Answer:
[tex]R=0.023m[/tex]
Explanation:
From the question we are told that:
Mass [tex]m=1.16*10^{-26}[/tex]
Potential difference [tex]V=523V[/tex]
Magnitude [tex]m=0.370 T[/tex]
Generally the equation for Velocity is mathematically given by
[tex]\frac{1}{2}mv^2=ev[/tex]
[tex]v=\frac{2ev}{m}[/tex]
[tex]v=\frac{2*1.6*10^{-19}*542}{1.16*10^{-26}}[/tex]
[tex]v=12.22*10^4m/s[/tex]
Generally the equation for Force is mathematically given by
[tex]F=qvBsin \theta[/tex]
Where
[tex]qVB=m\frac{v^2}{R}[/tex]
[tex]F=m\frac{v^2}{R}sin\theta[/tex]
Therefore
[tex]R=\frac{mv}{qB sin \theta}[/tex]
[tex]R=\frac{1.6*10^{-26}*12.2*10^{4}}{1.60*10^{-19}*0.394 sin 90}[/tex]
[tex]R=0.023m[/tex]
A motor is designed to operate on 117 V and draws a current of 17.7 A when it first starts up. At its normal operating speed, the motor draws a current of 2.78 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third normal speed.
Answer:
Resistance of the armature coil = 6.61 ohms
Back emf developed at normal speed = 98.62 V (Approx.)
Current drawn by the motor at one-third normal speed = 12.73 A
Explanation:
Given:
Potential difference V = 117 V
Current = 17.7 A
Motor drawn current = 2.78 A
Find:
Resistance of the armature coil
Back emf developed at normal speed
Current drawn by the motor at one-third normal speed
Computation:
A] Resistance of the armature coil R = V/ I
Resistance of the armature coil = 117 / 17.7
Resistance of the armature coil = 6.61 ohms
B] Back emf developed at normal speed = V- IR
Back emf developed at normal speed = 117 V - (2.78 A)(6.61 ohms)
Back emf developed at normal speed = 117 V - 18.37
Back emf developed at normal speed = 98.62 V (Approx.)
C] Current drawn by the motor at one-third normal speed = 17.7 A - (98.62/3)/(6.61 ohms)
Current drawn by the motor at one-third normal speed = 17.7 - 4.97
Current drawn by the motor at one-third normal speed = 12.73 A
An object whose weight is 100 lbf experiences a decrease in kinetic energy of 500 ft lbf and an increase in potential energy of 1500 ft lbf. The initial velocity and elevation of the object, each relative to the surface of the earth, are 40 ft/s and 30 ft, respectively. If g 5 32.2 ft/s2 , determine:
(a) the final velocity, in ft/s.
(b) the final elevation, in ft.
Answer:
a) [tex]v_2=35.60ft/sec[/tex]
b) [tex]h_2=45ft[/tex]
Explanation:
From the question we are told that:
Weight [tex]W=100lbf[/tex]
Decrease in kinetic energy [tex]dK.E= 500 ft lbf[/tex]
Increase in potential energy [tex]dP.E =1500 ft lbf.[/tex]
Velocity [tex]V_1=40[/tex]
Elevation [tex]h=30ft[/tex]
[tex]g=32.2 ft/s2[/tex]
a)
Generally the equation for Change in Kinetic Energy is mathematically given by
[tex]dK.E=\frac{1}{2}m(v_1^2-v_2^2)[/tex]
[tex]500=\frac{1}{2}*\frac{100}{32.2}(v_1^2-v_2^2)[/tex]
[tex]v_2=35.60ft/sec[/tex]
b)
Generally the equation for Change in Potential Energy is mathematically given by
[tex]dP.E=mg(h_2-h_1)[/tex]
[tex]1500=mg(h_2-h_1)[/tex]
[tex]h_2=45ft[/tex]
The kinetic theory of gases states that the kinetic energy of a gas is directly proportional to the temperature of the gas.
a. True
b. False
Answer:
true
Explanation:
The kinetic energy of a gas is directly proportional to the temperature of the gas.because temperature is the average kinetic energy of a substance
I hope this helps
Cho dòng điện xoay chiều trong sản xuất và sinh hoạt ở nước ta có tần số f = 50Hz. Tính chu kỳ T và tần số góc ω?
Answer:
T = 1/f = 1/50(s)
ω = 2πf = 100π (rad/s)
(vote 5 sao nhó :3 )
The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3. Express your answer numerically in joules.
The question is incomplete. The complete question is :
A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .
Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:
1. The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.
Solution :
Given :
[tex]A = 10 \ cm^2[/tex]
[tex]$=0.0010 \ m^2$[/tex]
d = 10 mm
= 0.010 m
Then, Capacitance,
[tex]$C=\frac{k \epsilon_0 A}{d}$[/tex]
[tex]$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$[/tex]
[tex]$C=2.655 \times 10^{12} \ F$[/tex]
[tex]$U_1 = \frac{1}{2}CV^2$[/tex]
[tex]$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_1=2.987 \times 10^{-10}\ J$[/tex]
Now,
[tex]$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$[/tex]
And
[tex]$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$[/tex]
In parallel combination,
[tex]$C_{eq}= C_k + C_{air}$[/tex]
[tex]$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$[/tex]
[tex]$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$[/tex]
[tex]$C_{eq} = 1.77 \times 10^{-12}\ F$[/tex]
Then energy,
[tex]$U_2 =\frac{1}{2} C_{eq} V^2$[/tex]
[tex]$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$[/tex]
[tex]$U_2=1.99 \times 10^{-10} \ J$[/tex]
b). Now the charge on the [tex]\text{capacitor}[/tex] is :
[tex]$Q=C_{eq} V$[/tex]
[tex]$Q = 1.77 \times 10^{-12} \times 15 V$[/tex]
[tex]$Q = 26.55 \times 10^{-12} \ C$[/tex]
Now when the capacitor gets disconnected from battery and the [tex]\text{dielectric}[/tex] is slowly [tex]\text{removed the rest}[/tex] of the way out of the [tex]\text{capacitor}[/tex] is :
[tex]$C_3=\frac{A \epsilon_0}{d}$[/tex]
[tex]$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$[/tex]
[tex]$C_3=0.885 \times 10^{-12} \ F$[/tex]
[tex]$C_3 = 0.885 \times 10^{-12} \ F$[/tex]
Without the dielectric,
[tex]$U_3=\frac{1}{2} \frac{Q^2}{C}$[/tex]
[tex]$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$[/tex]
[tex]$U_3=3.98 \times 10^{-10} \ J$[/tex]
Two forces A and B act at a point. If their resultant is [given by] (B - A) in the direction of B, then
A. A and B are equal
B. A is greater than B
C. the angle between A and B is 0°
D. the angle between A and B is 90°
E. the angle between A and B is 180°
Express 6revolutions to radians
Answer:
About 37.70 radians.
Explanation:
1 revolution = 2[tex]\pi[/tex] radians
∴ 6 revolutions = (6)(2[tex]\pi[/tex] radians)
6 revolutions = 37.6991 or ≈ 37.70 radians
Question
Name and write briefly on the international body
that Introduced si units .
Answer:
International System of Units. it was established in 1960 by the 11the General Conference on Weights and Measures
A cataract is a clouding or opacity that develops in the eye's lens, often in older people. In extreme cases, the lens of the eye may need to be removed. What effect would this have on someone?
a. He would become nearsighted.
b. He would become farsighted.
c. He would become neither nearsighted nor farsighted.
Answer:
b. He would become farsighted.
Explanation:
A cataract is defined as a medical condition where a person eyes becomes partially opaque and the person is not bale to see properly.
This is mainly caused due to aging or any injury in the eyes tissue which make up the lens of the eye.
It is the clouding of the lens of the eyes or opacity of the eyes. When treating cataract, in some cases the lens of the eyes are needed to be removed. This may lead to person becoming far sighted.
Therefore, the correct option is (b).
A charge is moving in a magnetic field that points to the
left
What direction can the charge move and experience no
magnetic force? Check all that apply.
O up
O down
Oleft
Oright
O into the screen
O out of the screen
Answer:
Magnetic Forces on Moving Charges. The magnetic force on a free moving charge is perpendicular to both the velocity of the charge and the magnetic field with direction given by the right hand rule.
The direction of the charge where it does not experience magnetic force is left and right. The correct option is C and D.
What is a magnetic field?A magnetic field is a region in space where a magnetic force can be observed. It is created by moving electric charges, such as electrons, and is characterized by the direction and strength of the force it exerts on other magnetic materials or moving charges.
Magnetic field lines are used to visualize the direction and strength of the magnetic field. They represent the path that a small magnetic north pole would follow if placed in the magnetic field. The direction of the magnetic field is given by the direction in which the north pole of a compass needle would point if placed in the field.
Magnetic flux is the measure of the strength of the magnetic field passing through a surface. It is given by the product of the magnetic field strength and the area of the surface, as well as the cosine of the angle between the magnetic field and the surface normal. The unit of magnetic flux is Weber (Wb).
Magnetic flux is important in many applications, such as electric motors and generators, where the interaction between the magnetic field and moving charges produces electrical energy. It is also used in magnetic imaging techniques, such as MRI, to visualize the internal structures of the human body.
Here in the question,
Options A (up), B (down), E (into the screen), and F (out of the screen) are all perpendicular to the direction of the magnetic field and so will experience a magnetic force.
Therefore, options C and D are correct i. e left and right.as they are parallel and anti-parallel to the direction of the magnetic field, respectively.
To learn about Ohm's law click:
brainly.com/question/29775285
#SPJ7
A 20 N south magnetic force pushes a charged particle traveling with a velocity of 4 m/s west through a 5 T magnetic field pointing downwards . What is the charge of the particle ?
Answer:
Charge of the particle is 1 coulomb.
Explanation:
Force, F:
[tex]{ \bf{F=BeV}}[/tex]
F is magnetic force.
B is the magnetic flux density.
e is the charge of the particle.
V is the velocity
[tex]{ \sf{20 = (5 \times e \times 4)}} \\ { \sf{20e = 20}} \\ { \sf{e = 1 \: coulomb}}[/tex]
what is science ? what qualities do we deal in deal in physic ?
science is all about the world around us
9. From this lab, we learn that the electric field and electric potential depend on both, the magnitude of the source charge (q), and the distance from the source charge (r). If we were to increase the magnitude of our source charge from 1 nC to 5 nC, then the magnitudes of the electric field and electric potential would be ____.(you can test this on the animation by dragging five 1 nC charges on top of each other and measuring E and V at a distance of 1 m)
Answer:
the electric field and the electric potential increase 5 times
Explanation:
The electric field created by a point charge is
E = k q / r²
in this case the charge changes from q₁ = 1 10⁺⁰ C to q₂ = 5 10⁻⁹ C
with the electric field is proportional to the charge
E₅ = 5 E₁
the electrical power for a point charge is
V = k q / r
as the electric power is proportional to the charge
V₅ = 5 V₁
consequently both the electric field and the electric potential increase 5 times
An electric field is produced by a charged object:
[tex]\to E = \frac{k q}{r^2}\\\\[/tex]
In this situation, the charge shifts:
[tex]\to q_1 = 1 10^{\circ}\ C \\\\ \to q_2 = 5 \times 10^{-9}\ C[/tex]
so with electric field remaining proportional to the charge
[tex]\to E_5 = 5 E_1[/tex]
The electrical power consumed for a point charge:
[tex]\to V = \frac{k q}{ r}[/tex]
since the electric power is related to the charge
[tex]\to V_5 = 5 V_1[/tex]
As a result, both the electric field as well as the electric potential increase by 5 times.
Learn more:
brainly.com/question/2017486
A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b) How long will it take for this person's temperature to return to 37°c if his mass is 90kg.
Answer:
(a) 209 Watt
(b) 4482.8 seconds
Explanation:
(a) P = 50×4.18
Where P = rate of heat loss in watt
P = 209 Watt
Applying,
Q = cm(t₁-t₂)................ Equation 1
Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.
From the question,
Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C
Constant: c = 3470 J/kg.K
Substtut these values into equation 1
Q = 90×3470(40-37)
Q = 936900 J
But,
P = Q/t.............. Equation 2
Where t = time
t = Q/P............ Equation 3
Given: P = 209 Watt, Q = 936900
Substitute into equation 3
t = 936900/209
t = 4482.8 seconds
Si un resorte de constante elástica 1300 n/m se comprime 12 cm ¿Cuanta energía almacena? Y si estira 12cm ¿Cuanta energía almacena?
La energía que almacena el resorte cuando se comprime y estira 12 cm es 9,4 J.
La energía potencial elástica del resorte se puede calcular con la siguiente ecuación:
[tex] E_{p} = \frac{1}{2}kx^{2} [/tex]
En donde:
k: es la constante del resorte = 1300 N/m
x: es la distancia de compresión o de elongación = 12 cm = 0,12 m
Dado que la energía es proporcional al cuadrado de la distancia recorrida por el resorte (x), la energía almacenada por el resorte durante la compresión será la misma que la energía almacenada por la elongación.
Por lo tanto, la energía almacenada es:
[tex]E_{p} = \frac{1}{2}kx^{2} = \frac{1}{2}1300 N/m*(0,12 m)^{2} = 9,4 J[/tex]
Entonces, la energía del resorte cuando se comprime y cuando se estira es la misma, a saber 9,4 J.
Para saber más sobre energía potencial visita este link: https://brainly.com/question/156316?referrer=searchResults
Espero que te sea de utilidad!
Answer:
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
Explanation:
La Energía Potencial Elástica almacenada por el resorte ([tex]U_{e}[/tex]), en joules, se calcula a partir de la Ley de Hooke, la definición de Trabajo y el Teorema del Trabajo y la Energía, cuya expresión se presenta abajo:
[tex]U_{e} = \frac{1}{2}\cdot k\cdot (x_{f}^{2}-x_{o}^{2})[/tex] (1)
Donde:
[tex]k[/tex] - Constante elástica del resorte, en newtons por metro.
[tex]x_{o}[/tex] - Posición inicial del resorte, en metros.
[tex]x_{f}[/tex] - Posición final del resorte, en metros.
Nótese que el resorte sin deformar tiene una posición de cero, la tensión tiene un valor positivo y la compresión, negativo.
Asumiendo que en ambos casos el resorte se encuentra inicialmente sin deformar, se reduce (1) a una forma de función par, es decir, una función que cumple con la propiedad de que [tex]f(x) = f(-x)[/tex], se encuentra que al comprimirse o estirarse en la misma medida almacena la misma cantidad de energía.
La cantidad de energía a almacenar es:
[tex]U_{e} = \frac{1}{2}\cdot \left(1300\,\frac{N}{m} \right)\cdot (0,12\,m)^{2}[/tex]
[tex]U_{e} = 9,360\,J[/tex]
Al comprimirse o estirarse 12 centímetros desde su posición sin deformar, el resorte almacena 9,360 joules.
In December of 2011 they announced that a planet has been discovered in a habitable zone around a
star! It has clouds! It has twice the radius of the earth, but with the same density as earth, about 5.515 × 10^3kg/m3
. Find the new acceleration of gravity on the surface of this planet.
Explanation:
The density of earth [tex]\rho_E[/tex] is given by
[tex]\rho_E = \dfrac{M_E}{\left(\frac{4\pi}{3}R_E^3\right)}[/tex]
and in terms of this density, we can write the acceleration due to gravity on earth as
[tex]g_E =G\dfrac{M_E}{R_E^2} = \dfrac{4\pi G}{3}\rho_ER_E[/tex]
Similarly, the acceleration due to gravity [tex]g_P[/tex] on this new planet is given by
[tex]g_P = G\dfrac{M_P}{R_P^2} = G\dfrac{\frac{4\pi}{3}R_p^3\rho_P}{R_P^2}[/tex]
[tex]\:\:\:\:\:= \dfrac{4\pi G}{3}\rho_PR_P[/tex]
We know that this planet has the same density as earth and has a radius 2 times as large. We can then rewrite [tex]g_P[/tex] as
[tex]g_P = \dfrac{4\pi G}{3}\rho_E(2R_E)[/tex]
[tex]\:\:\:\:\:= 2\left(\dfrac{4\pi G}{3}\rho_ER_E\right) = 2g_E[/tex]
[tex]\:\:\:\:\:= 2(9.8\:\text{m/s}^2) = 19.6\:\text{m/s}^2[/tex]
If car A passes car B, then car A must be
A. accelerating at a greater rate than car B.
B. moving faster than car B, but not necessarily accelerating
C. accelerating
D. moving faster than car B and accelerating more than car B
Answer:
B. moving faster than car B, but not necessarily accelerating
Explanation:
Velocity is the speed of something. So car A's velocity is greater than car B but does not mean car A is accelerating.
which watch is more preferable for the measurement of time among pendulum, quartz and atomic watch
Answer:
pendulum, quartz
Explanation:
i don't understand this, can someone help please??
Explanation:
N2 + H2 --> NH3
balance them:
N2 + 3 H2 --> 2 NH3
so if 6 moles of N2 react, 12 moles of NH3 will form.
(you have to look at the big number in front, in this case its N2 and 2 NH3, therefore the amount of N2 will produce double the amount of NH3 )
Three wires are connected at a branch point. One wire carries a positive current of 18 A into the branch point, and a second wire carries a positive current of 7 A away from the branch point. Find the current carried by the third wire into the branch point.
Answer:
The current in third branch is 11 A.
Explanation:
incoming current in one branch = 18 A
outgoing current in the other branch = 7 A
let the current in the third branch is i.
According to the Kirchoff's fist law in electricity
incoming current = out going current
18 = 7 + i
i = 11 A
The current in third branch is 11 A.
on a horizontal axis whose unit is the meter, a linear load ranging from 0 to 1 ma a linear load distribution = 2 nC / m.
determine the modulus of the electric field created by the previous loaded bar at the point A of abscissa 2m (we have to find the relation between l, which is the distance between the elementary bar and the point A and x which sweeps the segment [0: 1]
Answer:
The correct answer is - 8.99N/C
Explanation:
[tex]dE=k\dfrac{dq}{x^2}\\ dq=\lambda{dx}\\ \lambda=2nC/m\\ dq=2dxnC\\ dE=k\dfrac{2dx}{x^2}\\ E=2k\int_1^2\dfrac{dx}{x^2}\\ E=2k(\frac{-1}{x})_1^2=k\times10^{-9}N/C\\ E=8.99\times10^9\times10^{-9}N/C\\ E=8.99N/C\\dE=k[/tex]
a point object is 10 cm away from a plane mirror while the eye of an observer(pupil diameter is 5.0 mm) is 28 cm a way assuming both eye and the point to be on the same line perpendicular to the surface find the area of the mirror used in observing the reflection of the point
Answer:
1.37 mm²
Explanation:
From the image attached below:
Let's take a look at the two rays r and r' hitting the same mirror from two different positions.
Let x be the distance between these rays.
[tex]d_o =[/tex] distance between object as well as the mirror
[tex]d_{eye}[/tex] = distance between mirror as well as the eye
Thus, the formula for determining the distance between these rays can be expressed as:
[tex]x = 2d_o tan \theta[/tex]
where; the distance between the eye of the observer and the image is:
[tex]s = d_o + d_{eye}[/tex]
Then, the tangent of the angle θ is:
[tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex]
replacing [tex]tan \theta = \dfrac{R}{d_o+d_{eye}}[/tex] into [tex]x = 2d_o tan \theta[/tex], we have:
[tex]x = 2d_o \Big( \dfrac{R}{d_o+d_{eye}}\Big)[/tex]
[tex]x = 2(10) \Big( \dfrac{0.25}{10+28}\Big)[/tex]
[tex]x = 20\Big( \dfrac{0.25}{38}\Big) cm[/tex]
x = (0.13157 × 10) mm
x = 1.32 mm
Finally, the area A = π r²
[tex]A = \pi(\frac{x}{2})^2[/tex]
[tex]A = \pi(\frac{1.32}{2})^2[/tex]
A = 1.37 mm²
5. Steve is driving in his car to take care of some errands. The first errand has him driving to a location 2 km East and 6 km North of his starting location. Once he completes that errand, he drives to the second one which is 4 km East and 2 km South of the first errand. What is the magnitude of the vector that describes how far the car has traveled from its starting point, rounded to the nearest km?
Answer:
gshshs
Explanation:
hshsksksksbsbbshd
A solenoid has a length , a radius , and turns. The solenoid has a net resistance . A circular loop with radius is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance . At a time , the solenoid is connected to a battery that supplies a potential . At a time , what current flows through the outer loop
This question is incomplete, the complete question is;
A solenoid has a length 11.34 cm , a radius 1.85 cm , and 1627 turns. The solenoid has a net resistance of 144.9 Ω . A circular loop with radius of 3.77 cm is placed around the solenoid, such that it lies in a plane whose normal is aligned with the solenoid axis, and the center of the outer loop lies on the solenoid axis. The outer loop has a resistance of 1651.6 Ω. At a time of 0 s , the solenoid is connected to a battery that supplies a potential 34.95 V. At a time 2.58 μs , what current flows through the outer loop?
Answer:
the current flows through the outer loop is 1.3 × 10⁻⁵ A
Explanation:
Given the data in the question;
Length [tex]l[/tex] = 11.34 cm = 0.1134 m
radius a = 1.85 cm = 0.0185 m
turns N = 1627
Net resistance [tex]R_{sol[/tex] = 144.9 Ω
radius b = 3.77 cm = 0.0377 m
[tex]R_o[/tex] = 1651.6 Ω
ε = 34.95 V
t = 2.58 μs = 2.58 × 10⁻⁶ s
Now, Inductance; L = μ₀N²πa² / [tex]l[/tex]
so
L = [ ( 4π × 10⁻⁷ ) × ( 1627 )² × π( 0.0185 )² ] / 0.1134
L = 0.003576665 / 0.1134
L = 0.03154
Now,
ε = d∅/dt = [tex]\frac{d}{dt}[/tex]( BA ) = [tex]\frac{d}{dt}[/tex][ (μ₀In)πa² ]
so
ε = μ₀n [tex]\frac{dI}{dt}[/tex]( πa² )
ε = [ μ₀Nπa² / [tex]l[/tex] ] [tex]\frac{dI}{dt}[/tex]
ε = [ μ₀Nπa² / [tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
I = ε/[tex]R_o[/tex] = [ μ₀Nπa² / [tex]R_o[/tex][tex]l[/tex] ] [ (ε/L)e^( -t/[tex]R_{sol[/tex]) ]
so we substitute in our values;
I = [ (( 4π × 10⁻⁷ ) × 1627 × π(0.0185)²) / (1651.6 ×0.1134) ] [ ( 34.95 / 0.03154)e^( -2.58 × 10⁻⁶ / 144.9 ) ]
I = [ 2.198319 × 10⁻⁶ / 187.29144 ] [ 1108.116677 × e^( -1.7805 × 10⁻⁸ )
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.116677 × 0.99999998 ]
I = [ 1.17374 × 10⁻⁸ ] × [ 1108.11665 ]
I = 1.3 × 10⁻⁵ A
Therefore, the current flows through the outer loop is 1.3 × 10⁻⁵ A
Answer:
1.28 *10^-5 A
Explanation:
Same work as above answer. Needs to be more precise
What is cubical expansivity of liquid while freezing
Answer:
"the ratio of increase in the volume of a solid per degree rise of temperature to its initial volume" -web
Explanation:
tbh up above ✅
Answer:
cubic meter
Explanation:
Increase in volume of a body on heating is referred to as volumetric expansion or cubical expansion
what is simple machine?
Explanation:
Those tools that helps to make our work easier ,faster and more convenient in our daily life it is called simple Machine.
A political campaign manager must decide whether to emphasize television advertisements or letters to potential voters in a reelection campaign. Describe the production function for campaign votes.
A. Campaign managers produce campaign votes.
B. Reelection campaigns produce campaign votes.
C. Television advertisements and campaign votes produce letters to potential voters.
D. Television advertisements and letters to potential voters produce reelection campaigns.
E. Television advertisements and letters to potential voters produce campaign votes.
How might information about this function (such as the shape of the isoquants) help the campaign manager plan strategy?
A. If the marginal rate of technical substitution of television advertisements for letters to potential voters is constant, then the campaign manager should use a combination of the two inputs.
B. If television advertisements and letters to potential voters are perfect complements, then the campaign manager should use them in fixed proportions.
C. If television advertisements and letters to potential voters are perfect substitutes, then the campaign manager should use them in fixed proportions.
D. If the isoquant curves for television advertisements and letters to potential voters are convex, then the campaign manager should use only the cheaper input per vote.
E. If the isocost lines for television advertisements and letters to potential voters are convex, then the campaign manager should use a combination of the two inputs.
Answer:
First answer - (E)
Second answer - (B)
Explanation:
The trade-off here is between TV ADVERTISEMENTS and LETTERS TO POTENTIAL VOTERS. The campaign manager for the candidate who is running for reelection, is trying to decide which of the two factors he should use more of or emphasize. The production function for campaign votes can be simplified as
TVAD + LTPV = CV
This is the production function for campaign votes.
PART A
Describe the production function for campaign votes (in words).
ANSWER: (E)
Television advertisements and (or 'plus') letters to potential voters, produce (or 'equal') campaign votes.
PART B
How might information about this function (such as the shape of the isoquants) help the campaign manager plan strategy?
ANSWER: (B)
If television advertisements and letters to potential voters are perfect complements (complements are goods or actions that 'must' go together or be used together) then the campaign manager should use them in fixed proportions (e.g. in a ratio of 50:50).
A kind of variable that a researcher purposely changes in investigation is
Answer:
independent variable
Explanation:
A 2.0 kg puck is at rest on a level table. It is pushed straight north with a constant force of 5N for 1.50 s and then let go. How far does the puck move from rest in 2.25 s?
Answer:
d = 6.32 m
Explanation:
Given that,
The mass of a puck, m = 2 kg
It is pushed straight north with a constant force of 5N for 1.50 s and then let go.
We need to find the distance covered by the puck when move from rest in 2.25 s.
We know that,
F = ma
[tex]a=\dfrac{F}{m}\\\\a=\dfrac{5}{2}\\\\a=2.5\ m/s^2[/tex]
Let d is the distance moved in 2.25 s. Using second equation of motion,
[tex]d=ut+\dfrac{1}{2}at^2\\\\d=0+\dfrac{1}{2}\times 2.5\times (2.25)^2\\\\d=6.32\ m[/tex]
So, it will move 6.32 m from rest in 2.25 seconds.
Each rarefraction on a longitudinal wave correspond to what point on a transverse wave?
