The capacitance of a parallel-plate capacitor made of two 4.0 cm x 4.0 cm metal plates if the plates are separated by 5.0 mm is 0.073 F (or 73 nF).
The capacitance of a parallel-plate capacitor can be calculated using the formula:
C = ε_0 × A / d
where:
C is the capacitance in Farads
ε_0 is the vacuum permittivity constant (8.854 x 10⁻¹² F/m)
A is the area of each plate in square meters
d is the distance between the plates in meters
So, for the given parallel-plate capacitor:
A = 4.0 cm x 4.0 cm = 0.04 m x 0.04 m = 0.0016 m²
d = 5.0 mm = 5.0 x 10⁻³ m
C = ε_0 × A / d
C = 8.854 x 10⁻¹² F/m * 0.0016 m² / 5.0 x 10⁻³ m
C = 0.073 F (or 73 nF)
Therefore, the capacitance is 0.073 F (or 73 nF)
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18. in an early set of experiments (1911), millikan observed that the following measured charges, among others, appeared at different times on a single oil drop. what value of ele- mentary charge can be deduced from these data? a. 6.563 x b. 8.204 x c. 11.50 x d. 13.13 x e. 16.48 x 1o19c 10- 19 c 10- 19 c 10- 19 c 1o19c f. 18.08 x 10- 19 c g. 19.71 x 10- 19 c h. 22.89 x 10- 19 c i. 26.13 x 10.- 19 c
Based on Millikan's experiments and later studies, the most accurate value for the elementary charge is known to be 1.60 x 10^-19 C. Therefore, the correct answer is (g) 19.71 x 10^-19 C.
Robert Millikan performed experiments in 1911 to measure the charge on a single oil drop by measuring the change in its electric potential due to an applied electric field. He found that the charges on the oil drops were integer multiples of a fundamental unit of charge, which is now known as the elementary charge.
From the data in the question, we can deduce that the elementary charge is equal to one of the values listed in the answer choices.
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Three horizontal force are pulling on a ring, at ret. F1 i 6. 25 N at a 180 angle, and F2 i 8. 90 N at a 2430 direction. What i the y-component of F3?
According to the question of force, the y-component of F3 is -2.719 N.
What is force?
Force is a fundamental concept in physics that describes the interaction between two objects or systems. It is a vector quantity, meaning it has both magnitude and direction, and can be described mathematically. Force is a push or pull that can cause an object to accelerate, change its direction, or change its shape. Force is a function of mass and acceleration, as described by Newton's second law of motion.
The y-component of F3 can be calculated using the equation F3y = F1y + F2y. Since F1 is at a 180° angle, its y-component is 0. The y-component of F2 can be calculated using the equation F2y = F2 × sin(θ), where θ is the angle of F2. In this case, θ is 2430°, so the y-component of F2 is 8.90 × sin(2430°) = 8.90 × -0.3105 = -2.719 N. Therefore, the y-component of F3 is -2.719 N.
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Consider a force F = 80 N applied to a beam as shown in Fig. 8-37. The length of the beam is = 5.0 m, and 0 = 37°, so that x = 3.0m and y = 4.0 m. Of the following expressions, which ones give the correct torque produced by the force F around point P? (a) 80 N.(b) (80 N)(5.0 m).(c) (80 N)(5.0 m) (sin 37°).(d) (80 N)(4.0 m).(e) (80 N)(3.0 m).(f) (48 N)(5.0 m).(g) (48 N)(4.0 m) (sin 37°).
The math demonstrates that all three torques are equivalent.(80 N)(5.0 m)(sin 37∘)).(48 N)(5.0 m) (5.0 m).(80 N)(3.0 m) (3.0 m).
Three different approaches to express the torque are demonstrated by equations such. = rFsin. It can be the sine of the angle between them, the product of the force, the lever arm, and the response (c). It may be the result of the force and the part of the lever arm that is perpendicular to the force, as in answer (e). Another way to write it is as the sum of the force parallel to the lever arm and the lever arm, as in the following response (f). The math demonstrates that all three torques are equivalent.
(80 N)(5.0 m)(sin 37∘)).
(48 N)(5.0 m) (5.0 m).
(80 N)(3.0 m) (3.0 m).
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a point charge of 8.00×10−12 c is located at the center of a cubical gaussian surface. what is the electric flux φface through each face of the cube?
When a point charge of 8.00×10−12 c is located at the center of a cubical gaussian surface, 143.84 x 10^3 is the electric flux φface through each face of the cube.
The electric flux through a face of a cube is given by the product of the electric field and the area of the face. The electric field due to a point charge can be calculated using Coulomb's law:
=> E = k x Q / r^2
here,
E is electric field,
k is C constant (8.99 x 10^9),
Q is charge,
r is distance from the point charge to the surface,
In this case, the charge is
= 8.00 x 10^-12 C
and the distance from the point charge to the surface is equal to half the length of one side of the cube,
So, r = a/2
here,
a is length of a side of the cube.
The area of one face of the cube is a^2,
The electric flux of face is:
=> φface
= E x a^2
= k x Q / (a/2)^2 x a^2
= 2k x Q / a^2
Since there are six faces of the cube in the total electric flux is:-
=> φtotal
= 6 x φface
= 6 x 2k x Q / a^2
adding values, we get:
=> φtotal
= 6x2x8.99 x 10^9 Nm^2/C^2 x 8.00 x 10^-12 C / (a/2)^2
The electric flux of each face is :-
=> φface
= φtotal / 6
= 2 x 8.99 x 10^9 Nm^2/C^2 x 8.00 x 10^-12 C / a^2
= 143.84 x 10^3
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An electric field of constant magnitude ???? is tangent everywhere to a curved surface of area ????. What is the electric flux through the curved surface?
a) Zero
b) EA
c) –EA
d) 2EA
e) Not enough information
EA is the electric flux at the surface. because the electric flux is calculated by multiplying the electric field by the surface area it covers.
Because the surface area is known and the electric field intensity is constant, the electric flux can be computed as EA. The strength of an electric field flowing through a particular surface is measured by the electric flux. The strength of the electric field divided by the surface area through which it passes is used to compute it. The electric flux can be determined by by multiplying these two quantities if the surface area is known and the electric field magnitude is constant. Any closed surface surrounding a zero net charge has zero net electric flux. As a result, if we are aware of the net flux over a closed surface, we are also aware of the net charge contained.
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With the settings used in the simulation, you were unable to produce the 1st harmonic in either Part A or Part B. Why not? What specific changes to the simulation settings would enable you to see the 1st harmonic in each part? Write out your answer in a clear and well supported paragraph. Is the speed of the wave constant in Parts A and B? How did you determine this for each part? If the speed is not constant, how does it change as a function of harmonic number? Write out your answer in a clear and well supported paragraph.
insufficient data or simulation software constraints If the medium is heterogeneous or anisotropic, the wave's speed can change with harmonic number to produce the first harmonic in a simulation.
Numerous factors, such as improper simulation settings, software restrictions, or insufficient data, could prevent a simulation from producing the first harmonic. The simulation settings must be configured to accurately resolve the first harmonic frequency in order to produce the first harmonic in a simulation. This can be accomplished by changing parameters like the simulation's time step size, frequency range, or number of harmonics.
The medium through which the wave is propagated determines the wave's speed. travelling as well as the characteristics of that medium, such as its elastic modulus and density. The wave's speed is constant for each harmonic if the medium is homogeneous and isotropic. However, the wave's speed can vary with harmonic number if the medium is heterogeneous or anisotropic. In order to calculate the speed using the equation v = f,
where v is the speed, f is the frequency, and is the wavelength, the frequency and wavelength of the wave in Parts A and B must be measured.
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an object is placed 40.0 cm from a concave mirror of radius 20.0 cm. (a) find the location of the image. (b) what is the magnifi cation of the mirror? is the image real or virtual? is the image upright or inverted?
Provided that the concave mirror's given focal length is 20 cm long.The object is 40 cm away, which indicates that it is at the center of the curve.
When a concave mirror with a 20 cm focal length is put 40 cm away from the object? Provided that the concave mirror's given focal length is 20 cm long.The object is 40 cm away, which indicates that it is at the center of the curve.As a result, a real, inverted image that is the same size and precisely in the focus will be produced.As a result, the picture is created 10 centimeters in front of the mirror.The image's distance from the mirror is 40 cm, and the image is reversed at that distance.The image is 40 cm away from the viewer.To learn more about concave mirror refer
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when a 1.00-kg mass is suspended from a spring, the spring stretches by .050 m. if the suspended mass oscillates in simple harmonic motion, its period will be approximately
The required time period of the mass suspended from the spring is calculated to be 0.446 s.
The spring is said to be stretched by 0.050 m (x).
The mass suspended from the spring is 1 kg (m).
The suspended mass oscillates in simple harmonic motion.
Its time period T =?.
Let us find out the force constant of the spring.
k = F/x = m g/x = (1× 10)/0.050 = 200 N/m
Now, if the mass of 1 kg is suspended by the spring, then the period of oscillation, we know the formula,
T = 2π√(m/k)
where, m is mass
k is spring constant
Putting in the values,
T = 2π√(m/k) = 2π√(1/200) = 2π × 0.071 = 0.446 s
Thus, the required time period is calculated to be 0.446 s.
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A 650 KW power engine of a vehicle of
mass 1.5 * 105 Kg is rising on an inclined plane of inclination
1 in 100 with a constant speed of 60 km/hr. Find the frictional
force between the wheels of the vehicle and the plane.
The frictional force between the wheels of the vehicle and the plane is 14.7 N.
Find the frictional force between the wheels of the vehicle and the plane?Frictional Force = mgsin(angle)Frictional Force = (1.5 * 105 kg)(9.8 m/s2)sin(1°)Frictional Force = 1470 NThe frictional force between the wheels of the vehicle and the plane is given by the equation:Ff = μmgcosθwhere μ is the coefficient of friction, m is the mass of the vehicle, g is the acceleration due to gravity, and θ is the inclination of the plane.In this case, the mass of the vehicle is 1.5 * 105 Kg, the acceleration due to gravity is 9.8 m/s2, and the inclination of the plane is 1 in 100, or 0.01 radians.Therefore, the frictional force can be calculated as follows:Ff = μ * 1.5 * 105 * 9.8 * 0.01Ff = 14.7 NThis force is necessary to ensure that the vehicle maintains its constant speed of 60 km/hr while travelling up the inclined plane.This frictional force must be greater than the force needed to overcome the gravitational force of the vehicle, otherwise the vehicle will start to slow down.To learn more about The frictional force refer to:
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the threshold wavelength for tungsten is 2700 å. what is the maximum kinetic energy of photoelectrons produced by photons of wavelength 2200 å? a) 1.5 eV. b) 2.2 eV. c) 3.0 eV. d) 5.0 eV
The maximum kinetic energy of photoelectrons produced by photons of wavelength 2200 å is 5.0eV. Hence the correct option is (d).
The maximum kinetic energy of photoelectrons produced by photons of wavelength 2200 Å can be calculated using the following equation:
Kmax = hc/λ - W
Here Kmax is max. KE of photoelectrons, h is Planck's cons. (6.62 x 10^-34 Js), c is speed of photon i.e. light (2.998 x 10^8 m/s), λ is wavelength of incident photon (2200 Å), and W is work function of metal (the minimum amount of energy required to remove an electron from the metal).
Converting the wavelength from Å to m:
λ = 2200 x 10^-10 m
Using the above equation, we can calculate the maximum kinetic energy of photoelectrons produced by photons of wavelength 2200 Å:
Kmax = (6.62 x 10^-34)(2.998 x 10^8) / (2200 x 10^-10) - W
The value of W can be calculated from the threshold wavelength of tungsten, which is 2700 Å. The relationship between the threshold wavelength and the work function can be expressed as follows:
W = hc/λthreshold
Putting the value of threshold wavelength:
W = (6.62 x 10^-34)(2.998 x 10^8) / (2700 x 10^-10)
Inserting vale for W in equation for Kmax:
Kmax = (6.62 x 10^-34)(2.998 x 10^8) / (2200 x 10^-10) - (6.62 x 10^-34)(2.998 x 10^8) / (2700 x 10^-10)
Simplifying the expression, we get:
Kmax = 5.0 eV
Therefore, the maximum kinetic energy of photoelectrons produced by photons of wavelength 2200 Å is 5.0 eV (Option d).
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what is the rms current in an rl circuit when a 50.0 hz 120 v rms ac voltage is applied, where r = 1.20 kω , and l = 370 mh ?
The RMS current in an RL circuit when 50 hertz 120 volts RMS ac voltage is applied is 82.68 mA
RMS or root mean square current/voltage of the alternating current/voltage represents the d.c. current/voltage that dissipates the same amount of power as the average power dissipated by the alternating current/voltage. For sinusoidal oscillations, the RMS value equals peak value divided by the square root of 2.
Inductive reactance
XL= 2xpixfxL
=2pix50x0.37
XL=116.24 ohms
Impedance
Z= √[R^2+XL^2] =√[2900^2+116.24^2]
Z=2902.35 ohms
Rms current
I,rms=V,rms/Z
=240/2902.35
I,rms=0.08268 A or 82.68 mA.
Therefore , The RMS current in an RL circuit when 50 hertz 120 volts RMS ac voltage is applied is 82.68 mA
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A ga at a temperature of 50°C i under a preure of 700 MM Hg what will the preure be at 30°C auming the volume tay contant
The pressure will 657 MM Hg.
The relationship between pressure and temperature for a fixed volume of an ideal gas can be described by the Ideal Gas Law: PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
P1 = (700 mm Hg) x (101.3 kPa / 1 mm Hg) = 70,910 Pa
T1 = (50 + 273.15) K = 323.15 K
T2 = (30 + 273.15) K = 303.15 K
P2 = (70,910 Pa) x (303.15 K / 323.15 K) = 65,688 Pa
Therefore, the pressure at a temperature of 30°C while keeping the volume constant would be 65,688 Pa, which is equivalent to 657 mm Hg.
Since the volume is constant in this case, we can see that the pressure is directly proportional to the temperature. Therefore, if we decrease the temperature by 20°C, the pressure will also decrease by a factor of (T2/T1)(1-γ), where γ is the adiabatic index of the gas. In reality, the pressure will not decrease exactly as calculated using the Ideal Gas Law, since the gas is not an ideal gas and the volume may not remain constant. However, the Ideal Gas Law provides a useful approximation in many situations.
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Let the angle be the angle that the vector A makes with the +x-axis, measured counterclockwise from that axis. Find the
angle for a vector that has the following components.
Ax=1.50, Ay=1.30
The angle formed by the vector and the positive or counterclockwise of x-axis is 40.91°
What are Vectors ?The term "vector" is used informally to describe constituents of particular vector spaces or some values that cannot be described by a single integer.
A quantity or phenomena with independent qualities for both size and direction is called a vector. The word can also refer to a quantity's mathematical or geometrical representation. Velocity, momentum, force, electromagnetic fields, and weight are a few examples of vectors in nature.
Any vector with a two-dimensional direction may be conceived of as having two distinct parts. The impact of a single vector in a certain direction is described by its component.
The components are
Aₓ = 1.50
Ay = 1.30
We can find the angle of the vector by using the formula
[tex]tan\theta = \frac{A_{y} }{A_{x} }[/tex] = [tex]\frac{1.30}{1.50} =[/tex]
⇒[tex]\theta = tan\x^{-1}(1.3/1.5)[/tex]
⇒θ = 40.91°
The angle formed by the vector and the positive or counterclockwise of x-axis is 40.91°
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both segments of the wire are made of the same metal. current i1 flows into segment 1 from the left. how does current i1 in segment 1 compare to current i2 in segment 2?
The current i1 in segment 1 is equal to the current i2 in segment 2.
The current in each segment of a wire depends on the resistance of that segment. If both segments of the wire are made of the same metal, they have the same electrical resistance, and therefore the same current will flow through each segment.
The quantity of current that flows through a length of wire in an electrical circuit is controlled by Ohm's law, which states that the current in a circuit is directly proportional to the voltage across the circuit and inversely proportional to the resistance of the circuit.
If the voltage across the two segments of wire is the same and the resistance of each segment is the same, then the current flowing through each segment will also be the same.
This is because the same amount of electrical energy is being supplied to each segment, and since each segment has the same resistance, the same amount of current will flow through each segment in order to meet the same energy requirements.
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what did the magdeburg experiment demonstrate about air pressure? that atmospheric pressure on the earth's surface has tremendous force that the vacuum inside the two attached hemispheres has tremendous suction that horses are not intelligent enough to understand how to defeat atmospheric pressure that the horses could not get sufficient traction to pull the hemispheres apart
When the rims were greased and the air was blasted out, the Magdeburg experiment showed that air pressure existed and that a vacuum existed inside the sphere, which prevented teams of horses from pulling it apart.
Why do the hemispheres of Magdeburg adhere to one another?The two hemispheres are held together by the tremendous inward pressure from the surrounding atmosphere, which is there despite nearly no outward pressure from within.To show the strength of air pressure, they were utilized. The sphere held a vacuum and could not be dragged apart by teams of horses in a famous first public display in 1654 after the rims were greased and the air was pushed out.When the rims were greased and the air was blasted out, the Magdeburg experiment showed that air pressure existed and that a vacuum existed inside the sphere, which prevented teams of horses from pulling it apart.To learn more about hemispheres refer to:
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what is the value of 3.43 in a.m.u (1 a.m.u = 1.66 × 10^-27)
The center of the moon is 0.0026 AU (3.890 x 10⁸ m ) from the center of the Earth and has a relative mass of 0.0123. (a) Locate the cm of the Earth-Moon system with respect to the center of the Earth. (b) With respect to the center of the Moon. (c) Find the ratio of orbital velocities of the Earth and Moon about the cm of the system. (d) If the orbital velocity of the Moon is 1012 m/s , what is the approximate orbital velocity of the Earth about the cm of the Earth-Moon system? .
A) centre of mass will be 3.843 x 10⁸ m from the centre of moon
B) r₁ = 3.843 x 10⁸m
r₂ = 0.047 x 10⁸ m
C) Ve /Vm = 0.01226
A) Let the centre of moon be considered as point of origin.
therefore centre of mass = Mm x 0 + Me x 3.89x 10⁸/ Mm + Me
Mm = mass of moon
Me = mass of earth
Mm = 0.0123 Me
Centre of mass = 3.89 x 10⁸ / 1.0123 m
So centre of mass will be 3.843 x 10⁸ m from the centre of moon or it will be at a distance of 0.00256 Au from the centre of the moon.
B) For moon G Mm Me /r² = Mm Vm² /r²
For earth G Mm Me / r² = Me Ve² / r₂²
r₁ = 3.843 x 10⁸m
r₂ = 0.047 x 10⁸ m
C) Mm Vm² / r₁ = Me Ve² / r₂
(Ve/Vm)² = Mm r₂ / Me r₁
Ve/Vm =√ 0.047 x 0.0123/ 3.843
Thus Ve /Vm = 0.01226
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A) centre of mass will be 3.843 x 10⁸ m from the centre of moon B) r₁ = 3.843 x 10⁸m r₂ = 0.047 x 10⁸ m C) Ve /Vm = 0.01226 A) Let the centre of moon be considered as point of origin.
therefore centre of mass = Mm x 0 + Me x 3.89x 10⁸/ Mm + Me Mm = mass of moon Me = mass of earth Mm = 0.0123 Me Centre of mass = 3.89 x 10⁸ / 1.0123 m So centre of mass will be 3.843 x 10⁸ m from the centre of moon or it will be at a distance of 0.00256 Au from the centre of the moon. B) For moon G Mm Me /r² = Mm Vm² /r² For earth G Mm Me / r² = Me Ve² / r₂² r₁ = 3.843 x 10⁸m r₂ = 0.047 x 10⁸ m C) Mm Vm² / r₁ = Me Ve² / r₂ (Ve/Vm)² = Mm r₂ / Me r₁ Ve/Vm =√ 0.047 x 0.0123/ 3.843 Thus Ve /Vm = 0.01226 The center of the moon is 0.0026 AU (3.890 x 10⁸ m ) from the center of the Earth and has a relative mass of 0.0123. (a) Locate the cm of the Earth-Moon system with respect to the center of the Earth. (b) With respect to the center of the Moon. (c) Find the ratio of orbital velocities of the Earth and Moon about the cm of the system. (d) If the orbital velocity of the Moon is 1012 m/s , what is the approximate orbital velocity of the Earth about the cm of the Earth-Moon system
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how nonrenewable energy resources uses have changed over the past 30 years?
Answer:
Explanation:
The use of nonrenewable energy resources, such as coal, oil, and natural gas, has changed significantly over the past 30 years. Some key changes include:
Increased use of natural gas: The use of natural gas as an energy source has increased significantly in recent years, due to the development of new technologies that have made it easier and more cost-effective to extract and transport natural gas. This has led to a decrease in the use of coal as an energy source, as natural gas is cleaner and more efficient.Increasing use of renewable energy sources: Over the past 30 years, there has been a growing focus on renewable energy sources, such as wind, solar, and hydro power, as a way to reduce the dependence on fossil fuels and lower greenhouse gas emissions. The growth of renewable energy sources has been supported by government policies, technological advances and decreasing costs.Increase in the use of electric cars: Electric cars are becoming increasingly popular in recent years and this is expected to continue as the technology improves and the cost of batteries decreases. Electric cars are more energy-efficient and have lower emissions than traditional gasoline cars.Changes in oil consumption: The consumption of oil for transportation has decreased over the past 30 years due to more efficient vehicles and more use of biofuels. However, oil consumption for electricity generation and industrial use has remained relatively constant.Overall, the past 30 years have seen a shift away from traditional fossil fuels towards cleaner and more sustainable energy sources, driven by concerns about climate change and environmental degradation.
most swimmers find the pressure at a depth of about 10 ft painful to ears. what is the gauge pressure at this depth
The vapor pressure of water at room temperature is 0.2 psi absolute pressure (or -14.5 gauge pressure).
What is the definition of 10 foot water pressure?433 4.33 kilograms per square inch
A pressure of 10 X. 433 4.33 pounds per square inch would be exerted by ten feet of water. The same height-pressure relationship is true regardless of the area of the vertical liquid column. If the density of the liquid is known, the pressure exerted by it may be computed in the same way.
So the pressure due to the water column would be 43.2 psi at 100 feet deep. Add extra 14.5 psi to account for atmospheric pressure. And the total (absolute) pressure is around 57.7 psia
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Which of the following equations has four roots?
4x 3+8x-4=0
5x 4+x-7=0
(x-4)(x+2)(x-7)=0
The equation that has four roots is: (x-4)(x+2)(x-7)=0
How can I find the root of an equation?
To find where the function intersects the x-axis, set f(x)=0 and solve the equation for x. If the function is a linear function of degree 1, f(x)=mx+b and the x-intercept is the root of the equation found by solving the equation for x.
The equation that has four roots is: (x-4)(x+2)(x-7)=0
This is a polynomial equation of degree 3, and it has 4 roots x = 4, x = -2, and x = 7.
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a certain spring elongates 9.0 mm when it is suspended vertically and a block of mass m is hung on it. the natural angular frequency of this block-spring system:
The required natural frequency of the spring with certain elongation is calculated to be 5.3 hz.
Spring is an object which can be elongate or compress by some external force to store or release energy when ever needed.The natural frequency of the spring is the frequency at which it tends to vibrate without any external force intended.
The elongation of the spring is given as x = 9 mm
g is the acceleration due to gravity
The expression for the natural frequency according to Rayleigh's method is known as,
f = 1/2π √(g/x)
f = 1/2π √[9.8/(9×10⁻³)] = 5.3 hz
Thus, the required natural frequency is calculated to be 5.3 hz.
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consider an electric current, i, travelling through a circuit when it encounters a junction, splits into two branches a and b, and later rejoins back together. what parameter(s) are involved in determining the fraction of the original current, i, that travels path b?
The fraction of the original current that travels path b depends on the resistances of the two branches a and b, and the voltage across each branch.
The resistance of a conductor is proportional to the current flowing through it, and the voltage across a conductor is proportional to the power being supplied to it.
Ohm's law states that the voltage across a conductor is proportional to the current flowing through it and the resistance of the conductor, V = IR.
Therefore, the fraction of the original current that travels path b can be determined by the ratio of the resistance of path b to the total resistance of both paths a and b, and the voltage across path b to the total voltage across both paths.
The fraction of current that travels path b = (Resistance of path b / Total resistance) * (Voltage across path b / Total voltage)
In other words, the fraction of the original current that travels path b is determined by the impedance of path b relative to the impedance of both paths combined.
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given scientists' current understanding of the formation of the moon, how might the composition of the moon compare to that of the earth?
The composition of the moon is the same as that of the earth, both are made up of basalt.
The composition of the moon surface by weight is roughly 43.16% oxygen, 20.63% silicon, 19.45% magnesium, 10.35% iron, 3.36% calcium, 3% aluminum, 0.42% chromium, 0.18% titanium and 0.12% manganese.
The moon has only a very thin layer of atmosphere, so a layer of dust or a footprint can sustain their for centuries. And due to the absence of an atmosphere, heat is not held near the surface, so temperatures changes wildly. The temperatures of daytime on the sunny side of the moon reach 273° F (134°C), and in the night side, it gets as cold as -243° F(-153°C).
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a car starts from rest and accelerates with a constant acceleration of 1.00 m/s2 for 3.00 s. the car continues for 6.00 s at constant velocity. how far has the car traveled from its starting point?
The car will have traveled a total of 18 meters from its starting point. This can be calculated using the equation for displacement, s = ut + 0.5at^2, where u is the initial velocity (in this case 0 m/s), a is the acceleration (1.00 m/s2) and t is the time traveled (6.00 s).
Plugging in the given values into the equation, we get s = 0 + 0.51.006.00^2 = 18 m.
This equation can be used to calculate the displacement of an object over any period of time, as long as the acceleration and initial velocity are known. Additionally, it can be used to determine the average speed of the object over a given period of time.
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the driver of a car traveling at 25 m/s hits the gas pedal so that the car accelerates at 2 m/s2. find the speed of the car 5 s later
The speed of the car after 5 seconds will be 35 m/s, if the initial speed of the car is 25 m/s and acceleration is 2 m/s²
The speed of an object is defined as the distance travelled by the object in unit time. It is a scaler quantity.
Let the Initial speed of the car, u = 25 m/s
Time, t = 5 sec
Acceleration, a = 2 m/s²
Let the velocity of the car after 5 sec, = v
With the help of the first equation of motion, v = u + at
v = 25 + 2 × 5
v = 35 m/s.
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1.5 kg apple falls from a height of 2.0 m onto a spring scale. The scale’s spring constant is k = 1.5x105 N/m. At maximum compression, what is the reading on the scale?Group of answer choices3.0 x 103 N15 N2.1 x 103 N30 N1.5 x 103 N
The reading on the scale at its greatest compression is calculated to be 3,000 N.
The reading of the scale at the greatest compression is determined by applying the principle of conservation of energy.
The greatest compression of the scale occurs when the potential energy of the object at the given height falls on the scale.
Mathematically, the formula for potential energy of an object is given as,
P.E = m g h
where;
m is the mass of the object
g is acceleration due to gravity
h is the height of fall of the object
P.E = (1.5 kg × 9.8 × 2)
P.E = 29.4 J
P.E = ¹/₂ k x²
where;
x is the compression of the spring
k is the spring constant
x² = 2 P.E / k
x = √ ( 2 P.E / k )
x = √ ( 2 × 29.4 / 150000 )
x = 0.02 m
The reading of the scale is calculated as follows;
F = k x
F = 150,000 × 0.02
F = 3,000 N
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A woman of m= 62 kg stands on the ice. The contact area between her skate and the ice is A = 0.0041 m.
(a) Express the force that the person exerts on the ice, F, in terms of und g. You do not need to include the force of the column of air above her
(b) Calculate the numerical value of F in N
(c) Express the pressure on the ice from the person P in terms of F and A
(d) Calculate the numerical value of P in Pa
(a) 9.8 m/s^2 is acceleration due to gravity. (b) force is 607.6 N (c) Pressure on the ice from the person can be expressed as P = F/A (d) P is 147,757 Pa.
a) The force that the person exerts on the ice is given by:
F = m x g
where m = 62 kg is the mass of the person and g = 9.8 m/s^2 is the acceleration due to gravity. This is the force the person exerts on the ice due to her weight.
b) Numerical value of F can be calculated by multiplying the mass of the person by the acceleration due to gravity:
F = m x g = 62 kg x 9.8 m/s^2 = 607.6 N
c) Pressure on the ice from the person can be expressed as the force per unit area, or
P = F/A
where A = 0.0041 m^2 is the contact area between the skate and the ice.
d) Numerical value of P can be calculated by dividing the force by the area:
P = F/A = 607.6 N / 0.0041 m^2 = 147,757 Pa
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a force vector has a magnitude of 588 newtons and points at an angle of 37o below the positive x axis. what are (a) the x scalar component and (b) the y scalar component of the vector?
The force vector that has a magnitude of 588 newtons and points at an angle of 37 has the followings components:
v(x) =469.59 Nv(y) = 353.86 NWhat is a vector?It can be said to be a straight line described by a point (a) and (b) that has direction and sense.
To solve this exercise the formulas and procedures we will use are:
v(x) = v * cosine (angle)v(y) = v * sine (angle).Where:
v= magnitude of the vectorv(x) = component of the vector on the (x) axisv(y) = component of the vector on the (y) axisInformation about the problem:
angle = 37ºv = 588 Nv(x)= ?v(y)= ?Applying the formula of the component of the vector in the (x) axis we have:
v(x) = 588 N * cosine (37º)
v(x) =469.59 N
Applying the formula of the component of the vector in the (y) axis we have:
v(y) = 588 N * sine (37º)
v(y) = 353.86 N
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Explain what happens when a current flows in the low-current circuit (it is related to electromagnets)
Answer: A current flowing in a low-current circuit can generate an electromagnet. Electromagnets work by using an electric current to produce a magnetic field. When a current flows in a low-current circuit, it creates a magnetic field around the wire that is proportional to the amount of current flowing through the wire. This magnetic field can then be used to generate useful work, such as lifting a heavy object or generating a magnetic field for other purposes.
In a low-current circuit, the magnetic field produced is relatively weak compared to other types of electromagnets, but it still serves a purpose. For example, it can be used in relays, switches, and other electrical components that require a small amount of magnetic force to operate. Additionally, low-current circuits can be used to generate low-power signals or to create small, specialized electromagnets.
Overall, the generation of an electromagnet in a low-current circuit demonstrates the relationship between electrical current and magnetic fields, and highlights the importance of electromagnetism in a wide range of applications and technologies.
Explanation:
can generate an electromagnet.
MY
Explanation:
Why are Cepheid variable stars important to astronomers? How is the information they impart any more or less useful than using stellar parallax?
Cepheid variable stars are important to astronomers because their periods of variability are related to their absolute luminosity.
A Cepheid variable is a particular kind of variable star that pulses radially and undergoes changes in temperature and diameter. With a well defined, constant period and amplitude, its brightness varies. Cepheid variable stars are a unique class of variable star because they are hot, massive, and have a propensity to pulse radially. They also have a tendency to change in diameter and temperature.
An objective way to quantify the radiant power that a light-emitting device emits over time is luminosity. The total quantity of electromagnetic energy released by a star, galaxy, or other celestial object in a given amount of time is measured as luminosity in astronomy.
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