the energy stored at time t=0.3s is 0.0321 mJ (approx.).
A. Graph of the current vs time is as follows:
Given, current in 10μF capacitor is i(t)=5te^-³t mA.T
he plot of current with respect to time is shown in the following figure.
B. To find the voltage across as a function of time, we need to use the formula; v(t) = (1/C) ∫[0 to t] i(t) dt
Where, C = 10μ
F = 10^-5F
and i(t) = 5te^-³t mA
Voltage across as a function of time is;
v(t) = (1/C) ∫[0 to t] i(t) dtv(t)
= (1/10^-5) ∫[0 to t] 5te^-³t dt
taking the integration of the given expression we get;
v(t) = 1.67 - 1.67e^-³t Volts
To calculate the voltage value after t=30ms;
put the value of t = 30ms
= 0.03s in the above equation we get;
v(0.03) = 1.67 - 1.67e^-³(0.03)v(0.03)
= 1.662V (approx.)
Hence, voltage after t=30ms is 1.662V.
The graph of voltage vs time is shown in the following figure.
C. The energy stored in the capacitor is given by;
E(t) = (1/2) C v²(t)
put the value of capacitance and voltage found above, we get;
E(t) = (1/2) × 10^-5 × (1.67 - 1.67e^-³t)² Joules
The graph of energy stored vs time is shown in the following figure.
The energy stored at time t=0.3s is;
put the value of t = 0.3s in the above equation we get;
E(0.3) = (1/2) × 10^-5 × (1.67 - 1.67e^-³(0.3))²E(0.3)
= 0.0321 mJ (approx.)
To learn more on current :
https://brainly.com/question/25922783
#SPJ11
4. Given: \( \sigma=35 . \) \( \tau=35.7 \mathrm{lb} \mathrm{ft} \) \( r=0.0240 \mathrm{ft} \) \( F= \) ?
The torque required to twist the shaft is \(2420.57\; lb\; ft\).
The torque \(F\) required to twist the shaft can be calculated by the following formula,
\(F=\dfrac{Tr}{J}\) where, \(T\) is the torque applied to the shaft,\(r\) is the radius of the shaft, \(J\) is the polar moment of inertia.
The polar moment of inertia can be calculated as,
\(J=\dfrac{\pi d^{4}}{32}\) where, \(d\) is the diameter of the shaft.
The polar moment of inertia of the shaft is given by \(J=\dfrac{\pi d^{4}}{32}\)
We know that the radius of the shaft is given by \(r=0.0240\; ft\).
The diameter of the shaft is given by \(d=2r=2\times0.0240=0.0480\; ft\).
Therefore, \(d=0.0480\;ft\).
Substitute the values of \(T\) and \(r\) in the formula \(F=\dfrac{Tr}{J}\),\(\begin{aligned} F&=\dfrac{Tr}{J}\\ &=\dfrac{(35.7)\cdot(0.0240)}{\dfrac{\pi\cdot (0.0480)^{4}}{32}}\\ &=\dfrac{(35.7)\cdot(0.0240)\cdot(32)}{\pi\cdot (0.0480)^{4}}\\ &=2420.57\; lb \; ft \end{aligned}\)
Therefore, the torque required to twist the shaft is \(2420.57\; lb\; ft\).
To know more about torque please refer:
https://brainly.com/question/17512177
#SPJ11
A plain area (5 m by 5 m) is submerzed in water in such a way
that its centroid of area is at a depth of 41 from water surface.
Calculate the total force (in Newton) acting on the plan area.
Given, Length of the plain area = 5 m Breadth of the plain area = 5 m Centroid of area is at a depth of 41 m from water surface. The formula to calculate the total force acting on the plane area is given by:
Force = ρghA Where,
ρ = density of water
g = acceleration due to gravity
h = depth of centroid of plane area from water surface
A = area of plane area
The first step is to calculate the area of the plane area.
Area of plane area
= Length * Breadth
= 5 * 5
= 25 m²
Given, the depth of the centroid of the plane area from the water surface = 41 m The total force acting on the plane area can be calculated as follows:
Force = ρghA
= 1000 * 9.8 * 41 * 25
= 10,082,500 N
The total force acting on the plane area is 10,082,500 N, which is calculated using the formula Force = ρgh
To know more about Centroid visit:
https://brainly.com/question/31238804
#SPJ11
A fay of light strikes the midpoint of ore face of an equlangular (60
6
−634−600
6
) giass \{a) Trace the gath of the light ray throwgh the giass, and find the angles of incidance and refractian at each ourface. First surface: θ
inciatence
= 9
rufracsian
= Second surfoce: θ
incience
= 9
refration
= (o) If a whall fraction of light is also reflected at each surface, Find the agies of reflection at the suraces. θ
refeann
= - (second surface)
The path of the light ray through the glass and find the angles of incidence and refraction at each surface. Given that a ray of light strikes the midpoint of one face of an equilateral (60 degree) glass prism.
Let us consider the following diagram of the given problem: Since the ray is normal to the surface it does not bend at the entry point. So, θincidence = 0° for the first surface.The angle of incidence for the second surface of the prism is equal to the angle of refraction of the first surface. Since the first surface does not bend the light, θrefraction of the first surface = 0°.
Hence, θincidence = 0° for the second surface.Using Snell's law for the first surface of the prism, we get
;[tex]\frac{\sin\theta_i}{\sin\theta_r}=\frac{n_2}{n_1}[/tex]Here, [tex]\theta_i[/tex] = incidence angle, [tex]\theta_r[/tex] = refraction angle, [tex]n_1[/tex] = refractive index of air and [tex]n_2[/tex] = refractive index
of the glass prismWe know that the glass prism is made of equilateral glass.
Hence the refractive index for equilateral glass is 1.5. Using this value, we get:
[tex]\frac{\sin 30}{\sin\theta_r}=\frac{1.5}{1}[/tex][tex]\theta_r=19.47\degree[/tex]
For the second surface, the ray enters into the air from the glass. Hence, [tex]n_1[/tex] = 1 and [tex]n_2[/tex] = 1.5. Using Snell's law, we get
;[tex]\frac{\sin\theta_i}{\sin\theta_r}=\frac{n_2}{n_1}[/tex][tex]\frac{\sin\theta_i}{\sin 30}=\frac{1.5}{1}[/tex][tex]\sin\theta_i=0.75[/tex][tex]\theta_i=48.59\degree[/tex].
Thus, the angles of incidence and refraction at each surface are given as below:
First surface: [tex]\theta_{incidence}=0\degree[/tex] and [tex]\theta_{refraction}=19.47\degree[/tex]Second surface: [tex]\theta_{incidence}=48.59\degree[/tex] and [tex]\theta_{refraction}=0\degree[/tex]
The angle of reflection is equal to the angle of incidence. Hence, θreflection = θincidence. Thus, θreflection = 0° for the first surface and θreflection = 48.59° for the second surface.
To know more about refraction visit:
https://brainly.com/question/32684646
#SPJ11
What kind of heating systems involve circulation of the air in a room?
Heating systems that involve the circulation of air in a room are known as forced air heating systems.
Heating systems that involve the circulation of air in a room are known as forced air heating systems. These systems use a furnace or heat pump to generate heat, which is then distributed throughout the room or building using a network of ducts. The heated air is forced through the ducts by a blower or fan, allowing it to circulate and warm the space.
Forced air heating systems are commonly used in residential and commercial buildings due to their efficiency and ability to quickly heat large areas. They can be powered by various energy sources, including natural gas, electricity, or oil.
Learn more:About heating systems here:
https://brainly.com/question/26301591
#SPJ11
The kind of heating systems that involve circulation of the air in a room is the forced-air heating system. The forced-air heating system is a type of heating system that is found in many residential homes, commercial buildings and industrial applications.
It circulates the air in a room by using a fan or blower to distribute warm air throughout the building.An important component of a forced-air heating system is a furnace that generates heat and is located in a central location. The furnace heats up air and the warm air is then distributed through a network of ducts that run throughout the building.
The ducts are usually located in the walls, ceiling or floors of the building and they carry the warm air to the different rooms that require heating.In conclusion, a forced-air heating system involves circulation of the air in a room through the use of a furnace, fan or blower, and a network of ducts that distribute warm air throughout the building.
To know more about heating system visit:-
https://brainly.com/question/30651040
#SPJ11
1- Define the following: The polarizability - Polar molecules - Nonpolar molecules - Induced dipoles - Ferroelectric materials. 2- Deduce the Clausius-Mossotti equation. 3- Compute the polarizability of an atom, where the charge of the nucleus is (Ze) and the total charge of electrons (-Ze). 4- A point charge q is situated a large distance r from a neutral atom of polarizability a. Find the force of attraction between them. 5- Deduce the Langevin-Debye equation for polar molecules.
1- Polarizability: It is the tendency of a molecule or atom to become polarized when exposed to an electric field. Polar molecules: Molecules that have a positive or negative electrical charge at one end. Nonpolar molecules: Molecules that lack an electrical charge. Induced dipoles: When an electric field is applied to a nonpolar molecule, an induced dipole is formed.
Ferroelectric materials: Materials that exhibit spontaneous electric polarization in the absence of an electric field.
2- Clausius-Mossotti Equation
The Clausius-Mossotti equation can be expressed as:
(ε - 1) / (ε + 2) = (4πNa³α) / 3
The Clausius-Mossotti equation relates the dielectric constant (ε) of a substance to its polarizability (α). It provides a quantitative estimate of the polarizability of a molecule.
3- Computation of Polarizability
Polarizability of an atom can be computed using the following equation:
α = (1/6) × (e / ε₀) × (2a² + 3r²)
Where,α = polarizability of an atom
e = charge of the nucleus
r = distance between the electron and the nucleus
a = radius of the electron
ε₀ = permittivity of free space
4- Force of Attraction
The force of attraction (F) between a point charge (q) and a neutral atom of polarizability (a) can be computed using the following equation:
F = (q² / 4πε₀r²) × (α / 3)
When an electric field is applied to a nonpolar molecule, an induced dipole is formed. The induced dipole creates a temporary dipole, which creates an attractive force between the polar molecule and the point charge.
5- Langevin-Debye Equation
The Langevin-Debye equation can be expressed as:
(ε - ε₀) / (ε + 2ε₀) = 4πNpα / 3kT
The Langevin-Debye equation relates the dielectric constant (ε) of a substance to its polarizability (α), temperature (T), and particle density (Np). It is used to describe the behavior of polar molecules.
Therefore, the polarizability is the tendency of an atom or molecule to become polarized when exposed to an electric field. Polar molecules have a positive or negative electrical charge at one end while nonpolar molecules lack an electrical charge. Induced dipoles are formed when an electric field is applied to a nonpolar molecule. Ferroelectric materials exhibit spontaneous electric polarization in the absence of an electric field. The Clausius-Mossotti equation relates the dielectric constant (ε) of a substance to its polarizability (α). The polarizability of an atom can be computed using the formula.
The force of attraction (F) between a point charge (q) and a neutral atom of polarizability (a) can be computed using a formula. The Langevin-Debye equation relates the dielectric constant (ε) of a substance to its polarizability (α), temperature (T), and particle density (Np).
To know more about Polarizability, visit:
https://brainly.com/question/31842978
#SPJ11
3. Question 3 [25 marks] Consider the mass-spring system of Figure 3 where the masses of \( 2 m \) and \( m \) are bound to each other via a spring of stiffness \( k \) and connected to rigid walls vi
The mass-spring system is one of the classical examples of simple harmonic motion. A body undergoes simple harmonic motion if the force acting on the body is proportional to the displacement of the body from its equilibrium position and is directed towards the equilibrium position.
The system of masses and spring shown in Figure 3 is an example of a mass-spring system that can exhibit simple harmonic motion. In this system, there are two masses, one of mass 2m and the other of mass m, that are connected by a spring of stiffness k and are confined between two rigid walls. The two masses move along the x-axis with respect to their equilibrium positions, which is when the spring is unstretched and the forces on the masses are balanced.
The motion of the masses is governed by Hooke's Law, which states that the force exerted by the spring on each mass is proportional to the displacement of the mass from its equilibrium position and is directed towards the equilibrium position. The motion of the masses is periodic, with a period given by:
T=
\frac{2
\pi}{
\omega}=2
\pi
\sqrt{
\frac{3m}{k}}
In conclusion, the mass-spring system shown in Figure 3 is an example of a simple harmonic motion, with the motion of the masses being governed by Hooke's Law and the equations of motion being given by a second-order linear differential equation with constant coefficients. The frequency of oscillation and the period of the system are determined by the stiffness of the spring and the masses of the system.
To know more about proportional visit :
https://brainly.com/question/31548894
#SPJ11
L=65Ht=1 s A If you want current through it to be adjustable with a I second characteristic time constant, what is resistance of system in ohms? R= \ Omega (11\%) Problem 6: Two coils are placed close together in a physics lab to demonstrate Faraday's law of induction. A current of 5.5 A in one is switche in 2.5 ms, inducing an average 9 V emf in the other. What is their mutual inductance? Randomized Variables εave=9 Vt=2.5 msI=5.5 A a What is their mutual inductance in mH ? Problem 7: The inductance and capacitance in an LC circuit are 0.18mH and 4.5pF respectively. What is the angular frequency, in radians per second, at which the circuit oscillates? ω=∣
Problem 6: the mutual inductance is 4.1 mH.
Problem 7: the angular frequency of the LC circuit is 3
× 10¹² rad/s.
Problem 6:From Faraday's law of induction,
ε = - M(dI/dt),
Where ε is the average emf, M is the mutual inductance, and dI/dt is the rate of change of current.
dI/dt = 5.5 A/2.5 ms = 2200 A/sε = 9 V
Substituting all the values in the above equation, we get,
M = -ε/ (dI/dt) = -9/2200 = -0.0041H or -4.1 mH (taking negative sign as both the coils are opposite)
Therefore, the mutual inductance is 4.1 mH.
Problem 7: The formula for inductive reactance, Xl is given by the following equation:
Xl = 2πfL,
Where L is the inductance and f is the frequency.
Substituting the values of L and C, we get
Xl = 1/(2πfC)
We need to find the value of angular frequency, ω.
The formula for angular frequency, ω is given by the following equation,ω = 2πf.
Substituting the values of L and C in the above equation, we get,ω = 1/ √(LC)
Now, substituting the values of L and C, we get,
ω = 1/√(0.18 × 10⁻³ H × 4.5 × 10⁻¹² F)
ω = 1/√(0.81 × 10⁻²⁴)
ω = 3 × 10¹² rad/s
Therefore, the angular frequency of the LC circuit is 3
× 10¹² rad/s.
Learn more about inductance at
https://brainly.com/question/31960402
#SPJ11
A 3-phase I.M. operate at 400 Hz has a rated speed 10800 R.P.M, its speed at slip = 0.7 is equal to:
a) 3400 R.P.M.
b) 3600 R.P.M.
c) 3800 R.P.M.
d) None.
A 3-phase I.M. operate at 400 Hz has a rated speed 10800 R.P.M, its speed at slip = 0.7 is equal to: The correct answer is:
a) 3400 R.P.M. (Closest option to the calculated speed of 3240 R.P.M.)
To determine the speed of the 3-phase induction motor at slip = 0.7, we can use the formula:
Speed = Rated speed - (Slip × Rated speed)
Given:
Rated speed = 10800 R.P.M
Slip = 0.7
Substituting the values into the formula:
Speed = 10800 R.P.M - (0.7 × 10800 R.P.M)
= 10800 R.P.M - 7560 R.P.M
= 3240 R.P.M
Therefore, the speed of the 3-phase induction motor at slip = 0.7 is equal to 3240 R.P.M.
The correct answer is:
a) 3400 R.P.M. (Closest option to the calculated speed of 3240 R.P.M.)
To learn more about, speed, click here, https://brainly.com/question/6280317
#SPJ11
Lab #2: Isostasy
A) Purpose of the assignment:
This lab is meant to get you familiarized with the concept of
isostasy, which is invoked to explain how different topographic
heights can exist at the su
The purpose of Lab #2 is to introduce you to the concept of isostasy and its role in explaining variations in topographic heights.
Isostasy is the idea that the Earth's crust is in a state of equilibrium, with less dense materials, like continental crust, "floating" on denser materials, like the mantle. This equilibrium is maintained by the adjustment of material vertically in response to changes in the load on the crust.
For example, if there is a mountain range with a lot of material on top, it creates a downward force on the crust. In response, the crust will adjust by sinking deeper into the denser mantle to balance the load. Conversely, if material is eroded from the mountain range, the crust will rebound upward to maintain equilibrium.
This concept helps explain why different topographic heights can exist. The height of a landform is not solely determined by the elevation of the crust, but also by the density and thickness of the materials beneath it. So, variations in topography can be due to variations in crustal thickness and density.
In summary, Lab #2 aims to familiarize you with isostasy and its role in explaining topographic variations. By understanding this concept, you will gain insights into how the Earth's crust responds to changes in loads and the factors influencing topography.
Learn more about topographic from the following link:
https://brainly.com/question/24146311
#SPJ11
please explain in details
why Two coils are said to be mutually coupled if the magnetic flux
Ø emanating from one pass
through the other.
The principle of electromagnetic induction states that if there is a change in magnetic flux linking a coil, an electromotive force (emf) is induced in that coil. The magnitude of the induced emf is determined by the rate of change of the magnetic flux.
This forms the basis of electrical transformers. In an ideal transformer, all the flux in the primary winding links the secondary winding. In a practical transformer, however, the coupling between the windings may not be perfect. This is due to several factors such as leakage flux and poor core material.
Two coils are said to be mutually coupled if the magnetic flux Ø emanating from one passes through the other. For a perfect mutual coupling, all the flux in the primary coil passes through the secondary coil. In other words, if the coupling coefficient (k) is 1, then there is a perfect mutual coupling between the two coils.
When k is less than 1, there is a partial coupling between the two coils. The coupling coefficient k is defined as the ratio of the mutual inductance to the square root of the product of the individual inductances. Therefore, the greater the mutual inductance between two coils, the greater the coupling coefficient.
Learn more about electromagnetic induction from :
https://brainly.com/question/27613726
#SPJ11
IF I want to create a 12V DC solenoid lock. What are the mathematical modeling for it. Like how can I find the current, resistance, magnetic field, the force and whatever else is left. Please help me with proving all the equations and explanations.
To create a mathematical model for a 12V DC solenoid lock, we can consider various aspects such as the current, resistance, magnetic field, and force. Let's go through each one:
1. Current (I):
The current flowing through the solenoid can be determined using Ohm's Law:
I = V / R,
where V is the applied voltage (12V) and R is the resistance of the solenoid.
2. Resistance (R):
The resistance of the solenoid can be determined based on its physical characteristics, such as the length and cross-sectional area of the wire used. The resistance can be calculated using the formula:
R = ρ * (L / A),
where ρ is the resistivity of the wire material, L is the length of the wire, and A is the cross-sectional area of the wire.
3. Magnetic Field (B):
The magnetic field inside the solenoid can be calculated using Ampere's Law:
B = μ₀ * (N * I) / L,
where μ₀ is the permeability of free space, N is the number of turns in the solenoid, I is the current flowing through the solenoid, and L is the length of the solenoid.
4. Force (F):
The force exerted by the solenoid can be determined using the following equation:
F = B * (N * I) * A,
where B is the magnetic field strength, N is the number of turns, I is the current flowing through the solenoid, and A is the cross-sectional area of the solenoid.
Learn more about solenoid from :
https://brainly.com/question/1873362
#SPJ11
(10%) Problem 9: Several ice cubes (ϱi=0.9167 g/cm3) of total volume Vi=240 cm3 and temperature 273.15 K(0.000∘C) are put into a thermos containing Vt= 690 cm3 of tea at a temperature of 313.15 K, completely filling the thermos. The lid is then put on the thermos to close it. Assume that the density and the specific heat of the tea is the same as it is for fresh water (ϱw=1.00 g/cm3,c=4186 J/kgK) 33% Part (a) Calculate the amount of heat energy Qm in J needed to melt the ice cubes (Lf=334 kJ/kg). Qm=7.35∗10(4)Qm=7.350×104✓ Correct! 33\% Part (b) Calculate the equilibrium temperature TE in K of the final mixture of tea and water. TE=2.83∗10(2)TE=283.0∨ Correct! ▹≈33% Part (c) Calculate the magnitude of the total heat transferred QT in J from the tea to the ice cubes. QT=
The magnitude of the total heat transferred (QT) from the tea to the ice cubes is 1.74 × 105 J.
The equilibrium temperature of the final mixture of tea and water is 283.0 K. Part (c) The magnitude of the total heat transferred QT in J from the tea to the ice cubes is equal to the amount of heat energy (Q) m needed to melt the ice cubes plus the heat energy required to raise the temperature of the water and ice mixture from 0°C to the equilibrium temperature TE: QT = Q m + m water cΔT water where m water is the mass of water and ΔT water is the temperature change of water. Since ΔT water = TE - 273.15 K and using the equation for density ρ = m/V, we can write: m water = ρwater V water = 1.00 g/cm3 × 450 cm3 = 450 g. Therefore, QT = Q m + m water cΔTwater = 7.35 × 104 J + (450 g × 4186 J/kg K × (283.0 K - 273.15 K)) = 1.74 × 105 J. Therefore,
Part (a)The amount of heat energy Q m in J needed to melt the ice cubes can be calculated as follows: Q = m Lf Q = (240 cm3 × 0.9167 g/cm3) × (1 kg/1000 g) × (334 kJ/kg) = 7.35 × 104 J. Therefore, the amount of heat energy Q m needed to melt the ice cubes is 7.35 × 104 J. Part (b) The final temperature(T) of the mixture, TE can be calculated using the principle of energy conservation, which states that the amount of energy lost by the tea (or water) equals the amount of energy gained by the ice cubes during the melting process. The specific heat of water is 4186 J/kg K. Using the principle of energy conservation, we have: m water cΔTwater + m water Lf + m tea cΔTtea = 0where m water and m tea are the masses of water and tea, respectively; specific heat of water(c); latent heat of fusion of water(Lf); ΔTwater and ΔTtea are the temperature changes of water and tea, respectively. Since the system is insulated, we have: m water cΔTwater = - m tea cΔT tea using the equation for density ρ = m/V, we can write: m water = ρwater V water and m tea = ρtea V tea and the equation becomes: ρ water cΔT water V water = -ρtea cΔT tea V tea (ρwater cV water) ΔT water = -(ρtea c V tea)ΔTtea(1.00 g/cm3 × 690 cm3 × 4186 J/kg K) × (TE - 313.15 K) = -(0.9167 g/cm3 × 240 cm3 × 4186 J/kg K) × (TE - 273.15 K)Solving for TE, we get: TE = 283.0 K.
To know more about specific heat of water visit:
https://brainly.com/question/30551147
#SPJ11
An external force F moves a 4.50−kg box at a constant speed v up a frictionless ramp, as shown in the figure. The force acts in a direction parallel to the ramp. Calculate the work W done on the box by this force as it is pushed up the 5.00−m ramp to a height h=4.00 m. W= How does the work done on the box compare to the change in gravitational potential energy ΔUgrav that the box undergoes as it rises to its final height? W>ΔUgrav W=ΔUgrav W<ΔUgrav
The work done on the box is 220.5 Joules and the work done on the box is greater than the change in gravitational potential energy.
The work done on the box by the external force can be calculated using the formula,
W = Fd,
where
F is the magnitude of the force
d is the displacement.
In this case, the force is acting parallel to the ramp, so we can calculate the work done as the product of the force and the distance along the ramp.
Mass of the box (m) = 4.50 kg
Length of the ramp (d) = 5.00 m
Height (h) = 4.00 m
To calculate the work done, we need to determine the force acting on the box. Since the box is moving at a constant speed, the net force acting on it is zero. This means that the force exerted by the external force is equal in magnitude and opposite in direction to the gravitational force.
The gravitational force acting on the box can be calculated using the formula
F = mg,
where
m is the mass of the box
g is the acceleration due to gravity (approximately 9.8 m/s²).
F = (4.50 kg)(9.8 m/s²) = 44.1 N
Now, we can calculate the work done on the box:
W = Fd = (44.1 N)(5.00 m) = 220.5 J
So, the work done on the box is 220.5 Joules.
To compare the work done to the change in gravitational potential energy, we need to calculate the change in gravitational potential energy.
The change in gravitational potential energy can be calculated using the formula
ΔUgrav = mgh,
where
m is the mass of the box,
g is the acceleration due to gravity,
h is the change in height.
ΔUgrav = (4.50 kg)(9.8 m/s²)(4.00 m) = 176.4 J
Comparing the work done (220.5 J) to the change in gravitational potential energy (176.4 J), we can see that
W > ΔUgrav
This means that the work done on the box is greater than the change in gravitational potential energy.
learn more about gravitational potential energy
https://brainly.com/question/15896499
#SPJ11
Apoint charge of 870 nC is located on the nC as located at the origin and a second charge of 300 axis at a -1.75cm
The electric field at point P has a magnitude of 3.27x10⁵ N/C and is directed to the right.
The electric field due to a point charge can be calculated using Coulomb's law, which states that the electric field E at a distance r from a point charge q is given by E=kq/r², where k is Coulomb's constant.
In this scenario, a point charge of 870 nC is located at the origin, and a second charge of 300 nC is located at a distance of -1.75cm on the x-axis. We need to calculate the electric field at a point P located at a distance of 3.5 cm from the origin along the x-axis.
Let's begin by calculating the electric field at point P due to the charge of 870 nC. Using Coulomb's law, we have E₁=kq₁/r₁²where q₁=870 nC and r₁=3.5 cm=0.035 m Therefore, E₁=(9x10⁹ Nm²/C²)(870x10⁻⁹ C)/(0.035m)²=8.68x10⁴ N/C
Now let's calculate the electric field at point P due to the charge of 300 nC. Using Coulomb's law, we have E₂=kq₂/r₂² where q₂=300 nC and r₂=0.0175 m Therefore, E₂=(9x10⁹ Nm²/C²)(300x10⁻⁹ C)/(0.0175m)²=4.14x10⁵ N/C
Note that the electric field due to the charge of 300 nC is in the negative x-direction because the charge is to the left of point P. Therefore, the total electric field at point P is given by the vector sum of the electric fields due to the two charges: E=E₁+E₂=(-8.68x10⁴ N/C)+(4.14x10⁵ N/C)=3.27x10⁵ N/C
The electric field at point P has a magnitude of 3.27x10⁵ N/C and is directed to the right.
To know more about electric visit:
https://brainly.com/question/33513737
#SPJ11
Physics is in every action we take daily. Which examples do you see around you in your neighborhood and your home? Which laws of physics do you witness and experience every day? Begin your hunt and take pictures/videos of the laws of physics you see in action and upload them here.
Rubric:
Everyday Life: How do you use this physics discovery in your everyday life? Which equations covered in our physics course (in the chapters we covered) relate to this discovery? Do not include equations, formulas or concepts not included in our physics course. Provide examples by creating a physics problem in the same scientific style as the problems at the end of each Chapter in your ebook, or like in the Video Tutor. Solve your original creative new Physics Problem step-by-step with units of measurement consistency, framed final answer, and include a drawing or diagram.
Physics is a part of everything we do daily, whether we are driving a car, turning on a light switch, or even cooking. In this article, we will focus on the examples of physics that can be seen in our homes and neighborhoods. There are numerous instances of physics in our neighborhood and homes, and some of them are mentioned below:
Motion of the Sun: The sun rises in the east and sets in the west. This motion of the sun is related to the rotation of the earth around its axis.
Gravity: Gravity is what keeps our feet planted on the ground, and it is one of the fundamental forces of the universe. It pulls everything towards its center, keeping planets in orbit and keeping us grounded.
Inertia: When a car suddenly stops, the passengers continue to move forward due to their inertia. This is why we need seatbelts to keep us in place.
Friction: Friction is the force that opposes motion, and it is present everywhere. For example, when we walk, friction is what keeps our feet from slipping on the ground.
Magnetism: Magnetism is present in everyday life, such as in the magnets used to hold papers on the refrigerator or in the speakers in our phones.
Electricity: Electricity is used to power our homes and is present in everything from the lights we turn on to the chargers we use to charge our phones. In addition, it is used in appliances like refrigerators, televisions, and microwaves.
How do we use this physics discovery in our everyday life?
We use these physics principles in our everyday life to understand the world around us better. For example, we can understand how things move, why things fall, and how electricity works. By understanding these principles, we can create new technology and improve our quality of life. In addition, by understanding the laws of physics, we can create problems and equations to help us solve real-world problems. For instance, if we want to calculate the distance a car travels, we can use the equation distance = velocity x time.
Relation to equations in physics courses: The examples mentioned above relate to different physics concepts covered in the various chapters of the physics course. For example, the motion of the sun relates to the concept of circular motion, while gravity relates to the concept of forces. Furthermore, electricity and magnetism relate to the topics of electromagnetism and circuits in the physics course.
Creative new physics problem: The problem: A ball is thrown from a height of 20m with an initial velocity of 30 m/s at an angle of 30 degrees. What is the horizontal and vertical distance travelled by the ball before it hits the ground?
Solution: First, let us calculate the time taken by the ball to reach the ground. We can use the equation:
v = u + at
Where v = 0 m/s, u = 30 sin 30 m/s, and a = 9.8 m/s^2. We can rearrange this equation to get t = u/a. Substituting the values gives us: t = 1.94 s
Now, we can use the equations of motion to find the horizontal and vertical distance travelled by the ball. The equations of motion are: x = ut + (1/2) at^2 and v^2 = u^2 + 2ax
We can use these equations in the x and y directions separately.
Vertical direction: y = 20 m + uyt + (1/2) gt^2y = 20 + 30 sin 30 (1.94) - (1/2) (9.8) (1.94)^2y = 5.32 m
Horizontal direction: x = ux t + (1/2) axt^2x = 0 + 30 cos 30 (1.94) - (1/2) (0) (1.94)^2x = 27.87 m
Therefore, the horizontal distance travelled by the ball before it hits the ground is 27.87 m, while the vertical distance travelled by the ball before it hits the ground is 5.32 m.
To know more about friction visit:
https://brainly.com/question/28356847
#SPJ11
1. Explain the relationship between voltage and intensity in the:
R circuit
Circuit C
L Circuit
2. How does the theoretical value of the resonance frequency behave with relative to the experimental value? Calculate the percent error task.
3. Is the plot of Current vs. Frequency symmetrical about the resonance frequency? Explain your answer.
4. At the moment of resonance XL= XC and the circuit behaves as pure resistive. Using Ohm's Law, find the value of the endurance. Will that value be equal to 10 ohms? Explain why.
5. Summarize some technology applications that can have the RLC circuits
1. Explanation of the relationship between voltage and intensity in the following circuits:
R circuit:
The current and voltage are in phase with each other in a pure resistor circuit, where there is no inductance or capacitance. In a resistor circuit, the voltage is directly proportional to the current, as specified by Ohm's law.
Circuit C:
The capacitive circuit is one in which the voltage leads the current, with the current lagging behind the voltage by 90 degrees. The magnitude of the current decreases as the frequency increases, with the voltage remaining constant.
L Circuit:
The current in an inductive circuit lags behind the voltage, whereas the voltage leads the current. As the frequency of the source voltage increases, the magnitude of the current decreases, while the voltage remains constant.
2. The theoretical value of the resonant frequency is the frequency at which the reactive elements of the RLC circuit cancel each other out, resulting in a circuit that behaves as a purely resistive circuit.
The value obtained experimentally is compared to the theoretical value of the resonant frequency. The percentage difference between the theoretical and experimental values is referred to as the percent error in the measurement.
3. The plot of the current vs. frequency is symmetrical around the resonant frequency, with the maximum value of the current at the resonant frequency.
4. The circuit's behavior is purely resistive at resonance, with the inductive reactance (XL) being equal to the capacitive reactance (XC).
The impedance of the circuit is also purely resistive, and it is equal to the circuit's resistance (R). The value of the resistance can be calculated using Ohm's law, which is given by:
R = V / I
where V is the voltage and I is the current.
As a result, the resistance value will be equal to 10 ohms, and the circuit behaves like a pure resistive circuit at resonance.
5. RLC circuits are found in a variety of applications, including radio and television tuning circuits, acoustic filters, electronic oscillators, and power transmission lines. It is used in the following applications:
Resonant circuits in radio and television tuning Acoustic filters Electronic oscillators Power transmission line frequency filters in audio equipment and speakers LED light dimmers in lighting systems.
To know more about circuits visit:
https://brainly.com/question/12608491
#SPJ11
Measurements of the radioactivity of a certain isotope tell you that the decay rate decreases from 8280 decays per minute to 3100 decays per minute over a period of 5.00 days.
What is the half-life T1/2 of this isotope?
Express your answer numerically, in days, to three significant figures.
The half-life T1/2 of this isotope is 1.83 days if the decay rate decreases from 8280 decays per minute to 3100 decays per minute over a period of 5.00 days.
The half-life T1/2 of the isotope can be calculated using the formula given below:T1/2 = (t ln 2) / ln (N0 / Nt) where t is the time, N0 is the initial quantity, Nt is the final quantity, ln is the natural logarithm, and T1/2 is the half-life of the isotope. Let N0 be the initial quantity of the isotope, and Nt be the final quantity of the isotope. The decay rate decreases from 8280 decays per minute to 3100 decays per minute over a period of 5.00 days. Therefore, the initial quantity N0 can be expressed as:
N0 = 8280 decays per minute and the final quantity Nt can be expressed as: Nt = 3100 decays per minute
We know that the time t is 5.00 days. Substituting the given values in the above formula, we get:
T1/2 = (5.00 ln 2) / ln (8280 / 3100)T1/2 = 1.83 days
More on half-life: https://brainly.com/question/32245416
#SPJ11
Finding the work done in lifting a bucket.
A 6 lb bucket attached to a rope is lifted from the ground into the air by pulling in 16 ft of rope at a constant speed. If the rope weighs 0.9 lb/ft, how much work is done lifting the bucket and rope?
Find the work done in lifting the bucket (without the rope) 16 ft.
To find the work done in lifting the bucket without the rope, we can calculate the work done against the gravitational force. The work done in lifting the bucket (without the rope) 16 ft is approximately 126.722 Joules.
The work done against gravity is given by the formula: W = mgh
where W is the work done, m is the mass, g is the acceleration due to gravity, and h is the vertical distance.
In this case, we are given that the bucket weighs 6 lb and is lifted a vertical distance of 16 ft.
First, we need to convert the weight of the bucket from pounds (lb) to mass in the standard unit of kilograms (kg). The conversion factor is approximately 0.4536 kg/lb.
Mass of the bucket = 6 lb * 0.4536 kg/lb = 2.7216 kg
The acceleration due to gravity, g, is approximately 9.8 m/s^2.
The vertical distance, h, is given as 16 ft. We need to convert it to meters since the standard unit for distance is the meter. The conversion factor is approximately 0.3048 m/ft.
Vertical distance, h = 16 ft * 0.3048 m/ft = 4.8768 m
Now we can calculate the work done:
W = (2.7216 kg) * (9.8 m/s^2) * (4.8768 m)
W = 126.722 Joules
Therefore, the work done in lifting the bucket (without the rope) 16 ft is approximately 126.722 Joules.
To learn more about, gravitational force, click here, https://brainly.com/question/32684735
#SPJ11
An X-ray machine produces X-ray by bombarding a molybdenum ( Z=42 ) target with a beam of electrons. First, free electrons are ejected from a filament by thermionic emission and are accelerated by 25kV of potential difference between the filament and the target. Assume that the initial speed of electrons emitted from the filament is zero. For the calculation of characteristic X-ray, use σ=1 for the electron transition down to K shell (n=1) and σ=7.4 for the electron transition down to L shell (n=2). (a) What is the minimum wavelength of electromagnetic waves produced by bremsstrahlung? (6 pt) (b) What is the energy of the characteristic X-ray photon when an electron in n=4 orbital moves down to n=2 in the molybdenum target? ( 5 pt) (c) What is the frequency of the characteristic X-ray in part (b)? (2 pt) (d) What is the energy the characteristic X-ray photon when an electron in n=2 orbital moves down to n= 1 in the molybdenum target? ( 5 pt) (e) What is the frequency of the characteristic X-ray in part (d)? (2 pt)
(a) The minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.
Given, Initial speed of the emitted electrons, u = 0 m/s
Potential difference between the filament and target, V = 25 kV = 25,000 V
Charge of an electron, e = 1.6 × 10⁻¹⁹ C
Planck’s constant, h = 6.63 × 10⁻³⁴ Js
Speed of light, c = 3 × 10⁸ m/s
Electrons are accelerated by a potential difference between the filament and the target. The change in kinetic energy of the electron is equal to the work done by the electric field. The expression for the change in kinetic energy of the electron is given by
KE = eV … (1)
where
KE = kinetic energy of electron,
Ve = potential difference between the filament and the target, and e = charge of electron
The maximum kinetic energy of the electron is given by
KEmax = eV … (2)
where
KEmax = maximum kinetic energy of electron
When the accelerated electrons strike the target atoms, they slow down due to Coulombic interaction with the atomic nuclei. The kinetic energy lost by the electrons is emitted as electromagnetic radiation, called bremsstrahlung radiation.
The minimum wavelength of electromagnetic waves produced by bremsstrahlung radiation is given by
λmin = hc/KEmax … (3)
where
hc = Planck’s constant × speed of light
KEmax = maximum kinetic energy of electron
Substituting the given values in equation (2), we get
KEmax = eV= 1.6 × 10⁻¹⁹ C × 25,000
V= 4 × 10⁻¹⁵ J
Substituting the given values in equation (3), we get
λmin = hc/KEmax
= 6.63 × 10⁻³⁴ Js × 3 × 10⁸ m/s/4 × 10⁻¹⁵ J
= 0.491 nm
Therefore, the minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.(b) The energy of the characteristic X-ray photon when an electron in n = 4 orbital moves down to n = 2 in the molybdenum target is 0.63 keV
The minimum wavelength of electromagnetic waves produced by bremsstrahlung is 0.491 nm.
The energy of the characteristic X-ray photon when an electron in n=4 orbital moves down to n=2 in the molybdenum target is 0.63 keV.
The frequency of the characteristic X-ray in part (b) is 2.42 × 10¹⁸ Hz.
The energy the characteristic X-ray photon when an electron in n=2 orbital moves down to n= 1 in the molybdenum target is 17.4 keV.
The frequency of the characteristic X-ray in part (d) is 4.17 × 10¹⁸ Hz.
To know more about electromagnetic waves, visit:
https://brainly.com/question/29774932
#SPJ11
A 325-mm-diameter vitrified pipe is a m long, and by using the Hazen-Williams equation; determine the discharge capacity of this pipe if the head loss is 2.54 m and half full. a=[95+ (last digit of your id number / 2) ]m (20 POINTS) A=5=97,5
Discharge capacity of the given pipe is 12.57 m³/s.
The formula to calculate the discharge capacity of the pipe is given by;
Q = (C×π×d²/4)×(2gh)³
Here,
Q = Discharge capacity of the pipe
C = Hazen-Williams coefficient
π = 22/7
d = Diameter of the pipe
h = Head loss
g = Acceleration due to gravity (g = 9.81 m/s²)
We know that, the cross-sectional area of the pipe can be calculated by using the formula;
A = πd²/4
As the pipe is half full,
A = πd²/8
Also, the velocity of the flow in the pipe can be determined using the formula;
v = (2gh)^(1/2)
Putting the values in the formula, we get;
Q = C×A×vQ
= 130 × (πd²/8) × [(2gh)^(1/2)]Q
= 130 × (π/8) × (0.325 m)² × [(2 × 9.81 m/s² × 2.54 m)^(1/2)]Q
= 2.506 × (2 × 9.81 × 2.54)^(1/2) m³/sQ
= 2.506 × 5.018 m³/sQ
= 12.57 m³/s
Therefore, the discharge capacity of the given pipe is 12.57 m³/s.
Learn more about Discharge capacity from :
https://brainly.com/question/32293894
#SPJ11
There are no aurora on Venus because it
A. Lacks an ionosphere
B. Lacks atmospheric oxygen
C. Lacks a strong magnetic field
D. Lacks strong winds
The aurora is a natural light display in the sky, typically seen in high-latitude regions (around the poles). It is caused by the collision of charged particles from the sun with atoms in the Earth's atmosphere.
The aurora requires four things to appear:
Solar wind: The aurora is triggered by the solar wind, which is a stream of charged particles from the sun.
Earth's magnetic field: Earth's magnetic field guides the charged particles from the solar wind towards the poles, where they collide with atoms in the atmosphere and produce the aurora.
Atmosphere: The aurora is formed when charged particles from the solar wind collide with atoms in the Earth's atmosphere. These collisions release energy, which is typically seen as a light show.
Location: The aurora is typically seen in high-latitude regions (around the poles). This is because the Earth's magnetic field is strongest at the poles, which means that the solar wind particles are more likely to be guided there.
Venus does not have a strong magnetic field. This means that the solar wind particles are not guided towards the poles, and so they are unable to collide with atoms in the Venusian atmosphere and produce an aurora.
The magnetic field on Venus is around 20,000 times weaker than that on Earth. This is because Venus does not have a molten iron core, which is the source of Earth's magnetic field.
To know more about atmosphere visit:
https://brainly.com/question/11192430
#SPJ11
The rotating speed of a motor is 1440 RPM. What is the frequency (in Hz) of the peak in the vibration spectrum caused by rotor unbalance?
The frequency (in Hz) of the peak in the vibration spectrum caused by rotor unbalance is given by the equation: Frequency = (1/60) x RPM x No of Defects where RPM is the rotating speed of the motor and No of Defects is the number of unbalance defects.
Given RPM = 1440, we need to determine the frequency in Hz of the peak in the vibration spectrum caused by rotor unbalance. Frequency = (1/60) x RPM x No of Defects Frequency = (1/60) x 1440 x 1Frequency = 24 Hz
The frequency (in Hz) of the peak in the vibration spectrum caused by rotor unbalance is 24 Hz.
To know more about frequency visit:
https://brainly.com/question/29739263
#SPJ11
When your urinary bladder is full, the bladder pressure can reach up to 60 mm H2O. a Assuming that there is no height difference between your urinary bladder and where your urine comes out, calculate the speed at which your urine comes out. The density of urine is 1030 kg/m3 . b If the diameter of a urethra is 6 mm, estimate the volume flow rate of urine as it comes out in units of liters per second. If a full bladder constitutes 500 mL of urine, how long will it take you to remove all of the urine from your bladder? d Is the answer in c a realistic time for peeing? What should be added to make it more realistic? с
a)The speed at which urine comes out can be calculated using Bernoulli's equation, which relates the pressure of a fluid to its velocity. The equation is:P + (1/2)ρv² = constant
Where P is the pressure of the fluid, ρ is the density of the fluid, v is the velocity of the fluid, and the constant is the same at all points along the streamline. The constant can be neglected because the height difference is negligible. Therefore, at the bladder, P = 60 mmH2O (convert to Pa) and
ρ = 1030 kg/m³:60 mm H2O
= 60/1000 * 9.81 Pa
= 0.5886 PaThus,P + (1/2)ρv²
= 0.5886 PaRearranging this equation to solve for v gives:v = sqrt(2P/ρ)
= sqrt(2*0.5886/1030)
= 0.033 m/s
Answer: 0.033 m/s
b)The volume flow rate of urine can be calculated using the equation:Q = Avwhere Q is the volume flow rate, A is the cross-sectional area of the urethra, and v is the velocity of the urine found in part (a).
The diameter of the urethra is 6 mm, so the radius is 3 mm = 0.003 m:Area = πr²
= π(0.003)²
= 2.827E-5 m²
The volume flow rate is then:Q = Av = (2.827E-5 m²)(0.033 m/s)
= 9.32E-7 m³/s
To convert to L/s, divide by 1000:Q = 9.32E-7 m³/s ÷ 1000
= 9.32E-10 L/s
Answer: 9.32E-10 L/sc)If the bladder holds 500 mL of urine, it will take:Time = Volume flow rate⁻¹
= (500 mL) / (9.32E-10 L/s)
= 5.36E8 s (approx.)
Answer: 5.36E8 s, which is not a realistic time for peeing.
d)The answer in (c) is not a realistic time for peeing because it is several years. The units should be changed to minutes or seconds to make it more realistic. To make it more realistic, the person's rate of urine production should be taken into account. Most people produce urine at a rate of about 1-2 L per day, or 0.7-1.4 mL/min. Therefore, if a person has a full bladder, they should be able to empty it in less than a minute, assuming normal bladder function.
To know more about bladder pressure, visit:
https://brainly.com/question/32705790
#SPJ11
b. A tape measure is made of a particular material which has a linear thermal expansion coefficient of 20×10
−6
K
−1
. At −10
∘
C, using it you measure a piece of the material (which has a linear thermal expansion coefficient of 80×10
−6
K
−1
) to have a length of 10 m. What length would the tape measure say the piece of material has at 30
∘
C ? Assume the tape measure if marked to the mm level, and give your answer to that level of accuracy. [15 pts]
The length measurement of the material at `30°C` would be `10.040 m`.
Given that, A tape measure is made of a particular material which has a linear thermal expansion coefficient of `20×10^(-6)` K^(-1).At `-10°C`, using it you measure a piece of the material (which has a linear thermal expansion coefficient of `80×10^(-6)` K^(-1)) to have a length of `10 m`.
We need to find what length the tape measure would say the piece of material has at `30°C`.
Formula used: `∆L = Lα∆T` where, ∆L = Change in length L = Lengthα = Coefficient of linear expansion ∆T = Change in temperature
Length measurement of the material at `-10°C`, L₁ = `10 m`
Coefficient of linear expansion of the material, α₁ = `80×10^(-6)` K^(-1)
To find Length measurement at `30°C`
Coefficient of linear expansion of the tape measure, α₂ = `20×10^(-6)` K^(-1)
Change in temperature, ∆T = (`30°C`) - (`-10°C`) = `40°C`
Change in length, ∆L = Lα∆T = `10×80×10^(-6)×40 = 0.032 m`
Increase in length of the tape measure, ∆L₂ = L₂α₂∆T = `10×20×10^(-6)×40 = 0.008 m`
Total length at `30°C` = L + ∆L + ∆L₂ = `10 + 0.032 + 0.008 = 10.040 m`
Therefore, the length measurement of the material at `30°C` would be `10.040 m`.
To know more about linear expansion refer to:
https://brainly.com/question/28827009
#SPJ11
A monochromatic beam of light has a wavelength of 403 nm.
It is diffracted through a set of double slits, and produces a maxima of order 6 at an angle of 1.00 degrees.
What is the separation of the slits in m?
separation of slits is approximately 0.0013776 meters.
To find the separation of the slits, we can use the equation for the double-slit interference pattern:
dsinθ = mλ
where d is the separation between the slits, θ is the angle of maxima,
m is the order of the maxima, and λ is the wavelength of the light.
Given:
Wavelength, λ = 403 nm = 403 × 10^(-9) m
Angle of the maxima, θ = 1.00 degrees = 1.00 × π/180 radians
Order of the maxima, m = 6
Now, we can rearrange the equation to solve for d:
d = (mλ) / sinθ
Plugging in the values:
d = (6 × 403 × 10^(-9)) / sin(1.00 × π/180)
d ≈ 6 * (403 × 10^(-9) m) / sin(0.0175)
d ≈ 6 * (403 × 10^(-9) m) / 0.0175
d ≈ 0.0013776 m
Therefore, the separation of the slits is approximately 0.0013776 meters.
To Learn more about separation of the slits :
https://brainly.com/question/33715716
#SPJ11
Determine the net thermodynamic work (W) done by an engine in a cycle in which 17 moles of an ideal monatomic gas is compressed isothermally at 300 K and expanded isothermally at 554 K. The minimum and maximum volumes are 2 litres and 8 litres, respectively. The other two processes that complete the cycle are isovolumetric and can be ignored. O a.-5.0e4J O b. 5.9e4J O c.-1.1e5 J O d. 5.0e4J O e. 8.9e4J
The net thermodynamic work (W) done by the engine in the given cycle can be determined by calculating the work done during the isothermal compression and expansion processes.
The answer options are: a) -5.0e4J, b) 5.9e4J, c) -1.1e5J, d) 5.0e4J, and e) 8.9e4J.
In an isothermal process, the work done by or on the gas can be calculated using the equation W = nRT ln(V2/V1), where n is the number of moles of gas, R is the ideal gas constant, T is the temperature in Kelvin, and V2/V1 is the ratio of final volume to initial volume.
For the isothermal compression process, the temperature is 300 K and the volume changes from 8 liters to 2 liters. Plugging these values into the equation, we can calculate the work done during compression.
For the isothermal expansion process, the temperature is 554 K and the volume changes from 2 liters back to 8 liters. Using the same equation, we can calculate the work done during expansion.
The net work done by the engine in the cycle is the algebraic sum of the work done during compression and expansion. The sign of the work done depends on whether work is done on the gas (positive) or by the gas (negative).
To find the correct answer, calculate the work done during compression and expansion separately and then sum them up, considering the signs. The answer that matches the calculated net work will be the correct choice among the given options.
Learn more about The Thermodynamic work(W) from the given link:
https://brainly.com/question/2855564
#SPJ11
A nyicin rope (Y=2.909 Pa) has a length of 35:0 m and diameter of 22.0 mm, What force is needed to stretch this rope a length of 23.0 mm. 14
The force needed to stretch the nylon rope by 23.0 mm can be calculated using the formula:
Force = 2.909 Pa x Area x 0.023 m / 35.0 m
The force needed to stretch a nylon rope can be calculated using the formula:
Force = Young's modulus x Area x Change in length / Original length
In this case, the Young's modulus of nylon is given as 2.909 Pa, the original length is 35.0 m, and the change in length is 23.0 mm.
First, we need to convert the change in length from millimeters to meters. 23.0 mm is equal to 0.023 m.
Next, we need to calculate the area of the rope. The diameter is given as 22.0 mm, so the radius is half of that, which is 11.0 mm or 0.011 m. The area of the rope is then calculated using the formula for the area of a circle:
Area = [tex]\pi radius^2[/tex]
Once we have the area and the change in length in meters, we can substitute the values into the formula to calculate the force.
Learn more about Young's modulus here:
https://brainly.com/question/30139811
#SPJ11
An infinite surface charge density of -3n (/m² > Find charge located at -x-y plane (x=0) density everywhere.
When we talk about surface charge density (σ), we mean the amount of electric charge present per unit surface area. It is typically measured in coulombs per square meter (C/m2).
To determine the charge located at the -x-y plane (x=0), with a surface charge density of -3n C/m², we can use the following steps:Step 1: Determine the area of the plane We know that the plane is a 2D shape, and its area can be represented as:A = L x W where L is the length and W is the width.In this case, we have:L = ∞ (since it is infinite in one dimension)W = 1 (since it is a flat plane with width of 1)
Therefore, the area of the plane is:A = ∞ x 1
= ∞
Step 2: Calculate the total charge on the plane We can calculate the total charge Q on the plane by multiplying the surface charge density σ by the area A.Q = σ x AWe know that
σ = -3n C/m² and
A = ∞, so:
Q = -3n C/m² x ∞ = -∞ C
Therefore, the charge located at the -x-y plane (x=0) with a surface charge density of -3n C/m² is -∞ C.Therefore, the total charge on the plane is -∞ C.
To know more about electric charge visit:-
https://brainly.com/question/28457915
#SPJ11
In the design of a cam with the following characteristics
A slide follower moves a total slide height of 2"
At the beginning of the cycle, the follower is at rest between degrees 0° and 120°
Suffers a 2" elevation with cycloidal movement between 120° and 270° degrees
Suffers a 2" return with simple harmonic motion between 270° and 360° degrees
The diameter of the base circle is 2".
What is the height of the follower (from the center of rotation of the cam) at degree 60 of the cam?
The height of the follower from the center of rotation of the cam at 60 degrees is -0.83 units. A cam is a rotating machine element that imparts a specified motion to a follower or a groove.
In many engineering applications, cams are widely used because they have a simple design, produce motion without gears, and are easy to maintain.
Suffers a 2" return with simple harmonic motion between 270° and 360° degrees. The diameter of the base circle is 2".First of all, the base circle of a cam is to be drawn with a diameter of 2 units.
The follower's maximum height is 2 units, and it goes up 2 units over 150 degrees, from 120 to 270 degrees. From 0 to 120 degrees, the follower remains at 0 units of height.
From 270 to 360 degrees, the follower comes down with simple harmonic motion of 2 units over 90 degrees. This is shown in the diagram below:
The radius of the cam at 60 degrees can be found using the formula: RC = R cosθ + Hsinθ Where: RC is the radius of the cam at any angleθ is the angle H is the height of the cam, R is the radius of the base circle. The angle θ = 60 degrees.
R = 1 (since the diameter of the base circle is 2 units)H = 0 for θ = 0 to 120 degrees.
H = 2sin[(θ - 120)π /150] for θ
= 120 to 270 degrees H
= 2cos[(θ - 270)π /180] for θ
= 270 to 360 degrees
Substitute the values in the formula for the radius of the cam at 60 degrees. RC = R cosθ + HsinθR60
= 1 cos 60° + 2sin[(60 - 120)π /150]R60
= 0.5 + 2sin(240π /150)R60
= 0.5 - 1.33R60
= -0.83 units
Thus, the height of the follower from the center of rotation of the cam at 60 degrees is -0.83 units.
To know more about motion, refer
https://brainly.com/question/25951773
#SPJ11
Two carts mounted on an air track are moving toward one another. Cart 1has a speed of 1.1 m/s and a mass of 0.42 kg. Cart 2 has a mass of 0.71 kg. (a) If the total momentum of the system is to be zero, what is the initial speed (in m/s ) of Cart 2? (Enter a number.) m/s (b) Does it follow that the kinetic energy of the system is also zero since the momentum of the system is zero? Yes No (c) Determine the system's kinetic energy (in J) in order to substantiate your answer to part (b). (Enter a number.) J
a) initial speed of Cart 2 is approximately 0.651 m/s.
b) No, it does not follow that the kinetic energy of the system is also zero
c) system's kinetic energy is approximately 0.483 J.
(a) For initial speed of Cart 2, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision. Since the total momentum of the system is to be zero,
The momentum of an object is calculated by multiplying its mass by its velocity. So, we have:
Momentum of Cart 1 = Mass of Cart 1 * Velocity of Cart 1
Momentum of Cart 2 = Mass of Cart 2 * Velocity of Cart 2
Since the total momentum of the system is zero, we can set up the following equation:
0 = (0.42 kg * 1.1 m/s) + (0.71 kg * Velocity of Cart 2)
Solving for the velocity of Cart 2:
0 = 0.462 kg*m/s + (0.71 kg * Velocity of Cart 2)
-0.462 kg*m/s = 0.71 kg * Velocity of Cart 2
Velocity of Cart 2 = -0.462 kg*m/s / 0.71 kg
Velocity of Cart 2 ≈ -0.651 m/s
Therefore, the initial speed of Cart 2 is approximately 0.651 m/s.
(b) No, it does not follow that the kinetic energy of the system is also zero just because the momentum of the system is zero. Kinetic energy depends on the mass and velocity of an object, while momentum only considers the mass and velocity. Therefore, the kinetic energy can still be non-zero even if the momentum is zero.
(c) For system's kinetic energy, we can calculate the individual kinetic energies of Cart 1 and Cart 2, and then sum them up. The kinetic energy (KE) of an object is given by the formula:
KE = (1/2) * Mass * Velocity^2
The kinetic energy of Cart 1 is:
KE1 = (1/2) * 0.42 kg * (1.1 m/s)^2
The kinetic energy of Cart 2 is:
KE2 = (1/2) * 0.71 kg * (-0.651 m/s)^2
To find the total kinetic energy of the system, we add the individual kinetic energies together:
Total kinetic energy = KE1 + KE2
Total kinetic energy = 0.332 J + 0.151 J
Total kinetic energy = 0.483 J
Therefore, the system's kinetic energy is approximately 0.483 J.
To learn more about kinetic energy and Momentum :
https://brainly.com/question/12880315
#SPJ11
