Solution:
a) We know acceleration due to gravity, g = GM/r²
Differential change, dg/dr = -2GM/r³
Here, r = 50*Rh = 50*2GM/c² = 100GM/c ²
My height, h=dr = 1.7 m
Difference in gravitational acceleration between my head and my feet, dg = -10 m/s²
or, dg/dr = -10/1.7 = -2GM/(100GM/c²)³
or, 5.9*100³*G²*M² = 2c⁶
or, M = 0.59*c³/(1000G) = 2.39*1032 kg = [(2.39*1032)/(1.99*1030 )]Ms = 120*Ms
Mass of black hole which we can tolerate at the given distance is 120 time the mass of Sun.
b) This limit an upper limit ,we can tolerate smaller masses only.
The Mistuned Piano Strings Two identical piano strings of length 0.800 m are each tuned exactly to 480 Hz. The tension in one of the strings is then increased by 1.0%. If they are now struck, what is the beat frequency between the fundamentals of the two strings? SOLUTION Conceptualize As the tension in one of the strings is changed, its fundamental frequency changes. Therefore, when both strings are played, they will have different frequencies and beats be heard. Categorize We must combine our understanding of the waves model for strings with our new knowledge of beats.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The answer is
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
Explanation:
From the question we are told that
The first string has a frequency of [tex]f_1 = 230 Hz[/tex]
The period of the beat is [tex]t_{beat} = 0.99s[/tex]
Generally the frequency of the beat is
[tex]f_{beat} = \frac{1}{t_{beat}}[/tex]
Substituting values
[tex]f_{beat} = \frac{1}{0.99}[/tex]
[tex]= 1.01 Hz[/tex]
From the question
[tex]f_2 - f_1 = f_{beat}[/tex] for [tex]f_2[/tex] having a higher tension
So
[tex]f_2 - 230 = 1.01[/tex]
[tex]f_2 = 231.01Hz[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(231.01)^2}{(230)^2}[/tex]
[tex]T_2 = 1.008[/tex] % higher than [tex]T_1[/tex]
For [tex]f_2[/tex] having a lower tension
[tex]f_1 - f_2 = f_{beat}[/tex]
So
[tex]230 - f_2 = 1.01[/tex]
[tex]f_2 = 230 -1.01[/tex]
[tex]= 228.99[/tex]
From the question
[tex]\frac{f_2}{f_1} = \sqrt{\frac{T_2}{T_1} }[/tex]
[tex]\frac{T_2}{T_1} = \frac{f_2^2}{f_1^2}[/tex]
Substituting values
[tex]\frac{T_2}{T_1} = \frac{(228.99)^2}{(230)^2}[/tex]
[tex]T_2 = 0.99[/tex] % lower than [tex]T_1[/tex]
Determine the location of the center of mass of a "L" whose thin vertical and horizontal members have the same length L and the same mass M. Use the formal definition to find the x and y coordinates, and check your result by doing the calculation with respect to two different origins, one in the lower left corner at the intersection of the horizontal and vertical members and one at the top of the vertical member.
(a) Origin at the lower left
x = ?
y = ?
(b) Origin at the top of the vertical member
x = ?
y = ?
Answer:
a) x_{cm} = L / 2 , y_{cm}= L/2, b) x_{cm} = L / 2 , y_{cm}= L/2
Explanation:
The center of mass of a body is the point where all external forces are applied, it is defined by
[tex]x_{cm}[/tex] = 1 / M ∑ [tex]x_{i} m_{i}[/tex] = 1 /M ∫ x dm
[tex]y_{cm}[/tex] = 1 / M ∑ [tex]y_{i} m_{i}[/tex] = 1 / M ∫ y dm
where M is the total body mass
Let's calculate the center of mass of our L-shaped body, as formed by two rods one on the x axis and the other on the y axis
a) let's start with the reference zero at the left end of the horizontal rod
let's use the concept of linear density
λ = M / L = dm / dl
since the rod is on the x axis
dl = dx
dm = λ dx
let's calculate
x_{cm} = M ∫ x λ dx = λ / M ∫ x dx
x_{cm} = λ / M x² / 2
we evaluate between the lower integration limits x = 0 and upper x = L
x_{cm} = λ / M (L² / 2 - 0)
we introduce the value of the density that is cosntnate
x_{cm} = (M / L) L² / 2M
x_{cm} = L / 2
We repeat the calculation for verilla verilla
λ = M / L = dm / dy
y_{cm} = 1 / M ∫ y λ dy
y_{cm} = λ M y² / 2
[tex]y_{cm}[/tex] = M/L 1/M (L² - 0)
y_{cm}= L/2
b) we repeat the calculation for the origin the reference system is top of the vertical rod
horizontal rod
x_{cm} = 1 / M ∫λ x dx = λ/M x² / 2
we evaluate between the lower limits x = 0 and the upper limit x = -L
x_{cm} = λ / M [(-L)²/2 - 0] = (M / L) L² / 2M
x_{cm} = L / 2
vertical rod
y_{cm} = 1 / M ∫y dm
y_{cm} = λ / M ∫y dy
y_{cm} = λ / M y2 / 2
we evaluate between the integration limits x = 0 and higher x = -L
y_{cm} = (M / L) 1 / M ((-L)²/2 -0)
y_{cm} = L / 2
To find the center of mass of the 'L' structure, use the formal definition and consider the midpoint of each member. The x-coordinate is halfway between the ends and the y-coordinate is halfway between the top and bottom ends.
Explanation:To determine the location of the center of mass of the 'L' structure, we can use the formal definition of center of mass. Since the thin vertical and horizontal members have the same length and mass, the center of mass of each member is located at its midpoint. The x-coordinate of the center of mass is halfway between the x-coordinate of the left end and the x-coordinate of the right end of the 'L'. The y-coordinate of the center of mass is halfway between the y-coordinate of the bottom end and the y-coordinate of the top end of the 'L'.
(a) Origin at the lower left:
x = -L/4, y = L/4
(b) Origin at the top of the vertical member:
x = L/4, y = -L/4
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The centripetal acceleration might better be expressed as −ω2r⃗ (t)−ω2r→(t) because it is a vector. The magnitude of the centripetal acceleration is v2radial/Rvradial2/R. The magnitude of the centripetal acceleration is v2tangential/Rvtangential2/R. A particle that is going along a path with local radius of curvature RRR at speed sss experiences a centripetal acceleration −s2/R−s2/R. If you are in a car turning left, the force you feel pushing you to the right is the force that causes the centripetal acceleration. In these statements vradialvradialv_radial refers to the component of the velocity of an object in the direction toward or away from the origin of the coordinate system or the rotation axis. Conversely, vtangentialvtangentialv_tangential refers to the component of the velocity perpendicular to vradialvradialv_radial. Identify the statement or statements that are false.
Answer:
false b) a = v²(radial) / r and e)
Explanation:
Let's review given statement separately
a) centripetal acceleration
a = v² / r
linear and angular velocity are related
v = w r
we substitute
a = w²r
this acceleration is directed to the center of the circle, so the vector must be negative
a = - w r2
the bold are vector
True this statement
b) the magnitude is the scalar value
a = v² / r
where v is the tangential velocity, not the radial velocity, so this statement is
False
c) This is true
a = v²/ r
this speed is tangential
d) Newton's second law is
F = m a
if the acceleration is centripetal
F = m (- v² / r)
we substitute
F = m (- s² / R)
the statement is true
e) when the car turns to the left, the objects have to follow in a straight line, which is why
you need a force toward the center of the circle to take the curve.
Consequently there is no outward force
This statement is false
An infinite slab of charge of thickness 2z0 lies in the xy-plane between z=−z0 and z=+z0. The volume charge density rho(C/m3) is a constant. Part A Use Gauss's law to find an expression for the electric field strength inside the slab (−z0≤z≤z0). Express your answer in terms of the variables rho, z, z0, and constant ϵ0.
Final answer:
The electric field inside an infinite slab with a constant volume charge density is found using Gauss's law, resulting in an expression where the electric field is directly proportional to both the distance from the midplane and the charge density, and inversely proportional to the permittivity of free space.
Explanation:
Finding the Electric Field Inside an Infinite Slab Using Gauss's Law
To find the electric field strength inside an infinite slab of charge lying in the xy-plane and having a uniform volume charge density (ρ), we employ Gauss's law. Gauss's law states that the net electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). For an infinite slab of thickness 2z0, positioned between z = -z0 and z = +z0, we choose a Gaussian surface in the form of a rectangular box (pillbox) that extends symmetrically above and below the xy-plane. This choice is motivated by the symmetry of the slab and the desire to have an electric field that is either parallel or perpendicular to the faces of the box.
The charge enclosed by the Gaussian surface is given by Q = ρA(2z), where A is the cross-sectional area of the box parallel to the xy-plane and 2z is the extent of the box in the z-direction within the slab. Applying Gauss's law, the electric flux through the box is Ψ = EA = Q/ε0, where E is the magnitude of the electric field inside the slab and is directed along the z-axis. Simplifying, we find E = ρz/ε0. It follows that the electric field strength inside the slab, for -z0 ≤ z ≤ z0, is directly proportional to the distance z from the midplane and to the volume charge density ρ, and inversely proportional to the permittivity of free space ε0.
Using Gauss's law, the electric field strength inside the infinite slab of charge is found to be E = (ρ * z) / ε0.
To find the electric field strength inside an infinite slab of charge with uniform volume charge density ρ, we will use Gauss's law.
The slab extends from z = -z₀ to z = z₀ in the z-direction.
Step-by-Step Solution:
Consider a Gaussian surface in the form of a pillbox centered at a point within the slab at height z, with thickness 2z and cross-sectional area A.By symmetry, the electric field E points in the z direction and is perpendicular to the faces of the pillbox.The charge enclosed within this Gaussian surface is ρ * A * (z + z₀).Gauss's law states that the net electric flux through this surface is Φ = ∮E・dA = q_enclosed / ε₀.The electric flux through the pillbox is Φ = 2 * E * A, as the field is perpendicular to two faces of area A.Using Gauss's law: 2E * A = (ρ * A * (z + z₀)) / ε₀Simplifying, E = (ρ * (z + z₀)) / (2ε₀).The electric field strength inside the slab, for -z0 ≤ z ≤ z0, is given by:
E = (ρ * z) / ε₀
A toy car has a 1.5 A current, and its resistance is 2 . How much voltage does the car require?
Answer:
3 Volts
Explanation:
Using Ohm's Law
Voltage=current x resistance
V = I x R
V = 1.5 x 2
V = 3 V
A piece of glass has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and glass are 1.40 and 1.50, respectively, what is the thickness of the film if we see a bright reflection in the visible range
Answer:
200 nm
Explanation:
Refractive index of gasoline = 1.4
Wavelength = 560 nm
t = Thickness of film
m = Order = 1
Wavelength is given by
[tex]\lambda=\dfrac{560}{1.4}=400\ nm[/tex]
We have the relation
[tex]2t=m\lambda\\\Rightarrow t=\dfrac{m\lambda}{2}\\\Rightarrow t=\dfrac{1\times 400}{2}\\\Rightarrow t=200\ nm[/tex]
The thickness of the film is 200 nm
Problem 2: A string is wrapped around a uniform disk of mass M = 1.2 kg and radius R = 0.11 m. Attached to the disk are four low mass rods of radius b = 0.14 m, each with a small mass m = 0.4 kg at the end. The device is initially at rest on a nearly frictionless surface. Then you pull the string with a constant force F = 21 N for a time of ∆t=0.2 s. (Mom. of inertia of a disk I_(disk,CM)=1/2 MR^2 and point mass is I_(point mass)=mb^2.) Part 1. What is the angular speed of the apparatus? Part 2: What is the angle through which the apparatus turns in radians or degrees?
Answer:
the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]
the angle through which the apparatus turns in radians or degrees is : [tex]\theta = 68.54^0[/tex]
Explanation:
Given that :
mass of uniform disk M = 1.2 kg
Radius R = 0.11 m
lower radius (b) = 0.14 m
small mass (m) = 0.4 kg
Force (F) = 21 N
time (∆t) =0.2 s
Moment of Inertia of [tex]I_{disk, CM}[/tex] = [tex]\frac{1}{2}MR^2[/tex]
= [tex]\frac{1}{2}*1.2*0.11^2[/tex]
= 0.00726 kgm²
Point mass [tex]I_{point \ mass}[/tex] = mb²
But since four low rods are attached ; we have :
[tex]I_{point \ mass}[/tex] = 4 × mb²
= 4 × 0.4 (0.14)²
= 0.03136 kgm²
Total moment of Inertia = [tex]I_{disk, CM}[/tex] + [tex]I_{point \ mass}[/tex]
= (0.00726 + 0.03136) kgm²
= 0.03862 kgm²
Assuming ∝ = angular acceleration = constant;
Then; we can use the following kinematic equations
T = FR
T = 2.1 × 0.11 N
T = 2.31 N
T = I × ∝
2.31 = 0.03862 × ∝
∝ = [tex]\frac{2.31}{0.03862}[/tex]
∝ = 59.81 rad/s²
Using the formula [tex]\omega_f = \omega_i + \alpha \delta T[/tex] to determine the angular speed of the apparatus; we have:
[tex]\omega_f =0 + 59.81*0.2[/tex] since [tex]( w_i \ is \ at \ rest ; the n\ w_i = 0 )[/tex]
[tex]\omega_f =11.962 \ rad/s[/tex]
∴ the angular speed of the apparatus [tex]\omega_f =11.962 \ rad/s[/tex]
b) Using the formula :
[tex]\theta = \omega_i t + \frac{1}{2}*\alpha*(t)^2\\\\\theta = 0 *0.2 + \frac{1}{2}*59.81*(0.2)^2 \\ \\ \theta = 29.905 *(0.2)^2 \\ \\ \theta = 1.1962 rads \ \ \ ( to \ degree; \ we \ have) \\ \\ \theta = 68.54^0[/tex]
Thus, the angle through which the apparatus turns in radians or degrees is : [tex]\theta = 68.54^0[/tex]
To solve the problem, compute the angular acceleration using the torque and moment of inertia. Given the angular acceleration and the time, calculate the angular velocity. Finally, use the angular acceleration and time to compute the angle turned through.
Explanation:The first part of this problem involves calculating angular acceleration, which is the rate of change of angular velocity. Using the formula, angular acceleration (α) = Torque (τ) / Moment of Inertia (I), the torque can be calculated using the formula Torque = Force * Radius, and the moment of inertia is calculated using the given formulas for disk and point masses.
For the disk, I_(disk,CM) = 1/2 M R^2 and for the point masses, I_(point mass) = 4 * m * b^2. Summing these gives the total moment of inertia. The angular acceleration is then obtained by dividing the torque by the total moment of inertia.
Having the angular acceleration and the time, we can calculate the angular velocity (ω) using the formula ω = α * Δt.
For the second part, the angle through which the apparatus turns can be calculated using the formula θ = 0.5 * α * (Δt)^2, as the initial angular velocity was zero.
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At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until a circuit breaker trips at time t= 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.a.) Through what total angle did the wheel turn between t= 0 and the time it stopped? answer in radb.) At what time does the wheel stop? (secs)c.)What was the wheel's angular acceleration as it slowed down?Express your answer in radians per second per second.
Answer:
a) [tex]\Delta \theta = 617.604\,rad[/tex], b) [tex]\Delta t = 10.392\,s[/tex], c) [tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
Explanation:
a) The final angular speed at the end of the acceleration stage is:
[tex]\omega = 24\,\frac{rad}{s} + \left(35\,\frac{rad}{s^{2}}\right) \cdot (2.50\,s)[/tex]
[tex]\omega = 111.5\,\frac{rad}{s}[/tex]
The angular deceleration is:
[tex]\alpha = \frac{\omega^{2}-\omega_{o}^{2}}{2\cdot \theta}[/tex]
[tex]\alpha = \frac{\left(0\,\frac{rad}{s} \right)^{2}-\left(111.5\,\frac{rad}{s} \right)^{2}}{2\cdot (440\,rad)}[/tex]
[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
The change in angular position during the acceleration stage is:
[tex]\theta = \frac{\left(111.5\,\frac{rad}{s} \right)^{2}-\left(0\,\frac{rad}{s} \right)^{2}}{2\cdot \left(35\,\frac{rad}{s^{2}} \right)}[/tex]
[tex]\theta = 177.604\,rad[/tex]
Finally, the total change in angular position is:
[tex]\Delta \theta = 440\,rad + 177.604\,rad[/tex]
[tex]\Delta \theta = 617.604\,rad[/tex]
b) The time interval of the deceleration interval is:
[tex]\Delta t = \frac{0\,\frac{rad}{s} - 111.5\,\frac{rad}{s} }{-14.128\,\frac{rad}{s^{2}} }[/tex]
[tex]\Delta t = 7.892\,s[/tex]
The time required for the grinding wheel to stop is:
[tex]\Delta t = 2.50\,s + 7.892\,s[/tex]
[tex]\Delta t = 10.392\,s[/tex]
c) The angular deceleration of the grinding wheel is:
[tex]\alpha = -14.128\,\frac{rad}{s^{2}}[/tex]
Answer:
Ф_T =544
The wheel will stop after 10 s.
α_z = -11.15 rad/s^2
Explanation:
The angular acceleration is constant. Thus, we will apply the equations of rotation with constant angular acceleration model.
(a) In order to calculate the total angle, we will divide the entire interval from t = 0 to the time the wheel stops into two intervals.
From t = 0 to t = 2 s:
Ф-Ф_o =1/2(w_0z+w_z)t (1)
We will calculate w_z first:
w_z = w_0x +α_xt
w_z = 24 + (35)(2.5)
w_z = 111.5
Substitute w_x into Eq.1
Ф-Ф_o = 1/2(24+111.5)(2)
Ф-Ф_o = 136 rad
We can calculate it directly from the following equation:
Ф-Ф_o = w_0x*t+1/2a_xt^2
Ф-Ф_o = 24*2.5+1/2*35*2.5
Ф-Ф_o =103.75 rad
Thus, the total angle the wheel turned between t = 0 and the time it stopped:
Ф_T =103.75 +440
Ф_T =544
(b)
We will take the interval from when the circuit breaker trips until the wheel comes to a stop.
w_oz = 111.5 rad/s
wz = 0 the wheel will stop at the end. 1
Ф-Ф_o =1/2(w_0z+w_z)t
440 = 1/2(111.5+0)*t
t = 8s
Adding t of the first interval to t of the second interval :
t_T = 2 + 8 =10
The wheel will stop after 10 s.
(c)
We will take the interval from when the circuit breaker trips until the wheel comes to a stop.
w_oz = 111.5 rad/s
w_z = 0
w_x =w_oz+α_z*t
α_z = -11.15 rad/s^2
Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would require quantum mechanics to describe, that is, would show both particle and wave properties? Which objects would act like everyday objects, and be adequately described by classical mechanics? object quantum or classical? A virus with a mass of 9.4 x 10-17 g, 280. nm wide, moving at 0.50 µm/s. classical quantum A buckyball with a mass of 1.2 x 10-21 g, 0.7 nm wide, moving at 37. m/s. classical quantum A mosquito with a mass of 1.0 mg, 6.3 mm long, moving at 1.1 m/s. classical quantum A turtle with a mass of 710. g, 22. cm long, moving at 2.8 cm/s. classical quantum
All objects are described using classical mechanics, with the exception of the buckyball, which requires a quantum mechanical description due to its small mass and high speed that result in a significant de Brogli wavelength.
Explanation:Whether an object shows particle-wave duality, requiring quantum mechanics to describe, or behaves like everyday objects, described by classical mechanics, is determined by the de Broglie wavelength of the object. Calculating the de Broglie wavelength (λ) can be accomplished using the equation λ = h/mv, where h is Planck's constant and m and v are the object's mass and velocity, respectively.
For the virus, buckyball, mosquito, and turtle, the calculated de Broglie wavelengths are so small compared to their physical dimensions that quantum mechanical effects would be negligible. Hence, all these objects can be described using classical mechanics. The buckyball is the only exception. Owing to its tiny mass and high speed, its de Broglie wavelength becomes significant in relation to its size, requiring quantum mechanical description.
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Objects that are atomic or subatomic in size require quantum mechanics for their description, while larger objects can be adequately explained by classical mechanics. In the provided cases, a virus and a buckyball would require quantum mechanics, while a mosquito and a turtle would be described by classical mechanics.
Explanation:To determine whether an object would be adequately described by classical mechanics or require quantum mechanics, we need to consider its size, mass, and speed. Under the domain of quantum mechanics are typically objects that are atomic or subatomic in size, meaning they exhibit wave-particle duality due to their small size and significant motion.
1. A virus: Being atomic in size, a virus would require quantum mechanics for its description. Specifically, it would demonstrate both particle and wave properties due to its small size.
2. A buckyball: Given its subatomic mass, size, and significant speed, a buckyball falls into the quantum domain, displaying both particle and wave properties.
3. A mosquito and 4. A turtle : These are larger, macroscopic objects. Their motion and behavior can be adequately described using classical mechanics as they primarily exhibit particle properties, their wave-like characteristics being effectively obscured due to their larger mass and size. Their behaviors conform to the laws defined under classical mechanics and do not exhibit the quantum nature that smaller objects do.
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What attraction and repulsion behavior do the north and south poles of a magnet exhibit? Explain the left-hand rule for the flux direction in conductors. What is the most common source of power for most HVACR systems? What is the purpose for transformers? What is a solenoid valve?
Answer:
1.attraction and repulsion behavior do the north and south poles of a magnet exhibit?A.Like poles repel each other, and opposite poles attract.
2.Explain the left-hand rule for flux direction in conductors. Ans.The left-hand rule is used to determine the direction of rotation of the flux around the conductor
.3.What is the most common source of power for most HVACR systems?Ans.Alternating current is the most common source of power used for most HVACR systems.
4.What is the purpose of transformers?Ans..Transformers are used to increase or decrease incoming voltages to meet the requirements of the load.
5.What is a solenoid valve?Ans.Solenoid valves are electrically operated valves that open or close automatically by energizing or de-energizing a coil of wire.6.Is there an electrical connection between the primary and secondary windings of a transformer?A.There is no electrical connection between the primary and secondary windings.
Explanation:
The poles of a magnet attract and repel based on their polarity. The left-hand rule explains how magnetic fields interact with electric currents. Most HVACR systems use electricity, and transformers convert voltage levels to transmit power efficiently. A solenoid valve controls fluid or gas flow through electromagnetism.
Explanation:The north and south poles of a magnet exhibit a behavior where like poles repel and opposite poles attract. This means that north pole to north pole or south pole to south pole will repel, while north pole to south pole will attract.
The left-hand rule for flux direction, often used in electromagnetism, states that if you point your thumb in the direction of the current, the direction in which your fingers curl represents the direction of the magnetic field lines. This is significant in understanding the interaction of electric current with magnetic fields in conductors.
The most common source of power for most HVACR (Heating, Ventilation, Air Conditioning, and Refrigeration) systems is electricity. This power is used to run the system's various components including compressors, fans, and controls.
Transformers are used in electric power systems to convert voltages from one level to another. This is mainly used to transmit electricity over long distances with low energy losses and is essential in power grids.
A solenoid valve is an electromechanical device used to control the flow of a fluid or gas. When an electric current is passed through the coil, a magnetic field is created causing the valve to open or close.
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To practice Tactics Box 14.1 Identifying and analyzing simple harmonic motion. A complete description of simple harmonic motion must take into account several physical quantities and various mathematical relations among them. This Tactics Box summarizes the essential information needed to solve oscillation problems of this type. The position of a 60 g oscillating mass is given by x(t)=(2.0cm)cos(10t), where t is in seconds. Determine the velocity at t=0.40s. Assume that the oscillating mass described in Part A is attached to a spring. What would the spring constant k of this spring be? What is the total energy E of the mass described in the previous parts?
To solve for the velocity at t=0.40s of a mass in SHM, differentiate the position function with respect to time, and use the resulting velocity function. The spring constant can be calculated using Hooke's law, and the total energy can be found through the energy relationship in SHM.
Explanation:The student is asking about the characteristics of simple harmonic motion (SHM) related to a mass attached to a spring. To find the velocity at a specific time t for the mass undergoing SHM, we need to differentiate the position function x(t) with respect to time. The provided position function is x(t)=(2.0 cm)cos(10t), implying that the velocity function will be v(t)=-ω A sin(ωt), where ω is the angular frequency, and A is the amplitude. To calculate the spring constant k and total energy E, we will employ both Hooke's law, F = -kx, and the relationship between the total energy in SHM and the displacement from the equilibrium position, given by E = ½ k A².
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The velocity at [tex]t=0.40s[/tex] is [tex]0.1514m/s[/tex].
The spring constant is [tex]6N/m[/tex].
The total energy of the mass is 0.0012J.
Determining Velocity, Spring Constant, and Total Energy in Simple Harmonic Motion
The position of a 60 g oscillating mass is given by the function: [tex]x(t) = (2.0 cm)cos(10t)[/tex], where time t is in seconds. Here's how we solve for the velocity at [tex]t = 0.40 s[/tex], the spring constant k, and the total energy E of the system.
Step-by-Step Solution
Velocity at [tex]t = 0.40 s[/tex]:A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base of the hill. (a) At what rate was it rotating at the base of the hill? (b) What was the rotational kinetic energy then? (Ignore rolling friction and assume total mechanical energy is conserved). [Hint: Soccer ball is a thin walled h
Answer with Explanation:
We are given that
Diameter=d=22.6 cm
Mass,m=426 g=[tex]426\times 10^{-3} kg[/tex]
1 kg=1000 g
Radius,r=[tex]\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m[/tex]
1m=100 cm
Height,h=5m
[tex]I=\frac{2}{2}mr^2[/tex]
a.By law of conservation of energy
[tex]\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh[/tex]
[tex]\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh[/tex]
[tex]v=\omega r[/tex]
[tex]gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2[/tex]
[tex]\omega^2=\frac{6}{5r^2}gh[/tex]
[tex]\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s[/tex]
Where [tex]g=9.8m/s^2[/tex]
b.Rotational kinetic energy=[tex]\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J[/tex]
Rotational kinetic energy=8.35 J
g Show that the first derivative of the magnitude of the net magnetic field of the coils (dB/dx) vanishes at the mid- point P regardless of the value of s. Why would you expect this to be true from symmetry? (b) Show that the second derivative (d2 B/dx2 ) also vanishes at P, provided s R. This accounts for the uniformity of B near P for this particular coil separation.
Answer:
Hi Carter,
The complete answer along with the explanation is shown below.
I hope it will clear your query
Pls rate me brainliest bro
Explanation:
The magnitude of the magnetic field on the axis of a circular loop, a distance z from the loop center, is given by Eq.:
B = NμοiR² / 2(R²+Z²)³÷²
where
R is the radius of the loop
N is the number of turns
i is the current.
Both of the loops in the problem have the same radius, the same number of turns, and carry the same current. The currents are in the same sense, and the fields they produce are in the same direction in the region between them. We place the origin at the center of the left-hand loop and let x be the coordinate of a point on the axis between the loops. To calculate the field of the left-hand loop, we set z = x in the equation above. The chosen point on the axis is a distance s – x from the center of the right-hand loop. To calculate the field it produces, we put z = s – x in the equation above. The total field at the point is therefore
B = NμοiR²/2 [1/ 2(R²+x²)³÷² + 1/ 2(R²+x²-2sx+s²)³÷²]
Its derivative with respect to x is
dB /dx= - NμοiR²/2 [3x/ (R²+x²)⁵÷² + 3(x-s)/(R²+x²-2sx+s²)⁵÷² ]
When this is evaluated for x = s/2 (the midpoint between the loops) the result is
dB /dx= - NμοiR²/2 [3(s/2)/ (R²+s²/4)⁵÷² - 3(s/2)/(R²+s²/4)⁵÷² ] =0
independent of the value of s.
Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The length of each rod is 1.1 m, while the mass of each is 0.11 kg. One rod is held in place above the ground, and the other floats beneath it at a distance of 8.3 mm. Determine the current in the rods.
Answer:
I = 215.76 A
Explanation:
The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.
Fm = u_o*I^2*L/2*π*d
where I is the current in each rod, d = 0.0082 m is the distance 27rId
between each, L = 0.85 m is the length of each rod.
Fm = 4π*10^-7*I^2*1.1/2*π*0.0083
The mass of each rod is m = 0.0276 kg
F_m = mg
4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8
I = 215.76 A
note:
mathematical calculation maybe wrong or having little bit error but the method is perfectly fine
The current in the rods is approximately 132.75 A.
To determine the current in the rods, we need to use the formula for the force between two parallel current-carrying conductors, which is given by:
[tex]\[ F = \frac{\mu_0 I^2 L}{2 \pi d} \][/tex]
where:
F is the force between the two rods,
[tex]\( \mu_0 \)[/tex] is the permeability of free space [tex](\( 4\pi \times 10^{-7} \) T\·m/A)[/tex],
I is the current in each rod,
L is the length of the rods (1.1 m),
d is the distance between the rods (8.3 mm or [tex]\( 8.3 \times 10^{-3} \) m[/tex]).
The rods repel each other, and the force of repulsion must equal the gravitational force on the floating rod for it to levitate. The gravitational force is given by:
[tex]\[ F_g = m g \][/tex]
where:
m is the mass of the rod (0.11 kg),
g is the acceleration due to gravity (approximately 9.81 m/s²).
Setting the magnetic force equal to the gravitational force, we get:
[tex]\[ \frac{\mu_0 I^2 L}{2 \pi d} = m g \][/tex]
Solving for I, we have:
[tex]\[ I^2 = \frac{m g 2 \pi d}{\mu_0 L} \][/tex]
[tex]\[ I = \sqrt{\frac{m g 2 \pi d}{\mu_0 L}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \text{ kg} \times 9.81 \text{ m/s}^2 \times 2 \pi \times 8.3 \times 10^{-3} \text{ m}}{4\pi \times 10^{-7} \text{ T·m/A} \times 1.1 \text{ m}}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 2 \times 8.3 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]
[tex]\[ I = \sqrt{\frac{0.11 \times 9.81 \times 16.6 \times 10^{-3}}{4\pi \times 10^{-7} \times 1.1}} \][/tex]
[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{4\pi \times 10^{-7}}} \][/tex]
[tex]\[ I = \sqrt{\frac{1.77558 \times 10^{-2}}{1.256637 \times 10^{-6}}} \][/tex]
[tex]\[ I = \sqrt{1413.55} \][/tex]
[tex]\[ I \approx 132.75 \text{ A} \][/tex]
A surgical microscope weighing 200 lb is hung from a ceiling by four springs with stiffness 25 lb/ft. The ceiling has a vibration amplitude of 0.05mm at 2 Hz (a typical resonant frequency of a building). a) If there is no damping, how much transmitted vibration (amplitude of displacement) does the microscope experience
Answer:
If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
Explanation:
The motion of the ceiling is y = Y sinωt
y = 0.05 sin (2 π × 2) t
y = 0.05 sin 4 π t
K = 25 lb/ft × 4 sorings
K = 100 lb/ft
Amplitude of the microscope [tex]\frac{X}{Y}= [\frac{1+2 \epsilon (\omega/ W_n)^2}{(1-(\frac{\omega}{W_n})^2)^2+(2 \epsilon \frac{\omega}{W_n})^2}][/tex]
where;
[tex]\epsilon = 0[/tex]
[tex]W_n = \sqrt { \frac{k}{m}}[/tex]
= [tex]\sqrt { \frac{100*32.2}{200}}[/tex]
= 4.0124
replacing them into the above equation and making X the subject of the formula:
[tex]X =[/tex] [tex]Y * \frac{1}{\sqrt{(1-(\frac{\omega}{W_n})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]0.05 * \frac{1}{\sqrt{(1-(\frac{4 \pi}{4.0124})^2)^2})}}[/tex]
[tex]X =[/tex] [tex]5.676*10^{-3} \ mm[/tex]
Therefore; If there is no damping, the amount of transmitted vibration that the microscope experienced is = [tex]5.676*10^{-3} \ mm[/tex]
Consider two inductors, with the first inductor experiencing a current which is rising at twice the rate as that through the second inductor. Suppose the first inductor has half the EMF across it as the second inductor. How does the inductance of the first inductor compare to that of the second inductor?
Answer:
Inductance of first is one-4th that of the second inductor
Explanation:
See attached file
2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust the length of the air column in the tube. A tuning fork of frequency 330 Hz is sounded and held at the upper end as the tube is lifted out of the water until the first resonant sound is heard. The length of tube above the water level when the fundamental note sounds is 26 cm. a. Calculate the speed of sound in the air column in the tube. b. Determine another position for the tube that will result in a resonance with this tuning fork. c. Determine the first position (as we again lower the tube into the water) that will resonate with a 256 Hz tuning fork.
Answer:
Explanation:
This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows
l = λ / 4 where l is length of tube and λ is wave length.
here l = .26 m
λ = .26 x 4 = 1.04 m
frequency of sound = 330 Hz
velocity of sound = frequency x wave length
= 330 x 1.04
= 343.2 m /s
b )
Next overtone will be produced at 3 times the length
so next length of air column = 3 x 26
= 78 cm
c )
If frequency of sound = 256 Hz
wavelength = velocity / frequency
= 343.2 / 256
= 1.34 m
= 134 cm
length of air column for resonance
= wavelength / 4
134/4
= 33.5 cm
PROBLEM-10 GRAVITATIONAL POTINTIAL ENERGY In a typical professional downhill ski race, athletes start 820 m above where they cross the finish line. Assume that they start at rest, friction is negligible, and their mass is 80 kg. What is the magnitude of the change in gravitational potential energy of the Earth-athlete system over the course of the race? (in J )
Answer:
The magnitude of the change in gravitational potential energy of the earth-athlete system over the course of the race is 643,536Joules
Explanation
Potential energy is one of the form of mechanical energy and it is defined as the energy possessed by a body due to virtue of its position. When the body is under gravity, it possesses an energy called the gravitational potential energy.
Gravitational potential energy is expressed as shown:
GPE = mass × acceleration due to gravity × height
Given mass of athlete = 80kg
height covered = 820m
acceleration due to gravity = 9.81m/s
GPE = 80×9.81×820
GPE = 643,536Joules
Suppose the maximum safe intensity of microwaves for human exposure is taken to be 1.39~\mathrm{W/m^2}1.39 W/m 2 . If a radar unit leaks 10.0~\text{W}10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)
Answer:
0.763 m
Explanation:
Intensity I = power P ÷ area A of exposure (spherical area of propagation)
I = P/A
A = P/I
Power = 10.0 W
Intensity = 1.39 W/m^2
A = 10/1.39 = 7.19 m^2
Area A = 4¶r^2
7.19 = 4 x 3.142 x r^2
7.19 = 12.568r^2
r^2 = 7.19/12.568 = 0.57
r = 0.753 m
Find the charge q(t) on the capacitor and the current i(t) in the given LRC-series circuit. L = 1 h, R = 100 Ω, C = 0.0004 f, E(t) = 20 V, q(0) = 0 C, i(0) = 3 A q(t) = C i(t) = A
Find the maximum charge on the capacitor. (Round your answer to four decimal places.)
Answer:
The maximum charge around the capacitor is 0.03170189C
Explanation:
See attached file
In an RLC series circuit with given values for resistance, inductance, and capacitance, we can calculate the charge on the capacitor five cycles later and fifty cycles later using the equation q(t) = q(0) * e^(-(R/L)t)
Explanation:In an RLC series circuit with a resistance of 7.092 ohms, an inductance of 10 mH, and a capacitance of 3.0 µF, the charge on the capacitor can be calculated using the equation:
q(t) = q(0) * e^(-(R/L)t)
Given that the initial charge on the capacitor is 8.0 µC, we can calculate the charge five cycles later and fifty cycles later using this equation.
(a) To find the charge five cycles later, we need to find the time period of one cycle. The time period can be calculated using the formula:
T = 2π√(LC)
Once we have the time period, we can calculate the time it takes for five cycles to pass and substitute it in the equation to find the charge.
(b) To find the charge fifty cycles later, we can use the same approach as in part (a), but substituting the time it takes for fifty cycles to pass.
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You have a summer job working at a company developing systems to safely lower large loads down ramps. Your team is investigating a magnetic system by modeling it in the laboratory. The safety system is a conducting bar that slides on two parallel conducting rails that run down the ramp. The bar is perpendicular to the rails and is in contact with them. At the bottom of the ramp, the two rails are connected together. The bar slides down the rails through a vertical uniform magnetic field. The magnetic field is supposed to cause the bar to slide down the ramp at a constant velocity even when friction between the bar and the rails is negligible. Before setting up the laboratory model, your task is to calculate the constant velocity of the bar sliding down the ramp on rails in a vertical magnetic field as a function of the mass of the bar, the strength of the magnetic field, the angle of the ramp from the horizontal, the length of the bar which is the same as the distance between the tracks, and the resistance of the bar. Assume that all of the other conductors in the system have a much smaller resistance than the bar.
If the force due to the changing flux exactly cancells out the net force due to the combination of gravity and normal force, then the bare will cease to accelerate and instead move at a constant velocity. Please solve for this velocity algebraically.
Answer:
Explanation:
find the solution below
A steady stream of projectiles, with identical linear momenta, collides with a fixed target. The average force on the target is proportional to *
Answer:
The average force on the target is proportional to:
- The number of projectiles hitting the target.
- The mass of the projectiles.
- If the time increases or decreases
Explanation:
This is a problem that applies momentum and the amount of movement. Where this principle can be explained by the following equation:
m*v1 + Imp1_2 = m*v2
where:
∑F *Δt = Imp1_2 = impulse. [N*s]
m*v1 = mass by velocity before the impact [kg*m/s]
m*v2 = mass by velocity after the impact [kg*m/s]
When a problem includes two or more particles, each particle it can be considered separately and equation is written for every particle.
We clear the expression of force in the equation:
∑F *Δt = m*v2 - m*v1
In this equation if we have a different number of particles, given by the value n, we see that the equation is proportional to the number of particles
∑F *Δt = n*m*v2 - n*m*v1
Therefore the average force on the target is proportional to the number of projectiles hitting the target.
The Force F is also increased or decreased if the mass of the projectiles is changed. Therefore it is also proportional to the mass of the projectiles.
The Force F also changes if the time increases or decreases.
A ray of light passes from one material into a material with a higher index of refraction. Determine whether each of the following quantities increases, decreases, or remains unchanged. HINT (a) The ray's angle with the normal. increases decreases remains unchanged (b) The light's wavelength. increases decreases remains unchanged (c) The light's frequency. increases decreases remains unchanged (d) The light's speed. increases decreases remains unchanged (e) The photon energy. increases decreases remains unchanged
Answer:
a) the Angle te also decreases , b) decrease, c) unchanged , d) the speed decrease , e) unchanged
Explanation:
When a ray of light passes from one transparent material to another, it must comply with the law of refraction
n1 sin θ₁ = n2 sin θ₂
Where index v1 is for the incident ray and index 2 for the refracted ray
With this expression let's examine the questions
a) They indicate that the refractive index increases,
sin θ₂ = n₁ / n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
As m is greater than n1 the quantity on the right is less than one, the whole quantity in parentheses decreases so the Angle te also decreases
Answer is decrease
b) The wave velocity eta related to the wavelength and frequency
v = λ f
The frequency does not change since the passage from one medium to the other is a process of forced oscillation, resonance whereby the frequency in the two mediums is the same.
The speed decreases with the indicated refraction increases and therefore the wavelength decreases
λ = λ₀ / n
The answer is decrease
c) from the previous analysis the frequency remains unchanged
d) the refractive index is defined by
n = c / v
So if n increases, the speed must decrease
The answer is decrease
e) the energy of the photon is given by the Planck equation
E = h f
Since the frequency does not change, the energy does not change either
Answer remains unchanged
a) The Angle te also decreases, b) Decrease, c) Unchanged, d) The speed decrease, e) Unchanged
What is the Law of Refraction?
When a glimmer of light perishes from one translucent material to another, it must capitulate with the law of refraction
n1 sin θ₁ = n2 sin θ₂
Where index v1 is for the happening ray and index 2 is for the refracted ray With this expression let's discuss.
a) They indicate that the refractive index increases,
sin θ₂ = n₁ / n₂ sin θ₁
θ₂ = sin⁻¹ (n₁ /n₂ sin θ₁)
As m is more significant than n1 the quantity on the ownership is less than one, the whole quantity in parentheses diminishes so the Angle te also decreases
Solution is decrease
b) The wave velocity eta connected to the wavelength and frequency
v = λ f
The frequency does not modify since the passage from one medium to the other is a procedure of forced oscillation and resonance whereby the frequency in the two mediums is the same.
The speed decreases with the foreshadowed refraction increases and thus the wavelength decreases
So, λ = λ₀ / n
The answer is to decrease
c) from the earlier analysis the frequency remains unchanged
d) When the refractive index is defined by
n is = c / v
Then, if n increases, the speed must decrease
The solution is to decrease
e) When the energy of the photon is given by the Planck equation
E is = hf. Since When the frequency does not transform, the energy does not change either Solution is remains unchanged
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An inexperienced catcher catches a 126 km/h fastball of mass 160 g within 1.36 ms, whereas an experienced catcher slightly retracts his hand during the catch, extending the stopping time to 13.6 ms. What is the average force imparted to the two gloved hands during the catches of the inexperienced catcher
Answer:
4117.65 N
Explanation:
Speed of ball, u = 126 km/h = 35 m/s
Mass of ball, m = 160 g = 0.16 kg
Time interval, t = 1.36 ms = 0.00136 s
We can calculate the force as a measure of the momentum of the ball:
F = P/t
Momentum, P, is given as:
P = mv
Therefore:
F = (mv) / t
F = (0.16 * 35) / (0.00136)
F = 4117.65 N
The force imparted to the two glove d hands of the inexperienced catcher is 4117.65 N.
A heat conducting rod, 0.90 m long, is made of an aluminum section, 0.20 m long, and a copper section, 0.70 m long. Both sections have a cross-sectional area of 0.0004 m^2. The aluminum end and the copper end are maintained at temperatures of 30*C and 230*C respectively. The thermal conductivities of aluminum and copper are 205 and 385 W/m ? K, respectively. The temperature of the aluminum-copper junction in the rod, in *C, is closest to:
Answer:
is closest to 100*C temperature at the aluminum-copper junction
Explanation:
The expression for calculating the resistance of each rod is given by
[tex]R =\frac{ L}{ kA}[/tex]
Now; for Aluminium
[tex]R_{al} =\frac{ 0.20 }{ 205*0.0004}[/tex]
[tex]R_{al}[/tex] = 2.439
For Copper
[tex]R_{Cu}=\frac{0.70}{385*0.0004}[/tex]
[tex]R_{Cu} = 4.545[/tex]
Total Resistance [tex]R = R_{al} + R_{Cu}[/tex]
= 2.439 + 4.545
= 6.9845
Total temperature difference = 230*C + 30*C
= 200 *C
The Total rate of heat flow is then determined which is = [tex]\frac{ total \ temp \ difference}{total \ resistance }[/tex]
=[tex]\frac{200}{ 6.9845 }[/tex]
= 28.635 Watts
However. the temperature difference across the aluminium = Heat flow × Resistance of aluminium
= 28.635 × 2.349
= 69.84 *C
Finally. for as much as one end of the aluminium is = 30 *C , the other end is;
=30*C + 69.84*C
= 99.84 *C
which is closest to 100*C temperature at the aluminum-copper junction
As you may well know, placing metal objects inside a microwave oven can generate sparks. Two of your friends are arguing over the cause of the sparking, with one stating that the microwaves "herd" electrons into "pointy" areas of the metal object, from which the electrons jump from one part of the object to another. The other friend says that the sparks are caused by the photoelectric effect. Prove or disprove the latter idea using basic physics. Suppose the typical work function of the metal is roughly 4.910×10−19 J.4.910×10−19 J. Calculate the maximum wavelength in angstroms of the radiation that will eject electrons from the metal.
Answer:
[tex]4.048\times 10^{-7}\ m[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
E = Energy = [tex]4.91\times 10^{-19}\ J[/tex]
Wavelength ejected is given by
[tex]\lambda=\dfrac{hc}{E}\\\Rightarrow \lambda=\dfrac{6.626\times 10^{-34}\times 3\times 10^8}{4.91\times 10^{-19}}\\\Rightarrow \lambda=4.048\times 10^{-7}\ m[/tex]
The maximum wavelength in angstroms of the radiation that will eject electrons from the metal is [tex]4.048\times 10^{-7}\ m[/tex]
Suppose that a particle accelerator is used to move two beams of particles in opposite directions. In a particular region, electrons move to the right at 6020 m/s and protons move to the left at 1681 m/s. The particles are evenly spaced with 0.0476 m between electrons and 0.0662 m between protons. Assuming that there are no collisions and that the interactions between the particles are negligible, what is the magnitude of the average current in this region
Answer:
[tex]2.429783984\times 10^{-14}\ A[/tex]
Explanation:
Velocity of electron = 6020 m/s
Velocity of proton = 1681 m/s
Electron space = 0.0476 m
Proton space = 0.0662 m
e = Charge of particle = [tex]1.6\times 10^{-19}\ C[/tex]
Number of electrons passing per second
[tex]n_e=\dfrac{6020}{0.0476}\\\Rightarrow n_e=126470.588[/tex]
Number of protons passing per second
[tex]n_p=\dfrac{1681}{0.0662}\\\Rightarrow n_p=25392.749[/tex]
Current due to electrons
[tex]I_e=n_ee\\\Rightarrow I_e=126470.588\times 1.6\times 10^{-19}\\\Rightarrow I_e=2.0235\times 10^{-14}\ A[/tex]
Current due to protons
[tex]I_p=n_pe\\\Rightarrow I_p=25392.749\times 1.6\times 10^{-19}\\\Rightarrow I_p=4.06283984\times 10^{-15}\ A[/tex]
Total current
[tex]I=2.0235\times 10^{-14}+4.06283984\times 10^{-15}\\\Rightarrow I=2.429783984\times 10^{-14}\ A[/tex]
The average current is [tex]2.429783984\times 10^{-14}\ A[/tex]
For heat transfer purposes, an egg can be considered to be a 5.5-cm-diameter sphere having the properties of water. An egg that is initially at 4.3°C is dropped into boiling water at 100°C. The heat transfer coefficient at the surface of the egg is estimated to be 800 W/m2⋅K. If the egg is considered cooked when its center temperature reaches 74°C, determine how long the egg should be kept in the boiling water. Solve this problem using the analytical one-term approximation method. The thermal conductivity and diffusivity of the eggs can be approximated by those of water at room temperature to be k = 0.607 W/m·°C and α = 0.146×10−6 m2/s.
Using the analytical one-term approximation method, we can determine how long an egg should be kept in boiling water to reach a specific temperature at its center.
Explanation:In this problem, we are given the initial and final temperatures of an egg, as well as its diameter, the heat transfer coefficient, and the thermal conductivity and diffusivity of water. We are asked to find how long the egg should be kept in boiling water in order for its center temperature to reach a certain value.
To solve this problem, we can use the analytical one-term approximation method, which involves calculating the Biot number and the dimensionless temperature profile. By equating the temperature at the center of the egg to the desired final temperature, we can solve for the time required.
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The tub of a washing machine goes into its spin cycle, starting from rest and gaining angular speed steadily for 9.00 s, at which time it is turning at 7.00 rev/s. At this point, the lid of the washing machine is opened, and a safety switch turns it off. The tub then smoothly slows to rest in 13.0 s. Through how many revolutions does the tub rotate while it is in motion
Answer:
Explanation:
7 rev /s = 7 x 2π rad /s
angular velocity = 14π rad /s
Angular acceleration α = increase in angular velocity / time
= 14π - 0 / 9
α = 4.8844 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 4.8844 x 9²
= 197.8182 rad
2π n = 197.8182
n = 31.5 rotation
During acceleration , no of rotation made = 31.5
During deceleration : -----
deceleration =
Angular deceleration α = decrease in angular velocity / time
= 14π - 0 / 13
α = 3.3815 rad / s
θ = 1/2 α x t² θ is angle of rotation , t is time
= 1/2 x 3.3815 x 13²
= 285.736 rad
2π n = 285.736
n = 45.5 rotation
total rotation
= 45.5 + 31.5
= 77 .
A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is suspended from the bar 3/4ths of its length from the wall. The angle indicated is 60°. What is the tension in the cable? What is the horizontal force provided by the hinge?
Answer:
The tension is [tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]
The horizontal force provided by hinge [tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]
Explanation:
From the question we are told that
The mass of the beam is [tex]m_b =M[/tex]
The length of the beam is [tex]l = L[/tex]
The hanging mass is [tex]m_h = 3M[/tex]
The length of the hannging mass is [tex]l_h = \frac{3}{4} l[/tex]
The angle the cable makes with the wall is [tex]\theta = 60^o[/tex]
The free body diagram of this setup is shown on the first uploaded image
The force [tex]F_x \ \ and \ \ F_y[/tex] are the forces experienced by the beam due to the hinges
Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero
So
[tex]\sum F =0[/tex]
Now about the x-axis the moment is
[tex]F_x -T cos \theta = 0[/tex]
=> [tex]F_x = Tcos \theta[/tex]
Substituting values
[tex]F_x =T cos (60)[/tex]
[tex]F_x= \frac{T}{2} ---(1)[/tex]
Now about the y-axis the moment is
[tex]F_y + Tsin \theta = M *g + 3M *g ----(2)[/tex]
Now the torque on the system is zero because their is no rotation
So the torque above point 0 is
[tex]M* g * \frac{L}{2} + 3M * g \frac{3L}{2} - T sin(60) * L = 0[/tex]
[tex]\frac{Mg}{2} + \frac{9 Mg}{4} - T * \frac{\sqrt{3} }{2} = 0[/tex]
[tex]\frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}[/tex]
[tex]T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }[/tex]
[tex]T= \frac{11}{2\sqrt{3} } Mg[/tex]
The horizontal force provided by the hinge is
[tex]F_x= \frac{T}{2} ---(1)[/tex]
Now substituting for T
[tex]F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}[/tex]
[tex]Fx= \frac{11}{4\sqrt{3} } Mg[/tex]
The answer is:[tex]\frac{4Mg}{\sqrt{3}}.[/tex]
To solve this problem, we will analyze the forces acting on the uniform beam and apply the conditions for rotational and translational equilibrium. Here's the step-by-step solution:
1. Identify the forces acting on the beam:
- The weight of the beam, [tex]\( W_b = Mg \)[/tex], acting at the center of mass of the beam, which is at [tex]\( \frac{L}{2} \)[/tex] from the wall.
- The weight of the mass suspended from the bar, [tex]\( W_s = 3Mg \), acting at \( \frac{3L}{4} \)[/tex] from the wall.
- The tension in the cable, [tex]\( T \)[/tex], acting at an angle[tex]\( \theta = 60^\circ \)[/tex] with the horizontal.
- The horizontal force provided by the hi-nge,[tex]\( F_h \)[/tex].
2. Break down the tension in the cable into its horizontal and vertical components:
- The horizontal component of the tension,[tex]\( T_h = T \cos(60^\circ) \).[/tex]
- The vertical component of the tension,[tex]\( T_v = T \sin(60^\circ) \)[/tex].
3. Apply the condition for translational equilibrium in the vertical direction:
- The sum of the vertical forces must be zero:[tex]\( T_v + F_h = W_b + W_s \).[/tex]
- Substituting the expressions for the weights, we get: [tex]\( T \sin(60^\circ) = Mg + 3Mg \).[/tex]
- Simplifying, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex].
4. Solve for the tension [tex]\( T \)[/tex]:
[tex]- \( T = \frac{4Mg}{\sin(60^\circ)} \).- Since \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), we have \( T = \frac{4Mg}{\frac{\sqrt{3}}{2}} = \frac{8Mg}{\sqrt{3}} \).[/tex]
5. Apply the condition for rotational equilibrium about the hi-nge:
- The sum of the torques about the hi-nge must be zero.
- The torque due to the weight of the beam is [tex]\( \tau_b = W_b \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) \)[/tex].
- The torque due to the weight of the suspended mass is[tex]\( \tau_s = W_s \left(\frac{3L}{4}\right) = 3Mg \left(\frac{3L}{4}\right) \).[/tex]
- The torque due to the tension in the cable is[tex]\( \tau_T = T_v \left(\frac{L}{2}\right) \)[/tex] (since the tension acts at the end of the beam).
- Setting the sum of the torques to zero: [tex]\( \tau_T = \tau_b + \tau_s \).[/tex]
- Substituting the expressions for the torques, we get: [tex]\( T \sin(60^\circ) \left(\frac{L}{2}\right) = Mg \left(\frac{L}{2}\right) + 3Mg \left(\frac{3L}{4}\right) \)[/tex]
- We already know that [tex]\( T \sin(60^\circ) = 4Mg \)[/tex], so this equation is satisfied, confirming that our value for [tex]\( T \)[/tex] is correct.
6. Solve for the horizontal force [tex]\( F_h \)[/tex]:
- From the vertical equilibrium equation, [tex]\( T \sin(60^\circ) = 4Mg \)[/tex] , we have already found [tex]\( T \)[/tex].
- The horizontal component of the tension is [tex]\( T_h = T \cos(60^\circ) \)[/tex].
- Since there are no other horizontal forces, [tex]\( F_h = T_h \)[/tex].
- Substituting the value of[tex]\( T \), we get \( F_h = \frac{8Mg}{\sqrt{3}} \cos(60^\circ) \).[/tex]
- Since [tex]\( \cos(60^\circ) = \frac{1}{2} \), we have \( F_h = \frac{8Mg}{\sqrt{3}} \cdot \frac{1}{2} = \frac{4Mg}{\sqrt{3}} \).[/tex]
Therefore, the tension in the cable is [tex]\( \boxed{\frac{8Mg}{\sqrt{3}}} \)[/tex] and the horizontal force provided by the hi-nge is[tex]\( \boxed{\frac{4Mg}{\sqrt{3}}} \).[/tex]