Answers
Answer;
a 2.00-kg body will vibrate at 0.707Hz
Answer:-7.9
Explanation:
Related Questions
A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?
Answers
Answer:
50 cm long
When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.
An airplane flies 1,592 miles east from Phoenix, Arizona, to Atlanta, Georgia, in 3.68 hours.
What is the average velocity of the airplane? Round your answer to the nearest whole number.
Answers
Answer:
433
Explanation:
A metal sphere A of radius a is charged to potential V. What will be its potential if it is enclosed by a spherical conducting shell B of radius b and the two are connected by a wire?
Answers
Answer:
The potential will be Va/b
Explanation:
So Let sphere A charged Q to potential V.
so, V= KQ/a. ....(1
Thus, spherical shell B is connected to the sphere A by a wire, so all charge always reside on the outer surface.
therefore, potential will be ,
V ′ = KQ/b = Va/b... That is from .....(1), KQ=Va]
UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons
Answers
Answer:
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
Explanation:
First we calculate the energy of photon:
E = hc/λ
where,
E = Energy of Photon = ?
h = Plank's Constant = 6.626 x 10⁻³⁴ J.s
c = speed of light = 3 x 10⁸ m/s
λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m
Therefore,
E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)
E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)
E = 10.35 eV
Now, from Einstein's Photoelectric equation we know that:
Energy of Photon = Work Function + K.E of Electron
10.35 eV = 4.82 eV + K.E
K.E = 10.35 eV - 4.82 eV
K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J
The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".
Kinetic energyAccording to the question,
Speed of light, c = 3 × 10⁸ m/s
Wavelength, λ = 120 nm or,
= 1.2 × 10⁻⁷ m
Plank's Constant, h = 6.626 × 10⁻³⁴ J.s
Now,
The energy of photon will be:
→ E = [tex]\frac{hc}{\lambda}[/tex]
By substituting the values,
= [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]
= [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]
= 10.35 eV
By using Einstein's Photoelectric equation,
Energy of Photon = Work function + K.E
10.35 = 4.82 + K.E
K.E = 10.35 - 4.82
= 5.53 eV or,
= 8.85 × 10⁻¹⁹ J
Thus the response above is correct.
Find out more information about Kinetic energy here:
https://brainly.com/question/25959744
if a 1-m diameter sewer pipe is flowing at a depth of 0.4 m and has a flow rate of 0.15 m^3/s, what will be the flow rate when the pipe flows full?
Answers
Answer:
0.35 m³/s
Explanation:
When the pipe's depth is 0.4 m, the area of the circular segment is:
A = ½ R² (θ − sin θ)
The depth of the water is:
h = R (1 − cos(θ/2))
Solving for θ:
0.4 = 0.5 (1 − cos(θ/2))
0.8 = 1 − cos(θ/2)
cos(θ/2) = 0.2
θ/2 = acos(0.2)
θ = 2 acos(0.2)
θ ≈ 2.74 rad
The area is therefore:
A = ½ (0.5 m)² (2.74 − sin 2.74)
A = 0.338 m²
The cross-sectional area when the pipe is full is:
A = π (0.5 m)²
A = 0.785 m²
The flow velocity is constant:
v = v
Q / A = Q / A
(0.15 m³/s) / (0.338 m²) = Q / (0.785 m²)
Q = 0.35 m³/s
At what speed, as a fraction of c, will a moving rod have a length 65% that of an identical rod at rest
Answers
Answer:
v/c = 0.76
Explanation:
Formula for Length contraction is given by;
L = L_o(√(1 - (v²/c²))
Where;
L is the length of the object at a moving speed v
L_o is the length of the object at rest
v is the speed of the object
c is speed of light
Now, we are given; L = 65%L_o = 0.65L_o, since L_o is the length at rest.
Thus;
0.65L_o = L_o[√(1 - (v²/c²))]
Dividing both sides by L_o gives;
0.65 = √(1 - (v²/c²))
Squaring both sides, we have;
0.65² = (1 - (v²/c²))
v²/c² = 1 - 0.65²
v²/c² = 0.5775
Taking square root of both sides gives;
v/c = 0.76
The ceiling of your lecture hall is probably covered with acoustic tile, which has small holes separated by about 6.1 mm. Using light with a wavelength of 578 nm, how far could you be from this tile and still resolve these holes
Answers
Answer:
8.65x10^3m
Explanation:
See attached file
what is the average flow rate in of gasoline to the engine of a plane flying at 700 km/h if it averages 100.0 km/l
Answers
Answer:
1.94cm³/s
Explanation:
1L = 1000cm³
Ihr = 3600s
So
Using
Average flow rate
Fr= 1L/100Km x 700Km/1hr x 1hr/3600s x 1000cm³/ 1L
= 1.94cm³/s
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form _______. A. pure elements B. molecules C. metals D. the periodic table
Answers
Answer:
Atoms are the particles that all matter is made from. When two or more kinds of atoms combine, they form pure elements
The answer is option A
Answer:
its molecues
Explanation:
A homeowner purchases insulation for her attic rated at R-15. She wants the attic insulated to R-30. If the insulation she purchased is 10 cm thick, what thickness does she need to use
Answers
Answer:
she need to use 20 cm thick
Explanation:
given data
wants the attic insulated = R-30
purchased = 10 cm thick
solution
as per given we can say that
10 cm is for the R 15
but she want for R 30
so
R 30 thickness = [tex]\frac{30}{15} \times 10[/tex]
R 30 thickness = 20 cm
so she need to use 20 cm thick
3. El tambor de una lavadora que gira a 3 000 revoluciones por minuto (rpm) se acelera uniformemente hasta que alcanza las 6 000 rpm, completando un total de 12 revoluciones.
d. Determina la aceleración tangencial, centrípeta y la total en m.s-2 cuando el tambor a alcanzado los 60000 rpm
e. Explica lo que ocurre con la magnitud y dirección de los vectores aceleración tangencial, aceleración centrípeta, aceleración total, aceleración angular, velocidad angular cuando la lavadora ha girado desde 3000 rpm hasta 6000 rpm.
Answers
Answer:
d) α = 1693.5 rad / s² , a = 392.7 m / s² , a_total = α √(R² +1) ,
e) tan θ = a / α
Explanation:
This is an exercise in linear and angular kinematics.
We initialize reduction of all the magnitudes to the SI system
w₀ = 3000 rev / min (2π rad / 1rev) (1min / 60s) = 314.16 rad / s
w = 6000 rev / mi = 628.32 rad / s
θ = 12 rev = 12 rev (2π rad / 1 rev) = 75.398 rad
d) ask for centripetal, tangential and total acceleration.
Let's start by looking for centripetal acceleration, let's use the formula
w² = w₀² + 2 α θ
α = (w²- w₀²) / 2θ
we calculate
α = (628.32²2 - 314.16²) / 2 75.398
α = 1693.5 rad / s²
the quantity is linear and angular are related
the linear or tangential acceleration is
a = α R
where R is the radius of the drum
a = 1693.5 R
Unfortunately you do not give the radius of the drum for a complete calculation, but suppose it is a washing machine drum R = 20 cm = 0.20 m
a = 1693.5 0.20
a = 392.7 m / s²
the total acceleration is
a_total = √(a² + α²)
a_total = √ (α² R² + α²)
a_total = α √(R² +1)
e) The centripetal acceleration is directed towards the center of the movement is radial and its magnitude is constant
Tangential acceleration is tangency to radius and its value varies proportionally radius
the total accelracicon is the result of the vector sum of the two accelerations and their directions given by trigonometry
tan θ = a / α
the angular velocity increases linearly when with centripetal acceleration
A sinusoidal sound wave moves through a medium and is described by the displacement wave function s(x, t) = 1.99 cos(15.2x − 869t) where s is in micrometers, x is in meters, and t is in seconds. (a) Find the amplitude of this wave. µm (b) Find the wavelength of this wave. cm (c) Find the speed of this wave. m/s (d) Determine the instantaneous displacement from equilibrium of the elements of the medium at the position x = 0.050 9 m at t = 2.94 ms. µm (e) Determine the maximum speed of a element's oscillatory motion. mm/s
Answers
Answer:
a) A = 1.99 μm , b) λ = 0.4134 m , c) v = 57.2 m / s , d) s = - 1,946 nm ,
e) v_max = 1,739 mm / s
Explanation:
A sound wave has the general expression
s = s₀ sin (kx - wt)
where s is the displacement, s₀ the amplitude of the wave, k the wave vector and w the angular velocity, in this exercise the expression given is
s = 1.99 sin (15.2 x - 869 t)
a) the amplitude of the wave is
A = s₀
A = 1.99 μm
b) wave spectrum is
k = 2π /λ
in the equation k = 15.2 m⁻¹
λ = 2π / k
λ = 2π / 15.2
λ = 0.4134 m
c) the speed of the wave is given by the relation
v = λ f
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 869 / 2π
f = 138.3 Hz
v = 0.4134 138.3
v = 57.2 m / s
d) To find the instantaneous velocity, we substitute the given distance and time into the equation
s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
s = 1.99 sin (0.77368 - 2.55486)
remember that trigonometry functions must be in radians
s = 1.99 (-0.98895)
s = - 1,946 nm
The negative sign indicates that it shifts to the left
e) the speed of the oscillating part is
v = ds / dt)
v = - s₀(-w) cos (kx -wt)
the maximum speed occurs when the cosines is 1
v_maximo = s₀w
v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
let's reduce to mm / s
v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
v_max = 1,739 mm / s
a) A is = 1.99 μm , b) λ is = 0.4134 m , c) v is = 57.2 m / s , d) s is = - 1,946 nm, e) v_max is = 1,739 mm / s
Calculation of Wavelength
When A sound wave has the general expression is:
Then, s = s₀ sin (kx - wt)
Now, where s is the displacement, Then, s₀ is the amplitude of the wave, k the wave vector, and w the angular velocity, Now, in this exercise the expression given is
s is = 1.99 sin (15.2 x - 869 t)
a) When the amplitude of the wave is
A is = s₀
Thus, A = 1.99 μm
b) When the wave spectrum is
k is = 2π /λ
Now, in the equation k = 15.2 m⁻¹
Then, λ = 2π / k
After that, λ = 2π / 15.2
Thus, λ = 0.4134 m
c) When the speed of the wave is given by the relation is:
Then, v = λ f
Now, the angular velocity and frequency are related is:
w is = 2π f
Then, f = w / 2π
After that, f = 869 / 2π
Now, f = 138.3 Hz
Then, v = 0.4134 138.3
Thus, v = 57.2 m / s
d) Now, To find the instantaneous velocity, When we substitute the given distance and time into the equation
Then, s = 1.99 sin (15.2 0.0509 - 869 2.94 10⁻³)
After that, s = 1.99 sin (0.77368 - 2.55486)
Then remember that trigonometry functions must be in radians
After that, s = 1.99 (-0.98895)
Thus, s = - 1,946 nm
When The negative sign indicates that it shifts to the left
e) When the speed of the oscillating part is
Then, v = ds / dt)
Now, v = - s₀(-w) cos (kx -wt)
When the maximum speed occurs when the cosines is 1
Then, v_maximo = s₀w
After that, v_maximum = 1.99 869
v_maximo = 1739.31 μm / s
Now, let's reduce to mm / s
Then, v_maxio = 1739.31 miuy / s (1 mm / 103 mu)
Therefore, v_max = 1,739 mm / s
Finf more informmation about Wavelength here:
https://brainly.com/question/6352445
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Answers
Answer:
[tex]I=2.71\times 10^{-5}\ A[/tex]
Explanation:
A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
Let given is,
The diameter of a parallel plate capacitor is 6 cm or 0.06 m
Separation between plates, d = 0.046 mm
The potential difference across the capacitor is increasing at 500,000 V/s
We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :
[tex]C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}[/tex], r is radius
Let I is the displacement current. It is given by :
[tex]I=C\dfrac{dV}{dt}[/tex]
Here, [tex]\dfrac{dV}{dt}[/tex] is rate of increasing potential difference
So
[tex]I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A[/tex]
So, the value of displacement current is [tex]2.71\times 10^{-5}\ A[/tex].
A clown 2 m tall looks at himself in a full-length mirror (floor-to-ceiling). Where in the mirror must he look to see his feet?
Answers
Answer:
Around the center of the mirror
In a two-slit experiment, the slit separation is 3.34 ⋅ 10 − 5 m. The interference pattern is created on a screen that is 3.30 m away from the slits. If the 7th bright fringe on the screen is 29.0 cm away from the central fringe, what is the wavelength of the light?
Answers
Answer:
The wavelength is [tex]\lambda = 419 \ nm[/tex]
Explanation:
From the question we are told that
The distance of separation is [tex]d = 3.34 *10^{-5} \ m[/tex]
The distance of the screen is [tex]D = 3.30 \ m[/tex]
The order of the fringe is n = 7
The distance of separation of fringes is y = 29.0 cm = 0.29 m
Generally the wavelength of the light is mathematically represented as
[tex]\lambda = \frac{y * d }{ n * D}[/tex]
substituting values
[tex]\lambda = \frac{0.29 * 3.34*10^{-5} }{ 7 * 3.30}[/tex]
[tex]\lambda = 4.19*10^{-7}\ m[/tex]
[tex]\lambda = 419 \ nm[/tex]
A stationary coil is in a magnetic field that is changing with time. Does the emf induced in the coil depend
Answers
Answer:
Explanation:
The e.m.f induced in the coil depend on the following :
(a) No. of turns in the coil
(b) Cross-sectional Area of the coil
(c) Magnitude of Magnetic field
(d) Angular velocity of the coil
How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F
Answers
Answer:
Explanation:
To convert from °C to °F , the formula is
( F-32 ) / 9 = C / 5
F is reading fahrenheit scale and C is in centigrade scale .
F = 205 , C = ?
(205 - 32) / 9 = C / 5
C = 96°C approx .
Let us calculate the heat required .
Total heat required = heat required to heat up the ice at - 20 °C to 0°C + heat required to melt the ice + heat required to heat up the water at 0°C to
96°C.
= 5 x 2.04 x (20-0) + 5 x 336 + 5 x ( 96-0 ) x 4.2 kJ .
= 204 + 1680 + 2016
= 3900 kJ .
Describe at least two unique characteristics of the new space telescope.
Answers
Answer: Astro 1 will have a 10x larger field of view than the Hubble.
Explanation: The hubble will also be extremely light weight that way it can go further into space and the mission will be able to last a longer amount of time.
A stone is dropped from the bridge, it takes 4s to reach the water. what's the height of the bridge?
Answers
Explanation:
Using Equations of Motion :
[tex]s = ut + \frac{1}{2} g {t}^{2} [/tex]
Height = 0 * 4 + 4.9 * 16
Height = 78.4 m
Find the momentum of a particl with a mass of one gram moving with half the speed of light.
Answers
Answer:
129900
Explanation:
Given that
Mass of the particle, m = 1 g = 1*10^-3 kg
Speed of the particle, u = ½c
Speed of light, c = 3*10^8
To solve this, we will use the formula
p = ymu, where
y = √[1 - (u²/c²)]
Let's solve for y, first. We have
y = √[1 - (1.5*10^8²/3*10^8²)]
y = √(1 - ½²)
y = √(1 - ¼)
y = √0.75
y = 0.8660, using our newly gotten y, we use it to solve the final equation
p = ymu
p = 0.866 * 1*10^-3 * 1.5*10^8
p = 129900 kgm/s
thus, we have found that the momentum of the particle is 129900 kgm/s
Which of the following describes wavelength?
A.
the height of a wave
B.
the distance between crests of adjacent waves
C.
the distance a wave travels in a given amount of time
D.
the number of waves that pass a point in a given amount of time
Answers
A car moving east at 45 km/h turns and travels west at 30 km/h. What is the
magnitude and direction of the change in velocity?
mahalle 1.11
Answers
Explanation:
Change in Velocity = final velocity - initial velocity
Change in velocity = 30km/h - (- 45km/h )
= 75 km/h due west
A motorist traveling with a constant speed of 15 m/s (about 34 mi/h) passes a school-crossing corner, where the speed limit is 0 m/s (about 22 mi/h). Just as the motorist passes, a police officer on a motorcycle stopped at the comer starts off in pursuit with constant acceleration of 3.0 m/S2. (a) How much time elapses before the officer catches up with the motorist? (b) What is the officer's speed at that point? (c) What is the total distance each vehicle has traveled at that point? Please help me
Answers
Answer:
(a) 10 s
(b) 30 m/s
(c) 150 m
Explanation:
The motorist's position at time t is:
x = 15t
The officer's position at time t is:
x = ½ (3) t² = 1.5 t²
(a) When they have the same position, the time is:
15t = 1.5 t²
t = 0 or 10 s
(b) The officer's speed is:
v = 3t
v = 30 m/s
(c) The position is:
x = 15t = 150 m
A block of mass M rests on a block of mass M1 which is on a tabletop. A light string passes over a frictionless peg and connects the blocks. The coefficient of kinetic friction between the blocks and between M1 and the tabletop is the same. A force F pulls the upper block to the left and the lower block to the right. The blocks are moving at a constant speed.
Determine the mass of the upper block. (Express your answer to three significant figures.)
Answers
Answer:
M = F/3μ g - M₁/3
Explanation:
To solve this exercise we must use the equilibrium conditions translations
∑ F = 0
In the attachment we can see a free body diagram of each block
Block M (upper)
X axis
fr₁ + F₂ -F = 0
F = fr₁ + F₂ (1)
axis
N₁-W = 0
N₁ = Mg
the friction force has the formula
fr₁ = μ N₁
F = μ Mg + F₂
bottom block
X axis
F₂ - fr₁ - fr₂ = 0
F₂ = fr₁ + fr₂
Y axis
N - W₁ -W = 0
N = g (M + M₁)
we substitute
F₂ = μ Mg + μ (M + M1) g
F₂ = μ g (2M + M₁)
we substitute in 1
F = μ M g + μ g (2M + M₁)
F = μ g (3M + M₁)
we look for mass M
M = (F - μ g M₁)/ 3μ g
M = F/3μ g - M₁/3
the exercise does not have numerical data
The AB rope is fixed to the ground at its A end, and forms 30º with the vertical. Its other end is connected to two ropes by means of the B-ring of negligible weight. The vertical rope supports the E block and the other rope passes through the grounded articulated pulley C to join at its end to the 80 N weight block D. The inclined section of the BD rope forms 60º with the vertical one; determine the weight of the E block necessary for the balance of the system and calculate the tension in the AB rope.
Answers
Answer:
T = 80√3 N ≈ 139 N
W = 160 N
Explanation:
Sum of forces on B in the x direction:
∑F = ma
80 N sin 60° − T sin 30° = 0
T = 80 N sin 60° / sin 30°
T = 80√3 N
T ≈ 139 N
Sum of forces on B in the y direction:
∑F = ma
80 N cos 60° + T cos 30° − W = 0
W = 80 N cos 60° + T cos 30°
W = 40 N + 120 N
W = 160 N
a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?
Answers
Answer:
8.78 AmpsExplanation:
Given data:
power rating of the heater P= 1010 W
voltage of the heater V= 115 volts
current taken by the heater I= ?
We can apply the power formula to solve for the current in the heater
i.e P= IV
Making I the current subject of formula we have
I= P/V
Substituting our given data into the expression for I we have
I=1010/115= 8.78 A
Hence the current when the unit/heater is operating is 8.78 AmpInside the wall of a house, an L-shaped section of hot-water pipe consists of three parts: a straight, horizontal piece 28.0 cm long, an elbow, and a straight vertical piece ℓ = 159 cm long. A stud and a second-story floorboard hold the ends of this section of copper pipe stationary. Find the magnitude and direction of the displacement of the pipe elbow when the water flow is turned on, raising the temperature of the pipe from 18.0°C to 40.2°C. (The coefficient of linear expansion of copper is
Answers
Answer:
The magnitude and direction are
7.638×10-4m
80.01°
Explanation:
We know that the coefficient of linear expansion for copper = 16.6×10^-6 m/m-C
ΔT = 40.2 - 18.0 = 28.5 C°
The expansion of horizontal pipe length can be calculated as
= (0.28)(16.6×10^-6)(28.5) = 13247×10^-8
= 0.0001325 m
The expansion of vertical pipe length = (1.28)(16.6×10^-6)(28.5) = 60557×10^-8 = 0.000752229 m
horizontal displacement = 0.1325 mm
= 1.356×10^-4m
vertical displacement = 0.75223mm
=7.5223×10-4m
size of total displacement can be calculated as
√(x²+y²)
Where x and y are vertical and horizontal displacement respectively
= √(0.1325)²+(0.75223)² =
= 0.7638 mm
= 7.638×10-4m
Angle below horizontal = arctan Θ
= 0.75223/0.1325
=5.6772
= arctan (5.6772)
= 80.01°
Therefore, the the magnitude and direction of the displacement of the pipe elbow when the water flow is turned at (7.638×10-4m) 0.7638 mm and 80.01°
A spring is hung from the ceiling. When a block is attached to its end, it stretches 2.5 cm before reaching its new equilibrium length. The block is then pulled down slightly and released. What is the frequency of oscillation?
Answers
Answer:
0.99Hz
Explanation:
Using F= -mx ( spring force)
At equilibrium the gravitational force will be balanced by the spring force so mg= kx
K= mg/ 0.25 N/m
But
Frequency f= 1/2pi √g/0.25
Frequency is 0.99Hz
The block is pulled down slightly and released so, Frequency of oscillation is 3.15 Hz
Frequency of oscillation based problem:What information do we have?
Length starched = 2.5 cm
F = Kx
We know that
F = mg
So,
mg = Kx
K/m = g/x
[tex]f=\frac{1}{2\pi}\sqrt{\frac{g}{x} }\\f=\frac{1}{2\pi}\sqrt{\frac{9.8}{0.025} }[/tex]
Frequency of oscillation = 3.15 Hz
Find out more information about 'Oscillation'.
https://brainly.com/question/16016711?referrer=searchResults
The metal wire in an incandescent lightbulb glows when the light is switched on and stops glowing when it is switched off. This simple
process is which kind of a change?
OA a physical change
OB. a chemical change
OC. a nuclear change
OD
an ionic change
Answers
B. A chemical change
Explanation:
I'm guessing ?
An inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA. Part A What is the resistance RR of the inductor
Answers
i
CHECK COMPLETE QUESTION BELOW
inductor is connected to the terminals of a battery that has an emf of 12.0 VV and negligible internal resistance. The current is 4.96 mAmA at 0.800 msms after the connection is completed. After a long time the current is 6.60 mAmA.
Part A)What is the resistance RR of the inductor
PART B) what is inductance L of the conductor
Answer:
A)R=1818.18 ohms
B)L=1.0446H
Explanation:
We were given inductor L with resistance R , there is a connection between the battery and the inductor with Emf of 12V, we can see that the circuit is equivalent to a simple RL circuit.
There is current of 4.96mA at 0.8ms, at the end of the connection the current increase to 6.60mA,
.
a)A)What is the resistance RR of the inductor?
The current flowing into RL circuit can be calculated using below expresion
i=ε/R[1-e⁻(R/L)t]
at t=∞ there is maximum current
i(max)= ε/R
Where ε emf of the battery
R is the resistance
R=ε/i(max)
= 12V/(6.60*10⁻³A)
R=1818.18 ohms
Therefore, the resistance R=1818.18 ohms
b)what is inductance L of the conductor?
i(t=0.80ms and 4.96mA
RT/L = ⁻ln[1- 1/t(max)]
Making L subject of formula we have
L=-RT/ln[1-i/i(max)]
If we substitute the values into the above expresion we have
L= -(1818.18 )*(8.0*10⁻⁴)/ln[1-4.96/6.60)]
L=1.0446H
Therefore, the inductor L=1.0446H
