Answer:
entropy
Explanation:
A uniform disk with mass 43.9 kgkg and radius 0.280 mm is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 29.0 NN is applied tangent to the rim of the disk. Part A What is the magnitude vv of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.400 revolution
Answer:
1.36 m/s
Explanation:
I = ½mR²
τ = FR
α = τ/I = FR / (½mR²) = 2F/mR
a = Rα = 2F/m
s = θR
v² = u² + 2as
u = 0
v = √2as = √(2(2F/m)(θR)) = 2√(FθR/m)
v = 2√(29.0(0.400)(2π)(0.280) / 43.9) = 1.3636272...
A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?
Answer:
3.5 m/s
Explanation:
There are 3 forces on the cyclist:
Weight force mg pulling down,
Normal force N pushing up,
and friction force Nμ pushing left.
Sum of forces in the y direction:
∑F = ma
N − mg = 0
N = mg
∑F = ma
-Nμ = ma
-mgμ = ma
a = -gμ
a = -(10 m/s²)(0.60)
a = -6 m/s²
Velocity reached at end of 11 m:
v² = v₀² + 2aΔx
v² = (12 m/s)² + 2 (-6 m/s²) (11 m)
v = √12 m/s
v ≈ 3.5 m/s
A lab technician uses laser light with a wavelength of 650 nmnm to test a diffraction grating. When the grating is 42.0 cmcm from the screen, the first-order maxima appear 6.09 cmcm from the center of the pattern. How many lines per millimeter does this grating have?
Answer:
221 lines per millimetre
Explanation:
We know that for a diffraction grating, dsinθ =mλ where d = spacing between grating, θ = angle to maximum, m = order of maximum and λ = wavelength of light.
Since the grating is 42.0 cm from the screen and its first order maximum (m = 1) is at 6.09 cm from the center of the pattern,
tanθ = 6.09 cm/42.0 cm = 0.145
From trig ratios, cot²θ + 1 = cosec²θ
cosecθ = √((1/tanθ)² + 1) = √((1/0.145)² + 1) = √48.562 = 6.969
sinθ = 1/cosecθ = 1/6.969 = 0.1435
Also, sinθ = mλ/d at the first-order maximum, m = 1. So
sinθ = (1)λ/d = λ/d
Equating both expressions we have
0.1435 = λ/d
d = λ/0.1435
Now, λ = 650 nm = 650 × 10⁻⁹ m
d = 650 × 10⁻⁹ m/0.1435
d = 4529.62 × 10⁻⁹ m per line
d = 4.52962 × 10⁻⁶ m per line
d = 0.00452962 × 10⁻³ m per line
d = 0.00452962 mm per line
Since d = width of grating/number of lines of grating
Then number of lines per millimetre = 1/grating spacing
= 1/0.00452962
= 220.77 lines per millimetre
≅ 221 lines per millimetre since we can only have a whole number of lines.
A particle moves along the x axis. In order to calculate the torque on the particle, you need to know:
a. the rotational inertia of the particle
b. the velocity of the particle
c. the mass of the particle
d. the kinetic energy of the particle
e. the point about which the torque is to be calculated
Answer:
e. the point about which the torque is to be calculated
Explanation:
torque is the product of a force and a distance
the point about which the torque is calculated is required to know the distance.
None of the other terms are relevant as they refer to mass or its equivalent, and velocity. Force is not mentioned in any of them.
A particle moves along the x-axis. In order to calculate the torque on the particle, you need to know the point about which the torque is to be calculated. Therefore, option E is correct.
What is torque ?The rotating equivalent of force is torque. Depending on the subject of study, it is also known as the moment, moment of force, rotating force, or turning effect. It illustrates how a force can cause a change in the body's rotational motion.
Ancient Romans gave these necklaces the term "torque" by describing them as twisted and spiral screw-shaped using the Latin word "torquere," which also means "twisting" and "turning."
It's critical to realize that torque, which has to do with your motor's power in terms of rotational force, is not the same thing as speed. Find a motor with a top speed if you require more motor speed, and a motor with a motor torque that is maximized if you need more rotational force.
Thus, option E is correct.
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A 500 nm wavelength light illuminates a soap film with an index of refraction 1.33 to make it look bright. If the beam of light is incident normal on the film, what is the minimum thickness of the film
Answer:
t(min) = 94nm
Explanation:
The wavelength of the light incident is 500 nm.
The refractive index of the film is 1.33.
The minimum thickness of the soap film required for constructive interference.
The thickness of the film for the constructive interference is given by:
2*t= (m + 1/2) λ′
Now, λ′ = λ/μ = 500/1.33 = 376nm
The minimum thickness of the film
′t′ will be at m=0 :
2*t(min) = (0 + 1/2) 376
t(min) = 94nm
A/An is a type of blood cell that's also called a red blood cell. a) Jeukocyte O b) thrombocyte c) plasma d) erythrocyte
Answer:
red blood cell, also called erythrocyte
Explanation:
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How could a country benefit from making it into space?
Answer:
space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First what are the mass units?
Answer:
The mas unit is the the 'Kilogram' written as 'kg'
Volume is 10 L
Explanation:
The complete question is
If you have a density of 100 kg/L, and a mass of 1000 units, tell me the following: First, what are the mass units?
Second, what is the volume
mass units is the 'Kilogram', written as 'kg'
density = mass/volume = 100 kg/L
the mass = 1000 kg
volume = mass/density = 1000/100 = 10 L
What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from its siren? The speed of sound on this day is 345 m/s. Group of answer choices
Answer:
check photo for solve
Explanation:
What is the radiation pressure 1.5 m away from a 700 W lightbulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.
Answer:
3.30 x 10^-7 Pascal
Explanation:
distance r = 1.5 m
power P = 700 W
the radiation pressure is given as
Pr = P/A*c
where
area of the surface A = 4πr^2
calculate for A
speed of light is c = 3×10^8 m/s
plugging above values in equation above gives
Pr = 3.30 x 10^-7 Pascal
Determine the point of contraflexure
Answer:
The point of contraflexure (PoC) occurs where bending is zero and at the point of change between positive and negative (or between compression and tension). In a beam that is flexing (or bending), the point where there is zero bending moment is called the point of contraflexure.
Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?
Answer:
θ = 45.15°
Explanation:
We need to use the grating equation in this question. The grating equation is given as follows:
mλ = d Sin θ
where,
m = order number = 3
λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m
d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m
θ = angle from the axis = ?
Therefore,
(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ
Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)
Sin θ = 0.709
θ = Sin⁻¹(0.709)
θ = 45.15°
A diffraction grating has 6000 lines per centimeter ruled on it. What is the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe when the grating is illuminated with a beam of light of wavelength 500 nm
Explanation:
Hope it Will help he hsuejwoamxgehanwpalasmbwfwfqoqlmdbehendalmZbgevzuxwllw. yeh we pabdvddxhspapalw. X
The angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
What is diffraction?Waves spreading outward around obstructions are known as diffraction. Sound, electromagnetic radiation like light, X-rays, and gamma rays, as well as very small moving particles like atoms, neutrons, and electrons that exhibit wavelike qualities all exhibit diffraction.
Given:
The number of lines = 6000 per cm,
The Wavelength, λ = 500 nm = 500 × 10 ⁻⁹ m
Calculate the diffraction grating,
[tex]d = 1 / no\ of\ lines[/tex]
d = 10⁻² / 6000 m,
Calculate the second-order maxima angle and third-order maxima angle by the formula given below,
[tex]dsin\theta_1 = n_1 \lambda[/tex]
[tex]sin\theta_1 = n_1\lambda / d[/tex]
[tex]\theta _1 = sin^{-1}[2\times 500\times 10 ^{-9}/10^{-2}\times 6000][/tex]
θ₁ = sin⁻¹(0.6)
θ₁ = 36.87°
Similarly, for θ₂,
θ₂ = sin⁻¹(3 × 500 × 10 ⁻⁹ / 10⁻² × 6000)
θ₂ = sin⁻¹(0.9)
θ₂ = 64.16°
Calculate the separation as follows,
θ₂ - θ₁ = 64.16° - 36.87°
θ₂ - θ₁ = 27.29°
Therefore, the angular separation (in degrees) between the second and the third orders on the same side of the central bright fringe if the wavelength is 500 nm and A diffraction grating has 6000 lines per centimeter ruled on it, is 27.29°.
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Categorize each ray tracing statement as relating to ray 1, ray 2, or ray 3.
A. Drawn from the top of the object so that it passes through the center of the lens at the optical axis.
B. Drawn from the top of the object so that it passes through the focal point on the same side of the lens as the object.
C. Drawn parallel to the optical axis from the top of the object.
D. Ray bends parallel to the optical axis.
E. Ray bends so that it passes through the focal point on the opposite side of the lens as the object.
F. Ray does not bend.
Answer:
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
Explanation:
In this exercise you are asked to relate each with the answers
In general, in the optics diagram,
* Ray 1 is a horizontal ray that after stopping by the optical system goes to the focal point
* Ray 2 is a ray that passes through the intercept point between the optical axis and the system and does not deviate
* Ray 3 is a ray that passes through the focal length and after passing the optical system, it comes out horizontally.
With these statements, let's review the answers
statement 1 with answer C
statement 2 with answer F
statement 3 with answer B
Statement 1 with E
Statement 2 with A
Statement 3 with D
How much energy is required to accelerate a spaceship with a rest mass of 121 metric tons to a speed of 0.509 c?
Answer
1.07E22 Joules
Explanation;
We know that mass expands by a factor
=>>1/√[1-(v/c)²]
But v= 0.509c
So
1/√(1 - 0.509²)
=>>> 1/√(1 - 0.2591)
= >> 1/√(0.7409) = 1.16
But given that 121 tons is rest mass so 121- 1.16= 119.84 tons is kinetic energy
And we know that rest mass-energy equivalence is 9 x 10^19 joules per ton.
So Multiplying by 119.84
Kinetic energy will be 1.07x 10^22 joules
A wooden ice box has a total area of 1.50 m2 amd walls with an average thickness of 2.0 cm. The box contains ice at 0.0 oC. The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the shade of tree at 29 oC. (Assume the thermal conductivity of wood is 0.16 kJ/s m oC
Answer:
m = 9.1 x 10⁶ kg
Explanation:
First, we need to find the rate of heat transfer through the box to the ice. For this purpose, we use Fourier's Law of Heat Conduction:
Q = KA ΔT/L
where,
Q = Rate Of Heat Transfer = ?
K = Thermal Conductivity = 0.16 KW/m.°C = 160 W/m.°C
A = Area = 1.5 m²
ΔT = Difference in Temperature = 29°C - 0°C = 29°C
L = Thickness of wall = 2 cm = 0.002 m
Therefore,
Q = (160 W/m °C)(1.5 m²)(29°C)/(0.002 m)
Q = 3.48 x 10⁶ W
Now, we find the amount of heat transferred in one day to the ice:
q = Qt
where,
q = amount of heat = ?
t = time = (1 day)(24 h/1 day)(3600 s/1 h) = 86400 s
Therefore,
q = (3.48 x 10⁶ W)(8.64 x 10⁴ s)
q = 3 x 10¹¹ J
Now, for mass of ice melted in a day:
q = m H
m = q/H
where,
m = mass of ice melted in a day = ?
H = latent heat of fusion of ice = 3.3 x 10⁵ J/kg
Therefore,
m = (3 x 10¹¹ J)/(3.3 x 10⁵ J/kg)
m = 9.1 x 10⁶ kg
What would happen in a State if its citizens lack relevant knowledge, skills
and positive attitude?
Answer:
If the older generation is lacking, the younger generation would likely have knowledge, skill, or a positive attitude in some combination, but it is relative to the culture.
The simple reason is the desirability for genetic variation using recessive genes.
In other words, if the older generation lacks something, it tends to be something they don’t need, but something that will look good on young people. But mostly relative to the culture and education system.
Hope this helps
You have a horizontal grindstone (a disk) that is 95 kg, has a 0.38 m radius, is turning at 87 rpm (in the positive direction), and you press a steel axe against the edge with a force of 16 N in the radial direction.
(a) Assuming the kinetic coefficient of friction between steel and stone is 0.20, calculate the angular acceleration of the grindstone.
(b) How many turns will the stone make before coming to rest?
Answer:
Explanation:
The moment of inertia of the disk I = 1/2 m R² where R is radius of the disc and m is its mass .
putting the values
I = .5 x 95 x .38²
= 6.86 kg m²
n = 87 rpm = 87 / 60 rps
n = 1.45 rps
angular velocity ω = 2π n , n is frequency of rotation .
= 2 x 3.14 x 1.45
= 9.106 radian /s
frictional force = 16 x .2
= 3.2 N
torque created by frictional force = 3.2 x .38
= 1.216 N.m
angular acceleration = torque / moment of inertia
= - 3.2 / 6.86
α = - 0.4665 rad /s²
b ) ω² = ω₀² + 2 α θ , where α is angular acceleration
0 = 9.106² - 2 x .4665 θ
θ = 88.87 radian
no of turns = 88.87 / 2π
= 14.15 turns
A person, with his ear to the ground, sees a huge stone strike the concrete pavement. A moment later two sounds are heard from the impact: one travels in the air and the other in the concrete, and they are 0.50 s apart. The speed of sound in air is 343 m/s, and in concrete is 3000 m/s.
Required:
How far away did the impact occur?
Answer:
The distance is [tex]d = 193.6 \ m[/tex]
Explanation:
From the question we are told that
The time interval between the sounds is k[tex]t_1 = k + t_2[/tex] = 0.50 s
The speed of sound in air is [tex]v_s = 343 \ m/s[/tex]
The speed of sound in the concrete is [tex]v_c = 3000 \ m/s[/tex]
Generally the distance where the collision occurred is mathematically represented as
[tex]d = v * t[/tex]
Now from the question we see that d is the same for both sound waves
So
[tex]v_c t = v_s * t_1[/tex]
Now
So [tex]t_1 = k + t[/tex]
[tex]v_c t = v_s * (t+ k)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]3000 t = 343* (t+ 0.50)[/tex]
=> [tex]t = 0.0645 \ s[/tex]
So
[tex]d = 3000 * 0.0645[/tex]
[tex]d = 193.6 \ m[/tex]
The following equation is an example of
decay.
181
185
79
Au →
4
2
He+
Answer:
Alp decay.
Explanation:
From the above equation, the parent nucleus 185 79Au produces a daughter nuclei 181 77 Ir.
A careful observation of the atomic mass of the parent nucleus (185) and the atomic mass of the daughter nuclei (181) shows that the atomic mass of the daughter nuclei decreased by a factor of 4. Also, the atomic number of the daughter nuclei also decreased by a factor of 2 when compared with the parent nucleus as shown in the equation given above.
This simply means that the parent nucleus has undergone alpha decay which is represented with a helium atom as 4 2He.
Therefore, the equation is an example of alpha decay.
g A certain elevator cab has a total run of 195 m and a maximum speed is 306 m/min, and it accelerates from rest and then back to rest at 1.19 m/s2. (a) How far does the cab move while accelerating to full speed from rest
Answer:
About 23 meters
Explanation:
To do this, you'll want to apply one of the kinematic equations to find the time it takes for the cabin to reach max velocity from rest. (Use the max velocity as V_f and V_i=0)
Then, you can find the distance travelled during the acceleration by equating the acceleration to the change in distance of the time squared.
My work is in the attachment, comment if you have any questions.
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
Answer:
8.65x10^3m
Explanation:
See attached file
A rectangular coil lies flat on a horizontal surface. A bar magnet is held above the center of the coil with its north pole pointing down. What is the direction of the induced current in the coil?
Answer:
There is no induced current on the coil.
Explanation:
Current is induced in a coil or a circuit, when there is a break of flux linkage. A break in flux linkage is caused by a changing magnetic field, and must be achieved by a relative motion between the coil and the magnet. Holding the magnet above the center of the coil will cause no changing magnetic filed since there is no relative motion between the coil and the magnet.
?a wire is stretched 30% what is the percentage change in resistance
Answer:
The percentage change in resistance of the wire is 69%.
Explanation:
Resistance of a wire can be determined by,
R = (ρl) ÷ A
Where R is its resistance, l is the length of the wire, A its cross sectional area and ρ its resistivity.
When the wire is stretched, its length and area changes but its volume and resistivity remains constant.
[tex]l_{o}[/tex] = 1.3l, and [tex]A_{o}[/tex] = [tex]\frac{A}{1.3}[/tex]
So that;
[tex]R_{o}[/tex] = (ρ[tex]l_{o}[/tex]) ÷ [tex]A_{o}[/tex] = (ρ × 1.3l) ÷ ([tex]\frac{A}{1.3}[/tex])
= (1.3lρ) ÷ ([tex]\frac{A}{1.3}[/tex])
= [tex](1.3)^{2}[/tex] × [(ρl) ÷ A]
= 1.69R (∵ R = (ρl) ÷ A)
[tex]R_{o}[/tex] = 1.69R
Where [tex]R_{o}[/tex] is the new resistance, [tex]l_{o}[/tex] is the new length, and [tex]A_{o}[/tex] is the new area after stretching the wire.
The change in resistance of the wire = [tex]R_{o}[/tex] - R
= 1.69R - 1R
= 0.69R
The percentage change in resistance = [tex]\frac{0.69R}{R}[/tex] × 100
= 0.69 × 100
= 69%
The percentage change in resistance of the wire is 69%.
coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second later. assume g is 10m/s. how far below the top of the tower des coin 1 pass coin 2
The height below the tower at which coin 1 pass coin 2 is 89.04 m.
The given parameters:
height of the tower, h = 100 m
initial velocity of coin 1, v = 15 m/s
time spent in air by coin 1 before coin 2 was dropped = 2s
To find:
the height below the tower at which coin 1 passes coin 2Find the maximum height attained by coin 1 before falling to the ground:
[tex]v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m[/tex]
Find the time taken for coin 1 to fall to the ground:
Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m
[tex]t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s[/tex]
But the time taken for the coin 1 to reach 11.25 m above the tower:
[tex]t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s[/tex]
Total time spent by coin 1 before reaching ground with respect to coin 2:
time = (1.5 s + 4.72 s) - 2 s
time = 4.22 s
Note: the 2 s was subtracted to keep both coins at a fair starting time below the tower.
Find the total time taken for coin 2 to fall to the ground:
Height of coin 2 above the ground = 100 m
Total time taken by coin 2 before falling to the ground is calculated as:
[tex]t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 = 4.47s[/tex]
The time at which coin 1 will pass coin 2 is 4.22 s.
Find the height below the tower when the time is 4.22 s.
[tex]h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m[/tex]
Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.
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Two protons, A and B, are next to an infinite plane of positive charge. Proton B is twice as far from the plane as proton A. Which proton has the larg
Answer:
They both have the same acceleration
Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest wavelength in the series.
Answer and Explanation:
The computation of the shortest wavelength in the series is shown below:-
[tex]\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
Where
[tex]\lambda[/tex] represents wavelength
R represents Rydberg's constant
[tex]n_f[/tex] represents Final energy states
and [tex]n_i[/tex] represents initial energy states
Now Substitute is
[tex]1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )[/tex]
now we will put the values into the above formula
[tex]= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )[/tex]
[tex]= 1218888.889 m^{-1}[/tex]
Now we will rewrite the answer in the term of [tex]\lambda[/tex]
[tex]\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m[/tex]
So, the whole Paschen series is in the part of the spectrum.
A mass M slides downward along a rough plane surface inclined at angle \Theta\:Θ= 32.51 in degrees relative to the horizontal. Initially the mass has a speed V_0\:V 0 = 6.03 m/s, before it slides a distance L = 1.0 m down the incline. During this sliding, the magnitude of the power associated with the work done by friction is equal to the magnitude of the power associated with the work done by the gravitational force. What is the coefficient of kinetic friction between the mass and the incline?
Answer: μ = 0.8885
Explanation: Force due to friction is calculated as: [tex]F_{f} = \mu.N[/tex]
At an inclined plane, normal force (N) is: N = mgcosθ, in which θ=32.51.
Power associated with work done by friction is [tex]P=F_{f}.x[/tex]. The variable x is displacement the object "spent its energy".
Power associated with work done by gravitational force is P = mghcosθ, where h is height.
The decline forms with horizontal plane a triangle as draw in the picture.
To determine force due to friction:
[tex]F_{f}.x=mghcos(\theta)[/tex]
[tex]F_{f}=\frac{mghcos(\theta)}{x}[/tex]
Replacing force:
[tex]\frac{m.g.h.cos(\theta)}{x} = \mu.m.g.cos(\theta)[/tex]
[tex]\mu=\frac{h}{x}[/tex]
Calculating h using trigonometric relations:
[tex]sin(32.51) = \frac{h}{1}[/tex]
h = sin(32.51)
Coefficient of Kinetic friction is
[tex]\mu=\frac{sin(32.51)}{1}[/tex]
μ = 0.8885
For these conditions, coefficient of kinetic friction is μ = 0.8885.
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.
Answer:
433
Explanation:
An LR circuit consists of a 35-mH inductor, a resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed
Answer:
Current, I = 1.23 A
Explanation:
Given that,
Inductance, L = 35 mH
Resistance, R = 12 ohms
Potential difference, V = 18 V
We need to find current 5 ms after the switch is closed. Current in LR circuit is given by :
[tex]I=I_o(1-e^{-t/\tau })[/tex] ....(1)
Here,
[tex]I_o[/tex] is final current
[tex]I_o=\dfrac{V}{R}\\\\I_o=\dfrac{18}{12}=1.5\ A[/tex]
[tex]\tau[/tex] is time constant
[tex]\tau=\dfrac{L}{R}\\\\\tau=\dfrac{35\times 10^{-3}}{12}\\\\\tau=0.00291\ s[/tex]
So, equation (1) becomes :
[tex]I=1.5\times (1-e^{-5\times 10^{-3}/0.00291})\\\\I=1.23\ A[/tex]
So, after 5 ms the current in the circuit is 1.23 A.