconds, u = 4.73(cos40°i^ + sin40°j^), a = 1.93(cos200°i^ + sin200°j^)(i^,j^ are unit vectors in x and y directions respectively), s⃗ = 31.1(cos95°i^ + sin95°j^) + 4.73(cos40°i^ + sin40°j^) × 5.00 + 1/2 (1.93(cos200°i^ + sin200°j^)) × (5.00)². On solving we get,s⃗ = (-21.8i^ + 22.1j^) m.
Hence, the position vector of the snowmobile after 5.00 s is -21.8i^ + 22.1j^ m.
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Hubble's Law Hubble's law is the observation in physical cosmology that galaxies are moving away from Earth at speeds proportional to their distance. In other words, the farther they are, the faster they are moving away from Earth: v = H. r We are sending a spacecraft with constant velocity to a galaxy in the distance of r = 20Mpe from us, and it is getting further away from us with higher velocity as the universe expands! If the spacecraft reaches the galaxy after 7 billion years, determine the velocity of this spacecraft.
velocity of approximately 8.83 x 10^10 km/year. This means that the spacecraft's velocity will be higher than the calculated average velocity by the time it reaches the distant galaxy.
According to Hubble's law, galaxies are moving away from Earth at speeds proportional to their distance. If a spacecraft is sent to a galaxy located 20 million parsecs away and it takes 7 billion years to reach its destination, we can determine its velocity.
The velocity of the spacecraft can be calculated by dividing the distance traveled by the time taken. However, since the universe is expanding, the velocity of the spacecraft will increase due to the increasing separation between galaxies.
Hubble's law states that the velocity of a galaxy moving away from Earth is directly proportional to its distance. Mathematically, this can be expressed as v = H * r, where v is the velocity of the galaxy, H is the Hubble constant (representing the rate of the universe's expansion), and r is the distance between the galaxy and Earth.
In this case, the spacecraft is traveling to a galaxy located at a distance of r = 20 million parsecs. Given that it takes 7 billion years for the spacecraft to reach its destination, we can calculate its velocity.
First, we need to convert the distance from parsecs to a more standard unit, such as kilometers. Since 1 parsec is approximately equal to 3.09 x 10^13 kilometers, the distance can be calculated as 20 million parsecs * 3.09 x 10^13 km/parsec = 6.18 x 10^20 km.
Next, we divide the distance traveled (6.18 x 10^20 km) by the time taken (7 billion years or 7 x 10^9 years) to find the average velocity of the spacecraft. This gives us a velocity of approximately 8.83 x 10^10 km/year.
However, it's important to note that the spacecraft's velocity is not constant throughout its journey. Due to the expansion of the universe, the separation between galaxies increases over time.
Therefore, as the spacecraft travels, the velocity at which the galaxy it is heading towards is moving away from Earth also increases. This means that the spacecraft's velocity will be higher than the calculated average velocity by the time it reaches the distant galaxy.
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A man stands 10 m in front of a large plane mirror. How far must he walk before he is 5m away from his image? A. 10 cm B. 7.5 m C. 5 m D. 2.5 m
The man is 10m in front of a large plane mirror and we are to determine the distance he must walk before he is 5m away from his image.
The image formed by a plane mirror is a virtual image of the same size as the object and the distance between the object and its image is twice the distance of the object to the mirror.
The man’s distance to the mirror = 10m
Distance of man’s image to the mirror = 2 x 10 = 20m
Distance between man and his image = 20 - 10 = 10m To be 5m away from his image, he would need to walk half the distance between himself and the mirror.
Thus, he would need to walk a distance of 5m.
Option C 5 m is correct.
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Particle executes S.H.M. of period 12s and of amplitude 8cm. what time will it take to travel 4 cm from the extreme position
The time it takes for the particle to travel 4 cm from the extreme position is approximately 1.909 seconds (to three decimal places).
Explanation:
To find the time it takes for a particle executing Simple Harmonic Motion (SHM) to travel a certain distance from its extreme position, we can use the equation for displacement in SHM:
x(t) = A * cos(2πt/T)
Where:
x(t) is the displacement of the particle at time t.
A is the amplitude of the motion.
T is the period of the motion.
In this case, the amplitude is 8 cm and the period is 12 s.
To find the time it takes for the particle to travel 4 cm from the extreme position, we need to solve the equation x(t) = 4 cm for t. Let's do that:
4 = 8 * cos(2πt/12)
Divide both sides of the equation by 8:
0.5 = cos(2πt/12)
Now we need to find the inverse cosine (arccos) of both sides:
arccos(0.5) = 2πt/12
Using the inverse cosine function, we find that arccos(0.5) is equal to π/3 (or 60 degrees).
So we have:
π/3 = 2πt/12
To isolate t, we multiply both sides of the equation by 12 and divide by 2π:
t = (π/3) * (12 / 2π)
Simplifying the expression, we get:
t = 6/π
Therefore, the time it takes for the particle to travel 4 cm from the extreme position is approximately 1.909 seconds (to three decimal places).
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1. The south pole of a compass
a. points in the direction of Earth's magnetic field.
b. does not react to an electric current.
c. points toward a south magnetic pole.
d. points toward a north magnetic pole.
2. Electric current is a wire is
a. a flow of negative particles.
b. always clockwise if the charges are negative.
c. a flow of both positive and negative particles.
d. a flow of positive particles.
1. The south pole of a compass needle points toward a south magnetic pole.
2. Electric current in a wire is the flow of both positive and negative particles.
1. The south pole of a compass needle does not point towards the geographic south pole but actually points toward a south magnetic pole. This is because the Earth's magnetic field is generated by the movement of molten iron in its core. The magnetic field lines extend from the geographic north pole to the geographic south pole. Therefore, the south pole of a compass needle is attracted to the Earth's magnetic north pole, which acts as a magnetic south pole.
2. Electric current in a wire is the movement of electric charge. While historically, conventional current flow was defined as the movement of positive charges, it is now understood that electric current consists of the flow of both positive and negative charges. In most conductors, such as metals, the charge carriers are negatively charged electrons. However, there are also cases, such as in electrolytic solutions, where positive ions can contribute to the electric current. Hence, electric current in a wire can involve the movement of both positive and negative particles.
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