The capacitance per unit is 1.292 X 10⁻¹⁰.
The radius of inner part of the conductor = r = 2.6 X 10⁻³ m
The radius of outer part of the conductor = R = 4 X 10⁻³ m
Length of the capacitor = l = 2.1 X 10⁻³ m
Capacitance is purely geometrical property. It depends only on length, radius of conductor.
From the formula of cylindrical capacitor,
= C = (2πεl) / ln(R/r)
where ε = 8.85 X 10⁻¹²
But we need capacitance per unit length so,
= C/l = (2πε) / ln(R/r)
= C/l = (2 X 3.14 X 8.85 X 10⁻¹²) / ln ( 4 X 10⁻³ / 2.6 X 10⁻³)
= C/l = 55.58 X 10⁻¹² / 0.43
= C/l = 1.292 X 10⁻¹⁰
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In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression, where x is in centimeters and t is in seconds.x=(4.00cm)cos(3t+π/6)(a) At t=0, find the position of the piston.(b) What is its velocity?(c) What is its acceleration?(d) Find the period and amplitude of the motion.
The position of the piston is 3.785 cm. The velocity of the piston is -2.585 cm/s. The acceleration of the piston is −15.14 cm/s The period of piston is 3.14 s and the amplitude of piston is 4 cm.
It is a kind of periodic motion bounded between two extreme points. For example, Oscillation of Simple Pendulum, Spring-Mass System. The object will keep on moving between two extreme points about a fixed point is called the mean position (or) equilibrium position along any path.
(a) The equation for the piston's position is given as
= x=(4.00 cm)cos(3t+6π )
At t=0,
= x =(4.00 cm)cos(6π )
= x = 3.785 cm
(b) Differentiating the equation for position with respect to time gives us the piston's velocity:
= v = dt/dx = −(8.0 cm/s)sin(2t+ 6π )
At t=0,
= v = dt/dx = −(8.0 cm/s)sin(6π )
= v = -2.585 cm/s
(c) Differentiating again gives its acceleration:
= a = dv/ dt =−(16.0 cm/s 2 )cos(2t+ 6π )
At t=0,
= a = −(16.0 cm/s 2 )cos(6π )
= a = −15.14 cm/s
(d) The period of motion is = T
= T = ω/2π
= T = 2/2π
= T = 3.14 s
and,
The amplitude directly from the equation for x =
= A = 4 cm
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Which statement is a correct definition of a material’s strength?
A.
the ability to deflect sound
B.
the ability to resist radiation
C.
the ability to withstand stress
D.
the ability to resist magnetism
E.
the ability to expand with heat
Answer:
i think its stress (c) hope this helpa
Answer:
C. the ability to withstand stress
Explanation:
The strength of a material refers to its ability to withstand stress. Stress is a measure of the force applied to a material, and it can be either tensile (stretching) or compressive (squeezing). A material's strength is typically measured by its ability to resist deformation or failure under a given amount of stress.
two forces, f1 and f2, act on a particle. as a result the speed of the particle increases. which one of the following is not possible?two forces, f1 and f2, act on a particle. as a result the speed of the particle increases. a. The work done by F1 is positive, and the work done by F2 is zero b. The work done by each force is positive. c. The work done by F1 is positive, and the work done by F2 is negative. d. The work done by each force is negative. e. The work done by F1 is zero, and the work done by F2 is positive.
Each force produces a negative amount of effort (option d). Whenever there are two opposing forces, the net force will always be negative, no matter how strong one force is compared to the other. Reduced kinetic energy of the particle is indicated by a negative force.
How do forces work?When an object interacts with another object, it experiences a push or pull that is known as a force. Any instant two things connect, a force is transmitted to each one of them.
Force and example are defined.In daily life, forces can take many different forms. a bat's force on the ball is equal to its weight. the amount of pressure used by the hair brush to brush hair.
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a satellite of mass m is in an elliptical orbit around the earth, which has mass me and radius re. the orbit varies from a closest approach of distance a at point a to maximum distance of b from the center of the earth at point b. at point a, the speed of the satellite is vo. assume that the gravitational potential energy ug = 0 when masses are an infinite distance apart. Express your answers in terms of a, b, m, Me, Re, vo, and G.
a. Determine the total energy of the satellite when it is at A. b. What is the magnitude of the angular momentum of the satellite about the center of the Earth when it is at A? c. Determine the velocity of the satellite as it passes point B in its orbit.
The velocity of the satellite as it passes point B in its orbit is vᵇ = v₀a/b
The angular momentum of a inflexible item is described as the made of the moment of inertia and the angular velocity. it's miles analogous to linear momentum and is challenge to the essential constraints of the conservation of angular momentum principle if there's no outside torque at the item.
The belongings of any rotating object given by using moment of inertia times angular velocity. it's miles the assets of a rotating frame given by way of the fabricated from the instant of inertia and the angular speed of the rotating object.
(a) the total energy of the satellite when it is at A:-
E = PE + KE = - G*Me*M/a + 1/2mv₀²
(b) the magnitude of the angular momentum of the satellite about the center of the Earth when it is at A:-
L = mvr = mv₀a
(c) Conservation of angular momentum gives,
mv₀a = mvᵇb, or vᵇ = v₀a/b
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A 64 kg box is pushed at a constant speed up a frictionless plane for a distance of 18 m by a force parallel to the plane. If the plane makes an angle of 24 degrees above the horizontal, how much work was done?
1500 j,1200j,4600j,2900j,8800j
4600 j work is done.
To find the amount of work done, the equation for work must be used. This equation is Work = Force x Distance. Force is equal to the mass of the box multiplied by the acceleration due to gravity, and the distance is 18 meters. The angle of 24 degrees above the horizontal must also be taken into account, therefore the cosine of 24 degrees must be used. The mass of the box is 64 kg and the acceleration due to gravity is 9.8 m/s2. After plugging in all the values, the work done is 4600 J.
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suppose you observe a binary system containing a main-sequence star and a brown dwarf. the orbital period of the system is 1 year, and the average separation of the system is 1 au . you then measure the doppler shifts of the spectral lines from the main-sequence star and the brown dwarf, finding that the orbital speed of the brown dwarf in the system is 23 times greater than that of the main-sequence star.1. How massive is the brown dwarf?
Mdwarf = ? KG
The Mass of the Brown dwarf's star,M₂ = 4.132 * 10²⁷ kg
Let M₁ be the mass of the main-sequence star
M₂ be the Mass of the brown dwarf
v₁ be the speed of the main-sequence star
v₂ be the speed of the brown dwarf
The system's typical separation is 1 AU, where r₁ = r₂
The brown dwarf's orbital speed is 22 times that of the main-sequence star.
Therefore, v₂ = 22 v₁
As a main sequence star, the sun
The Mass of the main-sequence star is 2 * 10³⁰ kg
Centripetal force, F = Mv²/r
M₁v₁²/r₁² = M₂v₂²/r²
M₁v₁² = M₂v₂²
M₂/M₁ = v₁²/v₂²
M₂/M₁ = (1/22)²
M₂/M₁ = 0.002066
M₂ = 0.002066M₁
M₂ = 0.002066 * 2 * 10³⁰
M₂ = 4.132 * 10²⁷ kg
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through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 pa. the refinery needs the ethanol to be at a pressure of 2 atm (202600 pa) on a lower level. how far must the pipe drop in height in order to achieve this pressure? assume the velocity does not change. (hint: use the bernoulli equation. the density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. pay attention to units!)
The pipe must drop in height by 13.101 m in order to achieve this pressure.
Bernoulli stated that at whatever point along the fluid flow, the total amount of pressure, kinetic energy and potential energy per unit volume is the same.
P + 1/2 ρv² + ρ* g* h = constant
Given that, v = 1 m/s
P₁ = 101300 Pa
P₂ = 202600 Pa
ρ ethanol = 789 kg/m³
g = 9.8 m/s²
Now, let us write the bernoulli equation,
ρ*g*h₁ + P₁ = ρ*g*h₂ + P₂
P₁ - P₂ = ρ*g*h₂ - ρ*g*h₁
( h₂ - h₁) = (P₁ - P₂)/ρ*g
Δh = (P₁ - P₂)/ρ*g
Δh = -101300 / ( 789 * 9.8) = -13.101 m
Minus sign indicates that height must be reduced.
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cA proton and an electron are held a short distance apart and released from rest. The only nonnegligible forces on the particles are the electrostatic forces they exert on each other. Which of the following statements is true? The magnitude of the electrostatic force exerted on the proton by the electron is greater than that exerted on the electron by the proton. The magnitude of the acceleration of the proton is less than the magnitude of the acceleration of the electron The proton must move a larger distance to have the same magnitude change in potential energy as the electron. The proton must move the same distance to have the same change in speed as the electron.
The magnitude of the electrostatic force exerted on the proton by the electron is greater than that exerted on the electron by the proton.
What kind of force is electrostatic?
The Coulomb's inverse-square law, sometimes known as Coulomb's law, is an observational physical principle that measures the force exerted between two electrically charged particles that are stationary. Common names for the electric created between two charged objects at rest include electrostatic force and Coulomb force.
Electrostatic force, where is it?Between two charges that are separated by a distance, there is an electrostatic force. The size of each charges and the separation between them determine the strength of the electrostatic force. Two charges that are either positive or negative when placed together repel one another.
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1. which of the following is the speed of light? 300,000 km/sec 186,283 kilometers per second 186,283 meters per second 186,283 meters per hour 300,000 km/hr
Speed of light, is generally 299,792,458 m/s. The appropriate choice is "186,283 kilometers per second.
The speed of light is approx. 299,792,458 m/s. This is equal to approximately 186,282.397 miles per second or approximately 300,000 kilometers per second.
Therefore, the correct answer is "186,283 kilometers per second" or "300,000 km/sec". The other choices are incorrect because they are not the correct units for the speed of light. Visible light is an example of electromagnetic radiation that moves at the speed of light. Light and other electromagnetic waves appear to move instantly for many practical uses, but their finite speed has dramatic implications for vast distances and particularly sensitive measurements. The starlight that can be seen from Earth has been out there for a very long time, allowing people to view far-off objects and learn about the evolution of the universe. It can take minutes to hours for signals to reach from Earth to a spacecraft when corresponding with far-off space missions. The absolute shortest communication delay between computers, to computer memory, and within a CPU is fixed in computing at the speed of light.
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the electromagnetic spectrum displays electromagnetic waves in order of their wavelengths and frequencies. according to this diagram, what approximate range of wavelengths are visible to the human eye?
Gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, and radio waves are the components of the electromagnetic spectrum, listed in order of decreasing energy.
What are the 7 different wavelength types?The parts of the electromagnetic spectrum are designated as follows, from highest to lowest energy: gamma rays, X-rays, ultraviolet radiation, visible light, infrared radiation, & radio waves. The radio wave region of the electromagnetic spectrum includes microwaves.
What is the name of wavelength 1?When wavelength is expressed in metres, 1/ stands for the number of waves that can be found in a wave train that is one metre long, or for the number of waves that can be found in a wave train that is one centimeter long. The wavenumber of the spectrum line is this number.
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what's the effect of gravity on the balance?
Answer:
Explanation:
The lower your center of gravity, the easier it is to keep your balance. If you're sitting on a chair, you can lean over more than if you're standing up. With your center of gravity low, you can lean further to one side or the other without creating enough turning force to tip you over.
The 3.0 cm-diameter water line in the figure splits into two 1.0 cm-diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa.
What is the gauge pressure at point B?
At point B, the gauge pressure is 22 kPa.
Calculation:
Given gauge pressure is 50 kPa.
Diameter is 3.0cm
speed is 2.0m/s
the gauge pressure is 22 kPa.
How can you figure out a point's gauge pressure?Pg denotes gauge pressure, which is related to absolute pressure as follows: Pa is the local atmospheric pressure, and pg is equal to p - pa. Example: 32.0 psi is the tire pressure as measured by an automobile tire gauge. 14.2 psi is the atmospheric pressure in the area.
What is a point's gauge pressure?In order to calculate gauge pressure, atmospheric (ambient) pressure is subtracted from absolute pressure. A scenario where the absolute pressure is higher than the atmospheric pressure is referred to as positive overpressure, and a situation where the absolute pressure is lower than the atmospheric pressure is referred to as negative overpressure.
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Incomplete question:
The 3.0-cm diameter water line shown at right splits into two 1.0-cm diameter pipes. All pipes are circular and at the same elevation. At point A, the water speed is 2.0 m/s and the gauge pressure is 50 kPa. What is the gauge pressure at point B?
A)12 kPa B)22 kPa C)29 kPa D)38 kPa E)42 kPa
consider a rocket that has a specific heat ratio of 1.25 and a nozzle area ratio of 10:1. the stagnation pressure is 400 psi. using the plots below, estimate the design altitude for the rocket.
The points of the design altitude for the rocket, when the stagnation pressure is of 400 psi, using given plots will be :-
The Exit velocity and Pressure in the case of a rocket motor are 1290m/s and 8.1 atm respectively. The C* and thrust propellant calculations are based on certain formulas.
Exit velocity is v_e = sqrt((2 * C_p * 3000 K) / (1.25 - 1)) * sqrt((1.25 + 1) / 2) * (1 - (p_e / 60 atm) ^ ((1.25 - 1) / 1.25)) = 1290 m/s.Exit pressure is p_e = 60 atm * ((2 / (1.25 + 1)) * (1 - (1290 m/s / sqrt((1.25 + 1) / 2 * C_p * 3000 K)) / (1 - (p_e / 60 atm) ^ ((1.25 - 1) / 1.25)))) ^ (1.25 / (1.25 - 1)) = 8.1 atm. This is the pressure at the nozzle exit.To calculate c* for the propellant, we can use the equation c* = sqrt(1.25 * 8.314 J/mol-K * 3000 K / 18) = 654.3 m/s.To calculate the mass flow rate for the rocket, we find that m_dot = 60 atm * (pi * (5 cm / 2)^2 / 4) * 654.3 m/s / sqrt(3000 K) = 4.1 kg/s.To calculate the thrust of the rocket, we can use the equation that F= 4.1 kg/s * 1290 m/s + (8.1 atm * (pi * (30 cm / 2)^2 / 4) - 60 atm * (pi * (5 cm / 2)^2 / 4)) = 45600 N. This is the thrust produced by the rocket at sea level.To calculate the thrust coefficient of the rocket motor, we can use the equation C_F = (2 / (1.25 + 1)) * (1 - (1290 m/s / sqrt((1.25 + 1) / 2 * C_p * 3000 K)) / (1 - (8.1 atm / 60 atm) ^ ((1.25 - 1) / 1.25))) = 0.8. This is the thrust coefficient of the rocket motor.To calculate the effective exhaust velocity of the rocket motor, we can use the equation v_e, eff = v_e + g * Isp, where v_e, eff is the effective exhaust velocity, v_e is the exit velocity, g is the acceleration due to gravity, and Isp is the specific impulse. At sea level, g = 9.8 m/s^2.The specific impulse of the rocket motor is given by the equation Isp = (1.25 * 8.314 J/mol * K * 3000 K) / (9.8 m/s^2 * 18 g/mol) = 279 s.The effective exhaust velocity of the rocket motor is v_e, eff = 1290 m/s + 9.8 m/s^2 * 279 s = 36660 m/s. This is the effective exhaust velocity of the rocket motor when fired at sea level.To calculate the thrust of the rocket motor, we can use the equation F = 8.1 atm * 28.3 cm^2 * 1290 m/s = 3.1 * 10^6 N.To calculate the thrust coefficient, we can use the equation C_F = 3.1 * 10^6 N / (8.1 atm * 0.2 cm^2) = 0.16.To calculate the effective exhaust velocity of the rocket motor when fired in space, we can use the equation v_e = sqrt((2 * C_p * 3000 K) / (1.25 - 1)) * sqrt((1.25 + 1) / 2) * (1 - (0 / 60 atm) ^ ((1.25 - 1) / 1.25)) * sqrt(1 - ((18 / 18) ^ 2)) = 1290 m/s.To calculate the specific impulse of the rocket motor, we can use the equation I_sp = 1290 m/s / 9.8 m/s^2 = 131 s. This is the specific impulse of the rocket motor when fired in space.Therefore, these are the points of the design altitude for the rocket, when the stagnation pressure is of 400 psi, using given plots.
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[NOTE: THIS IS AN INCOMPLETE QUESTION. THE COMPLETE QUESTION IS: Consider a rocket motor equipped with a nozzle in which the stagnation conditions are 3000 K and 60 atm. The specific heat ratio is 1.25 and the molecular weight of the propellant is 18. The nozzle throat diameter is 5 cm and the nozzle exit diameter is 30 cm. a. Calculate the exit velocity b. Calculate the exit pressure c. Calculate c* for this propellant d. Calculate the mass flow rate for the rocket e. If the rocket is fired at sea level, calculate the thrust. f. If the rocket is fired at sea level, calculate the thrust coefficient g. If the rocket is fired at sea level, calculate the effective exhaust velocity h. If the rocket is fired at sea level, calculate the specific impulse. i. If the rocket is fired in space, calculate the thrust. j. If the rocket is fired in space, calculate the thrust coefficient k. If the rocket is fired in space, calculate the effective exhaust velocity I. If the rocket is fired in space, calculate the specific impulse]
On the surface of the moon, gravity is 1.6m/s^2. An astronaut has a mass of 90kg.
(a) Calculate his weight on earth.
(b) State his mass when on the moon.
(c) Calculate his weight on the moon.
Answer:
Hope this helps.
Explanation:
(a) To calculate the astronaut's weight on Earth, you can use the formula for weight, which is W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. On Earth, the acceleration due to gravity is approximately 9.8 m/s^2. So the astronaut's weight on Earth would be W = 90 kg * 9.8 m/s^2 = <<90*9.8=882>>882 N.
(b) The astronaut's mass when on the moon is the same as it is on Earth, which is 90 kg.
(c) To calculate the astronaut's weight on the moon, you can use the same formula as before, but with the acceleration due to gravity on the moon, which is 1.6 m/s^2. The astronaut's weight on the moon would be W = 90 kg * 1.6 m/s^2 = <<90*1.6=144>>144 N.
A 0.45 kg block oscillates back and forth along a straight line on a frictionless horizontal surface. Its displacement from the origin is given by
x = (12 cm)cos[(17 rad/s)t + p/2 rad]
(a) What is the oscillation frequency (in Hz)? (b) What is the maximum speed acquired by the block? (c) At what value of x does this occur? (d) What is the magnitude of the maximum acceleration of the block? (e) At what positive value of x does this occur? (f) What force, applied to the block by the spring, results in the given oscillation?
Give your answers in centimeter-based units, where applicable.
(a) The angular frequency is 2.7 Hz
(b) The maximum speed is 2.04 m/s
(c) The maximum value is occur at mean position
so the dispalcement is zero
(d) The maximum acceleration is 34.68 m/s^2.
(e) The acceleartion should be maximum at extrem positions
so the position x = A = 12 cm
( f ) The force on the block 15.6 N
Calculation :
The mass of the block m = 0.45 kg
The dispalcement of the block
x = 12cm cos [(17 rad/s) t + (π/2 rad)]
(a) The angular frequency
ω = 17 rad/s
2πf = 17 rad/s
then the frequency
f = 17 / 2π = 2.7 Hz
(b) The maximum speed
v max = Aω = (12 cm) (17 rad/s)
= 2.04 m/s
(c) The maximum value is occur at mean position
so the dispalcement is zero
(d) The maximum acceleration
a max = Aω^2
= (0.12 m) (17)^2 = 34.68 m/s^2
(e) The acceleartion should be maximum at extrem positions
so the position x = A = 12 cm
(f) The force on the block
F = ma = (0.45 kg) (34.68 )
= 15.6 N
Force or Energy Exercised or Applied: Cause of Movement or Change: Active Force. force of nature
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A circular saw is powered by a motor. When the saw is used to cut wood, the wood exerts a torque of 0.80N on the saw blade. If the blade rotates with a constant angular velocity of 20 rad/s the work done on the blade by the motor in 1.0 minis:
A. 0
B. 480 J
C. 960 J
D. 1400 J
J E. 1800 J
If the blade rotates with a constant angular velocity of 20 rad/s the work done on the blade by the motor in 1.0 min IS 960 J.
A torque is a force that causes rotation. Tells how effective force is at twisting or rotating an object.
Conditions for Equilibrium is that when FNet = 0 Translational EQ (Center of Mass) tNet = 0 Rotational EQ (True for any axis). Choose axis of rotation wisely to make problems easier, But as long as you’re consistent everything will be okay.
Gravitational Force Weight = mg, acts as force at center of mass, Torque about pivot due to gravity t = mgd, object not in static equilibrium.
As we know,
work = torque x angular displacement
θ = ωt
θ = 20 x 60
θ = 1200rad
hence,
work = 0.80 x 1200
W = 960J
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Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz 280 Hz b. 400 Hz 160 Hz 200 Hz e.180 Hz
160 Hz is the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.
According to this option c) is correct answer.
A string's fundamental frequency, succeeding frequencies, and two ends are defined as the sum of the entire depending on the number of harmonics. In other words, higher harmonics specify the frequencies that follow under the functions 2f, 3f, 4f, 5f, etc.
As a result, the higher harmonics would be:
1 x 80 Hz Equals 80 Hz (1st harmonic and Fundamental Frequency)
2 × 80Hz Equals 160Hz (2nd harmonic) (2nd harmonic)
3 x 80Hz Equals 240Hz (3rd harmonic)
4 × 80Hz = 320Hz (4th harmonic)
5 x 80Hz Equals 400Hz (5th harmonic)
As a result, the two higher harmonics of a string with an 80Hz fundamental frequency are 160Hz and 240Hz.
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I understand that the question you are looking for is:
Select the higher harmonics of a string fixed at both ends that has a fundamental frequency of 80 Hz.
a)280 Hz
b) 400 Hz
c) 160 Hz
d) 200 Hz
e) 180 Hz
A uniform conducting rod is fixed and allowed to pivot about its center, as shown in the figure. A uniform, positively charged conducting sphere is brought near the upper end of the rod, and the rod rotates clockwise. The rod is then reset to the position shown. The sphere is then brought to point X near the lower end of the rod, and the rod rotates counterclockwise. Which of the following could be true of the charge on the rod? Select two answers.
The rod is positively charged.
The rod is uncharged.
The rod has no net charge until the sphere is brought near it.
The rod is negatively charged.
The correct answers are: a) The rod is negatively charged and d) The rod is positively charged.
When the sphere is brought near the upper end of the rod and the rod rotates clockwise, this indicates that the sphere has a positive charge.
When the sphere is brought to point X near the lower end of the rod and the rod rotates counterclockwise, this also indicates that the sphere has a positive charge.
The other two options are not correct. The rod is not uncharged, because it has a charge when the sphere is brought near it. The rod also does not have no net charge until the sphere is brought near it, because it has a charge when the sphere is brought near it.
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which of the following statements is correct about mohr's circle. select all correct ones. group of answer choices mohr's circl is a graphical solution to state of stress that allows the evaluateof normal and shear stresses on an inclined plane (at an angle
An angle '[tex]\Theta[/tex]' on the circle Corresponds to an angle 2[tex]\Theta[/tex] on the part.
On the Mohr, the circle has a positive counterclockwise rotation, similar to the physical space convention for shear stress. is the clockwise shear stress and both are plotted. Mohr's circle is a graphical representation of the general state of stress at a point.
This is a graphical method for evaluating principal and maximum shear stresses. Normal and tangential stress in any plane. Soil mechanics use Mohr's circles to visualize the relationship between normal and shear stresses, and to estimate the maximum stress based on three or more soil samples taken from a site.
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In the engine of a locomotive, a cylindrical piece known as a piston oscillates in SHM in a cylinder head (cylindrical chamber) with an angular frequency of 180rev/min. Its stroke (twice the amplitude) is 0.76m. What is its maximum speed?
A cylindrical component known as a piston oscillates in the cylinder head (cylindrical chamber) of a locomotive's engine. 0.76 meters is the stroke, which is twice the amplitude. Its greatest speed is 7.2m/s.
We use v[tex]_{m}[/tex] =ωx[tex]_{m}[/tex]=2πfx[tex]_{m}[/tex], where the frequency is 180/(60s)=3.0Hz and the amplitude is half the stroke, or x[tex]_{m}[/tex]=0.38m. Thus,
v[tex]_{m}[/tex]=2π(3.0Hz)(0.38m)=7.2m/s.
A piston is a disc or short cylinder that fits snugly inside an engine cylinder and imparts motion to a pump or a liquid or gas by moving up and down against it. Pneumatic and hydraulic cylinders, reciprocating engines, reciprocating pumps, gas compressors, and other related devices all have pistons. It is the moving part that has a cylinder around it and piston rings that seal the gas tightness.
To transfer force from the expanding gas in the cylinder to the crankshaft via a piston rod and/or connecting rod is the function of a piston rod and/or connecting rod in an engine.
The processes of intake, compression, combustion, and exhaust are all four-stroke operations in automobile engines.
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which of the following statements is correct about mohr's circle. select all correct ones. group of answer choices mohr's circl is a graphical solution to state of stress that allows the evaluateof normal and shear stresses on an inclined plane (at an angle
Mohr's circle can be used to calculate the stresses on an oblique section if the stress condition on two planes and the angle between those planes are known. The correct option to this question is C.
An illustration of the transformation equations for plane stress problems is Mohr's circle. The relationships between the normal and shear loads operating on a stress element at any desired orientation can be visualized using this technique.
A Mohr's circle must have a diameter, and in order to draw the diameter, two straight line endpoints' coordinates are needed. Apply the principles on Planes A1 and B1 to obtain the stress locations for Mohr's circle diameter if the shear axis is positive upwards.
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Complete question: Consider the following statements:
Mohr's Circle is used to determine the stress on an oblique section of a body subjected to
1. direct tensile stress on one plane accompanied by a shear stress.
2. direct tensile stresses in two mutually perpendicular directions accompanied by a simple shear stress.
3. direct tensile stress in two mutually perpendicular directions.
4. A simple shear stress
Select the correct answe using the codes given below:
A 1 and 4
B 2 and 3
C 1, 2, 3 and 4
D 1, 2 and 3
if the ball is released from rest at a height of 0.91 m above the bottom of the track on the no-slip side, what is its angular speed when it is on the frictionless side of the track? assume the ball is a solid sphere of radius 2.1 cm and mass 0.14 kg .(figure 1)How high does the ball rise on the frictionless side?
Angular speed when it is on the frictionless side of the track is 89.7 rad/s
Given that, mass, m=0.14 kg
Ball is released. from rest. at a height. of, h= 0.83 m
Solid sphere of radius, R = 3.8 cm
=0.038 m
From the conservation of energy
ΔK = ΔU
1/2 mv2 + 1/2 Iω2 = mgh
Here, I=2/5MR2 , v=Rω
1/2 mv2 + 1/2 (2/5mR2)(v/R)2 =mgh
1/2 [v2 +2/5 v2] =gh
7/10 v2 = gh
0.7v2 = gh
v =√ [gh / (0.7) ]
=√ [(9.8 m/s2)(0.83 m) / (0.7) ]
= 3.408 m/s
Consequently, angular speed on the frictionless side of the rail,
ω = v/R
= (3.408 m/s)/(0.038 m)
= 89.7 rad/s
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ucy has three sources of sound that produce pure tones with wavelengths of 36 cm , 60 cm , and 92 cm . she also has an assortment of narrow tubes that are closed off at one end but open at the other. after experimenting, lucy notices that each tone will resonate with tubes of some lengths but not with others.for each wavelength, determine which tube lengths, if any, will resonate with sound of that wavelength.
The tubes with lengths of 15, 45, and 75 cm are found to resonate with this wavelength. Additionally, resonance is seen for tubes with lengths of 25 and 75 cm and 31 and 93 cm.
For this reason, at the place where the tube is closed, we have a node, and at the open point, we have a belly; in this case, the fundamental wave is = 4L. To determine the length of the tube that has resonance, we must establish the natural frequencies of the tubes.
The following resonance, known as the first harmonic, is 3 = 4L / 3.
The next fifth harmonic resonance is 5 = 4L / 5, and we can see that its general form is n= 4L / n, where n = 1, 3, and 5.
Let's apply these terms to the issue at hand.
Start with the wavelength that is the shortest.
Lam equals 60 cm.
Let's determine the length of the tube this harmonica provides.
L = λ n / 4
n = 1 L = 60 will yield the shortest tube length. 1/4\s L = 15 cm
For n = 3 L = 60 3/4\s L = 45 cm
For n = 5 L = 60 5/4\s L = 75 cm
For n = 7 L = 60 7/4\s L = 105cm
We can see that the harmonics 1, 3 and 5 of the tubes with lengths 15, 45, and 75 resonate with this wavelength.
b) λ = 100 cm
For n = 1 L = 100 1/4\s L = 25 cm
For n = 3 L = 100 3/4\s L = 75 cm
For n = 5 L = 100 5/4\s L = 125 cm
There is resonance in the fundamental and third ammonium frequencies for lengths of 25 and 75 cm. L = 124 1/4 L = 31 cm
In the case of the second resonance, L = 124 3/4 L = 93 cm.
For tubes with lengths of 31 and 93 cm, resonance occurs in the fundamental and third harmonics.
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A group of scientists discover a new, rare isotope and are able to store a small amount of it. They determine that the isotope is unstable and half of their sample will decay in 17.0 months. The scientists need a new laboratory to properly conduct measurements on the isotope. If a laboratory is built before half the sample decays there will still be enough of the isotope available for experiments. However, it will take 29.0 months to build the new lab. Fortunately, the scientists can quickly start construction and they have access to a spaceship that can travel at speeds approaching the speed of light c = 3.00×108 m/s. If they place their sample of the isotope on the spaceship, at what speed u must it travel in order for the new laboratory to be completed on Earth by the time half of the isotope on the spaceship decays? Assume that it took one month for the scientists to actually start construction and launch the spaceship so half the sample will remain in 16.0 months but it will still take 29.0 months to build the new lab.A) At what speed u must the spaceship travel for the scientists to have enough time to complete their laboratory by the time half of the isotope has decayed?
'u' must the spaceship travel for the scientists to have enough time to complete their laboratory by the time half of the isotope has decayed at speed of 2.416*10⁸ m/s .
Calculation :
[tex]29 = \frac{16}{1-\frac{c}{v} ^{2} }[/tex]
v = 0.805 * C
V = 0.805 * 3*10*⁸
= 2.146*10⁸ m/s
Isotopes are two or more types of atoms that have the same atomic number (number of protons in the nucleus) and position in the periodic table (and thus belong to the same chemical element), but that differ because of their nucleon number (mass number). The number of neutrons in the core is different. All isotopes of a particular element have approximately the same chemical properties, but they have different atomic masses and physical properties
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A block of mass 2kg starts from rest sliding down a rough inclined plane making an angle of 60⁰ with the horizontal length of plane in 8m it takes 4 second to return the bottom find the cofficent of kinetic friction
The force applied on the body that is on the inclined plane is given as, F=mgsinθ, F=2×9.8×sin30 =9.8N and f=μmgcosθ f=0.7×2×9.8cos30 =11.88N.
What is limiting friction?Any increase in the moving force will result in slippage; limiting friction only exists when the moving force and the force opposing motion are equal.
The limiting frictional force is independent of the area of contact and proportional to the normal reaction between the contacting surfaces.
The amount of friction that can be produced when two static surfaces come into touch. A motion will start as soon as the force applied to the two surfaces surpasses the limiting friction. The normal response force and the coefficient of limiting friction are combined to get the limiting friction for two dry surfaces.
Therefore, The force applied on the body that is on the inclined plane is given as, F=mgsinθ, F=2×9.8×sin30 =9.8N and f=μmgcosθ f=0.7×2×9.8cos30 =11.88N.
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9. a solid metal ball of radius 1.3 cm bearing a charge of 6.2 nc is located near to and along the midline of a hollow uniformly charged plastic rod of radius 1.9 cm and length 12.5 m bearing a uniformly distributed charge of 7.1 nc (see the figure) on its outer surface. the distance between the center of the rod and the center of the ball is 25.5 cm.
Since a metal ball is considered a solid sphere So the electric field inside the metal ball is given by E = (p/3∈₀)r
Where r is the distance from the center and rho is charge density. Hence at the center r = 0
Then electric field
E = (p/3∈₀) x 0 = 0 N/C
Hence electric field at the center of the metal ball is due to its own charge only being 0N/C.
An electrical property is associated with any point in space when some form of electric charge is present. The magnitude and direction of the electric field are represented by the value of E and are called field strength or electric field strength or simply electric field.
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A concave mirror with a focal length of 40 cm. How far should the mirror be
held from his face in order to give an image of two-fold magnification
The mirror should be held at a distance of 60cm
What is Magnification?Magnification is defined as the ratio of the height of image to the height of object. It can also be defined as the ratio of distance of image from mirror or lens to the distance of object from lens.
Magnification is given as v/u . where v is the image distance and u is the object distance.
therefore 2 = v/u
v = 2u
from the mirror formula
1/f = 1/v+ 1/u . where F is the focal length
represent 2u for v
1/f = 1/2u + I/u
1/f = 3/2u
1/40 = 3/2u
2u = 120
divide both sides by 2
u = 120/2
u = 60cm
therefore the object distance is 60cm. This means that to have a double sized image , the distance from the mirror and his face must be 60cm.
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Platinum is FCC and has a lattice constant of 0.39239 nm
Yes, that's correct. The FCC (face-centered cubic) crystal structure is a common structure for many metals, including platinum. The lattice constant is a measure of the distance between the atomic planes in a crystal structure. It is equal to the distance between two adjacent atoms in the lattice and is typically measured in nanometers (nm). The lattice constant for platinum is approximately 0.39239 nm.
Platinum is a dense, ductile, and malleable metal with a high melting point and good electrical and thermal conductivity. It is resistant to corrosion and is often used in catalytic converters, jewelry, and electronic components. It is also used in the production of chemicals, pharmaceuticals, and other industrial applications.
the figure shows the net force exerted on the satellite by the moon and the direction of the tangential velocity of the satellite at time t0. which of the following statements is true regarding the motion of the satellite?
The statement which is true regarding the motion of the satellite is D) The tangential velocity of the satellite does not remain constant.
Any object in orbit around the earth, the sun, or another large body is considered a satellite.
The satellite is able to maintain orbital motion even as gravity pulls it inward thanks to its tangential velocity. The necessary centripetal force is provided by gravitational attraction.
Newton was the first to say that a projectile launched with sufficient speed would actually orbit the earth.
The satellite moves faster when it is close to the earth and slows down while moving away to the top of the ellipse.
The question is incomplete. The complete question is ' A) The gravitational force exerted on the satellite by the moon will increase after time t0.
B) The satellite accelerates in a direction that is parallel to the direction of the tangential velocity.
C) The satellite will move toward the moon after time t0.
D) The tangential velocity of the satellite does not remain constant.'
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a) Traveling waves propagate with a fixed speed usuallydenoted as v(but sometimes c). The waves are called _______________ if their waveformrepeats every time interval T.
The fundamental relationship among frequency, wavelength, andvelocity is
v = f \lambda.
This relationship may be visualized as follows: In 1 s, f\cdot(1\;{\rm s}) cycles of the wave move past an observer. In thissame second the wavetrain moves a distance v\cdot(1\;{\rm s}).
c) If the velocity of the wave remains constant, then as thefrequency of the wave is increased, the wavelength___________.
d) The difference between the frequency f and the frequency omega is that fis measured in cycles per second or hertz (abbreviated Hz) whereasthe units for \omega are _______ per second.
e) Find an expression for the period of a wave T in terms of other kinematic variables.
f) What is the relationship between omega and f?
g) What is the simplest relationship between the angularwavenumber kand just one of the other kinematic variables?
The waves are called periodic wave if waveform repeats in time interval T. A wave is said to be periodic if its duration of transmission is infinity. In a periodic wave wavelength and frequency is regulated. In a given time interval the wave completes one cycle.
For a periodic wave the wave function is y(x,t) = A sin2(x -vt) .The term 2 scales the wave to the natural period of the sin function. The term A gives the amplitude of the wave which is the maximum displacement of the wave.
If the velocity of the wave remains constant, then as the frequency of the wave is increased, the wavelength decreases
The phase difference between any two points on a wave is found as follows:-At P1 the phase is 1 = kx1- t1 +,
At P2 the phase is 2 = kx2- t2 +.
So the phase difference is = 2-1 = k(x2 - x1)- (t2 - t1) The wave speed is given by:- v=k=2f2/=f
Examples of periodic wave can be seen in:-
1.Electromagnetic wave in optical fibre
2.Sound from a guitar string
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