The maximum current through an RLC circuit can be calculated using the following equation: I(max) = V/R, where V is the voltage applied across the circuit and R is the resistance of the circuit. Therefore, the answer is maximized.
An RLC circuit is an electrical circuit containing a resistor, an inductor, and a capacitor, which are the three most commonly used electronic components. When a sinusoidal voltage is applied to an RLC series circuit, an alternating current (AC) flows through it.
The current through an RLC circuit at resonance is maximized. Resonance can be described as the point at which the inductive reactance of a coil is equal to the capacitive reactance of a capacitor. At this point, the inductive reactance and capacitive reactance cancel out, resulting in a minimum impedance in the circuit and a maximum current flow.
The phase angle between the current and voltage in an RLC circuit at resonance is zero, indicating that they are in phase. At resonance, the RLC circuit's current is determined solely by the resistance of the circuit's resistor. The current in an RLC circuit at resonance is determined by the following equation:
I = V/R
Where, V is the voltage applied across the circuit, R is the resistance of the circuit, and I is the current flowing through the circuit. At resonance, the current through an RLC circuit is maximized.
To know more about voltage visit:
https://brainly.com/question/32002804
#SPJ11
A beam of electrons is accelerated from rest along the x-axis through a potential difference of 20.0 V. It is then directed at a single slit of width 1.00 x 10-4 m, and the width of the central maximum on a distant screen is measured to be Ay = 5.00x10-4 m. (a) Find the distance from the slit to the screen. [2] (b) What is the uncertainty Apy in the y-momentum of each electron striking this central maximum?
The distance from the slit to the screen is not provided in the given information, so it cannot be determined. The uncertainty in the y-momentum the central maximum is at least 2.65 × 10^-26 kg m/s.
B. Explanation:
(a) To find the distance from the slit to the screen, we can use the formula for the diffraction pattern from a single slit:
y = (λL) / (w)
where y is the width of the central maximum, λ is the de Broglie wavelength of the electrons, L is the distance from the slit to the screen, and w is the width of the slit.
We can rearrange the formula to solve for L:
L = (y * w) / λ
The de Broglie wavelength of an electron is given by the equation:
λ = h / p
where h is the Planck's constant (6.626 × 10^-34 J s) and p is the momentum of the electron.
The momentum of an electron can be calculated using the equation:
p = √(2mE)
where m is the mass of the electron (9.10938356 × 10^-31 kg) and E is the energy gained by the electron.
The energy gained by the electron can be calculated using the equation:
E = qV
where q is the charge of the electron (1.602 × 10^-19 C) and V is the potential difference through which the electrons are accelerated.
Substituting the given values:
E = [tex](1.602 ×*10^{-19} C) * (20.0 V) = 3.204 * 10^{-18} J[/tex]
Now we can calculate the momentum:
p = [tex]\sqrt{2} * (9.10938356 * 10^{-31 }kg) * (3.204 × 10^{-18 }J)) ≈ 4.777 * 10^{-23} kg m/s[/tex]
Substituting the values of y, w, and λ into the formula for L:
L = [tex]((5.00 ×*10^{-4 }m) * (1.00 * 10^{-4 }m)) / (4.777 ×*10^{-23 }kg m/s) = 1.047 * 10^{16} m[/tex]
Therefore, the distance from the slit to the screen is approximately 1.047 × 10^16 meters.
(b) The uncertainty in the y-momentum of each electron striking the central maximum, Apy, can be calculated using the uncertainty principle:
Apy * Ay ≥ h / (2Δx)
where Δx is the uncertainty in the position of the electron in the y-direction.
Since we are given the width of the central maximum Ay, we can take Δx to be half the width:
Δx = Ay / 2 = (5.00 × 10^-4 m) / 2 = 2.50 × 10^-4 m
Substituting the values into the uncertainty principle equation:
[tex]Apy \geq (5.00 * 10^{-4} m) ≥ (6.626 * 10^{-34 }J s) / (2 * (2.50 * 10^{-4} m))[/tex]
[tex]Apy \geq (6.626 * 10^{-34 }J s) / (2 * (2.50 * 10^{-4} m * 5.00 * 10^{-4} m))[/tex]
[tex]Apy \geq (6.626 * 10^{-34 }J s) / (2.50 * 10^{-8} m^2)[/tex]
[tex]Apy \geq 2.65 * 10^{-26} kg m/s[/tex]
Therefore, the uncertainty in the y-momentum of each electron striking the central maximum is at least 2.65 × 10^-26 kg m/s.
To learn more about diffraction click here:
brainly.com/question/12290582
#SPJ11
A beam of light reflects and refracts at point A on the interface between material 1 (n1 = 1.33) and material 2 (n2 = 1.66). The incident beam makes an angle of 40° with the interface. What is the angle of reflection at point A?
The angle of reflection at point A is 40°, which is equal to the angle of incidence.
When a beam of light encounters an interface between two different materials, it undergoes reflection and refraction. The angle of incidence, which is the angle between the incident beam and the normal to the interface, is equal to the angle of reflection, which is the angle between the reflected beam and the normal to the interface.
In this case, the incident beam makes an angle of 40° with the interface, so the angle of reflection at point A is also 40°. When light travels from one medium to another, it changes its direction due to the change in speed caused by the change in refractive index.
The law of reflection states that the angle of incidence is equal to the angle of reflection. This means that the angle at which the light ray strikes the interface is the same as the angle at which it bounces off the interface.
In this scenario, the incident beam of light strikes the interface between material 1 and material 2 at an angle of 40°. According to the law of reflection, the angle of reflection is equal to the angle of incidence, so the light ray will bounce off the interface at the same 40° angle with respect to the normal.
Learn more about reflection
brainly.com/question/30031394
#SPJ11
10 166 points ebook An ideal spring has a spring constant k 29.4 N/m. What is the amount of work that must be done to stretch the spring 0,660 m from its relaxed length?
The work done to stretch the spring by 0.660 m from its relaxed length is 6.38 J (approx).
Given: A spring has a spring constant k = 29.4 N/m and the spring is stretched by 0.660m from its relaxed length i.e initial length. We have to calculate the work that must be done to stretch the spring.
Concept: The work done to stretch a spring is given by the formula;W = (1/2)kx²Where,k = Spring constant,
x = Amount of stretch or compression of the spring.
So, the work done to stretch the spring is given by the above formula.Given: Spring constant, k = 29.4 N/mAmount of stretch, x = 0.660m.
Formula: W = (1/2)kx².Substituting the values in the above formula;W = (1/2)×29.4N/m×(0.660m)²,
W = (1/2)×29.4N/m×0.4356m²,
W = 6.38026 J (approx).
Therefore, the amount of work that must be done to stretch the spring by 0.660 m from its relaxed length is 6.38 J (approx).
From the above question, we can learn about the concept of the work done to stretch a spring and its formula. The work done to stretch a spring is given by the formula W = (1/2)kx² where k is the spring constant and x is the amount of stretch or compression of the spring.
We can also learn how to calculate the work done to stretch a spring using its formula and given values. Here, we are given the spring constant k = 29.4 N/m and the amount of stretch x = 0.660m.
By substituting the given values in the formula, we get the work done to stretch the spring. The amount of work that must be done to stretch the spring by 0.660 m from its relaxed length is 6.38 J (approx).
The work done to stretch a spring is an important concept of Physics. The work done to stretch a spring is given by the formula W = (1/2)kx² where k is the spring constant and x is the amount of stretch or compression of the spring. Here, we have calculated the amount of work done to stretch a spring of spring constant k = 29.4 N/m and an amount of stretch x = 0.660m. Therefore, the work done to stretch the spring by 0.660 m from its relaxed length is 6.38 J (approx).
To know more about Spring constant visit:
brainly.com/question/29975736
#SPJ11
What resistance R should be connected in series with an inductance L = 197 mH and capacitance C = 15.8 uF for the maximum charge on the capacitor to decay to 95.5% of its initial value in 72.0 cycles?
A resistance of approximately 2.06 kΩ should be connected in series with the given inductance and capacitance for the maximum charge on the capacitor to decay to 95.5% of its initial value in 72.0 cycles.
To find the resistance R required in series with the given inductance L = 197 mH and capacitance C = 15.8 uF, we can use the formula:
R = -(72.0/f) / (C * ln(0.955))
where f is the frequency of the circuit.
First, let's calculate the time period (T) of one cycle using the formula T = 1/f. Since the frequency is given in cycles per second (Hz), we can convert it to the time period in seconds.
T = 1 / f = 1 / (72.0 cycles) = 1.39... x 10^(-2) s/cycle.
Next, we calculate the angular frequency (ω) using the formula ω = 2πf.
ω = 2πf = 2π / T = 2π / (1.39... x 10^(-2) s/cycle) = 452.39... rad/s.
Now, let's substitute the values into the formula to find R:
R = -(72.0 / (1.39... x 10^(-2) s/cycle)) / (15.8 x 10^(-6) F * ln(0.955))
= -5202.8... / (15.8 x 10^(-6) F * (-0.046...))
≈ 2.06 x 10^(3) Ω.
Therefore, a resistance of approximately 2.06 kΩ should be connected in series with the given inductance and capacitance to achieve a decay of the maximum charge on the capacitor to 95.5% of its initial value in 72.0 cycles.
Learn more about circuit from the given link:
https://brainly.com/question/12608516
#SPJ11
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
Learn more about Work Done.
https://brainly.com/question/33261315
#SPJ11
How many electrons does carbon have? how many are valence electrons? what third-row element has the same number of valence electrons as carbon?
Carbon has 6 electrons. To determine the number of valence electrons, we need to look at the electron configuration of carbon, which is 1s² 2s² 2p². The third-row element that has the same number of valence electrons as carbon is silicon (Si).
In the case of carbon, the first shell (1s) is fully filled with 2 electrons, and the second shell (2s and 2p) contains the remaining 4 electrons. The 2s subshell can hold a maximum of 2 electrons, and the 2p subshell can hold a maximum of 6 electrons, but in carbon's case, only 2 of the 2p orbitals are occupied. These 4 electrons in the outermost shell, specifically the 2s² and 2p² orbitals, are called valence electrons. The electron configuration describes the distribution of electrons in the different energy levels or shells of an atom.
Therefore, carbon has 4 valence electrons. Valence electrons are crucial in determining the chemical properties and reactivity of an element, as they are involved in the formation of chemical bonds.
The third-row element that has the same number of valence electrons as carbon is silicon (Si). Silicon also has 4 valence electrons, which can be seen in its electron configuration of 1s² 2s² 2p⁶ 3s² 3p². Carbon and silicon are in the same group (Group 14) of the periodic table and share similar chemical properties due to their comparable valence electron configurations.
To learn more about, valence electrons, click here, https://brainly.com/question/31264554
#SPJ11
Carbon has 6 electrons in total, with 4 of them being valence electrons. Silicon is the third-row element that shares the same number of valence electrons as carbon.
Explanation:Carbon has 6 electrons in total. The electron configuration and orbital diagram for carbon are 1s²2s²2p¹, where the 1s and 2s orbitals are completely filled and the remaining two electrons occupy the 2p subshell. This means that carbon has 4 valence electrons.
The third-row element that has the same number of valence electrons as carbon is silicon (Si). Silicon also has 4 valence electrons.
Learn more about Electrons in Carbon here:https://brainly.com/question/33829891
#SPJ3
Consider a non-rotating space station in the shape of a long thin uniform rod of mass 8.85 x 10^6 kg and length 737 meters. Rocket motors on both ends of the rod are ignited, applying a constant force of F = 5.88 x 10^5 N to each end of the rod as shown in the diagram, causing the station to rotate about its center. If the motors are left running for 2 minutes and 37 seconds before shutting off, then how fast will the station be rotating when the engines stop? 1 1.62 rpm 2 0.65 rpm 3 2.59 rpm 4 3.11 rpm
The space station, has a mass of 8.85 x 10^6 kg and length of 737 meters. After running for 2 minutes and 37 seconds, the motors shut off, and the station will be rotating at approximately 1.62 rpm.
To determine the final rotational speed of the space station, we can use the principle of conservation of angular momentum.
The initial angular momentum (L_initial) of the space station is zero since it is initially at rest. The final angular momentum (L_final) can be calculated using the formula:
L_final = I × ω_final
where:
I is the moment of inertia of the space station
ω_final is the final angular velocity (rotational speed) of the space station
The moment of inertia of a uniform rod rotating about its center is given by:
[tex]I=\frac{1}{12} *m*L^{2}[/tex]
where:
m is the mass of the rod
L is the length of the rod
Substituting the given values:
m = 8.85 x [tex]10^{6}[/tex] kg
L = 737 m
[tex]I=\frac{1}{12} *(8.85*10^{6} )*737m^{2}[/tex]
Now, let's convert the time interval of 2 minutes and 37 seconds to seconds:
Time = 2 minutes + 37 seconds = (2 * 60 seconds) + 37 seconds = 120 seconds + 37 seconds = 157 seconds
The total torque (τ) exerted on the space station by the rocket motors is equal to the force applied (F) multiplied by the lever arm (r). Since the motors are applied at the ends of the rod, the lever arm is equal to half of the length of the rod:
r = [tex]\frac{L}{2} = \frac{737m}{2}[/tex] = 368.5 m
The torque can be calculated as:
τ = F × r
Substituting the given force:
F = 5.88 x [tex]10^{5}[/tex] N
τ = (5.88 x [tex]10^{5}[/tex] N) × (368.5 m)
Now, using the conservation of angular momentum, we equate the initial and final angular momenta:
L_initial = L_final
0 = I × ω_initial (initial angular velocity is zero)
0 = I × ω_final
Since ω_initial is zero, the final angular velocity is given by:
ω_final = τ ÷ I
Substituting the values of τ and I:
ω_final = [tex]\frac{(5.88 *10^{5}) *(368.5m)}{\frac{1}{12} *(8.858 *10^{6} kg)*(737m^{2}) }[/tex]
Calculating the final angular velocity:
ω_final ≈ 1.62 rad/s
To convert the angular velocity to revolutions per minute (rpm), we use the conversion factor:
1 rpm = [tex]\frac{2\pi rad}{60s}[/tex]
Converting ω_final to rpm:
ω_final_rpm = (1.62 rad/s) × [tex]\frac{60s}{2\pi rad}[/tex]
Calculating the final rotational speed in rpm:
ω_final_rpm ≈ 1.62 rpm
Therefore, the space station will be rotating at approximately 1.62 rpm when the engines stop.
The answer is 1) 1.62 rpm.
Learn more about velocity here:
https://brainly.com/question/30559316
#SPJ11
Part A A stone is thrown vertically upward with a speed of 15.6 m/s from the edge of a cliff 75.0 m high (Figure 1). How much later does it reach the bottom of the cliff? Express your answer to three significant figures and include the appropriate units. + OI? f Value Units Submit Request Answer - Part B What is its speed just before hitting? Express your answer to three significant figures and include the appropriate units. Value Units Submit Request Answer - Part What total distance did it travel? Express your answer to three significant figures and include the appropriate units. + 2 123 Figure 1 of 1 Value Units Submit Request Answer Provide Feedback
The stone reaches the bottom of the cliff approximately 4.20 seconds later. The speed just before hitting the bottom is approximately 40.6 m/s.
Part A: To find how much later the stone reaches the bottom of the cliff, we can use the kinematic equation for vertical motion. The equation is:
h = ut + (1/2)gt^2
Where:
h = height of the cliff (75.0 m, negative since it's downward)
u = initial velocity (15.6 m/s)
g = acceleration due to gravity (-9.8 m/s^2, negative since it's downward)
t = time
Plugging in the values, we get:
-75.0 = (15.6)t + (1/2)(-9.8)t^2
Solving this quadratic equation, we find two values for t: one for the stone going up and one for it coming down. We're interested in the time it takes for it to reach the bottom, so we take the positive value of t. Rounded to three significant figures, the time it takes for the stone to reach the bottom of the cliff is approximately 4.20 seconds.
Part B: The speed just before hitting the bottom can be found using the equation for final velocity in vertical motion:
v = u + gt
Where:
v = final velocity (what we want to find)
u = initial velocity (15.6 m/s)
g = acceleration due to gravity (-9.8 m/s^2, negative since it's downward)
t = time (4.20 s)
Plugging in the values, we get:
v = 15.6 + (-9.8)(4.20)
Calculating, we find that the speed just before hitting is approximately -40.6 m/s. Since speed is a scalar quantity, we take the magnitude of the value, giving us a speed of approximately 40.6 m/s.
Part C: To find the total distance traveled by the stone, we need to calculate the distance covered during the upward motion and the downward motion separately, and then add them together.
Distance covered during upward motion:
Using the equation for distance covered in vertical motion:
s = ut + (1/2)gt^2
Where:
s = distance covered during upward motion (what we want to find)
u = initial velocity (15.6 m/s)
g = acceleration due to gravity (-9.8 m/s^2, negative since it's downward)
t = time (4.20 s)
Plugging in the values, we get:
s = (15.6)(4.20) + (1/2)(-9.8)(4.20)^2
Calculating, we find that the distance covered during the upward motion is approximately 33.1 m.
Distance covered during downward motion:
Since the stone comes back down to the bottom of the cliff, the distance covered during the downward motion is equal to the height of the cliff, which is 75.0 m.
Total distance traveled:
Adding the distance covered during the upward and downward motion, we get:
Total distance = 33.1 + 75.0
Rounded to three significant figures, the total distance traveled by the stone is approximately 108 m.
To know more about distance:
https://brainly.com/question/13034462
#SPJ11
6. An electromagnetic wave travels in -z direction, which is -ck. What is/are the possible direction of its electric field, E, and magnetic field, B, at any moment? Electric field Magnetic field A. +E
For an electromagnetic wave traveling in the -z direction (opposite to the positive z-axis), the electric field (E) and magnetic field (B) are perpendicular to each other and to the direction of propagation.
Using the right-hand rule, we find that the electric field (E) will be in the +y direction. So, the correct answer for the electric field direction is:
A. +E (in the +y direction)
Since the magnetic field (B) is perpendicular to the electric field and the direction of propagation, it will be in the +x direction. So, the correct answer for the magnetic field direction is:
B. +x
Therefore, the correct answers are:
Electric field (E) direction: A. +E (in the +y direction)
Magnetic field (B) direction: B. +x
Learn more about electromagnetic wave here : brainly.com/question/29774932
#SPJ11
What is the position of the 2nd maxima for a double slit experiment with a slit width of d=20mm, if there is a laser of 500nm, with the screen 1m away from the slits?
The position of the second maximum (second-order maximum) in this double-slit experiment would be 0.05 mm.
How to find the the position of the second maximum (second-order maximum) in this double-slit experimentTo find the position of the second maximum (second-order maximum) in a double-slit experiment, we can use the formula for the position of the maxima:
[tex]\[ y = \frac{m \cdot \lambda \cdot L}{d} \][/tex]
Where:
- [tex]\( y \) is the position of the maxima[/tex]
- [tex]\( m \) is the order of the maxima (in this case, the second maximum has \( m = 2 \))[/tex]
-[tex]\( \lambda \) is the wavelength of the laser light (500 nm or \( 500 \times 10^{-9} \) m)[/tex]
-[tex]\( L \) is the distance from the slits to the screen (1 m)[/tex]
- [tex]\( d \) is the slit width (20 mm or \( 20 \times 10^{-3} \) m)[/tex]
Substituting the given values into the formula:
[tex]\[ y = \frac{2 \cdot 500 \times 10^{-9} \cdot 1}{20 \times 10^{-3}} \][/tex]
Simplifying the expression:
[tex]\[ y = \frac{2 \cdot 500 \times 10^{-9}}{20 \times 10^{-3}} \][/tex]
[tex]\[ y = 0.05 \times 10^{-3} \][/tex]
[tex]\[ y = 0.05 \, \text{mm} \][/tex]
Therefore, the position of the second maximum (second-order maximum) in this double-slit experiment would be 0.05 mm.
Learn more about experiment at https://brainly.com/question/25303029
#SPJ4
What is the value of the velocity of a body with a mass of 15 g that moves in a circular path of 0.20 min length? diameter and a centripetal force of 2 N acts: a. 5.34m/s b. 2.24m/s c. 2.54m d. 1.56Nm
The value of the velocity of a body with a mass of 15 g that moves in a circular path of 0.20 min length, diameter and a centripetal force of 2 N acts is 2.24 m/s.
The formula used to determine the value of velocity is:v = √(F * r / m)Where:
v = velocity
F = force (centripetal) applied to the mass
mr = radius of circular path
m = mass of the object
Now, substituting the given values in the formula:
V = √(F * r / m)
V = √(2 * 0.20 / 0.015)V = √26.67V = 2.24 m/s
Therefore, the answer is option b, 2.24 m/s.
To know more about path visit:
https://brainly.com/question/2047841
#SPJ11
Visible light shines upon a pair of closely-spaced thin slits. An interference pattern is seen on a screen located behind the slits. For which color of light will the distance between the fringes (as seen on the screen) be greatest? yellow-green green yellow
The distance between the fringes in an interference pattern, often referred to as the fringe spacing or fringe separation, is determined by the wavelength of the light used.
The greater the wavelength, the larger the fringe spacing.
Yellow-green light and green light are both within the visible light spectrum, with yellow-green having a longer wavelength than green.
Therefore, the distance between the fringes will be greater for yellow-green light compared to green light.
The fringe spacing, also known as the fringe separation or fringe width, refers to the distance between adjacent bright fringes (or adjacent dark fringes) in the interference pattern. It is directly related to the wavelength of the light used.
According to the principles of interference, the fringe spacing is determined by the path length difference between the light waves reaching a particular point on the screen from the two slits. Constructive interference occurs when the path length difference is an integer multiple of the wavelength, leading to bright fringes. Destructive interference occurs when the path length difference is a half-integer multiple of the wavelength, resulting in dark fringes.
Learn more about interference here : brainly.com/question/31857527
#SPJ11
Convert the orbital period of GJ 357 dfrom
days to seconds with the orbital radius given above, calculate
Kepler's constant for the Gliese 357 system in units of
s2 / m3.
The Kepler's constant for Gliese 357 system in units of s2 / m3 is:k = (4 * pi^2) / (G * 0.3 solar masses * (0.025 AU)^3) = 8.677528872262322 s^2
The steps involved in converting the orbital period of GJ 357 d from days to seconds, calculating Kepler's constant for the Gliese 357 system in units of s2 / m3:
1. Convert the orbital period of GJ 357 d from days to seconds. The orbital period of GJ 357 d is 3.37 days. There are 86,400 seconds in a day. Therefore, the orbital period of GJ 357 d in seconds is 3.37 days * 86,400 seconds/day = 291,167 seconds.
2. Calculate Kepler's constant for the Gliese 357 system in units of s2 / m3.Kepler's constant is a physical constant that relates the orbital period of a planet to the mass of the star it orbits and the distance between the planet and the star.
The value of Kepler's constant is 4 * pi^2 / G, where G is the gravitational constant. The mass of Gliese 357 is 0.3 solar masses. The orbital radius of GJ 357 d is 0.025 AU.
Therefore, Kepler's constant for the Gliese 357 system in units of s2 / m3 is: k = (4 * pi^2) / (G * 0.3 solar masses * (0.025 AU)^3) = 8.677528872262322 s^2 .
Learn more about keplers constant with the given link,
https://brainly.com/question/16705471
#SPJ11
All work/steps must be shown following the "Problem-Solving Procedure". Part II - Short Problems −4 points 1. Find the ' x ' and ' y ' components of the following vectors. a. F=67.9 N,38∘ b. v=8.76 m/s,−57.3∘ 2. Determine the 'polar coordinate' form of the following vector components. a. Ax=7.87 mAy=−8.43 m b. vx=−67.3 m/svy=−24.9 m/s
In problem 1, the x and y components of the vector F are found to be 50.19 N and 51.95 N, respectively. In problem 2, the polar coordinate form of vector A is determined to be 11.01 m at an angle of -48.92 degrees, while vector v is expressed as 76.46 m/s at an angle of -197.65 degrees.
In problem 1a, the vector force F, is given with a magnitude of 67.9 N and an angle of 38 degrees. To find the x and y components, we use the trigonometric functions cosine (cos) and sine (sin).
The x component is calculated as Fx = F * cos(θ), where θ is the angle, yielding Fx = 67.9 N * cos(38°) = 50.19 N. Similarly, the y component is determined as Fy = F * sin(θ), resulting in Fy = 67.9 N * sin(38°) = 51.95 N.
In problem 1b, the vector v is given with a magnitude of 8.76 m/s and an angle of -57.3 degrees. Using the same trigonometric functions, we can find the x and y components.
The x component is calculated as vx = v * cos(θ), which gives vx = 8.76 m/s * cos(-57.3°) = 4.44 m/s. The y component is determined as vy = v * sin(θ), resulting in vy = 8.76 m/s * sin(-57.3°) = -7.37 m/s.
In problem 2a, the vector components Ax = 7.87 m and Ay = -8.43 m are given. To express this vector in polar coordinate form, we can use the Pythagorean theorem to find the magnitude (r) of the vector, which is r = √(Ax^2 + Ay^2).
Substituting the given values, we obtain r = √((7.87 m)^2 + (-8.43 m)^2) ≈ 11.01 m. The angle (θ) can be determined using the inverse tangent function, tan^(-1)(Ay/Ax), which gives θ = tan^(-1)(-8.43 m/7.87 m) ≈ -48.92 degrees.
Therefore, the polar coordinate form of vector A is approximately 11.01 m at an angle of -48.92 degrees.In problem 2b, the vector components vx = -67.3 m/s and vy = -24.9 m/s are given.
Following a similar procedure as in problem 2a, we find the magnitude of the vector v as r = √(vx^2 + vy^2) = √((-67.3 m/s)^2 + (-24.9 m/s)^2) ≈ 76.46 m/s.
The angle θ can be determined using the inverse tangent function, tan^(-1)(vy/vx), resulting in θ = tan^(-1)(-24.9 m/s/-67.3 m/s) ≈ -197.65 degrees. Hence, the polar coordinate form of vector v is approximately 76.46 m/s at an angle of -197.65 degrees.
Learn more about force here ;
https://brainly.com/question/30507236
#SPJ11
If an applied force on an object acts antiparallel to the direction of the object's movement, the work done on by the applied force is: Negative Cannot be determined by the problem. Positive Zero
If an applied force on an object acts antiparallel to the direction of the object's movement, the work done by the applied force is negative.
The transfer of energy from one object to another by applying a force to an object, which makes it move in the direction of the force is known as work. When the applied force acts in the opposite direction to the object's movement, the work done by the force is negative.
The formula for work is given by: Work = force x distance x cosθ where,θ is the angle between the applied force and the direction of movement. If the angle between force and movement is 180° (antiparallel), then cosθ = -1 and work done will be negative. Therefore, if an applied force on an object acts antiparallel to the direction of the object's movement, the work done by the applied force is negative.
Learn more about work done here:
https://brainly.com/question/32263955
#SPJ11
A particle starts from the origin at t=0.0 s with a velocity of 8.1 i m/s and moves in the xy plane with a constant acceleration of (-9.3 i + 8.8 j)m/s2. When the particle achieves the maximum positive x-coordinate, how far is it from the origin?
When the particle achieves the maximum positive x-coordinate, it is approximately 4.667 meters away from the origin.
Explanation:
To find the distance of the particle from the origin when it achieves the maximum positive x-coordinate, we need to determine the time it takes for the particle to reach that point.
Let's assume the time at which the particle achieves the maximum positive x-coordinate is t_max. To find t_max, we can use the equation of motion in the x-direction:
x = x_0 + v_0x * t + (1/2) * a_x * t²
where:
x = position in the x-direction (maximum positive x-coordinate in this case)
x_0 = initial position in the x-direction (which is 0 in this case as the particle starts from the origin)
v_0x = initial velocity in the x-direction (which is 8.1 m/s in this case)
a_x = acceleration in the x-direction (which is -9.3 m/s² in this case)
t = time
Since the particle starts from the origin, x_0 is 0. Therefore, the equation simplifies to:
x = v_0x * t + (1/2) * a_x * t²
To find t_max, we set the velocity in the x-direction to 0:
0 = v_0x + a_x * t_max
Solving this equation for t_max gives:
t_max = -v_0x / a_x
Plugging in the values, we have:
t_max = -8.1 m/s / -9.3 m/s²
t_max = 0.871 s (approximately)
Now, we can find the distance of the particle from the origin at t_max using the equation:
distance = magnitude of displacement
= √[(x - x_0)² + (y - y_0)²]
Since the particle starts from the origin, the initial position (x_0, y_0) is (0, 0).
Therefore, the equation simplifies to:
distance = √[(x)^2 + (y)²]
To find x and y at t_max, we can use the equations of motion:
x = x_0 + v_0x * t + (1/2) * a_x *t²
y = y_0 + v_0y * t + (1/2) * a_y *t²
where:
v_0y = initial velocity in the y-direction (which is 0 in this case)
a_y = acceleration in the y-direction (which is 8.8 m/s² in this case)
For x:
x = 0 + (8.1 m/s) * (0.871 s) + (1/2) * (-9.3 m/s²) * (0.871 s)²
For y:
y = 0 + (0 m/s) * (0.871 s) + (1/2) * (8.8 m/s²) * (0.871 s)²
Evaluating these expressions, we find:
x ≈ 3.606 m
y ≈ 2.885 m
Now, we can calculate the distance:
distance = √[(3.606 m)² + (2.885 m)²]
distance ≈ 4.667 m
To know more about acceleration, visit:
https://brainly.com/question/2303856
#SPJ11
6. a. A space ship moves away from the earth at a speed of 0.850. The ship launches a small probe that moves with a speed of 0.780c with respect to the ship. From the point of view of the crew on the ship, the probe is moving back directly towards the earth. Compute the speed of the probe as seen from the earth. Is the probe moving towards the carth or away from the earth (as viewed from the earth)?
b. From the earth, the ship described above appears to have a length of 4.50 m. What would be the length of the ship as measured by the ship's crew?
(a.)The speed of the probe as seen from the earth is approximately 0.970c. (b.) The length of the ship as measured by the ship's crew is approximately 6.15 m.
a. To calculate the speed of the probe as seen from the earth, we need to use the relativistic velocity addition formula:
v' = (v + u) / (1 + (vu/c^2)),
where v' is the velocity of the probe as seen from the earth, v is the velocity of the ship (relative to the earth), u is the velocity of the probe (relative to the ship), and c is the speed of light.
v = 0.850c (speed of the ship relative to the earth),
u = 0.780c (speed of the probe relative to the ship).
Substituting the values into the formula:
v' = (0.850c + 0.780c) / (1 + (0.850c)(0.780c)/(c^2))
= (1.63c) / (1 + 0.663)
≈ 0.970c.
Therefore, the speed of the probe as seen from the earth is approximately 0.970c. Since the speed is greater than the speed of light, it implies that the probe is moving away from the earth (as viewed from the earth).
b. The length of the ship as measured by the ship's crew can be calculated using the relativistic length contraction formula:
L' = L * √(1 - (v^2/c^2)),
where L' is the length of the ship as measured by the crew, L is the length of the ship as measured by an observer at rest (in this case, the earth), v is the velocity of the ship (relative to the earth), and c is the speed of light.
L = 4.50 m (length of the ship as measured from the earth),
v = 0.850c (speed of the ship relative to the earth).
Substituting the values into the formula:
L' = 4.50 m * √(1 - (0.850c)^2/c^2)
= 4.50 m * √(1 - 0.7225)
= 4.50 m * √(0.2775)
≈ 6.15 m.
Therefore, the length of the ship as measured by the ship's crew is approximately 6.15 m.
a. The speed of the probe as seen from the earth is approximately 0.970c. The probe is moving away from the earth (as viewed from the earth).
b. The length of the ship as measured by the ship's crew is approximately 6.15 m.
To know more about probe , visit:
https://brainly.com/question/29244857
#SPJ11
4. The graph shows pulses A and B at time = 0 as they head toward each other. Each pulse travels at a constant speed of 1 square per second on a string which is 16 squares long. Show the resultant displacement of the string after 4 seconds has passed. Indicate the locations where constructive and destructive interference has occurred. (5 marks)
The resultant displacement of the string after 4 seconds is 4 squares long.
The given graph illustrates pulses A and B heading towards each other on a string, as shown below: The amplitude of each pulse is 1 square, and the string on which they travel is 16 squares long. Both pulses have a speed of 1 square per second.
Constructive interference occurs when two waves that have identical frequency and amplitude combine. As the amplitude of each pulse is the same and they have the same frequency, they will result in constructive interference when they meet. The distance between two consecutive points of constructive interference is equivalent to the wavelength.
Destructive interference occurs when two waves with the same frequency and amplitude, but opposite phases, meet. The distance between two consecutive points of destructive interference is equivalent to half a wavelength.
Therefore, we need to calculate the wavelength of the pulse, λ, in order to find where constructive and destructive interference occurs. The formula for the wavelength of a wave is as follows:
λ = v/f
whereλ = wavelength
v = velocity of the wave
f = frequency of the wave
Since the velocity of each pulse is 1 square per second, the formula becomes:
λ = 1/f. For the pulse shown in the diagram, f can be calculated by determining the time taken for the pulse to complete one cycle. Since the pulse has a speed of 1 square per second and an amplitude of 1 square, one cycle of the pulse is equivalent to twice the distance travelled by the pulse. As a result, one cycle of the pulse takes 2 seconds. Therefore, the frequency of the pulse is:f = 1/2 = 0.5 Hz
Substituting the value of f into the wavelength formula yields:
λ = 1/f = 1/0.5 = 2 squares
Resultant displacement after 4 seconds:
The pulses A and B have a combined wavelength of 2 squares and travel at a constant velocity of 1 square per second. As a result, the distance travelled by the pulses after 4 seconds can be calculated using the formula:
s = v/t
where v = velocity of waves = 1 square per second t = time = 4 seconds Substituting the values of v and t into the equation yields:s = 1 × 4 = 4 squares
Thus, the resultant displacement of the string after 4 seconds is 4 squares long.
The resultant displacement of the string after 4 seconds is 4 squares long, and constructive interference has occurred every 2 squares along the string while destructive interference has occurred halfway between the constructive interference points.
To know more about Constructive interference visit
brainly.com/question/31857527
#SPJ11
Use Gauss's Law to find the electric field inside and outside a solid metal sphere of radius R with charge Q.
Gauss's Law can be used to find the electric field inside and outside a solid metal sphere of radius R with charge Q.
Gauss's Law states that the electric flux through any closed surface is proportional to the total electric charge enclosed within the surface.
This can be expressed mathematically as:∫E.dA = Q/ε0
Where E is the electric field, A is the surface area, Q is the total electric charge enclosed within the surface, and ε0 is the permittivity of free space
total charge:ρ =[tex]Q/V = Q/(4/3 π R³)[/tex]
where ρ is the charge density, V is the volume of the sphere, and Q is the total charge of the sphere
.Substituting this equation into Gauss's Law,
we get:[tex]∫E.dA = ρV/ε0 = Q/ε0E ∫dA = Q/ε0E × 4πR² = Q/ε0E = Q/(4πε0R²)[/tex]
the electric field inside and outside the solid metal sphere is given by:
E =[tex]Q/(4πε0R²)[/tex]For r ≤ R (inside the sphere)
E = [tex]Q/(4πε0r²)[/tex]For r > R (outside the sphere)
:where r is the distance from the center of the sphere.
To know more about Gauss's Law visit:
https://brainly.com/question/13434428
#SPJ11
1. A steel bar of area 20mm² is under a force of 5000N, work out the stress. (3 marks)
Stress is a measure of the internal force experienced by a material due to an applied external force. To calculate the stress in the steel bar, we can use the formula: Stress = Force / Area. Therefore, the stress in the steel bar is 250,000,000 N/m² or 250 MPa (megapascals).
Given:
Force = 5000 N
Area = 20 mm²
First, we need to convert the area to square meters since the force is given in Newtons, which is the SI unit.
1 mm² = (1/1000)^2 m² = 1/1,000,000 m²
Area in square meters (A) = 20 mm² * (1/1,000,000 m²/mm²) = 0.00002 m²
Now we can calculate the stress:
Stress = Force / Area
Stress = 5000 N / 0.00002 m²
Stress = 250,000,000 N/m²
Therefore, the stress in the steel bar is 250,000,000 N/m² or 250 MPa (megapascals).
To learn more about, internal force, click here, https://brainly.com/question/32068975
#SPJ11
No radio antennas separated by d=272 m as shown in the figure below simultaneously broadcast identical signals at the same wavelength. A ar travels due north along a straight line at position x=1150 m from the center point between the antennas, and its radio receives the signals. ote: Do not use the small-angle approximation in this problem. (a) If the car is at the position of the second maximum after that at point O when it has traveled a distance y=400 morthward, what is the wavelength of the signals? x Return to the derivation of the locations of constructive interference in Young's double slit experiment. (b) How much farther must the car travel from this position to encounter the next minimum in reception? x You must work with the full trigonometric expressions for constructive and destructive interference because the angles are not small.
In this question, we determined the wavelength of the signals received by a car traveling due north along a straight line at position x = 1150 m from the center point between two radio antennas. We also determined the distance the car must travel from the second maximum position to encounter the next minimum in reception.
a)We have the distance between the antennas to be d = 272 m, the distance of the car from the center point of the antennas to be x = 1150 m and it has traveled a distance of y = 400 m to reach the second maximum point. We have to determine the wavelength of the signals.If we let θ be the angle between the line joining the car and the center point of the antennas and the line joining the two antennas. Let's denote the distance between the car and the first antenna as r1 and that between the car and the second antenna as r2. We have:r1² = (d/2)² + (x + y)² r2² = (d/2)² + (x - y)². From the diagram, we have:r1 + r2 = λ/2 + nλ ...........(1)
where λ is the wavelength of the signals and n is an integer. We are given that the car is at the position of the second maximum after that at point O, which means n = 1. Substituting the expressions for r1 and r2, we get:(d/2)² + (x + y)² + (d/2)² + (x - y)² = λ/2 + λ ...........(2)
After simplification, equation (2) reduces to: λ = (8y² + d²)/2d ................(3)
Substituting the values of y and d in equation (3),
we get:λ = (8 * 400² + 272²)/(2 * 272) = 700.66 m. Therefore, the wavelength of the signals is 700.66 m.
b)We have to determine how much farther the car must travel from the second maximum position to encounter the next minimum in reception. From equation (1), we have:r1 + r2 = λ/2 + nλ ...........(1)
where n is an integer. At a minimum, we have n = 0.Substituting the expressions for r1 and r2, we get:(d/2)² + (x + y)² + (d/2)² + (x - y)² = λ/2 ...........(2)
After simplification, equation (2) reduces to: y = (λ/4 - x²)/(2y) ................(3)
We know that the car is at the position of the second maximum after that at point O. Therefore, the distance it must travel to reach the first minimum is:y1 = λ/4 - x²/2λ ................(4)
From equation (4), we get:y1 = (700.66/4) - (1150²/(2 * 700.66)) = -112.06 m. Therefore, the car must travel a distance of 112.06 m from the second maximum position to encounter the next minimum in reception.
In this question, we determined the wavelength of the signals received by a car traveling due north along a straight line at position x = 1150 m from the center point between two radio antennas. We also determined the distance the car must travel from the second maximum position to encounter the next minimum in reception. We used the expressions for constructive and destructive interference for two coherent sources to derive the solutions.
To know more about wavelength visit:
brainly.com/question/31143857
#SPJ11
A water balloon is thrown straight down with an initial speed of 12.0 m 's from a second floor window, 5.00 m above ground level. With what speed v does the balloon strike the ground? Assume the effects of air resistance are negligible.
The water balloon will strike the ground, when it is thrown straight down with an initial speed of 12.0 m 's from a second floor window, 5.00 m above ground level, at a speed of 6.78 m/s.
To determine the speed at which the water balloon strikes the ground, we can use the kinematic equation for vertical motion:
v² = u² + 2as
Where: v is the final velocity (unknown), u is the initial velocity (12.0 m/s, downward), a is the acceleration due to gravity (-9.8 m/s², since the balloon is moving downward), s is the displacement (5.00 m, since the balloon is falling from a height of 5.00 m)
Substituting the given values into the equation:
v² = (12.0 m/s)² + 2(-9.8 m/s²)(5.00 m)
v² = 144 m²/s² - 98 m²/s²
v² = 46 m²/s²
Taking the square root of both sides:
v = √46 m/s
v = 6.78 m/s
Therefore, the water balloon will strike the ground with a speed of 6.78 m/s.
To learn more about speed: https://brainly.com/question/13943409
#SPJ11
A block is kept on horizontal table the table is undergoing simple harmonic motion of frequency 3Hz in a horizontal plane . the coefficient of static friction between block and the table surface is 0.72. find the maximum amplitude of the table at which the block does not slip on the surface.
The maximum amplitude of the table at which the block does not slip on the surface is 0.0727m.
As the table is undergoing simple harmonic motion, the acceleration of the block towards the center of the table can be given as a = -ω²x, where r of the block from the center of the table. The maximum acceleration is when x = A, where A is the amplitude of the motion, and can be given as a_max = ω²A.
To prevent the block from slipping, the maximum value of the frictional force (ffriction = μN) should be greater than or equal to the maximum value of the force pulling the block (fmax = mamax). Therefore, we have μmg >= mω²A, where m is the mass of the block and g is the acceleration due to gravity. Rearranging the equation, we get A <= (μg/ω²).
Substituting the given values, we get
A <= (0.729.8)/(2π3) = 0.0727m.
Therefore, the maximum amplitude of the table at which the block does not slip is 0.0727m.
To learn more about amplitude click brainly.com/question/3613222
#SPJ11
. You will need a partner. Run a tight figure-eight at increasing speed on a flat surface. Why is it difficult to run the figure-eight course at high speeds?
Running a figure-eight course at high speeds is difficult due to the increased centripetal force requirements, challenges in maintaining balance and coordination, the impact of inertia and momentum, and the presence of lateral forces and friction that can affect stability and control.
Running a figure-eight course at high speeds can be difficult due to the following reasons:
Centripetal force requirements: In order to make tight turns in the figure-eight pattern, a significant centripetal force is required to change the direction of motion. As the speed increases, the centripetal force required also increases, making it more challenging to generate and maintain that force while running.
Balance and coordination: Running a figure-eight involves sharp turns and changes in direction, which require precise balance and coordination. At higher speeds, it becomes more challenging to maintain balance and execute quick changes in direction without losing control.
Inertia and momentum: With increasing speed, the inertia and momentum of the runner also increase. This makes it harder to change directions rapidly and maintain control while transitioning between different parts of the figure-eight course.
Lateral forces and friction: During turns, lateral forces act on the runner, pulling them towards the outside of the turn. These lateral forces, combined with the friction between the feet and the ground, can make it difficult to maintain stability and prevent slipping or sliding, especially at higher speeds.
Overall, running a figure-eight course at high speeds requires a combination of physical strength, coordination, balance, and control. The increased demands on these factors make it challenging to execute the course smoothly and maintain stability throughout.
To learn more about Centripetal force visit : https://brainly.com/question/898360
#SPJ11
3) A proton is sent into a region of constant magnetic field, oriented perpendicular to the protons path. There the proton travels at a speed of 3 x 106m/s in a circular path of radius 20 cm. a) What is the magnitude of the magnetic field? b) What is the period? c) What is the value of the magnetic field, generated by the proton, at the center of the circular path?
Given,
Speed of the proton
v = 3x10⁶ m/s
The radius of the circular path
r = 20 cm
= 0.20 m
Here,
Force on the proton
F = qvB (B is the magnetic field and q is the charge of proton)
Centripetal force = Fq v
B = m v²/r
Substituting the value,
mv²/r = q v B
⇒ B = mv/qr
= (1.67 × 10⁻²⁷ × (3 × 10⁶)²)/(1.6 × 10⁻¹⁹ × 0.2)
= 1.76 × 10⁻⁴ T
Period, T = 2πr/v = 2 × 3.14 × 0.20/3 × 10⁶ = 4.19 × 10⁻⁷ s
The magnetic field generated by the proton at the center of the circular path
= B/2
= 1.76 × 10⁻⁴/2
= 0.88 × 10⁻⁴ T
Answer: a) 1.76 × 10⁻⁴ T;
b) 4.19 × 10⁻⁷ s;
c) 0.88 × 10⁻⁴ T
To know more about Centripetal force visit:
https://brainly.com/question/14021112
#SPJ11
A circular capacitor of radius ro = 5.0 cm and plate spacing d = 1.0 mm is being charged by a 9.0 V battery through a R = 10 Ω resistor. At which distance r from the center of the capacitor is the magnetic field strongest (in cm)?
The circular capacitor of radius ro = 5.0 cm and plate spacing d = 1.0 mm is being charged by a 9.0 V battery through a R = 10 Ω resistor. We are to determine the distance r from the center of the capacitor at which the magnetic field is strongest. By given information, we can determine that the magnetic field is strongest at a distance of r = 20 cm from the center of the capacitor.
The magnetic force is given by the formula
F = qvBsinθ
where,
q is the charge.
v is the velocity of the particle.
B is the magnetic field
θ is the angle between the velocity vector and the magnetic field vector. Since there is no current in the circuit, no magnetic field is produced by the capacitor. Therefore, the magnetic field is zero. The strongest electric field is at the center of the capacitor because it is equidistant from both plates. The electric field can be given as E = V/d
where V is the voltage and d is the separation distance between the plates.
Therefore, we have
E = 9/0.001 = 9000 V/m.
At the center of the capacitor, the electric field is given by
E = σ/2ε0, where σ is the surface charge density and ε0 is the permittivity of free space.
Therefore,
σ = 2ε0E = 2 × 8.85 × 10^-12 × 9000 = 1.59 × 10^-7 C/m^2.
At a distance r from the center of the capacitor, the surface charge density is given by
σ = Q/(2πrL), where Q is the charge on each plate, and L is the length of the plates.
Therefore, Q = σ × 2πrL = σπr^2L.
We can now find the capacitance C of the capacitor using C = Q/V.
Hence,
C = σπr^2L/V.
Substituting for V and simplifying, we obtain
C = σπr^2L/(IR) = 2.81 × 10^-13πr^2.Where I is the current in the circuit, which is given by I = V/R = 0.9 A.
The magnetic field B is given by B = μ0IR/2πr, where μ0 is the permeability of free space.
Substituting for I and simplifying, we get
B = 2.5 × 10^-5/r tesla.
At a distance of r = 20 cm from the center of the capacitor, the magnetic field is strongest. Therefore, the magnetic field is strongest at a distance of r = 20 cm from the center of the capacitor.
For further information on capacitors visit:
https://brainly.com/question/32648063
#SPJ11
Please explain steps for part A and what is the image distance,
di, in centimeters?
(11%) Problem 5: An object is located a distance do = 5.1 cm in front of a concave mirror with a radius of curvature r = 21.1 cm. 33% Part (a) Write an expression for the image distance, d;.
The image distance is 14.8 cm and it is virtual and upright. Image distance, di = -14.8 cm.
Part A: An expression for image distance, di The formula used to calculate the image distance in terms of the focal length is given as follows;
d = ((1 / f) - (1 / do))^-1
where;f = focal length do = object distance
So, we need to write an expression for the image distance in terms of the object distance and the radius of curvature, R.As we know that;
f = R / 2From the mirror formula;1 / do + 1 / di = 1 / f
Substitute the value of f in the above formula;1 / do + 1 / di = 2 / R Invert both sides; do / (do + di)
= R / 2di
= Rdo / (2do - R)
So, the expression for image distance is; di = Rdo / (2do - R)Substitute the given values;
di = (21.1 cm)(5.1 cm) / [2(5.1 cm) - 21.1 cm]
= -14.8 cm (negative sign indicates that the image is virtual and upright)
To know more about virtual visit:
https://brainly.com/question/31674424
#SPJ11
Express 18/4 as a fraction of more than 1
When expressed as a fraction of more than 1, 18/4 is equivalent to 4 and 1/2.
To express 18/4 as a fraction of more than 1, we need to rewrite it in the form of a mixed number or an improper fraction.
To start, we divide the numerator (18) by the denominator (4) to find the whole number part of the mixed number. 18 divided by 4 equals 4 with a remainder of 2. So the whole number part is 4.
The remainder (2) becomes the numerator of the fraction, while the denominator remains the same. Thus, the fraction part is 2/4.
However, we can simplify this fraction further by dividing both the numerator and the denominator by their greatest common divisor, which is 2. Dividing 2 by 2 equals 1, and dividing 4 by 2 equals 2. Therefore, the simplified fraction is 1/2.
Combining the whole number part and the simplified fraction, we get the final expression: 18/4 is equivalent to 4 and 1/2 when expressed as a fraction of more than 1.
To learn more about fractions
https://brainly.com/question/10354322
#SPJ8
Moving to another question will save this response. uestion 13 An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 mv's what is the frequency of the third ha
The frequency of the third harmonic of an organ pipe open at both ends with a length of 0.80 m and a velocity of sound in air of 340 m/s is 850 Hz. The correct option is C.
For an organ pipe open at both ends, the frequency of the harmonics can be determined using the formula:
fₙ = (nv) / (2L)
where fₙ is the frequency of the nth harmonic, n is the harmonic number, v is the velocity of sound, and L is the length of the pipe.
In this case, we want to find the frequency of the third harmonic, so n = 3. The length of the pipe is given as 0.80 m, and the velocity of sound in air is 340 m/s.
Substituting these values into the formula, we have:
f₃ = (3 * 340 m/s) / (2 * 0.80 m)
Calculating this expression gives us:
f₃ = 850 Hz
Therefore, the frequency of the third harmonic of the organ pipe is 850 Hz. Option C is correct one.
To know more about harmonics refer here:
https://brainly.com/question/28217835#
#SPJ11
Complete Question:
Moving to another question will save this response. uestion 13 An organ pipe open at both ends has a length of 0.80 m. If the velocity of sound in air is 340 mv's what is the frequency of the third harmonic of this pipe O 425 Hz O 638 Hz O 850 Hz 213 Hz
A 1100-kg automobile traveling at 15 m/s collides head-on with a 1800-kg automobile traveling at 10 m/s in the opposite direction. Is it possible to predict the velocities of the cars after the collision? Yes
No
Is it possible to predict the value that any pertinent physical quantity has immediately after the collision?
Yes, it is possiple to predict the total momentum. Yes, it is possiple to predict the sum of velocities.
No, it is impossiple to predict the value of any physical quantity.
1. Yes, the velocities of the cars after the collision can be predicted using conservation laws.
2. Yes, it is possible to predict the total momentum of the system immediately after the collision in an elastic collision.
1. Yes, it is possible to predict the velocities of the cars after the collision using the principles of conservation of momentum and kinetic energy. The collision between the two automobiles is an example of an elastic collision.
2. The pertinent physical quantity that can be predicted immediately after the collision is the total momentum of the system. In an elastic collision, the total momentum before the collision is equal to the total momentum after the collision.
Therefore, the correct answer to question 1 is "Yes," as the velocities of the cars can be predicted, and the correct answer to question 2 is "Yes, it is possible to predict the total momentum."
Learn more about velocity from this link:
https://brainly.com/question/80295
#SPJ11