Final answer:
For a step-up transformer with 22 primary turns and 800 secondary turns, to produce an output of 5300 V at 16 mA, the required input voltage is 145.75 volts and the input current needed is 0.5818 amperes.
Explanation:
Calculating Input Current and Voltage for a Step-Up Transformer
The student's question involves a step-up transformer with a known number of turns in the primary and secondary coils, a given secondary voltage, and a secondary current. To find the required input current and voltage, we can use the transformer equations that relate the primary and secondary sides of the transformer:
Primary voltage (VP) / Secondary voltage (VS) = Number of turns in the primary coil (NP) / Number of turns in the secondary coil (NS)
Primary current (IP) * Number of turns in the primary coil (NP) = Secondary current (IS) * Number of turns in the secondary coil (NS)
We're given:
NP = 22 turns
NS = 800 turns
IS = 16 mA = 0.016 A
VS = 5300 V
To find the input voltage VP:
VP = (NP / NS) * VS = (22 / 800) * 5300 V = 145.75 V
To find the input current IP:
IP = (NS / NP) * IS = (800 / 22) * 0.016 A = 0.5818 A
Therefore, the required input voltage is 145.75 volts, and the required input current is 0.5818 amperes.
How do the magnitudes of the inertial (the density times acceleration term), pressure, and viscous terms in the Navier-Stokes equation compare?
Answer:
The general equation of movement in fluids is obtained from the application, at fluid volumes, of the principle of conservation of the amount of linear movement. This principle establishes that the variation over time of the amount of linear movement of a fluid volume is equal to that resulting from all forces (of volume and surface) acting on it. Expressed in This equation is called the Navier-Stokes equation.
The equation is shown in the attached file
Explanation:
The derivative of velocity with respect to time determines the change in the velocity of a particle of the fluid as it moves in space. It also includes convective acceleration, expressed by a nonlinear term that comes from convective inertia forces). With this equation, Stokes studied the motion of an infinite incompressible viscous fluid at rest at infinity, and in which a solid sphere of radius r makes a rectilinear and uniform translational motion of velocity v. It assumes that there are no external forces and that the movement of the fluid relative to a reference system on the sphere is stationary. Stokes' approach consists in neglecting the nonlinear term (associated with inertial forces due to convective acceleration).
Problem 12.6 A hockey player hits a puck so that it comes to rest 10 s after sliding 100 ft on the ice. Determine (a) the initial velocity of the puck, (b) the coefficient of friction between the puck and the ice.
Answer:
a)The initial velocity of the puck is 20 ft/s.
b)The coefficient of friction is 0.062.
Explanation:
Hi there!
a)For this problem let's use the equations of position and velocity of an object moving in a straight line with constant acceleration:
x = x0 + v0 · t + 1/2 · a · t²
v = v0 + a · t
Where:
x = position of the puck after a time t.
x0 = initial position.
v0 = initial velocity.
a = acceleration.
t = time.
v = velocity of the puck at a time t.
Let's place the origin of the frame of reference at the point where the puck is hit so that x0 = 0.
We know that at t = 10 s the velocity of the puck is zero (v = 0) and its position is 100 ft (x = 100 ft):
100 ft = v0 · 10 s + 1/2 · a · (10 s)²
0 = v0 + a · 10 s
We have a system of two equations with two unknowns, so, we can solve the system.
Solving for v0 in the second equation:
0 = v0 + a · 10 s
v0 = -a · 10 s
Replacing v0 in the first equation:
100 ft = (-a · 10 s) · 10 s + 1/2 · a · (10 s)²
100 ft = -50 s² · a
100 ft / -50 s² = a
a = -2.0 ft/s²
Then the initial velocity of the puck will be:
v0 = -a · 10 s
v0 = -(-2.0 ft/s²) · 10 s
v0 = 20 ft/s
The initial velocity of the puck is 20 ft/s.
b) The friction force is calculated as follows:
Fr = N · μ
Where:
Fr = friction force.
N = normal force.
μ = coefficient of friction.
Since the only vertical forces acting on the puck are the weight of the puck and the normal force and since the puck is not being accelerated in the vertical direction, then, the normal force is equal to the weight of the puck. The weight (W) is calculated as follows:
W = m · g
Where "m" is the mass of the puck and "g" is the acceleration due to gravity (32.2 ft/s²).
Then the friction force can be calculated as follows:
Fr = m · g · μ
Since the acceleration of the puck is provided only by the friction force, then, due to Newton's second law:
Fr = m · a
Where "m" is the mass of the puck and "a" its acceleration. Then:
Fr = m · g · μ
Fr = m · a
m · g · μ = m · a
μ = a/g
μ = 2.0 ft/s² / 32.2 ft/s²
μ = 0.062
The coefficient of friction is 0.062.
The mass of the Sun is 2 × 1030 kg, the mass of the Earth is 6 × 1024 kg, and their center-to-center distance is 1.5 × 1011 m. Suppose that at some instant the Sun's momentum is zero (it's at rest). Ignoring all effects but that of the Earth, what will the Sun's speed be after 3 days? (Very small changes in the velocity of a star can be detected using the "Doppler" effect, a change in the frequency of the starlight, which has made it possible to identify the presence of planets in orbit around a star.)
Answer:
0.00461031264 m/s
Explanation:
G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²
M = Mass of the Earth = 6 × 10²⁴ kg
r = Distance between Earth and Sun = [tex]1.5\times 10^{11}\ m[/tex]
t = Time taken = 3 days
Acceleration is given by
[tex]a=\dfrac{GM}{r^2}\\\Rightarrow a=\dfrac{6.67\times 10^{-11}\times 6\times 10^{24}}{(1.5\times 10^{11})^2}\\\Rightarrow a=1.77867\times 10^{-8}\ m/s^2[/tex]
Velocity of the star
[tex]v=u+at\\\Rightarrow v=0+1.77867\times 10^{-8}\times 3\times 24\times 60\times 60\\\Rightarrow v=0.00461031264\ m/s[/tex]
The Sun's speed will be 0.00461031264 m/s
Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
Answer:
Explanation:
Under constant acceleration the average velocity of a particle is half the sum of its initial and final velocities. Is this still true if the acceleration is not constant? Explain.
A square is 1.0 m on a side. Point charges of +4.0 µC are placed in two diagonally opposite corners. In the other two corners are placed charges of +3.0 µC and -3.0 µC. What is the potential (relative to infinity) at the midpoint of the square?
Answer:
[tex]V = 1.44\times 10^{5}~V[/tex]
Explanation:
The electric potential can be found by using the following formula
[tex]V = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}[/tex]
Applying this formula to each charge gives the total potential.
[tex]V = V_1 + V_2 + V_3 + V_4\\V = \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{4\times 10^{-6}}{(\sqrt{2}/2)^2} + \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2} - \frac{1}{4\pi\epsilon_0}\frac{3\times 10^{-6}}{(\sqrt{2}/2)^2}\\V = \frac{16\times 10^{-6}}{4\pi\epsilon_0}\\V = 1.44\times 10^{5}~V[/tex]
Since the potential is a scalar quantity, it is safe to sum all the potentials straightforward. And since they all placed on the corners of a square, +3 and -3 μC charges cancel out each other.
A truck is passing over a bridge with a weight limit of 50,000 pounds. When empty, the front of the truck weighs 19,800 pounds and the back of the truck weighs 12,500 pounds. How much cargo (C), in pounds, can the truck carry and still be allowed to pass over the bridge?
Answer:
W = 17,700 lb
Explanation:
given,
Weight limit of the Bridge = 50,000 lb
Weight of empty truck = 19800 lb
Weight on the back of the truck = 12500 lb
Now,
Total weight of truck + cargo
= Weight of empty truck + Weight on the back of the truck
= 19800 + 12500
= 32300 lb
Weight of cargo which is still allowed.
W = weight limit - weight of the system at present
W = 50000 - 32300
W = 17,700 lb
Weight Truck can still carry is equal to W = 17,700 lb
An old light bulb draws only 54.3 W, rather than its original 60.0 W, due to evaporative thinning of its filament. By what factor is the diameter of the filament reduced, assuming uniform thinning along its length? Neglect any effects caused by temperature differences.
Answer:
The factor of the diameter is 0.95.
Explanation:
Given that,
Power of old light bulb = 54.3 W
Power = 60 W
We know that,
The resistance is inversely proportional to the diameter.
[tex]R\propto\dfrac{1}{D}[/tex]
The power is inversely proportional to the resistance.
[tex]P\propto\dfrac{1}{R}[/tex]
[tex]P\propto D^2[/tex]
We need to calculate the factor of the diameter of the filament reduced
Using relation of power and diameter
[tex]\dfrac{P_{i}}{P_{f}}=\dfrac{D_{i}^2}{D_{f}^2}[/tex]
Put the value into the formula
[tex]\dfrac{D_{i}^2}{D_{f}^2}=\dfrac{54.3}{60}[/tex]
[tex]\dfrac{D_{i}}{D_{f}}=0.95[/tex]
[tex]D_{i}=0.95 D_{f}[/tex]
Hence, The factor of the diameter is 0.95.
Answer:
Explanation:
Po = 60 W
P = 54.3 W
Let the initial diameter of the filament is do and the final diameter of the filament is d.
Let the voltage is V and the initial resistance is Ro and the final resistance is R.
The formula for power is given by
P = V²/R
The resistance of the filament is inversely proportional to the square of the diameter of the filament. As voltage is constant so the power is
Power α diameter²
So, initial power is
Po α do² ..... (1)
Final power is
P α d² ..... (2)
Divide equation (2) by equation (1), we get
P / Po = d² / do²
54.3 / 60 = d² / do²
d² / do² = 0.905
d = 0.95 d
Thus, the diameter of the filament is reduced to a factor of 0.95 .
What is the magnitude of the net force ∑ F on a 1.9 kg bathroom scale when a 74 kg person stands on it?
Answer:
725.2 N
Explanation:
Since it is not stated the scale, the person or both accelerated or experience weightlessness, the net force acting on the bathroom scale is the weight of the person acting downward as the person stands on the scale .
Weight = mass of a body × acceleration due to gravity
= 74 kg × 9.8 m/s²
= 725.2 N
An excited hydrogen atom emits light with a wavelength of 486.4 nm to reach the energy level for which n = 2. In which principal quantum level did the electron begin?
Answer:
The electron began in the quantum level of 4
Explanation:
Using the formula of wave number:
Wave Number = 1/λ = Rh(1/n1² - 1/n2²)
where,
Rh = Rhydberg's Constant = 1.09677 x 10^7 /m
λ = wavelength of light emitted = 486.4 nm = 486.4 x 10^-9 m
n1 = final shell = 2
n2 = initial shell = ?
Therefore,
1/486.4 x 10^-9 m = (1.09677 x 10^7 /m)(1/2² - 1/n2²)
1/4 - 1/(486.4 x 10^-9 m)(1.09677 x 10^7 /m) = 1/n2²
1/n2² = 0.06082
n2² = 16.44
n2 = 4
The principal quantum level in which the electron began is 5.
Given the following data:
Final transition = 2Wavelength = 486.4 nm = [tex]486.4 \times 10^{-9}\;m[/tex]Rydberg constant = [tex]1.09 \times 10^7\; m^{-1}[/tex]
To determine the principal quantum level in which the electron began, we would use the Rydberg equation:
Mathematically, the Rydberg equation is given by the formula:
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} -\frac{1}{n_i^2})[/tex]
Where:
[tex]\lambda[/tex] is the wavelength.R is the Rydberg constant.[tex]n_f[/tex] is the final transition.[tex]n_i[/tex] is the initial transition.Substituting the parameters into the formula, we have;
[tex]\frac{1}{486.4 \times 10^{-9}} = 1.09 \times 10^7(\frac{1}{2^2} -\frac{1}{n_i^2})\\\\\frac{1}{486.4 \times 10^{-9} \times 1.09 \times 10^7}=\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{5.3018} =\frac{1}{4} -\frac{1}{n_i^2}\\\\\frac{1}{n_i^2}=\frac{1}{4}-\frac{1}{5.3018}\\\\\frac{1}{n_i^2}=0.25-0.1886\\\\\frac{1}{n_i^2}=0.0614\\\\n_i^2=\frac{1}{0.0614} \\\\n_i=\sqrt{16.29} \\\\n_i=4.0[/tex]
Initial transition = 4.0
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Assume it takes 8.00 min to fill a 50.0-gal gasoline tank. (1 U.S. gal = 231 in.3) (a) Calculate the rate at which the tank is filled in gallons per second. .104 Correct: Your answer is correct. gal/s (b) Calculate the rate at which the tank is filled in cubic meters per second.
The volumetric rate or flow rate of a fluid is defined as the amount of the volume of a fluid circulating on a surface per unit of time. In this case we have units given initially: Gallons and minutes. For the first part we will convert the minutes to seconds, and we will obtain the flow rate under that measure. For the second case we will convert the gallons to cubic meters and obtain the desired value. Recall the following conversion rates,
[tex]1 min = 60s[/tex]
[tex]1 U.S Gal = 0.00378541178 m^3[/tex]
If the flow rate is defined as the volume by time, the flow rate with the given values is
[tex]Q = \frac{V}{t}[/tex]
[tex]Q = \frac{50Gal}{8min}[/tex]
[tex]Q = 6.25 Gal/min[/tex]
PART A ) Converting to Gal/seconds, we have,
[tex]Q = 6.25 \frac{Gal}{min}(\frac{1min}{60s})[/tex]
[tex]Q = 0.10416Gal/s[/tex]
PART B) Converting Gal/seconds to [tex]m^3/s[/tex]
[tex]Q = 0.104116\frac{Gal}{s} (\frac{0.00378541178 m^3}{1 Gal})[/tex]
[tex]Q = 3.941*10^{-4}m^3/s[/tex]
The rate of filling the gasoline tank is 0.104 gallons per second or approximately 0.000383 cubic meters per second.
Explanation:To calculate the rate at which the gasoline tank is being filled, we need to first convert the given quantities into the relevant units. Given that the gasoline tank is 50.0 gallons and it takes 8.00 minutes to fill it, the flow rate is 50/480 (since 8 min = 480 s) = 0.104 gallons/second. Since 1 US gallon = 231 cubic inches, this gives us a flow rate of 0.104 x 231 = 23.5 cubic inches per second.
To convert this to the rate in cubic meters per second, we use the fact that 1 inch = 0.254 cm and 1 cubic meter = 1,000,000 cubic cm. Therefore, 23.5 cubic inches = 23.5 x 0.254^3 cubic meters = approximately 0.000383 cubic meters per second (m^3/s).
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Under what conditions does the magnitude of the average velocity equal the average speed?
Final answer:
The magnitude of the average velocity equals the average speed when the direction of motion doesn't change and the speed is constant. These conditions are met in straightforward travel without directional change or speed variations. For round trips or trips involving direction changes, the average speed may differ from the magnitude of the average velocity.
Explanation:
The conditions under which the magnitude of the average velocity equals the average speed occur when the motion does not involve a change in direction. The average speed is calculated by dividing the total distance traveled by the elapsed time, while the magnitude of the average velocity is the total displacement divided by the elapsed time. For these two quantities to be the same, the direction of travel must remain constant, meaning there is no reversal or change in direction.
When you take a road trip and do not change direction, and your speed is consistent, then your average speed is equal to the magnitude of the average velocity. However, if you ended up back at your starting point, despite having moved, your displacement would be zero, and hence so would your average velocity, even though your average speed is greater than zero.
If you're calculating the ratio of the total distance as shown on the car's odometer to the time of the trip, you're calculating average speed. The speedometer of a car measures instantaneous speed, not velocity, because it does not provide information about direction.
a large parallel plate capacitor has plate seperation of 1.00 cm and plate area of 314 cm^2. The capacitor is connected across a voltage of 20.0 V and has air betweeen the plates. How much work is done on the capacitor as the plate seperation is inceased to 2.00 cm?
Answer:
[tex]W = -2.76\times 10^{-9}~J[/tex]
Explanation:
The work done on the capacitor is equal to the difference in potential energy stored in the capacitor in two different cases.
The potential energy is given by the following formula:
[tex]U = \frac{1}{2}CV^2[/tex]
where C can be calculated using the plate separation and area.
[tex]C = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.01} = 3.14\epsilon[/tex]
Therefore, the potential energy in the first case is
[tex]U = \frac{1}{2}3.14\epsilon (20)^2 = 628\epsilon[/tex]
In the second case:
[tex]C_2 = \epsilon\frac{A}{d} = \epsilon\frac{0.0314}{0.02} = 1.57\epsilon\\U = \frac{1}{2}C_2 V^2 = \frac{1}{2}1.57\epsilon (20)^2 = 314\epsilon[/tex]
The permittivity of the air is very close to that of vacuum, which is 8.8 x 10^-12.
So, the difference in the potential energy is
[tex]W = U_2 - U_1 = \epsilon(314 - 628) = -314 \times 8.8 \times 10^{-12} = -2.76\times 10^{-9}~J[/tex]
An electron is accelerated eastward at 1.06 109 m/s2 by an electric field. Determine the magnitude and direction of the electric field.
Answer:
Electric field, [tex]E=6.02\times 10^{-3}\ N/C[/tex] to the west direction.
Explanation:
Given that,
Acceleration of the electron, [tex]a=1.06\times 10^9\ m/s^2[/tex] (eastwards)
We need to find the magnitude and direction of the electric field. From Newton's law and electrostatic force,
ma = qE
[tex]E=\dfrac{ma}{q}[/tex]
[tex]E=\dfrac{9.1\times 10^{-31}\times 1.06\times 10^9}{1.6\times 10^{-19}}[/tex]
[tex]E=6.02\times 10^{-3}\ N/C[/tex]
The direction of electric field is in opposite direction of the acceleration of the electron. So, the electric field is acting in west direction.
The magnitude of the electric field accelerating an electron eastward is 6.04 × 10⁴ N/C, and its direction is westward since the electron has a negative charge.
Explanation:To determine the magnitude and direction of the electric field that accelerates an electron eastward, we can use the formula F = qE, where F is the force, q is the charge of the electron, and E is the electric field strength. The force exerted on the electron can be found using Newton's second law, F = ma, where m is the mass of the electron (9.11 × 10⁻³¹ kg) and a is the acceleration (1.06 × 10⁹ m/s²). Thus, F = (9.11 × 10⁻³¹ kg)(1.06 × 10⁹ m/s²) = 9.66 × 10⁻¹¹ N. We then solve for E by rearranging the formula to E = F/q, with q being the charge of an electron (-1.60 × 10⁻¹¹ C). Thus, E = (9.66 × 10⁻¹¹ N) / (-1.60 × 10⁻¹¹ C) = -6.04 × 10⁴ N/C. The negative sign indicates that the electric field points westward since it accelerates the electron eastward, and the electron has a negative charge.
In reaching her destination, a backpacker walks with an average velocity of 1.20 m/s, due west. This average velocity results, because she hikes for 5.63 km with an average velocity of 2.33 m/s due west, turns around, and hikes with an average velocity of 0.374 m/s due east. How far east did she walk (in kilometers)?
Answer:
[tex] x_2 = 648.46\ m[/tex]
Explanation:
given,
Average velocity due west = 1.20 m/s
case 1
Distance moved in west, x₁ = 5.63 km
speed due west, v₁ = 2.33 m/s
Case 2
Distance moved in east = x₂
speed due east, v₂ = 0.374 m/s
total distance = x₁ + x₂ = 5.62
total time = t₁ + t₂ = 5630/2.33 + x₂/0.374
now,
[tex]average\ velocity = \dfrac{total\ distance}{total\ time}[/tex]
[tex]-1.20= \dfrac{-5630 + x_2}{\dfrac{5630}{2.33}+\dfrac{x_2}{0.374}}[/tex]
negative sign is used because we want the distance in east but velocity is in west.
[tex] - 2900 - 3.21 x_2 = -5630 + x_2[/tex]
[tex]-4.21 x_2 = -2730[/tex]
[tex] x_2 = 648.46\ m[/tex]
The distance she walked in east is equal to 648.46 m
If an electric current of 8.50 A flows for 3.75 hours through an electrolytic cell containing copper-sulfate (CuSO4) solution, then how much copper is deposited on the cathode (the negative electrode) of the cell? (Copper ions carry two units of positive elementary charge, and the atomic mass of copper is 63.5 g/mol.)
Answer:
75.5g
Explanation:
From the ionic equation, we can write
[tex]CU^{2+}+SO^{2-}_{4}\\[/tex]
next we find the number of charge
Note Q=it
for i=8.5A, t=3.75 to secs 3.75*60*60=13500secs
hence
[tex]Q=8.5*13500\\Q=114750C[/tex]
Since one faraday represent one mole of electron which equal 96500C
Hence the number of mole produced by 114750C is
114750/96500=1.2mol
The mass of copper produced is
[tex]mol=\frac{mass}{molar mass} \\mass=mole*molar mass\\mass=1.2*63.5\\mass=75.5g[/tex]
Hence the amount of copper produced is 75.5g
Answer:
75.5 g.
Explanation:
Dissociation equation:
Cu2+ + SO4^2- --> CuSO4
Q = I * t
Where,
I = current
= 8.5 A
t = time
= 3.75 hours
= 13500 s
Q = 8.5 * 13500
= 114750 C
1 faraday represent one mole of electron which equal 96500C
Number of mole of Cu2+
= 114750/96500
= 1.19 mol.
Mass = number of moles * molar mass
= 1.19 * 63.5
= 75.5 g.
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. The properties of steam at 220°C are given as follows: vf = 0.001190 m3/kg and vg = 0.08609 m3/kg.
Answer:
a) P = 2319.6[kPa]; b) 2.6%
Explanation:
Since the problem data is not complete, the following information is entered:
A 1.78-m3 rigid tank contains steam at 220°C. One-third of the volume is in the liquid phase and the rest is in the vapor form. Determine (a) the pressure of the steam, and (b) the quality of the saturated mixture.
From the information provided in the problem we can say that you have a mixture of liquid and steam.
a) Using the steam tables we can see (attached image) that the saturation pressure at 220 °C is equal to:
[tex]P_{sat} =2319.6[kPa][/tex]
[tex]v_{f}=0.001190[m^{3}/hr]\\v_{g}=0.08609[m^{3}/hr]\\[/tex]
b) Since the specific volume of the gas and liquid is known, we can find the mass of each phase using the following equation:
[tex]m_{f}=\frac{V_{f} }{v_{f} } \\m_{g}=\frac{V_{g} }{v_{g} } \\where:\\V_{f}=volume of the fluid[m^3]\\v_{f}=specific volume of the fluid [m^3/kg]\\[/tex]
We know that the volume of the fluid is equal to:
[tex]V_{f}=1/3*V_{total} \\V_{total}=1.78[m^3]\\[/tex]
Now we can find the mass of the gas and the liquid.
[tex]m_{f}=\frac{1/3*1.78}{0.001190} \\m_{f}=498.6[kg]\\m_{g}=\frac{2/3*1.78}{0.08609}\\m_{g}=\ 13.78[kg][/tex]
The total mass is the sum of both
[tex]m_{total} =m_{g} + m_{fluid} \\m_{total} = 498.6 + 13.78\\m_{total} = 512.38[kg][/tex]
The quality will be equal to:
[tex]x = \frac{m_{g} }{m_{T} }\\ x= \frac{13.78}{512.38} \\x = 0.026 = 2.6%[/tex]
A moving electron passes near the nucleus of a gold atom, which contains 79 protons and 118 neutrons. At a particular moment the electron is a distance of 7.5 × 10−9 m from the gold nucleus. (a) What is the magnitude of the electric force exerted by the gold nucleus on the electron?
Answer:
[tex]F=3.2345*10^{-10}N[/tex]
Explanation:
Given data
Distance r=7.5×10⁻⁹m
Charge of electron -e= -1.6×10⁻¹⁹C
Charge of proton e=1.6×10⁻¹⁹C
To find
Electric force F
Solution
From Coulombs law we know that:
[tex]F=K\frac{q_{1}q_{2} }{r^{2} }[/tex]
q₁ is charge of electron
q₂ is the charge of gold nucleus which contains 79 positively charge protons and 118 neutral neutrons.
The Charge of single proton e=1.6×10⁻¹⁹C
79 proton charge q₂=79×1.6×10⁻¹⁹=1.264×10⁻¹⁷C
So
[tex]F=\frac{1}{4\pi *8.85*10^{-12} } \frac{-1.6*10^{-19}*1.264*10^{-17}}{(7.5*10^{-9})^{2} }\\ F=3.2345*10^{-10}N[/tex]
The magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].
What is electric force?Electric force is the force of attraction or repulsion between two charged particles. It can be given as,
[tex]F=K\dfrac{q_1\times q_2}{r^2}[/tex]
Here [tex]k[/tex] is coulomb's constant, [tex]q[/tex] is charge on the objects and [tex]r[/tex] is the distance between two objects.
Given information-
The number of proton in gold atom is 79.
The number of neutrons in gold atom is 118.
The distance of the electron from the nucleus is [tex]7.5 \times 10^{-9} \rm m[/tex].[tex]r[/tex]
a) The magnitude of the electric force exerted by the gold nucleus on the electron-The charge on electron is [tex]-1.6\times 10^{-19} C[/tex] and the charge on the proton is [tex]1.6\times 10^{-19} C[/tex].
Put the values in the above equation as,
[tex]F=8.98\times10^9\times\dfrac{(79\times1.6\times10^{-19})\times1.6\times10^{-19}}{(7.5\times10^{-19})^2}\\F=3.235\times10^{-10}\rm N[/tex]
Hence the magnitude of the electric force exerted by the gold nucleus on the electron [tex]3.235\times10^{-10}\rm N[/tex].
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Is the electric-field magnitude between the plates larger or smaller than that for the original capacitor?
Answer:
The magnitude of the electric field will decrease
Explanation:
The capacitance of a parallel plate capacitor having plate area A and plate separation d is C=ϵ0A/d.
Where ϵ0 is the permittivity of free space.
A capacitor filled with dielectric slab of dielectric constant K, will have a new capacitance C1=ϵ0kA/d
C1=K(ϵ0A/d)
C1=KC
Where C is the capacitance with no dielectric.
The new capacitance is k times the capacitance of the capacitor without dielectric slab.
This implies that the charge storing capacity of a capacitor increases k times that of the capacitor without dielectric slab.
The charge stored in the original capacitor Q=CV
The charge stored in the original capacitor after inserting dielectric Q1=C1V1
The law of conservation of energy states that the energy stored is constant:
i.e Charge stored in the original capacitor is same as charge stored after the dielectric is inserted.
Q = Q1
CV = C1V1
CV = C1V1 -------2
We derived C1=KC. Inserting this into equation 2
CV = KCV1
V1 = (CV)/KC
V/K
This implies the voltage decreases when a dielectric is used within the plate.
The relationship between electric field and potential voltage is a linear one
V= Ed
Therefore the electric field will decrease
Final answer:
The capacitance of a parallel-plate capacitor with non-uniform electric field due to increased plate separation will be less than the ideal scenario calculated by C = € A/d.
Explanation:
When the separation of the plates in a parallel-plate capacitor is not small enough to maintain a uniform electric field, the field lines will bulge outwards at the edges. This bulging implies that some of the field lines that emanate from one plate do not reach the opposite plate, reducing the effective area over which the field is acting. Consequently, this non-uniform field means that the capacitance of the capacitor will be less than the ideal calculation using the formula C = € A/d, where C is the capacitance, € is the permittivity of the medium between the plates, A is the plate area, and d is the separation between the plates.
A solid nonconducting sphere of radiusRcarries a chargeQdistributed uniformly throughout itsvolume. At a certain distancer1(r1< R) from the center of the sphere, the electric field has magnitudeE.If the same chargeQwere distributed uniformly throughout a sphere of radius 2R, the magnitude of theelectric field at the same distancer1from the center would be equal to______
Answer:
[tex]E' = \frac{1}{8} E[/tex]
Explanation:
Given data:
first case
Distance of electric field from center of sphere is r_1 <R
Electric field at r_1< R
[tex]E = \frac{kQr_1}{R^3}[/tex]
second case
Distance of electric field from centre of sphere is r_1 < 2R
Electric field at r_1< 2R
[tex]E' = \frac{kQr_1}{8R^3}[/tex]
so, we have
[tex]E' = \frac{1}{8} E[/tex]
What is the magnitude of a point charge that would create an electric field of 1.18 N/C at points 0.822 m away?
Answer:
q = 8.85 x 10⁻¹¹ C
Explanation:
given,
Electric field, E = 1.18 N/C
distance, r = 0.822 m
Charge magnitude = ?
using formula of electric field.
[tex]E = \dfrac{kq}{r^2}[/tex]
k is the coulomb constant
[tex]q= \dfrac{Er^2}{k}[/tex]
[tex]q= \dfrac{1.18\times 0.822^2}{9\times 10^9}[/tex]
q = 8.85 x 10⁻¹¹ C
The magnitude of charge is equal to q = 8.85 x 10⁻¹¹ C
100 kg of R-134a at 280 kPa are contained in a piston-cylinder device whose volume is 8.672 m3. The piston is now moved until the volume is one-half its original size. This is done such that the pressure of the R-134a does not change. Determine the final temperature and the change in the total internal energy of the R-134a. (Round the final answers to two decimal places.)
Answer:
∆u =-111.8 kJ/kg
T_fin=-10.09℃
Explanation:
note:
solution is attached in word file due to some technical issue in mathematical equation. please find the attached documents.
The acceleration of a bus is given by ax(t) = αt, where α = 1.2 m/s3. (a) If the bus’s velocity at time t = 1.0 s is 5.0 m/s, what is its velocity at time t = 2.0 s? (b) If the bus’s position at time t = 1.0 s is 6.0 m, what is its position at time t = 2.0 s? (c) Sketch ay-t, vy-t , and x-t graphs for the motion.
Answer:
(a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
Explanation:
Given that,
[tex]\alha=1.2\ m/s^3[/tex]
Time t = 1.0 s
Velocity = 5.0
The Acceleration equation is
[tex]a_{x(t)}=\alpha t[/tex]
We need to calculate the velocity
Using formula of acceleration
[tex]a=\dfrac{dv}{dt}[/tex]
On integrating
[tex]\int_{v_{0}}^{v}{dv}=\int_{0}^{t}{a dt}[/tex]
Put the value into the formula
[tex]v-v_{0}=1.2\int_{0}^{t}{t dt}[/tex]
[tex]v-v_{0}=0.6t^2[/tex]
[tex]v=v_{0}+0.6t^2[/tex]
Put the value into the formula
[tex]v_{0}=5.0-0.6\times(1.0)^2[/tex]
[tex]v_{0}=4.4\ m/s[/tex]
We need to calculate the velocity at 2.0 sec
Put the value of initial velocity in the equation
[tex]v=4.4+0.6\times(2.0)^2[/tex]
[tex]v=6.8\ m/s[/tex]
(b). If the bus’s position at time t = 1.0 s is 6.0 m,
We need to calculate the position
Using formula of velocity
[tex]v=\dfrac{dx}{dt}[/tex]
On integrating
[tex]\int_{x_{0}}^{x}{dx}=\int_{0}^{t}{v dt}[/tex]
[tex]x_{0}-x=\int_{0}^{t}{v_{0}dt}+\int_{0}^{t}{0.6 t^2}[/tex]
[tex]x_{0}-x=v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x=x_{0}+v_{0}t+\dfrac{0.6}{3}t^3[/tex]
[tex]x_{0}=6-4.4\times1-\dfrac{0.6}{3}\times1^3[/tex]
[tex]x=1.4\ m[/tex]
The position at t = 2.0 s
[tex]x=1.4+4.4\times2.0+\dfrac{0.6}{3}\times2^3[/tex]
[tex]x=11.8\ m[/tex]
Hence, (a). The velocity of bus at 2.0 sec is 6.8 m/s.
(b). The position of bus at 2.0 s is 11.8 m.
(c). [tex]a_{y}-t[/tex], [tex]v_{y}-t[/tex] and x-t graphs
When a sound wave moves through a medium such as air, the motion of the molecules of the medium is in what direction (with respect to the motion of the sound wave)? Group of answer choicesa. Perpendicularb. Parallalc. Anit-paralleld. Both choices B and C ara valid
Answer:
Both choices B and C are valid
Explanation:
Sound wave are Mechanical wave. Air (or viscus fluid) is the medium of propagation. Sound is produced by the back and forth vibration of the object. Consider the vibration of object is from left to right then this back and forth vibrations of object displaces the molecules of the medium both rightward and leftward (to and fro) to the direction of the energy transport forming compression and rarefaction. This shows that the motion of the molecules of the medium is both parallel (and anti-parallel) to the direction of the sound wave propagation.
Final answer:
The motion of the particles of a medium in a sound wave is parallel to the direction of the wave motion.
Explanation:
A sound wave is a longitudinal wave, which means that the motion of the particles of the medium is parallel to the direction of the wave motion. In other words, the particles of the medium vibrate or oscillate back and forth in the same direction that the sound wave is traveling. This can be compared to compressing and stretching a coiled spring. As the wave propagates through the medium, it creates zones of compression and rarefaction, causing the air molecules to move in the same direction as the sound wave.
A pair of identical 10-cm-diameter circular rings face each other. The distance between the rings is 20.0 cm . The rings each have a charge of + 20.0 nC . What is the magnitude of the electric field at the center of either ring?
Answer:
The magnitude of electric field at the center of each ring is 129.96 N/C
Explanation:
As per the question:
The diameter of the ring , d = 10 cm = 0.1 m
Radius, [tex]r = \frac{d}{2} = \frac{0.1}{2} = 0.05\ m[/tex]
Separation between the rings, d = 20.0 cm = 0.20 m
Charge on a ring, q = +20 nC = [tex]20\times 10^{- 9}\ C[/tex]
Now,
The electric field at the center of either ring is given by:
[tex]E = \frac{1}{4\pi \epsilon_{o}}\frac{qd}{(d^{2} + r^{2})^{\frac{3}{2}}}[/tex]
where
[tex]\frac{1}{4\pi \epsilon_{o}} = 9\times 10^{9}[/tex]
Thus
[tex]E = 9\times 10^{9}\times \frac{20\times 10^{- 9}\times 0.20}{(0.20^{2} + 0.05^{2})^{\frac{3}{2}}}[/tex]
E = 129.96 N/C
An electron is brought from rest infinitely far away to rest at point P located at a distance of 0.033 m from a fixed charge q. That process required 111 eV of energy from an eternal agent to perform the necessary work.
Answer:
The potential at point P is -111 Volt.
Explanation:
Given that,
Distance = 0.033 m
Work = 111 ev
Suppose what is the potential at point P?
We need to calculate the potential at point P
Using formula of potential
[tex]W=qV[/tex]
[tex]V=\dfrac{W}{q}[/tex]
Where, W = work
q = charge
Put the value into the formula
[tex]V= \dfrac{111\times1.6\times10^{-19}}{-1.6\times10^{-19}}[/tex]
[tex]V=-111\ V[/tex]
Hence, The potential at point P is -111 Volt.
The question is about the process of bringing an electron from rest to rest near a fixed charge q at a specific distance, and the energy required for this process. The answer explains the equation for the energy required per unit charge and the concept of electron volts (eV) as an energy unit.
Explanation:The question is asking about the process of bringing an electron from rest at an infinite distance to rest at a point located a distance of 0.033 m from a fixed charge q using an external agent. The process requires 111 eV of energy to perform the necessary work. The equation for the energy required per unit charge is given, which relates the potential energy to the charge and distance. The concept of electron volts (eV) as an energy unit is also mentioned.
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A bird is flying due east. Its distance from a tall building is given by x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. What is the instantaneous velocity of the bird when t = 8.00s?
Answer:
3.76 m/s
Explanation:
Instantaneous velocity: This can be defined as the velocity of an object in a non uniform motion. The S.I unit is m/s.
v' = dx(t)/dt..................... Equation 1
Where v' = instantaneous velocity, x = distance, t = time.
Given the expression,
x(t) = 28.0 m + (12.4 m/s)t - (0.0450 m/s³)t³
x(t) = 28 + 12.4t - 0.0450t³
Differentiating x(t) with respect to t.
dx(t)/dt = 12.4 - 0.135t²
dx(t)/dt = 12.4 - 0.135t²
When t = 8.00 s.
dx(t)/dt = 12.4 - 0.135(8)²
dx(t)/dt = 12.4 - 8.64
dx(t)/dt = 3.76 m/s.
Therefore,
v' = 3.76 m/s.
Hence, the instantaneous velocity = 3.76 m/s
Final answer:
To find the bird's instantaneous velocity at t = 8.00s, we differentiate its position function to get v(t) = 12.4 m/s - 0.135 m/s^2 × t^2, then substitute t = 8.00s to find v(8.00) = 3.76 m/s east.
Explanation:
The question asks for the instantaneous velocity of a bird flying due east when t = 8.00s, given the position function x(t) = 28.0 m + (12.4 m/s)t – (0.0450 m/s3)t3. To find the instantaneous velocity, we need to differentiate the position function with respect to time (t) to get the velocity function, v(t).
First, let's differentiate x(t):
Derivative of 28.0 m is 0 since it's a constant.
Derivative of (12.4 m/s)t is 12.4 m/s, as the derivative of t is 1.
Derivative of (-0.0450 m/s3)t3 is -0.135 m/s2 × t2, using the power rule for derivatives.
So, the velocity function is v(t) = 12.4 m/s - 0.135 m/s2 × t2. To find the instantaneous velocity at t = 8.00s, we plug in t = 8.00 into the velocity function:
v(8.00) = 12.4 m/s - 0.135 m/s2 × (8.002)
Calculating this gives us:
v(8.00) = 12.4 m/s - 0.135 m/s2 × 64.00 = 12.4 m/s - 8.64 m/s = 3.76 m/s
Therefore, the instantaneous velocity of the bird when t = 8.00s is 3.76 m/s east.
If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
Answer:
The magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
Explanation:
As data is incomplete here, so by seeing the complete question from the search the data is
vx_0=1.1 x 10^6
ax=0 As acceleration is zero in the horizontal axis so
Equation of motion in horizontal direction is given as
[tex]s_x=v_x_0 t[/tex]
[tex]t=\frac{s_x}{v_x}\\t=\frac{2 \times 10^{-2}}{1.1 \times 6}\\t=1.82 \times 10^{-8} s[/tex]
Now for the vertical distance
vy_o=0
than the equation of motion becomes
[tex]s_y=v_y_0 t+\frac{1}{2} at^2\\s_y=\frac{1}{2} at^2\\0.5 \times 10^{-2}=\frac{1}{2} a(1.82 \times 10^{-8})^2\\a=3.02 \times 10^{13} m/s^2[/tex]
Now using this acceleration the value of electric field is calculated as
[tex]E=\frac{F}{q}\\E=\frac{ma}{q}\\E=\frac{m_ea}{q_e}\\[/tex]
Here a is calculated above, m is the mass of electron while q is the charge of electron, substituting values in the equation
[tex]E=\frac{9.1\times 10^{-31} \times 3.02 \times 10^{13} }{1.6 \times 10^{-19}}\\E=171.76 N/C[/tex]
So the magnitude of the electric field be 171.76 N/C so that the electron misses the plate.
If the electron misses the upper plate, the magnitude of the electric field is equal to 171.88 Newton per coulomb.
Given the following data:
Distance = 2 cm to m = 0.02 meter.Vertical speed = [tex]1.6 \times 10^6[/tex] m/sVertical distance = 1 cm = [tex]\frac{0.01}{2} = 0.005\;m[/tex]Scientific data:
Mass of electron = [tex]9.1 \times 10^{-31}\;kg[/tex]Charge of electron = [tex]1.6 \times 10^{-19}\;C[/tex]To calculate the magnitude of the electric field:
First of all, we would determine the time taken by this electron to travel through the plates.
Time in the vertical direction.Mathematically, time is given by this formula:
[tex]Time = \frac{distance}{speed} \\ \\ Time = \frac{0.02}{1.10 \times 10^6} \\ \\ Time = 1.82 \times 10^{-8}\;m/s[/tex]
Next, we would find the acceleration of the electron in the vertical direction by using this formula:
[tex]a=\frac{2y}{t^2} \\ \\ a=\frac{2 \times 0.005}{(1.82 \times 10^{-8})^2}\\ \\ a=\frac{0.01}{3.31 \times 10^{-16}}\\ \\ a=3.02 \times 10^{13}\;m/s^2[/tex]
The formula for electric field.Mathematically, the electric field is given by this formula:
[tex]E=\frac{ma}{q}[/tex]
Where:
q is the charge.a is the acceleration.m is the mass.Substituting the given parameters into the formula, we have;
[tex]E=\frac{9.1 \times 10^{-31} \times 3.02 \times 10^{13}}{1.6 \times 10^{-19}}\\ \\ E=\frac{2.75 \times 10^{-17}}{1.6 \times 10^{-19}}[/tex]
Electric field, E = 171.88 N/C.
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Complete Question:
An electron is projected with an initial speed v0 = [tex]1.6 \times 10^6[/tex] m/s into the uniform field between the parallel plates. The distance between the plates is 1 cm and the length of the plates is 2 cm. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. E = N/C
(a) If the electron just misses the upper plate as it emerges from the field, find the magnitude of the electric field.
A firefighter who weighs 712 N slides down a vertical pole with an acceleration of 3.00 m/s 2 ,directed downward.What are the (a) magnitude and (b) direction (up or down) of the vertical force on the firefighter from the pole and the (c) magnitude and (d) di- rection of the vertical force on the pole from the firefighter
Answer:
494.262996942 N
Upward
494.262996942 N
Downward
Explanation:
W = Weight of the firefighter = 712 N
g = Acceleration due to gravity = 9.81 m/s²
Mass is given by
[tex]m=\dfrac{W}{g}\\\Rightarrow m=\dfrac{712}{9.81}\\\Rightarrow m=72.5790010194\ kg[/tex]
Force is given by
[tex]T-W=-ma\\\Rightarrow T=W-ma\\\Rightarrow T=712-72.5790010194\times 3\\\Rightarrow T=494.262996942\ N[/tex]
The force on the firefighter is 494.262996942 N
directed upward
On the pole the force will be the same 494.262996942 N
But the direction will be downward
Final answer:
The magnitude of the vertical force on the firefighter from the pole is determined by the net force required for the given downward acceleration. The direction of the force the firefighter applies to the pole is downward while the reaction force from the pole is upward, with the magnitude being the same for both forces.
Explanation:
The question relates to the physics concept called Newton's second law of motion, which states that the acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the object's mass. The law is usually expressed by the equation F = ma, where F is the net force, m is the mass, and a is the acceleration.
To find the magnitude of the vertical force on the firefighter from the pole, we need to take into account both the downward force due to gravity (which is the firefighter's weight) and the additional force needed to produce the acceleration. If the firefighter weighs 712 N and has an acceleration of 3.00 m/s2 downward, we can calculate the force exerted on the pole using F = ma. The mass (m) of the firefighter can be obtained by dividing the weight (W) by the acceleration due to gravity (g), m = W/g. From F = ma, we get F = (W/g)a. Inserting the given values, we have F = (712 N/9.8 m/s2) × 3.00 m/s2. The resulting force is less than the weight of the firefighter because the net force is reduced by the downward acceleration. The direction of the force exerted by the firefighter on the pole is downward since the firefighter is sliding down.
The reaction force exerted on the firefighter by the pole, according to Newton's third law, is equal in magnitude but opposite in direction to the force exerted by the firefighter on the pole. So, the magnitude would be the same, and the direction would be upward.
A particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the origin, at t = 0 and moves to the right. The amplitude of its motion is 2.00 cm, and the frequency is 1.50 Hz(a) show that the position of the particle is given by x = (2.00 cm) sin(3.00πt) (b) the maximum speed and the earliest time (t > 0) at which the particle has this speed, (c) the maximum
acceleration and the earliest time (t > 0) at which the particle has this acceleration, and (d) the total distance traveled between t = 0 and t = 1.00 s.
Answer:
(a). [tex]x=2.00\sin(3.00\pi t)[/tex], hence proved
(b). The maximum speed is 18.8 cm/s.
(c). The maximum acceleration is 177.65 cm/s².
(d). The total distance is 12 cm.
Explanation:
Given that,
Amplitude = 2.00 cm
Frequency = 1.50 Hz
Given equation of position of the particle is
[tex]x=2.00\ sin(3.00\pit)[/tex]
(a). show that the position of the particle is given by
[tex]x=2.00\sin(3.00\pi t)[/tex]
We know the general equation of S.H.M
[tex]x=A\sin(\omega t[/tex]...(I)
At t =0, x = 0
On differentiating equation (I)
[tex]v=\dfrac{dx}{dt}[/tex]
[tex]v=A\omega\cos(\omega t)[/tex]
At t = 0, the particle moving to the right
[tex]V=A\omega[/tex] > 0
Given statement is true.
The equation of position is
[tex]x=A\sin(\omega t)[/tex]
here, [tex]\Omega= 2\pi f[/tex]
Put the value in the equation
[tex]x=2.00\sin(2\times1.50\pi t)[/tex]
[tex]x=2.00\sin(3.00\pi t)[/tex]
Hence proved.
(b). We need to calculate the maximum speed
[tex]V=A\omega\cos(\omega t)[/tex]....(II)
At t = 0,
[tex]V_{max}=A\omega[/tex]
Put the value into the formula
[tex]V_{max}=2.00\times2\pi\times1.50[/tex]
[tex]V_{max}=6\pi[/tex]
[tex]V_{max}=18.8\ cm/s[/tex]
(c). We need to calculate the maximum acceleration
Using equation (II)
[tex]V=A\omega\cos(\omega t)[/tex]
On differentiating
[tex]a=\dfrac{dV}{dt}[/tex]
[tex]a=-A\omega^2\sin(\omega t)[/tex]
[tex]a_{max}\ when\ \sin(\omega t)\ is\ -1[/tex]
[tex]a_{max}=-A\omega^2\times-1[/tex]
[tex]a=A\omega^2[/tex]
[tex]a_{max}=2\times(3\pi)^2\approx 177.65 cm/s^2[/tex]
(d). We need to calculate the total distance traveled between t = 0 and t = 1.00 s
Using equation (II)
[tex]V=A\omega\cos(\omega t)[/tex]
On integration
[tex]\int{V}=\int_{t}^{t'}{A\omega\cos(\omega t)}[/tex]
Put the vale into the formula
[tex]\int{V}=\int_{0}^{1}{A\omega\cos(\omega t)}[/tex]
[tex]D=\int_{0}^{1}|6\pi\cos\left(3\pi t\right)|dt[/tex]
[tex]D=12\ cm[/tex]
Hence, (b). The maximum speed is 18.8 cm/s.
(c). The maximum acceleration is 177.65 cm/s².
(d). The total distance is 12 cm.
A particle in simple harmonic motion can be described by the equation x = (2.00 cm)sin(3.00πt). The maximum speed and maximum acceleration occur when the particle is at its maximum displacement from the equilibrium position. The total distance traveled between t = 0 and t = 1.00 s is 4.00 cm.
Explanation:The position of the particle in simple harmonic motion is given by the equation x = (2.00 cm)sin(3.00πt), where x represents the displacement from the equilibrium position and t represents time in seconds.
(b) The maximum speed of the particle is equal to the amplitude of the motion multiplied by the angular frequency, vmax = (2.00 cm)(2π)(1.50 Hz). The earliest time at which the particle reaches this maximum speed is when it passes through its equilibrium position, t = 0.
(c) The maximum acceleration of the particle is equal to the amplitude of the motion multiplied by the angular frequency squared, amax = (2.00 cm)(2π)(1.50 Hz)2. The earliest time at which the particle reaches this maximum acceleration is when it is at its maximum displacement from the equilibrium position, which occurs at t = 0.
(d) To find the total distance traveled between t = 0 and t = 1.00 s, we calculate the area under the velocity versus time graph. Since the particle is in simple harmonic motion, the velocity varies sinusoidally, and the total distance traveled is equal to two times the amplitude of the motion, dtotal = 2(2.00 cm) = 4.00 cm.
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A series RL circuit with L = 3.00 H and a series RC circuit with C = 3.00 F have equal time constants. If the two circuits contain the same resistances R, (a) what is the value of R and (b) what sit the time constant?
Answer:
(a) R = 1Ω
(b) τ = 3
Explanation:
The time constants of the given circuits are as follows
[tex]\tau_{RL} = \frac{L}{R}\\\tau_{RC} = RC[/tex]
If the two circuits have equal time constants, then
[tex]\frac{L}{R} = RC\\R^2 = \frac{L}{C} = \frac{3}{3} = 1\\R = 1\Omega[/tex]
Therefore, the time constant in any of the circuits is
[tex]\tau = RC = 3[/tex]
(a). Value of resistance R is 1 ohm.
(b). The Value of time constant will be 3 second.
The time constant is defined as, time taken by the system to reach at 63.2% of its final value.
Time constant in RL circuit is, [tex]=\frac{L}{R}[/tex]
Time constant in RC circuit is, [tex]=RC[/tex]
Since, in question given that both circuit have same time constant.
So, [tex]RC=\frac{L}{R}\\\\R^{2}=\frac{L}{C}\\\\R=\sqrt{\frac{L}{C} }[/tex]
substituting L = 3H and C = 3F in above expression.
[tex]R=\sqrt{\frac{3}{3} }=1ohm[/tex]
Time constant = [tex]RC=1*3=3s[/tex]
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