Answers
A wave is considered to be transverse if its oscillations run counterclockwise to the wave's direction of advance. A longitudinal wave, on the other hand, moves in the direction of its oscillations. Transverse waves include water waves.
A waveform signal's wavelength, which is the distance between two identical locations (adjacent crests) in the succeeding cycles, determines whether it is sent through space or via a wire. This length is typically defined in wireless systems in metres (m), centimetres (cm), or millimetres (mm).
The largest displacement or distance made by a point on a wave or vibrating body relative to its equilibrium position is its amplitude. It is equal to one-half of the vibration path.
The crest and trough of a wave, respectively, are its highest and lowest surface portions. The wave height is the vertical distance between the peak and trough. The wavelength is the horizontal separation between two consecutive crests or troughs.
The horizontal line at the wave's centre stands in for balance. A period is the length of time it takes to complete a cycle, which includes travelling from one peak to another, from one trough to another, or from one equilibrium point to another (both equilibrium points same direction).
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Related Questions
true or false? a liter is a metric unit for area
Answers
A iter is defined as 1 cubic decimeter. It is the volume that fits inside a cube whose side measures 10 centimeters:
Therefore, the statement is false because a liter is a metric unit for volume.
if the dolphin is moving horizontally when it goes through the hoop how high above the water is the center of the hoop
Answers
We are given that a dolphin moves describing a projectile motion. This can be represented in the following graph of position vs time:
Since the dolphin moves horizontally as he goes through the hoop this means that the hoop is at the maximum height of the motion. The maximum height of a projectile motion is given by:
[tex]h_{\max }=\frac{v^2\sin ^2\theta}{2g}[/tex]Where:
[tex]\begin{gathered} h_{\max }=\text{ max}imum\text{ height} \\ v=velocity_{} \\ \theta=\text{ initial angle} \\ g=\text{ acceleration of gravity} \end{gathered}[/tex]Now, we plug in the values:
[tex]h_{\text{max}}=\frac{(10\frac{m}{s})^2(\sin (41))^2}{2(9.8\frac{m}{s^2})}[/tex]Solving the operations:
[tex]h_{\max }=2.2m[/tex]Therefore, the hoop is at 2.2 meters above the water.
7.0 J of work is done to draw a bowstring back. The bow launches an arrow with a mass of 0.09 kg straight upward.
(a) What is the arrow's kinetic energy as it leaves the bow? (Round your answer to one decimal place.)
J
(b) What is the arrow's speed? (Round your answer to one decimal place.)
m/s
(c) What maximum height does the arrow reach? (Round your answer to one decimal place.)
m
Answers
The final answer is
(a) The KE = 3.88 * 10
(b) The speed of the arrow is 2.93 * 10
(c) Height which the arrow attains is 4.38 * 10
When the arrow is drawn from a bowstring back then there will be some work done. The pulling of the bowstring will have a distance moved and the energy used to do this.
Given,
Work done = 7 J
Mass of the arrow = 0.09 kg
The computations for a and b are the following:
To calculate the Kinetic energy,Work done = force x distance
7.0 J = force x 0.09
force = 7/0.09=77.77 = 7.77 x 10
force = m. a = 0.09 x a
77.77 / 0.09 = a = 864.1 = 8.64*10²
KE = 0.5 x m x v²
= 0.5 x 0.09 x 864.1
= 3.88 * 10
The speed of the arrow can be calculated asspeed 38.8/(0.5 x 0.09) = v² = 862 sq-root
Hence, v = 2.93*10¹ m/s is the answer
Height that the arrow reaches29.3² = 2gh
= 29.3²/(2 x 9.8) = 858.49 / (2 x 9.8) = 43.8 m
h = 43.8 m or 4.38 * 10¹ is the answer
Therefore, we can conclude that the kinetic energy of the arrow that leaves bowstring is 3.88*10 with the speed of 29.3 m/s and it reaches the height of 43.8 m.
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An object on a horizontal, frictionless surface is attached to a spring, displaced, and then released. If it isdisplaced 0.12m from its equilibrium position and released after 0.8s its displacement is found to be 0.12m onthe opposites side, and passed the equilibrium position once during the interval. Find:
Answers
Given:
The maximum displacement from the equilibrium position is A = 0.12 m.
Half of the time period is
[tex]T_{\frac{1}{2}}=\text{ 0.8 s}[/tex]To find the amplitude, time period, and frequency.
Explanation:
Amplitude is the maximum displacement from the equilibrium position.
Thus, the amplitude is A = 0.12 m.
One time period is the time taken from maximum displacement on one side(say A) to maximum displacement on the opposite side and back to the maximum displacement on the same side(A).
Thus, the time period is
[tex]\begin{gathered} T=2T_{\frac{1}{2}} \\ =2\times0.8 \\ =1.6\text{ s} \end{gathered}[/tex]The frequency will be
[tex]\begin{gathered} f=\frac{1}{T} \\ =\frac{1}{1.6} \\ =0.625\text{ Hz} \end{gathered}[/tex]An electron is in an infinite one dimension well that is 8.9 nm wide. What is the ground state energy of the electron?
Answers
The ground state energy of the electron is 2.23 x 10⁻¹⁷ J.
What is the ground state energy of the electron?The ground state energy of the electron is calculated by applying the formula for energy of photons.
E = hf
E = hc/λ
where;
h is Planck's constantc is speed of lightλ is the wavelengthE = (6.626 x 10⁻³⁴ x 3 x 10⁸) / (8.9 x 10⁻⁹)
E = 2.23 x 10⁻¹⁷ J
Thus, the ground state energy of the electron is determined by applying the principle or formula for energy of a single photon at the given wavelength of 8.9 nm.
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An object is launched at a speed of 20 m/s. If it rises to a height of 12 m, at what angle was it launched? ________° above the +x direction
Answers
Given:
The initial speed of the object is,
[tex]u=20\text{ m/s}[/tex]The maximum height is,
[tex]H=12\text{ m}[/tex]To find:
The angle of launch above the X-axis
Explanation:
The maximum height above the X-axis is,
[tex]H=\frac{u^2sin^2\theta}{2g}[/tex]Here, the angle above the X-axis is,
[tex]\theta[/tex]Substituting the values we get,
[tex]\begin{gathered} 12=\frac{(20)^2sin^2\theta}{2\times9.8} \\ sin^2\theta=\frac{12\times2\times9.8}{400} \\ sin\theta=0.7668\text{ \lparen taking positive value only\rparen} \\ \theta=50.1\degree \end{gathered}[/tex]Hence, the angle of launch is,
[tex]50.1\degree[/tex]Identify as many different ways as you can for giving energy to a basketball? (Select all that apply)
Answers
To answer this question we need to remember each kind of energy:
• Potential energy is the energy held by an object because of its position relative to other objects.
,• Kinetic energy is the energy held by an object due to is motion.
,• Internal energy is the energy due to the movement of the molecules of the object.
With this in mind we conclude that the following are ways of giving energy to a basketball:
• You can give a basketball kinetic energy by pushing on it with your hand, as in throwing or dribbling.
• You can give a basketball kinetic energy by spinning it on your finger.
• You can give a basketball potential energy by lifting it upward with your hand, as when shooting a free throw.
,• You can give a basketball internal energy by heating it.
The initial earthquake had a Richter scale reading of 3.24. The aftershock was 7470 times as strong as the initial earthquake. What was the Richter scale reading of the aftershock?(a) 7.11(b) 6.48(c) 4.32(d) 3.87
Answers
In order to determine the Richter scale reading of the aftershock, use the following formula:
[tex]M_2-M_1=\log (\frac{I_2}{I_1})[/tex]Where I1 and I2 are the intensity of both earthquake and its aftershock, M1 and M2 are the magnitude on the Richter Scale for earthquake and aftershock.
In this case, you have:
I1 = intensity earthquake
I2 = 7470*I1
M1 = 3.24
M2 = ?
solve the equation above for M2 and replace the values of the given parameters:
[tex]undefined[/tex]Consider a fluid of density 3.43 g⋅cm−3 flowing through a pipe of varying cross-section. The diameter of the pipe in one section is 9.1 cm, while the diameter in a second section is 12.6 cm. When the diameter of the pipe is 9.1 cm, the flow speed of the fluid is 339 cm⋅s−1 and the pressure is 2.93 × 105 Pa.A)Calculate the flow speed (in m⋅s−1) of the fluid when the diameter of the pipe is 12.6 cm. B)Calculate the pressure (in × 105 Pa) when the pipe has a diameter of 12.6 cm
Answers
Given that the pipe has varying cross-sections.
The diameter of one section is d1 = 9.1 cm and the diameter of second section is d2 = 12.6 cm.
Also, the fluid has the density,
[tex]\rho=3.43gcm^{-3}[/tex]The area of the cross-section for the first section is
[tex]\begin{gathered} A_1=\frac{\pi(d1)^2}{4} \\ =\frac{\pi(9.1)^2}{4}cm^2 \end{gathered}[/tex]The area of the cross-section for the second section is
[tex]\begin{gathered} A_2=\frac{\pi(d2)^2}{4} \\ =\frac{\pi(12.6)^2}{4}cm^2 \end{gathered}[/tex]The flow speed for the first section is v1 = 339 cm s^-1
The flow speed for the second section will be v2.
(a) The flow speed for the second section can be calculated as
[tex]\begin{gathered} A_1v1=A_2_{}v2 \\ v2=\frac{A_1v1}{A_2} \\ =\frac{\pi(9.1)^2\times339\times4}{4\times\pi\times(12.6)^2} \\ =\text{ 176.82 cm/s} \\ =1.7682\text{ m/s} \end{gathered}[/tex](b) The pressure for first section is p1 = 2.93 x 10^5 Pa
The pressure for the second section will be p2.
The pressure for the second section can be calculated by the formula,
[tex]\begin{gathered} p2=p1+\frac{1}{2}\rho\mleft\lbrace(v1)^2-(v2\mright)^2\} \\ =2.93\times10^5+\frac{1}{2}\times3.43\mleft\lbrace(339)^2-(176.82)^2\mright\rbrace \\ =4.36\text{ }\times10^5\text{ Pa} \end{gathered}[/tex]A 5.0 kg block moves in a straight line on a horizontal frictionless surface under the influence of a force that varies with position as shown in the figure. The scale of the figure's vertical axis is set by F = 20.0 N. How much work is done by the force as the block moves from the origin to x = 8.0 m?
Answers
The work done by the force to move the object from the origin to x = 8 m is 160 J.
The mass of the block, m = 5 kg
The block moves in a straight line on a frictionless surface.
A force of F = 20 N is acting on the object.
The work done by the force when the block moves from one place to another can be defined by the formula,
W = Fd where F is the force and d is the distance covered.
Now, we have F = 20 N and;
d = x = 8 m
Therefore, the work done by the object will be:
W = Fx
W = 20 × 8
W = 160 joules
The work done by the force to move the object to x = 8 m is 160 J.
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but 9 volt battery is connected to a 4 ohm resistor and the 5 Ohm resistor as shown in the diagram. How much current flows through the 4 ohm resistor ?
Answers
Given data
The resistance of the first resistor is R1 = 4 ohm
The resistance of the second resistor is R2 = 5 ohm
The potential difference of the battery is V = 9 V
The resistors are connected in series. The expression for the equivalent resistance is given as:
[tex]\begin{gathered} R=R_1+R_2_{} \\ R=4\text{ }\Omega+5\text{ }\Omega \\ R=9\Omega \end{gathered}[/tex]The expression for the current in the 4-ohm resistor is given as:
[tex]\begin{gathered} I=\frac{V}{R} \\ I=\frac{9\text{ V}}{9\text{ }\Omega} \\ I=1\text{ A} \end{gathered}[/tex]Thus, the magnitude of the current flows through the 4-ohm resistor is 1 A.
Instructions: 1) write out the question 2) work out the solution 3) Explain in words how you would know to do that1. What is the momentum of a 5.0 g bullet with a velocity of 500 m/s?
Answers
Given data:
The mass of bullet is m=5.0 g.
The velocity fo bullet is v=500 m/s.
The formula for the momentum is given by,
[tex]p=mv[/tex]Substitute the given values in above equation,
[tex]\begin{gathered} p=(5g\times\frac{1kg}{1000\text{ g}})(\frac{500m}{s}) \\ p=\frac{2.5kgm}{s} \end{gathered}[/tex]Thus, the momentum of the bullet is 2.5 kgm/s.
According to the definition of momentum, "It is the product of mass and velocity". Therefore, the momentum of moving body can be calculated by multiplying the mass of the body and velocity at which the body is moving.
So, that is how we know how to calculate the momentum of the body.
Thermal equilibrium implies that:A:the state of restB:absolute zero temperatureC:the maximum temperatureD:equilibrium temperature
Answers
Explanation
Heat is the flow of energy from a high temperature to a low temperature. When these temperatures balance out, heat stops flowing
so, after some time both regions reach thermal equilibrium and no more energy is transfered.so we can conclude that
thermal equilibrium inplies that there is equilibrium temperature
D.equilibrium temperature
I hope this helps you
A car travels 400 km in the first 4.5 hours of a trip. It stops for an hour and then travels final 300 km in 2.5 hours. Find the average speed of the car.
Answers
Given data:
Distance traveled by car in t_1=4.5 hr is s_1=400 km.
Distance traveled by car in t_2=1 hr is s_2=0 km (as the car was stopped).
Distance traveled by car in t_3=2.5 hr is s_3=300 km.
The average speed is given as,
[tex]\begin{gathered} v_{avg}=\frac{\text{ total distance traveled}}{\text{total time taken}} \\ =\frac{s_1+s_2+s_3}{t_1+t_2+t_3} \end{gathered}[/tex]Substitute all known values,
[tex]\begin{gathered} v_{avg}=\frac{(400\text{ km})+(0\text{ km})+(300\text{ km})}{(4.5\text{ hr})+(1\text{ hr})+(2.5\text{ hr})} \\ =87.5\text{ km/h} \end{gathered}[/tex]Therefore, the average speed of the car is 87.5 km/h.
Calculate the kinetic energy of a 3.15 kg bowling ball rolling down the lane at 2.5 m/s. Include the units.
Answers
Given:
The mass of the ball is: m = 3.15 kg
The velocity of the ball is: v = 2.5 m/s
To find:
The kinetic energy of the ball.
Explanation:
The kinetic energy is the energy that a particle has when is is in motion. For the particle of mass m moving with velocity v, the kinetic energy KE is given as:
[tex]KE=\frac{1}{2}mv^2[/tex]Substitute the values in the above expression, we get:
[tex]\begin{gathered} KE=\frac{1}{2}\times3.15\text{ kg}\times(2.5\text{ m/s\rparen}^2 \\ \\ KE=\frac{1}{2}\times3.15\text{ kg}\times6.25\text{ m}^2\text{/s}^2 \\ \\ KE=9.84\text{ kg.m}^2\text{/s}^2 \\ \\ KE=9.84\text{ J} \end{gathered}[/tex]Final answer:
The kinetic energy of the ball is 9.84 J.
Which part of the rock cycle can only occur because of thermal convection?
1. Magma changing to igneous rock
2. Igneous rock changing to metamorphic rock
3. Metamorphic rock changing to sedimentary rock
4. Sediment changing to sedimentary rock
Answers
Answer:
magma changing to igneous rock
Explanation:
In rock formation, there are several processes involved which could include sedimentation, melting, crystallization, etc but there is a part of the rock cycle that can only occur because of thermal convection which is magma changing to igneous rock.
This occurs when the heat from inside the earth and cooler temperature moves the rock in its liquid state which is known as convection and can also be defined as a cycle of heat transfer from where the hot material rises and the cool material condenses.
jerome pitches a baseball of mass 0.300 kg. the ball arrives at home plate with a speed of 40.0 m/s and is batted straight back to jerome with a return speed of 52.0 m/s. what is the magnitude of change in the ball's momentum?
Answers
The change in the momentum of the ball is 3.6 Kgm/s.
What is momentum?The term momentum has to do with the product in the velocity of a body and mass of the body. We have to recall at this point that rate of change of momentum is directly related to the impressed force and this is in accordance with the Newton second law of motion.
Now we have to look at the few pieces of information that we can be able to glean from the question;
Mass of the object = 0.300 kg
Initial speed of the object = 40.0 m/s
Final speed of the object = 52.0 m/s
Given that the change in the velocity of the object is given by;
m( v - u)
m = Mass of the object
v = Final speed of the object
u = Initial speed of the object
change in the velocity of the object = 0.300(52 - 40)
= 3.6 Kgm/s
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A spring with a spring constant of 52N/m sits on a desk. The spring is 34cm long. A block of mass 0.12kg is placed on top of the spring. How high above the desk does the block rest?
Answers
Answer:
H = 31.7 cm
Explanation:
Given:
k = 52 N/m - Spring rate
L = 34 cm = 0.34 m - Spring length
m = 0.12 kg
g = 9.8 m/s²
____________
H - ?
Block weight:
F = m*g = 0.12·9.8 ≈ 1.18 N
According to Hooke's law:
F = k·ΔL
The spring is compressed by:
ΔL = F / k = 1.18 / 52 ≈ 0.023 м
Height of the block above the table:
H = L -ΔL = 0.34 - 0.023 = 0.317 m or H = 31.7 cm
A hydroelectric plant takes energy from water and turns it into electrical energy.What are the transformations of energy in the water molecules that are used in theprocess of generating electricity this way?The water particles initially have kinetic energy due to their motion. This kineticenergy is transformed into potential energy due to the position of the water andthen this energy is used to produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into thermal energy and then this energy is used toproduce electricity.The water particles initially have chemical energy due to the bonds in water. Thischemical energy is transformed into thermal and then the thermal energy is usedto produce electricity.The water particles initially have potential due to their position. This potentialenergy is transformed into kinetic energy due to the motion of energy and thenthe kinetic energy is used to produce electricity.
Answers
Answer:
The last option.
Explanation:
The principle of hydroelectric dams is the following:
Basically, we utilizie the energy of the falling wat
Which of the following is needed for an electric circuit to work?A. Fusion sourceB. Nuclear sourceC. Chemical sourceD. Energy sourceVoltage is a measure of the ________ it gives to the current.A. AccelerationB. SpeedC. PullD. Push
Answers
We will have the following:
The "part" needed for an electric circuit is an energy source.
Voltage is the measure of the acceleration it gives to the current.
What do you diagram to analyze orbital motion ?
Answers
The diagram to analyze the orbital motion can be shown as,
Here, a is the acceleration of moon and v is the speed.
The above diagram indicates the orbital motion of the moon around the earth. The moon is more towards the earth than the sun due to larger gravity of earth and at the same time the moon has its velocity that tends moon to move. Therefore, the moon has balanced gravitational and centripetal force to keep in an uniform orbital motion.
Every time a cycle of convection currents occur, the temperature of the water ________.A. IncreasesB. Stays the sameC. DecreasesD. Goes to zero
Answers
The answer to this question is B
The reason the answer is B is because of how the cycle works. In the beggining, water is heated up and evaporates into the atmosphere as water vapor. Then the water cools down and condensates, coming back down. The water then ends up back to the same temperature as the earth. So in the end, the entire cycle doesn't add or remove heat from the water.
If there is a source voltage of 12 volts with a 2.5 volt drop over the LED, and an intended current of 100 mA, what size resistor should be used?
Answers
Given:
A voltage source has a voltage of 12 V.
A voltage drop over the LED is 2.5 V.
The current in the circuit is 100 mA.
To find:
The resistance of the resistor should be used.
Explanation:
LED is a forward-biased PN junction that emits light. The voltage drop over the LED is 2.5 V.
The voltage source has a voltage of 12 V.
Hence, the total voltage in the circuit = 12 V - 2.5 V = 9.5 V
The resistance of the circuit in the forward biased is almost equal to zero. Let the resistance in the circuit be R.
Thus, the total resistance in the circuit = 0 + R = R
The current in the circuit is = 100 mA = 0.1 A
The resistance R of the circuit can be calculated as:
[tex]R=\frac{V}{I}[/tex]Here, V is the voltage in the circuit and I is the current in the circuit.
Substituting the values in the above equation, we get:
[tex]\begin{gathered} R=\frac{9.5\text{ V}}{0.1\text{ A}} \\ \\ R=95\text{ }\Omega \end{gathered}[/tex]Final answer:
Hence, a resistor of 95 Ω should be used in the circuit.
25. A student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force?
Work = change in energy
Fd = 1/2 mv^2
F = 1/2 x 120 x 5^2 / 10
F = 150 N
Answers
If a student cycles along a level road at a speed of 5.0 m / s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m, then the average braking force would be 150 Newtons.
What is power?The rate of doing work is known as power. The Si unit of power is the watt.
Power =work / time
Work done by the braking force = change in kinetic energy
F × s = 1/2 × m × v²
F = 0.5 x 120 x 5² / 10
F = 150 Newtons
Thus, the average braking force would be 150 Newtons.
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Alpha decay results in the emission of aPositively charged helium ionPhotonNegatively charged electronNeutrino
Answers
An alpha particle is a Helium nucleus that is emitted during alpha decay, this meant that alpha decay results in the amission of a Positively charged helium ion.
A Curve in a road has 60m radius The angle of bank of the road is 4.7° Find the maximum speed of car can have without skidding. If the Coefficient of Static Friction between tyres and road is 0.8.
Answers
Answer:
An unbanked curve has a radius of 60m. The maximum speed at which a car can make a turn if the coefficient of static friction is 0.75, is.
Explanation:
The train above is traveling at a constant velocity because the forces acting on it are in equilibrium. Therefore, the missing force must have a magnitude (blank) of newtons to the (blank).
Answers
The missing force has a magnitude of 800 N to the right
Explanation:The forces acting on the train are in equilibrium.
This means that the sum of all the forces acting in the right direction equals the sum of all the forces acting in the left direction
Let the missing force be represented by F
1700 = 900 + F
F = 1700 - 900
F = 800 N
Therefore, the missing force has a magnitude of 800 N to the right
You exert a force of 5.3 N on a book to slide it across a table. If you do 2.5 J of work in the process, how far did the book move?
Answers
We will have the following:
[tex]2.5J=5.3N\cdot x\Rightarrow x=\frac{2.5J}{5.3N}[/tex][tex]\Rightarrow x=\frac{25}{53}m\Rightarrow x\approx0.47m[/tex]So, the book moved 25/53 meters, that is approximately 0.47 meters.
A spring with spring constant 40 N/m is compressed .1m past it natural length. A mass of .5kg is attached to the spring. A. What is the elastic potential energy stored in the spring?B. The spring is released. What is the speed of the masses as it reaches the natural length of the spring?
Answers
Given data
*The given spring constant is k = 40 N/m
*The given compressed length is x = 0.1 m
*The given mass is m = 0.5 kg
(a)
The formula for the elastic potential energy stored in the spring is given as
[tex]U_p=\frac{1}{2}kx^2[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} U_p=\frac{1}{2}(40)(0.1)^2 \\ =0.2\text{ J} \end{gathered}[/tex]Hence, the elastic potential energy stored in the spring is 0.2 J
(b)
The formula for the speed of the masses is given by the conservation of energy as
[tex]\begin{gathered} U_p=U_k \\ \frac{1}{2}kx^2=\frac{1}{2}mv^2 \\ v=x\sqrt[]{\frac{k}{m}} \end{gathered}[/tex]Substitute the values in the above expression as
[tex]\begin{gathered} v=(0.1)\sqrt[]{\frac{40}{0.5}} \\ =0.89\text{ m/s} \end{gathered}[/tex]Hence, the speed of the masses as it reaches the length of the spring is v = 0.89 m/s
Three bulbs of resistance 100. Ω, 200, Ω and 300 Ω are connected in parallel to a 120. V DC power supply. Draw the diagram and find thea) current in each bulb b) current drawn from the power supplyc) total power drawn power supply d) the net resistance of all bulbs
Answers
Let's use the formula for electric current.
[tex]I=\frac{V}{R}[/tex]Where V is the power supply 120 V, and R is the resistance. Let's find the current in each bulb.
[tex]\begin{gathered} I=\frac{120V}{100\Omega}=1.20A \\ I=\frac{120V}{200\Omega}=0.6A \\ I=\frac{120V}{300\Omega}=0.4A \end{gathered}[/tex](a) The current in each bulb is 1.20A, 0.6A, and 0.4A, respectively.(b) (c) The diagram of the circuit isTo find the net resistance, we use the following formula.
[tex]\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}+\frac{1}{R_3}[/tex]Let's use the given magnitudes.
[tex]\begin{gathered} \frac{1}{R}=\frac{1}{100\Omega}+\frac{1}{200\Omega}+\frac{1}{300\Omega} \\ \frac{1}{R}=\frac{6+3+2}{600\Omega} \\ \frac{1}{R}=\frac{11}{600\Omega} \\ R=\frac{600}{11}\Omega \\ R\approx54.55\Omega \end{gathered}[/tex]Therefore, the net resistance of all bulbs is 54.55 ohms.