Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed
Answer:
Final energy = Uf = initial energy × d₂/d₁
Explanation:
Energy is the ability to do work.
capacitor is an electronic device that store charges
where
V is the potential difference
d is the distance of seperation between the two plates
ε₀ is the dielectric constant of the material used in seperating the two plates, e.g., paper, mica, glass etc.
A = cross sectional area
U =¹/₂CV²
C =ε₀A/d
C × d=ε₀A=constant
C₂d₂=C₁d₁
C₂=C₁d₁/d₂
charge will 'q' remains same in the capacitor, if the capacitor was disconnected from the electric potential source (v) before the separation of the plates was replaced
Energy=U =(1/2)q²/C
U₂C₂ = U₁C₁
U₂ =U₁C₁ /C₂
U₂ =U₁d₂/d₁
Final energy = Uf = initial energy × d₂/d₁
Two ships of equal mass are 109 m apart. What is the acceleration of either ship due to the gravitational attraction of the other? Treat the ships as particles and assume each has a mass of 39,000 metric tons. (Give the magnitude of your answer in m/s2.)
Answer:
The acceleration is [tex]a = 2.190 *10^{-7} \ m/s^2[/tex]
Explanation:
From the question we are told that
The distance of separation of the ship is [tex]r= 109 \ m[/tex]
The mass of each ship is [tex]M = 39,000 \ metric\ tons =39,000 * 1000 = 3.9 *10^{7}\ kg[/tex]
The gravitational force of attraction exerted on each other is mathematically represented as
[tex]F_g = \frac{ GMM}{r^2}[/tex]
Where G is the gravitational constant with value
substituting values
[tex]F_g = \frac{ 6.674 30 * 10^{-11} (3.9 *10^{7})^2}{109^2}[/tex]
[tex]F_g = 8.54 \ N[/tex]
This force can also be mathematically represented as
[tex]F_g = M * a[/tex]
=> [tex]a = \frac{F_g}{M}[/tex]
substituting values
[tex]a = \frac{8.544}{3.9 *10^{7}}[/tex]
[tex]a = 2.190 *10^{-7} \ m/s^2[/tex]
A 285-kg object and a 585-kg object are separated by 4.30 m. (a) Find the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object placed midway between them.
Answer:
The magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Explanation:
Given;
first object with mass, m₁ = 285 kg
second object with mass, m₂ = 585 kg
distance between the two objects, r = 4.3 m
The midpoint between the two objects = r/₂ = 4.3 /2 = 2.15 m
Gravitational force between the first object and the 42 kg object;
[tex]F = \frac{GMm}{r^2}[/tex]
where;
G = 6.67 x 10⁻¹¹ Nm²kg⁻²
[tex]F = \frac{6.67*10^{-11} *285*42}{2.15^2} \\\\F = 1.727*10^{-7} \ N[/tex]
Gravitational force between the second object and the 42 kg object
[tex]F = \frac{6.67*10^{-11} *585*42}{2.15^2} \\\\F = 3.545*10^{-7} \ N[/tex]
Magnitude of net gravitational force exerted on 42kg object;
F = 3.545x 10⁻⁷ N - 1.727 x 10⁻⁷ N
F = 1.818 x 10⁻⁷ N
Therefore, the magnitude of the net gravitational force exerted by these objects on a 42.0-kg object is 1.818 x 10⁻⁷ N
Each side of a metal plate is illuminated by light of different wavelengths. The left side is illuminated by light with λ0 = 500 nm and the right side by light of unknown λ. Two electrodes A and B provide the stopping potential for the ejected electrons. If the voltage across AB is VAB=1.2775 V, what is the unknown λ?
Answer:
The wavelength is [tex]\lambda = 1029 nm[/tex]
Explanation:
From the question we are told that
The wavelength of the left light is [tex]\lambda_o = 500 nm = 500 *10^{-9} \ m[/tex]
The voltage across A and B is [tex]V_{AB } = 1.2775 \ V[/tex]
Let the stopping potential at A be [tex]V_A[/tex] and the electric potential at B be [tex]V_B[/tex]
The voltage across A and B is mathematically represented as
[tex]V_{AB} = V_A - V_B[/tex]
Now According to Einstein's photoelectric equation the stopping potential at A for the ejected electron from the left side in terms of electron volt is mathematically represented as
[tex]eV_A = \frac{h * c}{\lambda_o } - W[/tex]
Where W is the work function of the metal
h is the Planck constant with values [tex]h = 6.626 *10^{-34} \ J \cdot s[/tex]
c is the speed of light with value [tex]c = 3.0 *10^{8} \ m/s[/tex]
And the stopping potential at B for the ejected electron from the right side in terms of electron volt is mathematically represented as
[tex]eV_B = \frac{h * c}{\lambda } - W[/tex]
So
[tex]eV_{AB} = eV_A - eV_B[/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - W - [\frac{h * c}{\lambda } - W][/tex]
=> [tex]eV_{AB} = \frac{h * c}{\lambda_o } - \frac{h * c}{\lambda }[/tex]
=> [tex]\frac{h * c}{\lambda } = \frac{h * c}{\lambda_o } -eV_{AB}[/tex]
=> [tex]\frac{1}{\lambda } =\frac{1}{\lambda_o } - \frac{ eV_{AB}}{hc}[/tex]
Where e is the charge on an electron with the value [tex]e = 1.60 *10^{-19} \ C[/tex]
=> [tex]\frac{1}{\lambda } = \frac{1}{500 *10^{-9} } - \frac{1.60 *10^{-19} * 1.2775}{6.626 *10^{-34} * 3.0 *10^{8}}[/tex]
=> [tex]\frac{1}{\lambda } = 9.717*10^{5} m^{-1}[/tex]
=> [tex]\lambda = 1.029 *10^{-6} \ m[/tex]
=> [tex]\lambda = 1029 nm[/tex]
An air-filled capacitor is formed from two long conducting cylindrical shells that are coaxial and have radii of 42 mm and 74 mm. The electric potential of the inner conductor with respect to the outer conductor is -308 V ( = 1/4πε0 = 8.99 × 10^9 N · m^2/C^2).
The maximum energy density of the capacitor is closest to:_______
Correct answer is 2.7 x 10^-3 J/m3
I hope that helps ! <33
The maximum energy density of the capacitor is closest to: 2.7 x 10^-3 J/m3.
What is meant by the energy density of a capacitor?Energy density is defined as the total energy per unit volume of the capacitor. Since, Now, for a parallel plate capacitor, A × d = Volume of space between plates to which electric field E = V / d is confined. Therefore, Energy is stored per unit volume.
How do you calculate energy density?All Answers (14) Energy density is equal to 1/2*C*V2/weight, where C is the capacitance you computed and V should be your nominal voltage (i.e 2.7 V). Power Density is V2/4/ESR/weight, where ESR is the equivalent series resistance.
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The Atwood machine consists of two masses hanging from the ends of a rope that passes over a pulley. The pulley can be approximated by a uniform disk with mass p=7.95 kg and radius p=0.89 m. The hanging masses are L=32.0 kg and R=17.8 kg. Calculate the magnitude of the masses' acceleration and the tension in the left and right ends of the rope, L and R , respectively.
Answer:
Acceleration(a) = 2.588 m/s²
TL = 230.784 N
TR = 220.5 N
Explanation:
Given:
M = 7.95 kg
mL = 32 kg
mR = 17.8 kg
g = 9.8 m/s²
Find:
Acceleration(a)
TL
TR
Computation:
Acceleration(a) = [(mL - mR)g] / [mL + mR + M/2]
Acceleration(a) = [(32 - 17.8)9.8] / [32 + 17.8 + 7.95/2]
Acceleration(a) = [139.16] / [53.775]
Acceleration(a) = 2.588 m/s²
TL = mL(g-a)
TL = 32(9.8-2.588)
TL = 230.784 N
TR = mR(g+a)
TR = 17.8(9.8+2.588)
TR = 220.5 N
While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor RRR will
Answer:
The induced current through resistor R will
b) flow from a to b
Explanation:
The image is shown below, and the full question is written down as
The two solenoids in the figure are coaxial and fairly close to each other. While the resistance of the variable resistor in the left-hand solenoid is decreased at a constant rate, the induced current through the resistor R will
a) Flow from b to a
b) flow from a to b
c) be zero because the rate is constant.
From the image, the current in the left hand solenoid flows from the positive terminal of the battery to the negative terminal in an anticlockwise direction by convention.
Varying a rheostat causes a change in the resistance of electricity through the solenoid, and a changing current through a solenoid will induce current to flow through another solenoid placed nearby. Therefore, the left-hand solenoid induces a current flow on the right-hand solenoid.
Since the current in the left-hand solenoid flows in an anticlockwise direction, then it will have an equivalent magnetic polarity of a north pole on a magnet.
Also remember that Lenz law states that the induce current acts in such a way as to oppose the motion, or action producing it.
In this case, the induced current in the right-hand solenoid will act as to repel the left-hand solenoid away from itself. The only way is by the right-hand solenoid also having a north pole equivalent magnetic pole on it since like poles repel each other. This means that the induced current in the right-hand solenoid will flow in an anticlockwise manner too, from a to b.
Imagine that while you and a passenger are in a deep-diving submersible in the North Pacific near Alaska’s Aleutian Islands, you encounter a long, narrow depression on the ocean floor. Your passenger asks whether you think it is a submarine canyon, a rift valley, or a deep-ocean trench. How would you respond? Explain your response.
Answer:
I would say its a deep ocean trench
Explanation:
This is because deep ocean trenches are found at the deepest part of the ocean and also at Pacific ocean margins or Rim where subduction usually occurs and Aleutian islands are part of the Pacific Rim
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage across its terminals. (a) At the moment contact is made with the battery the voltage across the capacitor is
Answer:
(a) D. Zero.
(b) C. Equal to the battery's terminal voltage.
Explanation:
The question is incomplete, see the complete question for your reference and information.
A resistor and a capacitor are connected in series across an ideal battery having a constant voltage
across its terminals. At the moment contact is made with the battery
(a) the voltage across the capacitor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero.
(b) the voltage across the resistor is
A) equal to the battery's terminal voltage.
B) less than the battery's terminal voltage, but greater than zero.
C) equal to the battery's terminal voltage.
D) zero
A RC circuit is a circuit that is composed of both resistors and capacitors connect to a source of current or voltage.
basically when a voltage source is applied to an RC circuit, the capacitor, C charges up through the resistance, R
An electric car uses a 45-kW (160-hp) motor. If the battery pack is designed for 340V, what current would the motor need to draw from the battery? Neglect any energy losses in getting energy from the battery to the motor.
Answer:
Current = 132.35 A
The motor needs to draw 132.35 Amperes current from the battery.
Explanation:
The formula of electric power is given as follows:
Power = (Voltage)(Current)
Current = Power/Voltage
In this question, we have:
Power = 45 KW = 45000 W
Voltage of Battery Pack = 340 V
Current needed to be drawn = ?
Therefore,
Current = 45000 W/340 V
Current = 132.35 A
The motor needs to draw 132.35 Amperes current from the battery.
A light wave with an electric field amplitude of E0 and a phase constant of zero is to be combined with one of the following waves. Which of these combinations produces the greatest intensity?
a. Wave A has an amplitude of E0 and a phase constant of zero.
b. Wave B has an amplitude of E0 and a phase constant of π.
c. Wave C has an amplitude of 2E0 and a phase constant of zero.
d. Wave D has an amplitude of 2E0 and a phase constant of π.
e. Wave E has an amplitude of 3E0 and a phase constant of π.
Answer:
the greatest intensity is obtained from c
Explanation:
An electromagnetic wave stagnant by the expression
E = E₀ sin (kx -wt)
when two waves meet their electric fields add up
E_total = E₁ + E₂
the intensity is
I = E_total . E_total
I = E₁² + E₂² + 2E₁ E₂ cos θ
where θ is the phase angle between the two rays
Let's examine the two waves
in this case E₁ = E₂ = E₀
I = Eo2 + Eo2 + 2 E₀ E₀ coasts
I = E₀² (2 + 2 cos θ )
I = 2 I₀ (1 + cos θ )
let's apply this expression to different cases
a) In this case the angle is zero therefore the cosine is worth 1 and the intensity is I_total = 4 I₀
b) cos π = -1 this implies that I_total = 0
c) the cosine is 1,
I = E₀² + 4E₀² + 2 E₀ (2E₀) cos θ
I = E₀² (5 +4 cos θ)
I = E₀² 9
I = 9 Io
d) in this case the cos pi = -1
I = E₀² (5 -4)
I = I₀
e) we rewrite the equation
I = E₀² + 9 E₀² + 2 E₀ (3E₀) cos θ
I = Eo2 (10 + 6 cos θ)
cos π = -1
I = E₀² (10-6)
I = 4 I₀
the greatest intensity is obtained from c
The combination that has the greatest intensity is C. Wave C has an amplitude of 2E0 and a phase constant of zero.
What is an amplitude?An amplitude simply means the variable that meaures the change that occur in a single variable. It's the maximum diatance moved.
In this case, the combination that has the greatest intensity is Wave C since it has an amplitude of 2E0 and a phase constant of zero.
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An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
a. The radius of the circular path the electron travels
b. The magnitude of the electron's acceleration inside the field
c. The time the electron is in the magnetic field
d. The magnitude of the net force acting on the electron inside the field
Answer:
Explanation:
For circular path in magnetic field
mv² / R = Bqv ,
m is mass , v is velocity , R is radius of circular path , B is magnetic field , q is charge on the particle .
a )
R = mv / Bq
If v is changed to 2v , keeping other factors unchanged , R will be doubled
b )
magnitude of acceleration inside field
= v² / R
= Bqv / m
As v is doubled , acceleration will also be doubled
c )
If T be the time inside the magnetic field
T = π R / v
= π / v x mv / Bq
= π m / Bq
As is does not contain v that means T remains unchanged .
d )
Net force acting on electron
= m v² / R = Bqv
Net force = Bqv
As v becomes twice force too becomes twice .
So a . b , d are correct answer.
A parallel combination of a 1.01 μF capacitor and a 2.93 μF capacitor is connected in series to a 4.75 μF capacitor. This three‑capacitor combination is connected to a 16.3 V battery. Determine the charge on each capacitor.
Answer:
A.16.5x10^-6C
B. 47.5x10^-6C
C.77x10^-6C
Explanation:
Pls see attached file
You are moving at a speed 2/3 c toward Randy when shines a light toward you. At what speed do you see the light approaching you
Answer:
The speed of light will be c=3x10^8m/s
Explanation:
This is the same as the speed of light because your speed does not affecttje speed of light so you will see the light approaching you at the same speed of light c
Search Results Web results A car of mass 650 kg is moving at a speed of 0.7
Answer:
W = 1413.75 J
Explanation:
It is given that,
Mass of car, m = 650 kg
Initial speed of the car, u = 0.7 m/s
Let a man pushes the car, increasing the speed to 2.2 m/s, v = 2.2 m/s
Let us assume to find the work done by the man. According to the work energy theorem, work done is equal to the change in kinetic energy.
[tex]W=\dfrac{1}{2}m(v^2-u^2)\\\\W=\dfrac{1}{2}\times 650\times ((2.2)^2-(0.7)^2)\\\\W=1413.75\ J[/tex]
So, the work done by the car is 1413.75 J.
We repeat the experiment from the video, but this time we connect the wires in parallel rather than in series. Which wire will now dissipate the most heat?
Both wires will dissipate the same amount of heat.
A. The Nichrome wire (resistance 2.7)
B. The copper wire (resistance 0.1)
Answer: B. The copper wire (resistance 0.1)
Explanation: When resistance is in parallel, voltage (V) is the same but current is different for every resistance. Current (i) is related to voltage and resistance (R) by Ohm's Law
i = [tex]\frac{V}{R}[/tex]
So, since both wires are in parallel, they have the same voltage but because the copper wire resistance is smaller than Nichrome wire, the first's current will be bigger.
Every resistor in a circuit dissipates electrical power (P) that is converted into heat energy. The dissipation can be found by:
P = [tex]i^{2}*R[/tex]
As current for copper wire is bigger than nichrome, power will be bigger and it will dissipate more heat.
In conclusion, the copper wire will dissipate more heat when connected in parallel.
A long wire carrying a 5.8 A current perpendicular to the xy-plane intersects the x-axis at x=−2.3cm. A second, parallel wire carrying a 3.0 A current intersects the x-axis at x=+2.3cm.
Required:
a. At what point on the x-axis is the magnetic field zero if the two currents are in the same direction?
b. At what point on the x-axis is the magnetic field zero if the two currents are in opposite directions?
Answer:
a) v r = 0.7318 cm , b) r = 7.23 cm
Explanation:
The magnetic field generated by a wire carrying a current can be found with Ampere's law
∫ B. ds = μ₀ I
the length of a surface circulates around the wire is
s = 2π r
where r is the point of interest of the calculation of the magnetic field
B = μ₀ I / 2π r
In this exercise we have two wires, write the equation of the magnetic field of each one
wire 1 I = 5.8 A
B₁ = μ₀ 5,8 / 2π r₁
wire 2 I = 3.0 A
B₂ = μ₀ 3/2π r₂
the direction of the field is given by the rule of the right hand, the thumb indicates the direction of the current and the other fingers the direction of the magnetic field
Let's apply these expressions to our case
a) the two streams go in the same direction
using the right hand rule for each wire we see that between the two wires the magnetic fields have opposite directions so there is some point where the total value is zero
B₁ - B₂ = 0
B₁ = B₂
μ₀ 5,8 / 2π r₁ = μ₀ 3 / 2π r₂
5.8 / r₁ = 3 / r₂
5.8 r₂ = 3r₁
the value of r is measured from each wire, therefore
r₁ = 2.3 + r
r₂ = 2.3 -r
we substitute
5.8 (2.3 - r) = 3 (2.3 + r)
r (3 + 5.8) = 2.3 (5.8 - 3)
r = 2.3 2.8 / 8.8
r = 0.7318 cm
b) the two currents have directional opposite
with the right hand rule in the field you have opposite directions outside the wires
suppose it is zero on the right side where the wire with the lowest current is
B₁ = B₂
5.8 / r₁ = 3 / r₂
5.8 r₂ = 3 r₁
r₁ = 2.3 + r
r₂ = r - 2.3
5.8 (r - 2.3) = 3 (2.3 + r)
r (5.8 -3) = 2.3 (3 + 5.8)
r = 2.3 8.8 / 2.8
r = 7.23 cm
An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
A. The time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Answer:
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels
Explanation:
The radius of the electron's motion in a uniform magnetic field is given by
[tex]R = \frac{MV}{qB}[/tex]
where;
m is the mass of the electron
q is the charge of the electron
B is the magnitude of the magnetic field
V is speed of the electron
R is the radius of the electron's
Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.
The magnitude of the electron's acceleration inside the field is given by;
[tex]a_c = \frac{V^2}{R}[/tex]
where;
[tex]a_c[/tex] is centripetal acceleration of electron
Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.
The time the electron is in the magnetic field is given by;
[tex]T = \frac{2\pi M}{qB}[/tex]
The time of electron motion will not change
The magnitude of the net force acting on the electron inside the field will not change;
[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]
Therefore, the correct options are "C" and "D"
A department store expects to have 225 customers and 20 employees at peak times in summer. Determine the contribution of people to the total cooling load of the store. The average rate of heat generation from people doing light work is 115 W, and 70% of it is in sensible form.
Answer:
The contribution of people to the cooling load of the store is 19722.5 W
Explanation:
Total amount of customers = 225
Total amount of employees = 20
Total amount of people in the store at that instant n = 245 people
Average rate of heat generation Q = 115 W
percentage of these heat generated that is sensible heat = 70%
Sensible heat raises the surrounding temperature. Latent heat only causes a change of state.
The total heat generated by all the people in the store = n x Q
==> 245 x 115 = 28175 W
but only 70% of this heat is sensible heat that raises the temperature of the store, therefore, the contribution of people to the cooling load of the store = 70% of 28175 W
==> 0.7 x 28175 = 19722.5 W
How much heat is needed to melt 2.5 KG of water at its melting point? Use Q= mass x latent heat of fusion.
Answer:
Q = 832 kJ
Explanation:
It is given that,
Mass of the water, m = 2.5 kg
The latent heat of fusion, L = 333 kJ/kg
We need to find the heat needed to melt water at its melting point. The formula of heat needed to melt is given by :
Q = mL
[tex]Q=2.5\ kg\times 333\ kJ/kg\\\\Q=832.5\ kJ[/tex]
or
Q = 832 kJ
So, the heat needed to melt the water is 832 kJ.
At time t = 1, a particle is located at position (x, y) = (5, 2). If it moves in the velocity field F(x, y) = xy − 1, y2 − 11 find its approximate location at time t = 1.02.
Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
The rotor of a gas turbine is rotating at a speed of 7000 rpm when the turbine is shut down. It is observed that 3.5 minutes is required for the rotor to coast to rest. Assuming uniformly accelerated motion, determine the number of revolutions that the rotor executes before coming to rest. Hint: there will be a large number of rotations.
Answer:
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
Explanation:
Given that rotor of gas turbine is decelerating at constant rate, it is required to obtained the value of angular acceleration as a function of time, as well as initial and final angular speeds. That is:
[tex]\dot n = \dot n_{o} + \ddot n \cdot t[/tex]
Where:
[tex]\dot n_{o}[/tex] - Initial angular speed, measured in revolutions per minute.
[tex]\dot n[/tex] - Final angular speed, measured in revolutions per minute.
[tex]t[/tex] - Time, measured in minutes.
[tex]\ddot n[/tex] - Angular acceleration, measured in revoiutions per square minute.
The angular acceleration is now cleared:
[tex]\ddot n = \frac{\dot n - \dot n_{o}}{t}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]t = 3.5\,min[/tex], the angular acceleration is:
[tex]\ddot n = \frac{0\,\frac{rev}{min}-7000\,\frac{rev}{min} }{3.5\,min}[/tex]
[tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex]
Now, the final angular speed as a function of initial angular speed, angular acceleration and the change in angular position is represented by this kinematic equation:
[tex]\dot n^{2} = \dot n_{o}^{2} + 2\cdot \ddot n \cdot (n-n_{o})[/tex]
Where [tex]n[/tex] and [tex]n_{o}[/tex] are the initial and final angular position, respectively.
The change in angular position is cleared herein:
[tex]n-n_{o} = \frac{\dot n^{2}-\dot n_{o}^{2}}{2\cdot \ddot n}[/tex]
If [tex]\dot n_{o} = 7000\,\frac{rev}{min}[/tex], [tex]\dot n = 0\,\frac{rev}{min}[/tex] and [tex]\ddot n = -2000\,\frac{rev}{min^{2}}[/tex], the change in angular position is:
[tex]n-n_{o} = \frac{\left(0\,\frac{rev}{min} \right)^{2}-\left(7000\,\frac{rev}{min} \right)^{2}}{2\cdot \left(-2000\,\frac{rev}{min^{2}} \right)}[/tex]
[tex]n-n_{o} = 12250\,rev[/tex]
The rotor of the gas turbine rotates 12250 revolutions before coming to rest.
A 0.500-V potential difference is maintained across a 1.50-m length of tungsten wire that has a cross-sectional area of 0.800 mm2. What is the current in the wire
Answer:
5.95 A
Explanation:
From the question
R = ρL/A..................... Equation 1
Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.
Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m
Substitute these values into equation 1
R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶
R = 0.084 Ω.
Finally, using Ohm law,
V = IR
Where V = Voltage, I = current
Make I the subject of the equation
I = V/R............... Equation 2
I = 0.5/0.084
I = 5.95 A
3. According to Hund's rule, what's the expected magnetic behavior of vanadium (V)?
O A. Ferromagnetic
O B. Non-magnetic
C. Diamagnetic
O D. Paramagnetic
Answer:
Diamagnetic
Explanation:
Hunds rule states that electrons occupy each orbital singly first before pairing takes place in degenerate orbitals. This implies that the most stable arrangement of electrons in an orbital is one in which there is the greatest number of parallel spins(unpaired electrons).
For vanadium V ion, there are 18 electrons which will be arranged as follows;
1s2 2s2 2p6 3s2 3p6.
All the electrons present are spin paired hence the ion is expected to be diamagnetic.
Answer:
its paramagnetic
Explanation:
i took this quiz
the efficiency of a carnot cycle is 1/6.If on reducing the temperature of the sink 75 degrees celcius ,the efficiency becomes 1/3,determine he initial and final temperatures between which the cycle is working.
Answer:
450°C
Explanation: Given that the efficiency of Carnot engine if T₁ and T₂ temperature are initial and final temperature .
η = 1 - T2 / T1
η = 1/6 initially
when T2 is reduced by 65°C then η becomes 1/3
Solution
η = 1/6
1 - T2 / T1 = 1/6 [ using the Formula ]........................(1)
When η = 1/3 :
η = 1 - ( T2 - 75 ) / T1
1/3 = 1 - (T2 - 75)/T1.........................(2)
T2 - T1 = -75 [ because T2 is reduced by 75°C ]
T2 = T1 - 75...........................(3)
Put this in (2) :
> 1/3 = 1 - ( T1 - 75 - 75 ) / T1
> 1/3 = 1 - (T1 - 150 ) /T1
> (T1 - 150) / T1 = 1 - 1/3
> ( T1 -150 ) / T1 = 2/3
> 3 ( T1 - 150 ) = 2 T1
> 3 T1 - 450 = 2 T1
Collecting the like terms
3 T1- 2 T1 = 450
T1 = 450
The temperature initially was 450°C
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s2 until it runs out of fuel at an altitude of 850 m . After this point, its acceleration is that of gravity, downward.
Answer:
v = 73.75 m/s
Explanation:
It is given that,
A rocket rises vertically, from rest, with an acceleration of 3.2 m/s² until it runs out of fuel at an altitude of 850 m.
Let us assume we need to find the velocity of the rocket when it runs out of fuel.
Let v is the final speed. Using the third equation of kinematics as :
[tex]v^2-u^2=2as[/tex]
u = 0
[tex]v=\sqrt{2as} \\\\v=\sqrt{2\times 3.2\times 850}\\\\v=73.75\ m/s[/tex]
So, the velocity of the rocket when it runs out of the fuel is 73.75 m/s
A stone with a mass m is dropped from an airplane that has a horizontal velocity v at a height h above a lake. If air resistance is neglected, the horizontal distance R from the point on the lake directly below the point of release to the point where the stone strikes the water is given by which formula? R=v(2h/g)2 None of these are correct. R = 2mv sqrt(2h/g) R = v sqrt(2h/g) R=(1/2)gt2
Answer:
R = v √(2h / g)
Explanation:
This exercise can be solved using the concepts of science, projectile launching
let's calculate the time it takes to get to the water
y = y₀ +[tex]v_{oy}[/tex] t - ½ g t²
as the stone is skipped the vertical speed is zero
y = y₀ - ½ g t²
for y=0
t = √ (2y₀ / g)
the horizontal distance it covers in this time is
R = v₀ₓ t
R = v₀ₓ √(2 y₀ / g)
let's call the horizontal velocity as v and the height is h
R = v √(2h / g)
Question 8
A spring is attached to the ceiling and pulled 8 cm down from equilibrium and released. The
damping factor for the spring is determined to be 0.4 and the spring oscillates 12 times each
second. Find an equation for the displacement, D(t), of the spring from equilibrium in terms of
seconds, t.
D(t) =
Can someone please help me ASAP?!!!!
Answer: D(t) = [tex]8.e^{-0.4t}.cos(\frac{\pi }{6}.t )[/tex]
Explanation: A harmonic motion of a spring can be modeled by a sinusoidal function, which, in general, is of the form:
y = [tex]a.sin(\omega.t)[/tex] or y = [tex]a.cos(\omega.t)[/tex]
where:
|a| is initil displacement
[tex]\frac{2.\pi}{\omega}[/tex] is period
For a Damped Harmonic Motion, i.e., when the spring doesn't bounce up and down forever, equations for displacement is:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex] or [tex]y=a.e^{-ct}.sin(\omega.t)[/tex]
For this question in particular, initial displacement is maximum at 8cm, so it is used the cosine function:
[tex]y=a.e^{-ct}.cos(\omega.t)[/tex]
period = [tex]\frac{2.\pi}{\omega}[/tex]
12 = [tex]\frac{2.\pi}{\omega}[/tex]
ω = [tex]\frac{\pi}{6}[/tex]
Replacing values:
[tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex]
The equation of displacement, D(t), of a spring with damping factor is [tex]D(t)=8.e^{-0.4t}.cos(\frac{\pi}{6} .t)[/tex].
An object with a mass of 5.5 kg is allowed to slide from rest down an inclined plane. The plane makes an angle of 30o with the horizontal and is 72 m long. The coefficient of friction between the plane and the object is 0.35. The speed of the object at the bottom of the plane is:_________.
a. 24 m/s.
b. 11 m/s.
c. 15 m/s.
d. 5.3 m/s.
e. 17 m/s.
Answer:
The speed will be "16.67 m/s".
Explanation:
The given values are:
Distance
= 72 m
Angle
= 30°
Acceleration
= [tex]g(sin \theta-ucos \theta)[/tex]
= [tex](9.8\times sin30^{\circ}) - (0.53\times cos30^{\circ})[/tex]
= [tex]1.929 \ m/s^2[/tex]
Let the speed be "v".
⇒ [tex]v^2=u^2+2as[/tex]
⇒ [tex]v^2=0(2\times 1.929\times 72)[/tex]
⇒ [tex]v^2=277.226[/tex]
⇒ [tex]v=\sqrt{277.776}[/tex]
⇒ [tex]v=16.67 \ m/s[/tex]
Two particles, of charges q1 and q2, are separated by a distance d on the x-axis with q1 at the origin and q2 in the positive direction. The net electric field due to the particles is zero at x = d/4. With V = 0 at infinity, locate (in terms of d) any point on the x-axis (other than infinity) at which the electric potential due to the two particles is zero.
Answer:
No point on the x-axis
Pls see attached file