Answer:
It depends on the rate at which the current through it is changing.
Explanation:
As per the Faraday's law, the induced emf is given by :
[tex]\epsilon=-L\dfrac{di}{dt}[/tex]
Where
L is the inductance of the inductor
[tex]\dfrac{di}{dt}[/tex] is the rate of change of current
So, the self-induced emf of an inductor depends on the rate at which the current through it is changing. Hence, the correct option is (d).
A peg is located a distance h directly below the point of attachment of the cord. If h = 0.760 L, what will be the speed of the ball when it reaches the top of its circular path about the peg?
Answer:
Explanation:
Given
Pivot is at h=0.76 L
Where L is the length of String
Conserving Energy at A and B
[tex]mgL=\frac{1}{2}mu^2[/tex]
where u=velocity at bottom
[tex]u=\sqrt{2gL}[/tex]
After coming at bottom the ball completes the circle with radius r=L-0.76 L
Suppose v is the velocity at the top
Conserving Energy at B and C
[tex]\frac{1}{2}mu^2=mg(2r)+\frac{1}{2}mv^2[/tex]
Eliminating m
[tex]u^2=4r+v^2[/tex]
[tex]v^2=u^2-4\cdot gr[/tex]
[tex]v^2=2gL-4g(L-0.76L)[/tex]
[tex]v^2=1.04gL[/tex]
[tex]v=\sqrt{1.04gL}[/tex]
A Machine part is undergoing SHM with a frequency of 5.00 Hz and amplitude 1.80 CM, how long does it take the part to go from x=0 to x=-1.80 cm?
It takes 0.05 seconds for a part undergoing simple harmonic motion with a frequency of 5.00 Hz to move from x=0 to x=-1.80 cm.
Explanation:The student is asking how long it takes for a machine part undergoing simple harmonic motion (SHM) with a frequency of 5.00 Hz and amplitude of 1.80 cm to move from its equilibrium position (x=0) to its maximum negative displacement (x=-1.80 cm).
In SHM, the period (T) is the time required for one complete cycle of the motion. The period can be calculated using the formula T = 1/f, where f is the frequency. For a frequency of 5.00 Hz, the period is T = 1/5.00 Hz = 0.20 s. In half of a period, the part moves from one extreme to the other, passing through the equilibrium position at the quarter-period. Thus, to go from x=0 to x=-1.80 cm, it takes a quarter of this period, or (1/4)T = 0.05 s.
Therefore, it takes 0.05 seconds for the part to travel from x=0 to x=-1.80 cm while undergoing SHM.
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A skater has rotational inertia 4.2 kg.m2 with his fists held to his chest and 5.7 kg.m2 with his arms outstretched. The skater is spinning at 3.0 rev/s while holding a 2.5-kg weight in each outstretched hand; the weights are 76 cm from his rotation axis. If he pulls his hands in to his chest, so they’re essentially on his ro- tation axis, how fast will he be spinning?
Answer:
6.13428 rev/s
Explanation:
[tex]I_f[/tex] = Final moment of inertia = 4.2 kgm²
I = Moment of inertia with fists close to chest = 5.7 kgm²
[tex]\omega_i[/tex] = Initial angular speed = 3 rev/s
[tex]\omega_f[/tex] = Final angular speed
r = Radius = 76 cm
m = Mass = 2.5 kg
Moment of inertia of the skater is given by
[tex]I_i=I+2mr^2[/tex]
In this system the angular momentum is conserved
[tex]L_f=L_i\\\Rightarrow I_f\omega_f=I_i\omega_i\\\Rightarrow \omega_f=\dfrac{I_i\omega_i}{I_f}\\\Rightarrow \omega_f=\dfrac{(5.7+2\times 2.5\times 0.76^2)3}{4.2}\\\Rightarrow \omega_f=6.13428\ rev/s[/tex]
The rotational speed will be 6.13428 rev/s
Final answer:
To solve this problem, we need to use the principle of conservation of angular momentum. The skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.
Explanation:
To solve this problem, we need to use the principle of conservation of angular momentum. According to this principle, the initial angular momentum of the skater with his arms extended is equal to the final angular momentum when his hands are brought to his chest.
The initial angular momentum is given by the equation:
Li = Iiωi
where Li is the initial angular momentum, Ii is the initial moment of inertia, and ωi is the initial angular velocity. The final angular momentum is given by:
Lf = Ifωf
where Lf is the final angular momentum, If is the final moment of inertia, and ωf is the final angular velocity.
Since the skater is pulling his hands in to his chest, his moment of inertia decreases, and his angular velocity increases. We can set up the following equation:
Iiωi = Ifωf
Substituting the given values, we have:
(5.7 kg.m2)(2π(3.0 rev/s)) = (4.2 kg.m2)ωf
Solving for ωf, we find that the skater will be spinning at approximately 8.0 rev/s when his hands are brought to his chest.
As a comet approaches the Sun, it arcs around the Sun more rapidly. Why?
Answer:
Explanation:
According to Kepler's law a radius vector joining planet and sun swept equal area in equal interval of time thus it can be applied for comets.
when a comet is nearer to sun it has to swept more area so its velocity is more nearer to the sun ,
The basics of this formula comes from conservation of angular momentum thus comet moves faster when it approaches the sun.
A gas in a cylinder is held at a constant pressure of 1.80 * 105 Pa and is cooled and compressed from 1.70 m3 to 1.20 m3 . The internal energy of the gas decreases by 1.40 * 105 J.
(a) Find the work done by the gas.
(b) Find the absolute value |Q| of the heat flow into or out of the gas, and state the direction of the heat flow.
(c) Does it matter whether the gas is ideal? Why or why not
This question is wrong because of not correct values.The Correct question is
A gas in a cylinder is held at a constant pressure of 1.80×10⁵Pa and is cooled and compressed from 1.70m³ to 1.20m³. The internal energy of the gas decreases by 1.40×10⁵J.
(a) Find the work done by the gas.
(b) Find the absolute value of the heat flow, |Q|, into or out of the gas, and state the direction of the heat flow.
(c) Does it matter whether the gas is ideal? Why or why not?
Answer:
(a) W= -9×10⁴J
(b) |Q|=2.3×10⁵
(c) It does not matter whether gas is ideal or non-ideal
Explanation:
Given
V₁=1.70m³
V₂=1.20m³
p=1.8×10⁵ pa
ΔU= -1.40×10⁵J
For (a) work done by gas
[tex]W=p(V_{2}-V_{1} )\\W=1.8*10^{5}(1.2-1.7)\\ W=-9*10^{4}J[/tex]
For (b) Heat flow |Q|
|Q|=ΔU+W
[tex]Q=(-1.4*10^{5}) +(-9*10^{4})\\Q=-2.3*10^{5}J\\So\\|Q|=|-2.3*10^{5}|\\|Q|=2.3*10^{5}J[/tex]
For (c) part
It does not matter whether the gas is ideal or not because the first law of thermodynamics which applied in our solution could applied to any material ideal or non ideal
An engineer has an odd-shaped 13.5 kg object and needs to find its rotational inertia about an axis through its center of mass. The object is supported on a wire stretched along the desired axis. The wire has a torsion constant κ = 0.618 N·m. If this torsion pendulum oscillates through 28 cycles in 58.1 s, what is the rotational inertia of the object?
Answer:
I = 0.0674 kg.m²
Explanation:
given,
mass = 13.5 Kg
torsion constant = k = 0.618 N.m
number of cycle = 28
time = 58.1 s
Time of one cycle
[tex]T = \dfrac{58.1}{28}[/tex]
[tex]T =2.075\ s[/tex]
we know,
[tex]T = 2\pi\sqrt{\dfrac{I}{k}}[/tex]
[tex]I = k (\dfrac{T}{2\pi})^2[/tex]
[tex]I =0.618\times \dfrac{T^2}{4\pi^2}[/tex]
[tex]I =0.618\times \dfrac{2.075^2}{4\pi^2}[/tex]
I = 0.0674 kg.m²
the rotational inertia of the object is equal to I = 0.0674 kg.m²
A miniature spring-loaded, radio-controlled gun is mounted on an air puck. The gun's bullet has a mass of 7.00 g, and the gun and puck have a combined mass of 150 g. With the system initially at rest, the radio controlled trigger releases the bullet causing the puck and empty gun to move with a speed of 0.530 m/s. What is the bullet's speed?
Answer:
Explanation:
Given
mass of bullet [tex]m=7 gm[/tex]
mass of gun and Puck [tex]M=150 gm[/tex]
speed of gun and Puck is [tex]v=0.53 m/s[/tex]
Let speed of bullet be u
conserving Momentum
initial momentum=Final Momentum
[tex]0=Mv+mu[/tex]
[tex]u=-\frac{M}{m}v[/tex]
[tex]u=-\frac{150}{7}\times 0.53=-11.35 m/s[/tex]
negative sign indicates that bullet is moving in opposite direction
Using the law of conservation of momentum, it's possible to calculate that the bullet moves in the opposite direction of the gun and puck at a speed of approximately 113.6 m/s.
Explanation:This question represents a practical example of the conservation of momentum. According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces act on it. As the system (gun and bullet) is initially at rest, the total momentum is zero, and it remains zero even after the bullet is fired.
Let's denote the bullet's velocity as v. The momentum of the bullet after firing is its mass times its velocity, or 0.007kg * v. The momentum of the gun and puck is their combined mass times their velocity, or 0.150kg * 0.530m/s. As the total momentum should remain zero, these two quantities must be equal and opposite. Thus, 0.007kg * v = - 0.150kg * 0.530m/s. Solving for v, we find that the bullet's speed is approximately -113.6 m/s. The negative sign indicates that the bullet moves in the opposite direction of the gun and puck.
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An ideal gas goes through the following two-step process. 1) The container holding the gas has a fixed volume of 0.200 m3 while the pressure of the gas increases from 3.00×105 Pa to 4.00×105 Pa . 2) The container holding the gas is then compressed to a volume of 0.120 m3 while maintaining a constant pressure of 4.00×105 Pa .A) What is the total work done by the gas for this two-step process?
Answer:
[tex]W=-3.2\times 10^4\ J[/tex]
Explanation:
Given:
Process 1:
Volume of ideal gas is constant, [tex]V_1_i=0.2\ m^3[/tex]Initial pressure, [tex]P_1_i=3\times 10^5\ Pa[/tex]Final pressure, [tex]P_1_f=4\times 10^5\ Pa[/tex]Process 2:
Pressure of ideal gas is constant, [tex]P_2_f=4\times 10^5\ Pa[/tex]Final volume, [tex]V_2_f=0.12\ m^3[/tex]We know that the work done by an ideal gas is given as:
[tex]W=P\times (V_f-V_i)[/tex]
Now for process 1:
[tex]W_1=0\ J[/tex]
∵there is no change in volume in this process.
For process 2:
[tex]W_2=4\times 10^5\time (0.12-0.2)\ J[/tex]
[tex]W_2=-3.2\times 10^4\ J[/tex]
∵Negative , sign indicates that the work is being done on the gas here since the gas is being compressed.
Hence the total work done by the gas during this two step process is :
[tex]W=W_1+W_2[/tex]
[tex]W=0-3.2\times 10^4[/tex]
[tex]W=-3.2\times 10^4\ J[/tex] is the work done by the gas.
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.At what speed v should an archerfish spit the water to shoot down an insect floating on the water surface located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60∘ above the water surface.
Answer:
Explanation:
Here is the full question and answer,
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.
When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.
Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.
Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?
Answer
A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.
Calculate the speed of the archer fish.
The time of the flight of spitted water is,
[tex]t = \frac{{2v\sin \theta }}{g}[/tex]
Substitute [tex]9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\[/tex]
Spitted water will travel [tex]0.80{\rm{ m}}[/tex] horizontally.
Displacement of water in this time period is
[tex]x = vt\cos \theta[/tex]
Substitute [tex]\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}[/tex] for [tex]t\rm 60^\circ[tex] for [tex]\theta[/tex] and [tex]0.80{\rm{ m}}[/tex] for x in above equation.
[tex]\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.
Calculate the height of insect above the surface.
Vertical component of the velocity is,
[tex]{v_v} = v\sin \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
Horizontal component of the velocity is,
[tex]{v_h} = v\cos \theta[/tex]
Substitute [tex]3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for v and [tex]60^\circ[/tex] for [tex]\theta[/tex] in above equation.
[tex]\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\[/tex]
When horizontal [tex]({0.60\;{\rm{m}}}[/tex] distance away from the fish.
The time of flight for distance (d) is ,
[tex]t = \frac{d}{{{v_h}}}[/tex]
Substitute [tex]0.60\;{\rm{m}}[/tex] for d and [tex]1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_h}[/tex] in equation [tex]t = \frac{d}{{{v_h}}}[/tex]
[tex]\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\[/tex]
Distance of the insect above the surface is,
[tex]s = {v_v}t + \frac{1}{2}g{t^2}[/tex]
Substitute [tex]2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}[/tex] for [tex]{v_v}[/tex] and [tex]0.3987{\rm{ s}}[/tex] for t and [tex]- 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}[/tex] for g in above equation.
[tex]\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\[/tex]
The 8 kg block is then released and accelerates to the right, toward the 2 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 8 kg block. Find the speed of the 8 kg block just before it collides with the 2 kg block. Answer in units of m/s.
Answer:
3.258 m/s
Explanation:
k = Spring constant = 263 N/m (Assumed, as it is not given)
x = Displacement of spring = 0.7 m (Assumed, as it is not given)
[tex]\mu[/tex] = Coefficient of friction = 0.4
Energy stored in spring is given by
[tex]U=\dfrac{1}{2}kx^2\\\Rightarrow U=\dfrac{1}{2}\times 263\times 0.7^2\\\Rightarrow U=64.435\ J[/tex]
As the energy in the system is conserved we have
[tex]\dfrac{1}{2}mv^2=U-\mu mgx\\\Rightarrow v=\sqrt{2\dfrac{U-\mu mgx}{m}}\\\Rightarrow v=\sqrt{2\dfrac{64.435-0.4\times 8\times 9.81\times 0.7}{8}}\\\Rightarrow v=3.258\ m/s[/tex]
The speed of the 8 kg block just before collision is 3.258 m/s
The speed of the 8 kg block before the collision is found by considering the work done by friction and the initial potential energy in the spring and using the conservation of energy principle to solve for the speed.
Explanation:To find the speed of the 8 kg block just before it collides with the 2 kg block, we must calculate the motion of the block under the influence of friction before the collision occurs.
First, we determine the force of friction using the coefficient of friction (μ) and the normal force (N), which in this case equals the weight of the block (m × g) since the surface is horizontal. So, the force of friction is given by Fₙ = μ × m × g = 0.4 × 8 kg × 9.8 m/s² = 31.36 N.
Since there is friction, work done by friction must be considered to find the velocity. The work done by friction (Wₙ) is the force of friction (Fₙ) multiplied by the distance the block moves before colliding with the second block (d), which is not given in this problem. But if we had the distance, we could use the work-energy principle, where the net work done on an object is equal to the change in its kinetic energy (KE).
Assuming no other forces act on the block, then the work done by friction is equal to the initial potential energy stored in the spring. Using Hooke's Law, the potential energy (PE) in the spring is given by PE = ½ kx², where k is the spring constant and x is the compression distance. This energy gets converted entirely into the kinetic energy of the 8 kg block minus the losses due to friction.
Using the conservation of energy:
PE - Wₙ = ½ m v²
½ kx² - Fₙ × d = ½ m v²
We solve this equation for v, the speed of the block, if we know the distance (d).
A 946.4 kg ( 2083 lb ) car is moving at 14.8 m / s ( 33.0 mph ) . Calculate the magnitude of its momentum.
Answer:
14009. 72 kgms^-1
Explanation:
Momentum is the product of an objects mass and velocity
The calculated value of the magnitude of the car's momentum be 14006.72 kg.m/s or 68739 lb.mph.
What is momentum?Momentum is a property of a body moving in linear motion, It is product of mass and velocity. As velocity is a vector quantity, momentum is also a vector quantity with magnitude and direction. SI unit of momentum be kg.m/s.
Given parameters:
Mass of the car: m = 946.4 kg. = 2083 lb
Speed of the car: v = 14.8 m/s. = 33.0 mph
So, momentum of the car: p = mv = 946.4 × 14.8 kg.m/s = 14006.72 kg.m/s.
or p = mv = 2083 lb×33.0 mph = 68739 lb.mph.
Hence, the magnitude of its momentum be 14006.72 kg.m/s or 68739 lb.mph.
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You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.
Answer:
45 N, 30N ,20N
Explanation
torque = perpendicular force× distance
perpendicular force = torque/ distance
1) perpendicular force = 18/0.400 = 45 N
2) perpendicular force = 18/0.600 = 30N
3) perpendicular force = 18/0.900 = 20N
We have that for the Question " the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque" it can be said that
F=45NF=30NF=20N
From the question we are told
You want to apply 18 N ⋅ m of torque to tighten a nut, using a wrench with a length of either 0.400 m, 0.600 m, or 0.900 m. For each wrench, determine the perpendicular force that you would have to apply to the end of the wrench in order to produce the desired torque.
Generally the equation for the Force is mathematically given as
t=F*d
a)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.4}\\\\[/tex]
F=45N
b)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\\F=\frac{18}{0.6}\\\\[/tex]
F=30N
c)
[tex]t=F*d\\\\F=\frac{t}{d}\\\\F=\frac{18}{0.6}\\\\[/tex]
F=20N
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You set out to reproduce Thomson’s e/m experiment with an accelerating potential of 150 V and adeflecting electric field of magnitude 6.0 ×106 N/C(a) At what fraction of the speed of light do the electrons move?(b) What magnetic-field magnitude will yield zero beam deflection?(c) With this magnetic field, how will the electron beam behave if you increase the acceleratingpotential above 150 V?
Answer:
a. v' = 24.22 x 10 ⁻³
b. β = 0.825 T
Explanation:
V = 150 V , E = 6.0 x 10 ⁶ N / C
a.
¹/₂ * m * v² = e * V
v = √ 2 * e / m * V
v = √ 2 * 1.76 x 10 ¹¹ * 150 v = √ 5.28 x 10 ¹³
v = 7.26 x 10 ⁶ m /s
v' / c = 7.26 x 10 ⁶ m /s / 3.0 x 10 ⁸ = 24.22 x 10 ⁻³
b.
β = E / V
β = 6.0 x 10 ⁶ / 7.266 x 10 ⁶
β = 0.825 T
c.
When the increasing the accelerating potential speed it doesn't change the up wired the electric force and the magnetic force the electron be beat down more .
You have 10 ohm and a 100 ohm resistor in parallel. You place this equivalent resistance in series with an LED, which is rated to have a voltage drop of 1.83V at 20mA. What voltage should you connect to this circuit to provide 20mA of current for the LED? What would the current be, if you connected to a 3V battery? Draw all of these diagrams before attempting the calculations.
Answer:
Approximately [tex]\rm 2.0\; V[/tex].
Approximately [tex]\rm 30 \; mA[/tex]. (assumption: the LED here is an Ohmic resistor.)
Explanation:
The two resistors here [tex]R_1= 10\; \Omega[/tex] and [tex]R_2= 100\; \Omega[/tex] are connected in parallel. Their effective resistance would be equal to
[tex]\displaystyle \frac{1}{\dfrac{1}{R_1} + \dfrac{1}{R_2}} = \frac{1}{\dfrac{1}{10} + \dfrac{1}{100}} = \frac{10}{11} \; \Omega[/tex].
The current in a serial circuit is supposed to be the same everywhere. In this case, the current through the LED should be [tex]20\; \rm mA = 0.020\; \rm A[/tex]. That should also be the current through the effective [tex]\displaystyle \rm \frac{10}{11} \; \Omega[/tex] resistor. Make sure all values are in standard units. The voltage drop across that resistor would be
[tex]V = I \cdot R = 0.020 \times \dfrac{10}{11} \approx 0.182\; \rm V[/tex].
The voltage drop across the entire circuit would equal to
the voltage drop across the resistors, plusthe voltage drop across the LED.In this case, that value would be equal to [tex]1.83 + 0.182 \approx 2.0\; \rm V[/tex]. That's the voltage that needs to be supplied to the circuit to achieve a current of [tex]20\; \rm mA[/tex] through the LED.
Assuming that the LED is an Ohmic resistor. In other words, assume that its resistance is the same for all currents. Calculate its resistance:
[tex]\displaystyle R(\text{LED}) = \frac{V(\text{LED})}{I(\text{LED})}= \frac{1.83}{0.020} \approx 91.5\; \Omega[/tex].
The resistance of a serial circuit is equal to the resistance of its parts. In this case,
[tex]\displaystyle R = R(\text{LED}) + R(\text{Resistors}) = 91.5 + \frac{10}{11} \approx 100\; \Omega[/tex].
Again, the current in a serial circuit is the same in all appliances.
[tex]\displaystyle I = \frac{V}{R} = \frac{3}{100} \approx 0.030\; \rm A = 30\; mA[/tex].