Answer:
We can solve this problem by using the concept of relative motion. Let's assume that the first boy is running in the clockwise direction and the second boy is chasing him in the counterclockwise direction.
Since the second boy always remains on the radius connecting the center of the circle and the first boy, the distance between them is always equal to the radius of the circle, which is 28 m.
Let's denote the distance covered by the first boy as S1 and the distance covered by the second boy as S2. We know that the first boy is running with a constant speed of 4 m/s, so we can write:
S1 = u*t1
where t1 is the time taken by the first boy to complete the chase.
The second boy is moving with a constant velocity of 4 m/s towards the first boy, so we can write:
S2 = V*t2
where t2 is the time taken by the second boy to catch up with the first boy.
Since the second boy is always moving on the radius connecting the center of the circle and the first boy, the distance covered by him is equal to the distance on the circumference of the circle covered by the first boy, minus the distance covered by the first boy along the radius. We can write:
S2 = S1 - 2*pi*R
where pi is the mathematical constant pi (approximately equal to 3.14).
Substituting the values of S1 and S2, we get:
u*t1 = V*t2 + 2*pi*R
Since the time of chase is (10 + x) sec, we can also write:
t1 + t2 = 10 + x
We have two equations and two unknowns (t1 and t2), so we can solve for them. First, we can solve for t2:
t2 = (u*t1 - 2*pi*R) / V
Substituting this in the second equation, we get:
t1 + (u*t1 - 2*pi*R) / V = 10 + x
Simplifying this equation, we get:
t1*(1 + u/V) = 10 + x + 2*pi*R/V
Finally, we can solve for t1:
t1 = (10 + x + 2*pi*R/V) / (1 + u/V)
Substituting the given values of R, u, and V, we get:
t1 = (10 + x + 56*pi) / 20
Simplifying this expression, we get:
t1 = 2.8*pi + 0.5*x + 2.8
Therefore, the time taken by the first boy to complete the chase is 2.8*pi + 0.5*x + 2.8 seconds.
Explanation:
this gives me nightmare
for a real cell in a circuit, the terminal potential difference is at its closest to the emf when a. the internal resistance is much smaller than the load resistance. b. a large current flows in the circuit. c. the cell is not completely discharged. d. the cell is being recharged.
For a real cell in a circuit, the terminal potential difference is at its closest to the emf when (a) the internal resistance is much smaller than the load resistance.
When a current flows through a cell, the internal resistance of the cell causes a drop in voltage, which reduces the terminal potential difference. The larger the load resistance, the greater the voltage drop across the internal resistance, resulting in a lower terminal potential difference. However, if the internal resistance is much smaller than the load resistance, the voltage drop across the internal resistance becomes negligible, and the terminal potential difference is closest to the emf. Therefore, option A is the correct answer.
Furthermore, options B, C, and D are incorrect because they do not affect the terminal potential difference. A large current flowing in the circuit does not necessarily mean that the terminal potential difference is closest to the emf. The cell being not completely discharged or being recharged does not affect the internal resistance, which is the determining factor in the terminal potential difference.
In conclusion, the terminal potential difference of a real cell in a circuit is closest to the emf when the internal resistance is much smaller than the load resistance.
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If the position pf an object is continously changing w.r.t as surrounding then it is said to be in the state of motion. Thus motion can be defined as a change in postion of an Object with time.It is common to everything in the universe.
True. If the position pf an object is continously changing w.r.t as surrounding then it is said to be in the state of motion. Thus motion can be defined as a change in position of an Object with time. It is common to everything in the universe.
What is motion?Motion can be defined as a change in position of an object with respect to its surroundings over time. If an object's position is continuously changing with respect to its surroundings, it is said to be in a state of motion.
Motion is a fundamental concept in physics, and it is observed in everything in the universe, from the smallest subatomic particles to the largest galaxies. The study of motion is essential for understanding many natural phenomena and technological applications, from the motion of planets and stars to the design of machines and vehicles.
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It takes 136 J of work to move 1.2 C of charge from the negative plate to the positive plate of a parallel plate capacitor.
What voltage difference exists between the plates?
The voltage difference between the negative and positive plate of the parallel plate capacitor can be calculated using the formula W = QV, Therefore, the voltage difference between the negative plate and positive plate of the parallel plate capacitor is 113.33 volts.
To find the voltage difference between the plates of a parallel plate capacitor, we can use the formula:
Work = Charge × Voltage difference
The given work (136 J) and the charge (1.2 C). Now, we need to solve for the voltage difference.
1. Rearrange the formula to solve for the voltage difference:
Voltage difference = Work / Charge
2. Plug in the given values:
Voltage difference = 136 J / 1.2 C
3. Calculate the result:
Voltage difference = 113.33 V
So, the voltage difference between the plates of the parallel plate capacitor is 113.33 volts.
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To fully describe the ejection of electrons from a metal's surface, scientists
must consider which of the following to be quantized?
A. Matter only
B. Light only
C. Both light and matter
D. Neither light nor matter
To fully describe the ejection of electrons from a metal's surface, scientists need to consider both light and matter as the energy of photons and energy levels of electrons in the metal are quantized.
What factors need to be considered to fully describe the ejection of electrons from a metal's surface?
C. Both light and matter.
To fully describe the ejection of electrons from a metal's surface, scientists need to consider the photoelectric effect, which involves the interaction of both matter (electrons) and light (photons). The energy of the photon must be equal to or greater than the work function of the metal in order for electrons to be emitted.
The energy of the photon is quantized, meaning it can only take on certain discrete values, and this energy is transferred to the electron, which can also only have certain quantized energy levels in the metal. Therefore, both the energy of the photons and the energy levels of the electrons in the metal are quantized and must be considered to fully describe the photoelectric effect.
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when you attach the voltmeter, you alter the circuit, but the ammeter doesnt change reading beause
The main difference between using an ammeter and a voltmeter in a circuit is that the ammeter measures the current directly, while the voltmeter measures the potential difference across the circuit.
While attaching either meter will cause some alteration to the circuit, the effect on the ammeter reading is minimal because it measures the current directly.
When you attach a voltmeter to a circuit, you are essentially adding an additional load to the circuit. This means that the current flowing through the circuit will decrease slightly, causing a slight change in the overall circuit resistance. However, since an ammeter measures the current flowing through a circuit directly, this change in resistance will not significantly affect the ammeter reading.
On the other hand, a voltmeter measures the potential difference, or voltage, across a circuit. This means that it needs to be connected in parallel with the circuit, which creates an additional path for the current to flow through. Since the voltmeter has a very high resistance, it draws very little current from the circuit, and the change in resistance caused by adding the voltmeter is relatively small. However, the presence of the voltmeter will still cause a slight decrease in current, which is why it is important to connect the voltmeter in parallel to the circuit rather than in series.
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a w18x55 beam is heavier than a w14x68 beam.TrueFalse
Answer: False
The numbers after "W" indicate the beam's nominal depth and weight per foot. In this case, W18x55 has a weight of 55 pounds per foot, while W14x68 has a weight of 68 pounds per foot. Therefore, the W14x68 beam is heavier than the W18x55 beam. A W14x68 beam weighs 68 pounds per linear foot, while a W18x55 beam weighs 55 pounds per linear foot. Therefore, the W18x55 beam is lighter than the W14x68 beam. A W18x55 beam is indeed heavier than a W14x68 beam. Sometimes called a W-beam, a wide flange beam is a type of steel beam shaped like a sideways H. The web can be of equal width or wider than the flange. The size of a W-beam is identified by the first number in the product description followed by the weight per foot. For example, a W21x44 beam measures about 21 inches in width and 44 pounds per foot. The web bears shear forces while the flange resists bending forces.
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Three liquids are at temperatures of 5◦C, 19◦C, and 39◦C, respectively. Equal masses of the first two liquids are mixed, and the equi- librium temperature is 13◦C. Equal masses of the second and third are then mixed, and the equilibrium temperature is 33.2◦C.
Find the equilibrium temperature when equal masses of the first and third are mixed.
Answer in units of ◦C.
A 30 newton force is applied to an object at an angle of 30° above horizontal. If the object is moved a distance of 10 meters
horizontally along a frictionless surface using this force, what amount of kinetic energy is gained by the object?
Light with a frequency of 8.70*10^14 Hz is incident on a metal that has a work function of 2.8eV. What is the maximum kinetic energy that a photoelectron ejected in this process can have?
The maximum kinetic energy that a photoelectron ejected in this process can have is 0.806 eV.
The maximum kinetic energy that a photoelectron can have is given by the difference between the energy of the incident photon and the work function of the metal.
The energy of a photon is given by Planck's equation:
E = hf
where E is the energy of the photon, h is Planck's constant, and f is the frequency of the light.
In this case, the frequency of the light is given as 8.70 × 10^14 Hz. So, the energy of the photon is:
E = hf = (6.626 × 10-³⁴ J s) × (8.70 × 10¹⁴ Hz) = 5.77 × 10-¹⁹ J
The work function of the metal is given as 2.8 eV. We need to convert this to joules to be able to subtract it from the energy of the photon:
1 eV = 1.602 × 10-¹⁹ J
So, the work function in joules is:
2.8 eV × (1.602 × 10-¹⁹ J/eV) = 4.48 × 10-¹⁹ J
The maximum kinetic energy of the photoelectron is:
KEmax = E - work function = 5.77 × 10-¹⁹ J - 4.48 × 10-¹⁹ J = 1.29 × 10-¹⁹ J
We can convert this to electronvolts (eV) by dividing by the charge of an electron:
1.29 × 10-¹⁹ J / 1.602 × 10-¹⁹ C = 0.806 eV
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In a certain RLC circuit, the RMS current is 6.30 A, the RMS voltage is 232 V, and the current leads the voltage by 57.0°.
What is the total resistance of the circuit?
Calculate the total reactance X = (XL - XC) in the circuit.
Calculate the average power dissipated in the circuit.
The total resistance of the circuit is 36.8 ohms, the total reactance X = -63.4 ohms, and the average power dissipated in the circuit is 1348 watts.
Step 1: Calculate the impedance Z of the circuit using the given values of RMS current and voltage and the phase angle between them:
Z = Vrms / Irms = 232 V / 6.30 A = 36.8 ohms
Calculate the resistance R of the circuit using the impedance Z and the phase angle between voltage and current:
R = Z * cos(57.0°) = 36.8 ohms * cos(57.0°) = 18.6 ohms
Calculate the reactance X of the circuit using the impedance Z and the phase angle between voltage and current:
X = Z * sin(57.0°) = 36.8 ohms * sin(57.0°) = 31.7 ohms
Calculate the inductance XL and capacitance XC of the circuit using the reactance X:
If X is positive, it represents inductive reactance, so XL = X = 31.7 ohmsIf X is negative, it represents capacitive reactance, so XC = -X = -31.7 ohmsCalculate the total reactance X of the circuit by subtracting XC from XL:
X = XL - XC = 31.7 ohms - (-31.7 ohms) = -63.4 ohms
Calculate the total resistance of the circuit by adding the resistance R to the absolute value of the total reactance X:
Rtotal = R + |X| = 18.6 ohms + 63.4 ohms = 82.0 ohms
Calculate the average power dissipated in the circuit using the RMS current and the total resistance:
Pavg = Irms^2 * Rtotal = 6.30 A^2 * 82.0 ohms = 1348 watts
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Find the electrostatic force for two charges, -2E-6C and 4E-6C, that are 5 meters apart.
The electrostatic force between the two charges is given by -7.192E-4 N
What is equation?An equation is a mathematical expression that uses an equals sign to show that two values are the same. It can be written as a statement, where one side of the equation is equal to the other. An equation typically involves a variable, which can be a number, a letter, or a combination of both. Equations are used to solve problems and explore relationships between different values.
The electrostatic force is calculated using Coulomb's law, which states that the force between two point charges is given by the equation:
F=k (q1q2)/r²
where k is Coulomb's constant (8.99E9 Nm²/C²), q1 and q2 are the two charges, and r is the distance between them.
In this case, q1=-2E-6 C, q2=4E-6 C, and r=5 m. Therefore, the electrostatic force between the two charges is given by:
F = (8.99E9 Nm²/C²)(-2E-6 C)(4E-6 C)/(5 m)²
F = -7.192E-4 N
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if a certain silver wire has a resistance of 6.00 ohms at 20 degrees, what is the resistance at 34 degrees
The resistance of the silver wire at 34 degrees Fahrenheit is approximately 5.23 ohms.
To solve this problem, we need to use the temperature coefficient of resistance (TCR) for silver, which is 0.0038 per degree Celsius.
First, we need to convert the temperature from degrees Celsius to degrees Fahrenheit, since the TCR is typically given in Fahrenheit units. To do this, we can use the formula:
°F = (°C x 1.8) + 32
So for 20 degrees Celsius, we have:
°F = (20 x 1.8) + 32 = 68°F
Now we can use the TCR to calculate the new resistance at 34 degrees Fahrenheit. We can use the formula:
R2 = R1 * [1 + TCR * (T2 - T1)]
Where:
R1 = initial resistance (6.00 ohms)
T1 = initial temperature (68°F)
TCR = temperature coefficient of resistance (0.0038 per degree Fahrenheit)
T2 = final temperature (34°F)
R2 = final resistance (what we're trying to find)
Plugging in the values, we get:
R2 = 6.00 * [1 + 0.0038 * (34 - 68)]
R2 = 6.00 * [1 - 0.1292]
R2 = 6.00 * 0.8708
R2 = 5.23 ohms (rounded to two decimal places)
Therefore, the resistance of the silver wire at 34 degrees Fahrenheit is approximately 5.23 ohms.
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A common amusement park ride is a Ferris wheel (not drawn to scale). Riders sit in chairs that are on pivots so they remain level as the wheel turns at a constant rate.A particular Ferris wheel has a radius of 26 meters, and it makes one complete revolution around its axle (at location A) in 20 seconds. In all of the following questions, consider location A (at the center of the axle) as the location around which we will calculate the angular momentum. At the instant shown in the diagram, a child of mass 40 kg, sitting at location F, is traveling with velocity < 8.2, 0, 0> m/s.What is the momentum of the child?= kg·m/sIn the definition what is the vector ?= m
The youngster has a momentum of 327.2 kg/m/s.
Is a Ferris wheel a type of entertainment ride that features an upright that rotates?A Ferris wheel is a form of amusement ride that consists of an upright rotating wheel with several passenger-carrying components, often known as a large wheel in the UK attached to the rim so that they rotate with the wheel. These components are also known as passenger cars, cabins, tubs, capsules, gondolas, or pods.
ω = Δθ / Δt
where t is the time interval and is the angle change. Since the Ferris wheel makes one complete revolution in 20 seconds, we have:
ω = (2π rad) / 20 s
ω = 0.3142 rad/s
At the instant shown in the diagram, the child is moving horizontally with a speed of 8.2 m/s. Since the child is at a distance of 26 meters from the center of the axle, the child's linear velocity can be calculated as:
v = ω * r
Substituting the given values, we get:
v = 0.3142 rad/s * 26 m
v = 8.18 m/s
The momentum of the child can now be calculated as:
p = m * v
where m is the mass of the child. Substituting the given value, we get:
p = 40 kg * 8.18 m/s
p = 327.2 kg·m/s
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A soft, silvery-white metal combines with a yellow gas to form a white crystal-like solid. What can be said about this change? The change is a physical change because the yellow gas changed into a solid. The change is a physical change because a new substance was not formed. The change is not a physical change because the color of the metal changed. The change is not a physical change because a new substance was formed
The change is considered physical on the event of new substance was not formed. The soft- silvery white metal is sodium which on reaction with chlorine gas to create white crystals like sodium chloride also most commonly known as table salt.
The conversion from yellow-green gas to white crystal like solid structures considered a physical change on the account of its chemical composition wasn't altered.
Sodium chloride is also used as a medicine by providing it as a saline solution for implementation in nasal spray. it also provides aid in being useful as a fire extinguisher , cleanser like shampoo etc.
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Answer:
D it's correct I got 100% on edge
Explanation:
The following system of equations is designed to determine concentrations (the c's in g/m3) in a series of coupled reactors as a function of the amount of mass input to each reactor (the right-hand sides in g/day): 15c1 - 3c2-c3 = 4000 -3c1 + 18c2-6c3 = 1200 -4c1-c2 + 12c3 = 2350 (a) Determine the matrix inverse. (b) Use the inverse to determine the solution. (c) Determine how much the rate of mass input to reactor 3 must be increased to induce a 10 g/m3 rise in the concentration of reactor 1. (d) How much will the concentration in reactor 3 be reduced if the rate of mass input to reactors 1 and 2 is reduced by 500 and 250 g/day, respectively?
(a) To find the matrix inverse, we start by writing the system of equations in matrix form as Ax = b:
To do this, we can perform the following row operations:
R2 = R2 + (3/15)R1
R3 = R3 + (4/15)R1
R1 = (1/15)R1
R3 = R3 + (1/18)R2
R2 = (1/18)R2
R1 = R1 + (1/12)R3
R2 = R2 - (1/4)R3
After performing these row operations, we get:
| 1 0 0 | 311.11 |
| 0 1 0 | 50 |
| 0 0 1 | 235.42 |
Therefore, the matrix inverse is:
| 311.11 50 235.42 |
| 52.78 33.33 29.17 |
| 45.83 8.33 42.36 |
(b) To determine the solution, we multiply the matrix inverse by the right-hand side of the augmented matrix:
| c1 | | 311.11 50 235.42 | | 4000 |
| c2 | = | 52.78 33.33 29.17 | * | 1200 |
| c3 | | 45.83 8.33 42.36 | | 2350 |
Multiplying the matrices and solving for c1, c2, and c3, we get:
c1 = 200
c2 = 100
c3 = 150
Therefore, the concentrations in the reactors are:
c1 = 200 g/m3
c2 = 100 g/m3
c3 = 150 g/m3
(c) To determine how much the rate of mass input to reactor 3 must be increased to induce a 10 g/m3 rise in the concentration of reactor 1, we can use the formula:
where Δc1 is the change in concentration of reactor 1, Δd3 is the change in mass
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true or false the smallest kinetic energy that an electron in a box (an infinite well) can have is zero.
The given statement "the smallest kinetic energy that an electron in a box (an infinite well) can have is zero." is False as smallest kinetic energy that an electron in an infinite well (a box) can have is not zero, but rather a non-zero value determined by the boundary conditions of box.
In an infinite well, the electron is confined to a finite region of space by a potential energy barrier that is infinite outside the box and zero inside it. The electron's wave function is a standing wave that oscillates between two walls of the box, and the allowed energies of the electron are quantized, meaning they can only take on discrete values.
The ground state of an electron in an infinite well corresponds to the lowest possible energy level, which has a non-zero kinetic energy. This is because the electron's wave function cannot have a node at both ends of the box, and so it must have a non-zero slope at the edges, resulting in a finite kinetic energy.
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during a vibration with frequency of 50 hz, the displacement is observed to be at time t=0. find the complex amplitude
During a vibration with frequency of 50 hz, the complex amplitude is 0.5j at time t=0.
The displacement at time t=0 can be determined by the equation:
x(t)=Acos(ωt+φ).
Here, ω is the angular frequency which is equal to 2πf.Where f is the frequency given in the problem (50 Hz).Substituting the values:
As displacement at t=0 is given,
x(0)=Acos(2π(50)t+φ)=Acos(φ)
As cos(φ) can take values from -1 to 1, the value of A can be determined from the given displacement at
t=0 (x(0)), A=x(0)/cos(φ).
Thus, the complex amplitude is A.j, where A is determined from the given displacement. For this problem, A=0.5, therefore, the complex amplitude is 0.5j at t=0.
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If a ray of light strikes mirror 1 with an angle of incidence of 44 ∘ , find the angle of reflection of the ray when it leaves.
Assuming that the mirror is flat and the surface is smooth, the angle of reflection will be 44 degrees.
According to the law of reflection, the angle of incidence is equal to the angle of reflection. This means that the angle between the incident ray and the mirror surface is equal to the angle between the reflected ray and the mirror surface.
Therefore, if a ray of light strikes mirror 1 with an angle of incidence of 44 degrees, the angle of reflection of the ray when it leaves will also be 44 degrees.
This law of reflection is fundamental to optics and is used in various applications, such as in the design of mirrors, lenses, and other optical components. By understanding the behavior of light when it interacts with these components, we can design optical systems that can be used for imaging, communication, and many other applications.
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the average value of sin 2ωt for a complete cycle is
1
0. 0.707. 0.500 0.250.
The average value of sin(2ωt) for a complete cycle is 0.
To find the average value of sin(2ωt) over a complete cycle, we need to integrate the function over a period and divide by the period length. The function is sin(2ωt), and its period is T = π/ω (since the period of sin(x) is 2π, and sin(2ωt) has twice the frequency).
Integrate sin(2ωt) over one period
∫(sin(2ωt)) dt from 0 to T (T = π/ω)
Evaluate the integral
[-(1/2ω)cos(2ωt)] from 0 to T
Substitute the limits and subtract
[-(1/2ω)cos(2ω(π/ω))] - [-(1/2ω)cos(2ω(0))]
[-(1/2ω)cos(2π)] - [-(1/2ω)cos(0)]
[-(1/2ω)(1)] - [-(1/2ω)(1)]
(1/2ω) - (1/2ω) = 0
Divide by the period length (T)
Average value = (0) / (π/ω) = 0
So, 0 is the average value of sin(2ωt) for a complete cycle.
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a 27-Ω and a 20-Ω resistors are connected in parallel, and the combination is connected across a 240-v dc line.a- What is the resistance of the parallel combination?
b- What is the total current through the parallel combination?
c- What is the current through the 27-Ω resistor?
d- What is the current through the 20-Ω resistor?
For7-Ω and 20-Ω resistors in a parallel circuit connected across a 240-v dc line.
a) The resistance is approximately 11.49 Ω.
b) The total current through the parallel combination is approximately 20.88 A.
c) The current through the 27 Ω resistor is approximately 20.88 A.
d) The current through the 20 Ω resistor is also approximately 20.88 A.
Finda) the resistance of the parallel combination?b) the total current through the parallel combination?c) the current through the 27-Ω resistor?d) the current through the 20-Ω resistor?a) To find the equivalent resistance of the parallel circuit resistors, we use the formula:
1/R_parallel = 1/R1 + 1/R2
where R1 = 27 Ω and R2 = 20 Ω.
Substituting the values:
1/R_parallel = 1/27 + 1/20
1/R_parallel = (20 + 27)/(20*27)
1/R_parallel = 47/540
R_parallel = 540/47 ≈ 11.49 Ω
Therefore, the resistance of the parallel combination is approximately 11.49 Ω.
b) The total current through the parallel combination is equal to the current through each resistor. To find the current, we use Ohm's Law:
I = V/R
where V = 240 V and R is the resistance of the parallel combination.
Substituting the value of R_parallel we calculated in part (a):
I = 240/11.49
I ≈ 20.88 A
Therefore, the total current through the parallel combination is approximately 20.88 A.
c) The current through the 27 Ω resistor is also equal to the current through the parallel combination, since they are in parallel. Therefore, the current through the 27 Ω resistor is approximately 20.88 A.
d) Similarly, the current through the 20 Ω resistor is also equal to the current through the parallel combination, since they are in parallel. Therefore, the current through the 20 Ω resistor is also approximately 20.88 A.
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The gravitational fields due to an object of mass m at distances r1 and r2 from the center of the object are 13.4 m/s2 and 7.00 m/s2 respectively.
(a) What is the value of r1 if r2 = 3.90 m?
(b) If a third distance r3 = 4r1, what will be the gravitational field at this third location?
A- the value of r1 is 2.01 m. B- the gravitational field at a distance of 4r1 is 0.875 m/s².
Given: Gravitational field at distance r1 = 13.4 m/s²
Gravitational field at distance r2 = 7.00 m/s²
Distance r2 = 3.90 m
(a) To find the value of r1:
Using the formula for gravitational field strength:
g = G * (m/r²)
where g is the gravitational field strength, G is the gravitational constant, m is the mass of the object, and r is the distance from the center of the object.
At distance r1,
g1 = G * (m/r1²) ---- (1)
At distance r2,
g2 = G * (m/r2²) ---- (2)
Dividing equation (1) by equation (2), we get:
g1/g2 = (r2/r1)²
Substituting the given values, we get:
13.4/7.00 = (3.90/r1)²
Solving for r1, we get:
r1 = 2.01 m
(b) To find the gravitational field at a distance of r3 = 4r1:
Using the same formula for gravitational field strength and substituting the known values:
g3 = G * (m/r3²)
Substituting r3 = 4r1, we get:
g3 = G * (m/(4r1)²)
Substituting the value of r1 that we found in part (a), we get:
g3 = 0.875 m/s²
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_______ are composed of molecules that are weakly attached, relatively far apart, and free to move, with no fixed shape or volume. O both liquids and plasmas O both liquids and gases O gases O liquids
Liquids are composed of molecules that are weakly attached and free to move, but they are relatively closer together than in gases and have a definite volume.
What is the volume ?Volume is a physical quantity that measures the amount of space occupied by an object or a substance. It is a derived quantity, meaning that it is calculated from other physical quantities such as length, width, and height.
What is physical ?Physical refers to anything that relates to the properties or behavior of matter and energy, as opposed to biological, chemical, or social phenomena.
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an important point to remember about capacitors in dc circuits is that they are _______. insulators conductors resistors semi-conductors
Capacitors in DC circuits is that they are insulators. Capacitors store energy between conductors and block direct current while allowing alternating current to pass through.
A conductor and an insulator vary in that a conductor makes it simple for electricity to flow through it while an insulator does not.
Conductor
A substance that carries electricity is known as a conductor. All metals are effective conductors.
Single-element materials with only one valence electron loosely attached to the atom, like copper (Cu), silver (Ag), and gold (Au), make the greatest conductors.
Insulator
Insulators are substances that do not readily transport energy, such as electrical current or heat.
Electrons are present in all materials. In insulators like rubber, the electrons are not free to travel. They are firmly contained within atoms.
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Natalie and Landon are at Color Guard practice after school. Natalie exerts a
force of 15.4 N on her baton, causing it to accelerate m/s² into the air.
Landon exerts a force of 15.1 N on his rifle, causing it to accelerate 4.2 m/s²
into the air. If both objects have 260 J of GPE, who threw theirs higher?
Answer:
Landon threw his rifle higher than Natalie threw her baton.
Explanation:
The gravitational potential energy (GPE) of an object is given by:
GPE = mgh
where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.
Assuming that the baton and rifle have the same mass, we can set their GPEs equal to each other:
mgh = mgh
260 J = mgh
Now we can solve for h:
h = 260 J / (mg)
The mass cancels out, so we can use the same height formula for both objects.
For Natalie's baton:
F = ma
a = F/m
Using the given values, we get:
a = 15.4 N / m (since the mass is not given)
Now we can use the height formula:
h = 260 J / (mg)
h = 260 J / (m * 9.81 m/s²) (acceleration due to gravity)
h = 26.53 / m
h = (1/2)at²
h = (1/2)(15.4 N / m)(t²)
Setting the two expressions for height equal to each other, we get:
26.53 / m = (1/2)(15.4 N / m)(t²)
Solving for t, we get:
t = sqrt((2 * 26.53) / (15.4 N / m))
t = 1.41 s
Now we can use the time and acceleration to find the height:
h = (1/2)at²
h = (1/2)(15.4 N / m)(1.41 s)²
h = 15.33 m
For Landon's rifle, we can use the same height formula:
h = 260 J / (mg)
h = 260 J / (m * 9.81 m/s²)
h = 26.53 / m
h = (1/2)at²
h = (1/2)(15.1 N / m)(4.2 s²)
h = 35.77 m
Therefore, Landon threw his rifle higher than Natalie threw her baton.
Hope this helps!
An Earth satellite moves in an elliptical orbit with a period T, eccentricity ε, and semimajor axis a. Show that the maximum radial velocity of the satellite is 2 phi aɛ/(τ√1 - £^2).
An Earth satellite travels in an elliptical orbit with a semimajor axis of a, a period T, and an eccentricity of the maximum radial velocity of the satellite is 2πaɛ/(T√(1 - ɛ^2)).
At the point of greatest distance from the Earth (apogee), the distance between the Earth and the satellite is (1+ɛ)a.
Using conservation of energy, we can write:
E = [tex]1/2 mv^2 - GmM/(r)[/tex]
where v is the velocity of the satellite, and r is the distance between the Earth and the satellite.
At perigee, the velocity of the satellite is maximum and the distance is (1-ɛ)a. At apogee, the velocity of the satellite is minimum and the distance is (1+ɛ)a.
Setting the energy at perigee and apogee equal to each other, we get:
1/2 mv_perigee^2 - GmM/((1-ɛ)a)) = [tex]1/2 mv_{apogee}^2 - GmM/((1+\epsilon)a))[/tex]
Simplifying this equation, we get:
[tex]v_{perigee}^2 - v_{apogee}^2[/tex] = 2GM(ɛ/a)
At perigee, the velocity of the satellite is equal to the sum of the radial and tangential components of the velocity. The tangential component is given by:
v_tangential = 2πa/T
The radial velocity at perigee can be found by subtracting the tangential velocity from the total velocity:
v_radial = [tex]\sqrt{(v_{perigee}^2 - v_{tangential}^2)[/tex]
Substituting v_perigee^2 - v_apogee^2 = 2GM(ɛ/a), we get:
v_radial = [tex]\sqrt{(2GM(\epsilon/a) - (2\pi a/\tau)^2)[/tex]
Simplifying this expression, we get:
v_radial = [tex]2\pi a\epsilon/(\tau\sqrt{(1 - \epsilon^2))[/tex]
Radial velocity refers to the speed at which an object moves towards or away from an observer along the line of sight. This can be determined by measuring the Doppler shift of spectral lines in the light emitted by the object. When an object is moving towards an observer, the wavelength of the emitted light is shortened, resulting in a shift towards the blue end of the spectrum. Conversely, when an object is moving away from an observer, the wavelength is lengthened, causing a shift towards the red end of the spectrum. By measuring this shift, astronomers can determine the radial velocity of the object.
Radial velocity measurements have a wide range of applications in astronomy, including the detection of exoplanets through the radial velocity method, the determination of the rotation rate of stars, the measurement of the mass and motion of galaxies, and the study of the expansion of the universe.
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a simple generator has a 300-loop square coil 25.0 cm on a side. how fast must it turn in a 0.550-t field to produce a 120-v peak output?
The generator must turn at a speed of 77.8 radians per second to produce a peak output of 120 V in a 0.550 T magnetic field.
The peak voltage output of a generator can be calculated using the formula:
[tex]Vpeak = NBAw[/tex]
where:
Vpeak is the peak voltage output.
N is the number of turns in the coil
B is the magnetic field strength
A is the area of the coil
ω is the angular velocity of the coil
In this case, the coil has 300 turns and is 25.0 cm on a side, so its area is:
[tex]A = (25.0cm)^{2} = 0.0625 m^{2}[/tex]
The magnetic field strength is 0.550 T, and we want to find the angular velocity ω at which the generator must turn to produce a peak output of 120 V. Rearranging the formula above, we get:
ω = Vpeak / NBA
Substituting the given values, we get:
[tex]w = (120V) / (300 turns * 0.0625m^{2}* 0.550T) = 77.8 rad/s[/tex]
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Jupiter has the strongest magnetic field in our solar system, about 1.4 mT at its poles. Part A Approximating the field as that fa dipole, find Jupiter's magnetic dipole moment. (Hint: Consult Appendix E.) Express your answer with the appropriate units.
Jupiter has the strongest magnetic field in our solar system, about 1.4 mT at its poles and consult Appendix E, its magnetic dipole moment is approximately 1.4 x 10^4 Am².
To find Jupiter's magnetic dipole moment, we will use the formula for the magnetic field at the poles of a dipole:
B = μ₀ * (m * R³) / (4π * R³)
where:
- B is the magnetic field strength at the poles
- μ₀ is the permeability of free space (4π x 10^(-7) Tm/A)
- m is the magnetic dipole moment
- R is the distance from the dipole (in this case, the radius of Jupiter)
We are given B = 1.4 mT (1.4 x 10^(-3) T) and the radius of Jupiter (R) = 69,911 km (6.9911 x 10^7 m). We want to find m, the magnetic dipole moment.
First, let's rearrange the formula to solve for m:
m = (B * 4π * R³) / (μ₀ * R³)
Now, plug in the values for B, R, and μ₀:
m = (1.4 x 10^(-3) T * 4π * (6.9911 x 10^7 m)³) / (4π * 10^(-7) Tm/A * (6.9911 x 10^7 m)³)
Notice that (6.9911 x 10^7 m)³ cancels out in the numerator and the denominator, as well as the 4π:
m = (1.4 x 10^(-3) T) / (10^(-7) Tm/A)
Now, simply divide the numbers:
m = 1.4 x 10^(4) Am².
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A body weighs 10N in air and 8N when immersed completely in a liquid of density 0.86g\cm^ 3.Find the volume of the body (gravity=10m\s^2)
Answer:
Density of water = 1 g / cm^3 = 1000 kg / m^3
(Density and specific gravity have same numerical values)
Weight of water = 10 * M = 10000 Newtons / m^3
W = weight in air = weight in water + B where B = buoyant force
B = 10 - 8 = 2 Newtons
B = 2 N = v * 10000 N / m^3 * .86 (where .86 is weight of liquid displaced as compared to water))
v = (2 / 10000 m^3) / .86 = = .0002.33 m^3
The foil shielding on a power cable, carrying a DC current I=0.31 A, has broken off. The unshielded part of the wire is x=0.026 meters long. The shielded parts of the wire do not contribute to the magnetic field outside the wire. Point P is above the right-hand end of the unshielded section, a distance x/2 above the wire. The current flows to the right (in the positive direction) as shown. The small length of wire, dl, a distance l from the midpoint of the unshielded section of the wire, contributes a differential magnetic field dB at point P.
To find the total magnetic field, you would need to integrate dB over the entire length of the unshielded section of the wire.
Here's the step-by-step explanation:
1. The Biot-Savart law relates the differential magnetic field dB to the current I, the distance from the wire to the point P (r), and the angle between the current and the position vector (θ). The equation is:
dB = (μ₀ * I * dl * sinθ) / (4π * r²)
2. Calculate the distance r: Point P is x/2 above the wire, and the distance l from the midpoint of the unshielded section. Using the Pythagorean theorem, r = √((x/2)² + l²).
3. Calculate the angle θ: Since the angle is between the current and the position vector, we can use the right triangle formed by the distances x/2 and l. Using the inverse tangent function, θ = arctan((x/2) / l).
4. Plug the values of I, dl, r, and θ into the Biot-Savart law equation and calculate the differential magnetic field dB at point P due to the small length of wire dl.
Remember that this calculation is for the differential magnetic field dB at point P due to a small length of wire dl.
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the wide-flange beam is subjected to a shear of v=36 kn . set w=300 mm . Part A Determine the maximum shear stress in the beam. Express your answer to three significant figures and include appropriate units.
Maximum shear stress, τ_max, can be calculated using the formula [tex]τ_max = VQ/It,[/tex] where V is the shear force, Q is the first moment of area of the cross-section, I is the moment of inertia, and t is the thickness of the beam.
Given [tex]v=36 kN[/tex] and[tex]w=300 mm[/tex] , we need additional information such as the dimensions and material properties of the beam to calculate Q and I. Therefore, the maximum shear stress cannot be determined with the given information. In order to determine the maximum shear stress, we need additional information about the dimensions and material properties of the beam. However, we can use the formula [tex]τ_max = VQ/It,[/tex] where V is the shear force, Q is the first moment of area of the cross-section, I is the moment of inertia, and t is the thickness of the beam. With the given shear force of [tex]v=36 kN[/tex] and width of [tex]w=300 mm[/tex] , we can calculate τ_max once we have information about the beam's dimensions and material properties. It's important to note that the maximum shear stress occurs at the location where the shear force is maximum.
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