A sound wave travelling in water at 10.2 m/s has a wavelength of 1.5 m.

Answers

Answer 1

Answer:

Frequency of sound wave in water = 6.8 Hz

Explanation:

Given:

Speed of sound wave in water = 10.2 m/s

The wavelength of the sound wave in water = 1.5 m

Find:

Frequency of sound wave in water

Computation:

Frequency = Speed of wave / Wavelength

Frequency of sound wave in water = Speed of sound wave in water / The wavelength of the sound wave in water

Frequency of sound wave in water = 10.2 / 1.5

Frequency of sound wave in water = 6.8 Hz


Related Questions

Pls help! Tysm!!! :D

Answers

Answer:

Slowing a bike by applying the brkaes is an xmaple of acceleration

Explanation:

Select all the correct answers.
What happens when a ray of light moves from alr to water?

Its speed changes.
its color changes.
Its direction changes.
It's absorbed by water.​

Answers

Answer:

it changes it's direction

Explanation:

When light travels from air into water, it slows down, causing it to change direction slightly. This change of direction is called refraction. When light enters a more dense substance (higher refractive index), it 'bends' more towards the normal line

Answer:

Hope this helps!

Explanation:

When light travels from air into water, it slows down, causing it to change direction slightly. This change of direction is called refraction. When light enters a more dense substance (higher refractive index), it 'bends' more towards the normal line.

A toy is undergoing SHM on the end of a horizontal spring with force constant 310 N/m . When the toy is 0.120 m from its equilibrium position, it is observed to have a speed of 3 m/s and a total energy of 3.6 J .
1.) Find the mass of the toy.
2.) Find the amplitude of the motion.
3.) Find the maximum speed obtained by the object during its motion.

Answers

Answer:

1) the mass of the toy is 0.304 kg

2) the amplitude of the motion is 0.1524 m

3) the maximum speed obtained by the object during its motion is 4.87 m/s

Explanation:

Given the data in the question;

force constant of spring k = 310 N/m

position of toy x = 0.120 m

speed v = 3 m/s

Total energy E = 3.6 J

let m represent mass of toy.

1.) Find the mass of the toy.

we know that; The total energy of the system equals the sum of kinetic energy of the toy and the potential energy of the spring;

E = [tex]\frac{1}{2}[/tex]kx² + [tex]\frac{1}{2}[/tex]mv²

we substitute

3.6 = ( [tex]\frac{1}{2}[/tex] × 310 × (0.120)² ) + ( [tex]\frac{1}{2}[/tex] × m × (3)² )

3.6 = 2.232 + 4.5m

3.6 - 2.232 = 4.5m

1.368 = 4.5m

m = 1.368 / 4.5

m = 0.304 kg

Therefore, the mass of the toy is 0.304 kg

2) Find the amplitude of the motion.

we know that;

E = [tex]\frac{1}{2}[/tex]kA²

where A is the amplitude of the motion,

we substitute

3.6 = [tex]\frac{1}{2}[/tex] × 310 × A²

3.6 = 155 × A²

A² = 3.6 / 155

A² = 0.0232258

A = √0.0232258

A = 0.1524 m

Therefore, the amplitude of the motion is 0.1524 m

3) Find the maximum speed obtained by the object during its motion;

we know that;

E = [tex]\frac{1}{2}[/tex]m[tex]v_{max[/tex]²

where [tex]v_{max[/tex] is the maximum speed

so we substitute

3.6 = [tex]\frac{1}{2}[/tex] × 0.304 × [tex]v_{max[/tex]²

3.6 = 0.152 × [tex]v_{max[/tex]²

[tex]v_{max[/tex]² = 3.6 / 0.152

[tex]v_{max[/tex]² = 23.6842

[tex]v_{max[/tex]² = √23.6842

[tex]v_{max[/tex]² = 4.8664 ≈ 4.87 m/s

Therefore, the maximum speed obtained by the object during its motion is 4.87 m/s

If you build a circuit with 3 D batteries, what is the voltage of the circuit?

A) 3 volts
B) 4.5 volts
C) 3 Amps
D) 4.5 Ohms

Answers

Answer:

a

Explanation:

Answer:

d

Explanation:

A uniform rod of length 50cm and mass 0.2kg is placed on a fulcrum at a distance of 40cm from the left end of the rod. At what distance from the left end of the rod should a 0.6kg mass be hung to balance the rod?
a.48 cm
b. 50 cm
c. 45 cm
d. the rod can not be balanced with this mass.
e.42 cm
show your work. NO LINKS. ​

Answers

Answer:

x = 45 cm

Explanation:

Given that,

The length of a rod, L = 50 cm

Mass, m₁ = 0.2 kg

It is at 40cm from the left end of the rod.

We need to find the distance from the left end of the rod should a 0.6kg mass be hung to balance the rod.

The centre of mass of the rod is at 25 cm.

Taking moments of both masses such that,

[tex]15\times 0.2=x\times 0.6\\\\x=\drac{3}{0.6}\\\\x=5\ cm[/tex]

The distance from the left end is 40+5 = 45 cm.

Hence, at a distance of 45 cm from the left end it will balance the rod.

What are the sources of error in density of liquid experiment?​

Answers

The most common reasons for density errors include using the wrong or inaccurate instruments and not accounting for temperature changes. There are many different instruments you can use to measure liquid volume such as kitchen measuring cups, beakers, graduated cylinders and volumetric pipettes.

We have that the sources of error in density of liquid experiment is

From the Question we are told to find

Sources of error in density of liquid experiment

It is important to note that  common reasons for  error in density of liquid experiment

using the inaccurate instruments temperature changes overlook

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A person pulls a box across the floor with a rope. The rope makes an angle of 40 degrees tot he horizontal, and a total of 125 newtons of force is applied. How much work is done if the box is pulled for 25 meters?

Answers

Answer:

The angle formed of the rope with the surface = 40°

Force applied = 125Newtons

The displacement covered by the box =25metres

W= FDcos theta

[125×40×cos(40°) ] Joules

= [ (3125×0.76604444311)]Joules

= 2393.88888472 joules(ans)

Hope it helps

An initially uncharged sphere is on an insulated stand and isolated in a chamber. The charge on the sphere is monitored as a beam of monochromatic light shines on the sphere. Initially nothing happens. The wavelength of the light is slowly decreased. When the wavelength reaches a certain value, a positive charge is suddenly measured on the sphere. The wavelength is then held constant, and the charge continues to increase at a constant rate. The intensity of the beam is then increased without the wavelength being changed, and the rate of increase of the charge becomes greater.

Required:
a. In a coherent paragraph-length response, describe the cause of the charge on the sphere and the changes in the observations about the charge, in terms of physics principles.
b. An electron in the chamber is moving with speed 2×10^5m/s when it collides with a positron (a particle identical to an electron except for the sign of its charge) moving with the same speed in the opposite direction. The particles annihilate each other. How much energy is released due to the annihilation?
c. In another experiment, a beam of electrons with uniform wavelength λe is incident on a slit, where the width of the slit is much larger than λe. A detector is placed near the slit, but no diffraction pattern is observed. What change should be made that would result in a diffraction pattern? Indicate why this change is the one needed.

Answers

Answer:

a)   E = K + Φ, b)  ΔE = 1.64 10⁻¹³ J.,  c)    λ = a

Explanation:

a) In this case it is an example of the photoelectric effect that was correctly described by Einstein assuming that the light ray is composed of a series of particles called photons, each one with an energy given by the Planck equation

        E = h f

         c = λ f

         

substituting

          E = h c /λ

We can see from this equation that as the wavelength of the ray decreases the energy of each photon increases, the moment arrives that the energy is sufficient to remove an electron from the sphere, thus leaving an unbalanced positive charge, this description explains why the positive charge appears on the sphere; the minimum wavelength to remove an electron is

           E = K + Φ

if K = 0

           E =Φ

where fi is the work function of the material.

When the intensity of the ray increases according to Eintein's description, the number of photons increases, so if the number of photons increases, the number of shocks and the number of electrons expelled increases, therefore the unbalanced positive charge also increases.

b) the energy released in collision is the sum of the energy of each particle

   

for the electron

          E = K + m c² = (pc) ² + (m c²) ²

where the moment is

          p = γ m u

          γ = [tex]\sqrt{1- (\frac{u}{c})^2 }[/tex]

          γ = [tex]\sqrt {1- ( \frac{2 \ 10^5}{3 \ 10^8 })^2 } = \sqrt{1- (6.67 \ 10 ^{-4)^2 }[/tex]

          γ ≅ 1

in this case since the speed of the particles is much less than the speed of light,

            E = (m u) ² + (m c²) ²

            E = m² (u² + c²)

            E = 9.1 10⁻³¹ [(2 10⁵) ² + (3 10⁸) ²

            E = 9.1 10⁻³¹   9 10¹⁶

            E = 8.2 10⁻¹⁴ J

the positron has an energy of equal magnitude, so when the two particles annihilate the energy change is

            ΔE = 2E

            ΔE = 2 (8.2 10⁻¹⁴)

            ΔE = 1.64 10⁻¹³ J.

c) the expression that describes the diffraction process is

            a sin θ = m λ

            sin  θ = m λ/ a

where a is the width of the slit and m in diffraction order

The greatest value that the sine function can have is 1

               1 = m λ / a

               λ = a / m

therefore we can see that to see the diffraction phenomenon the width of the slit must be greater than or equal to the wavelength

Part D- Isolating a Variable with a Coefficient In some cases, neither of the two equations in the system will contain a variable with a coefficient of 1, so we must take a further step to isolate it. Let's say we now have • 30 + 4D = 5 • 2C + 5D = 2 t to None of these terms has a coefficient of 1. Instead, we'll pick the variable with the smallest coefficient and isolate it. Move the term with the lowest coefficient so that it's alone on one side of its equation, then divide by the coefficient. Which of the following expressions would result from that process?
1. D= 5/4 - 3/4C
2. C= 5/3 - 4/3D
Part E- Now that you have one of the two variables in Part D isolated, use substitution to solve for the two variables. You may want to review the Multiplication and Division of Fractions and Simplifying an Expression Primers. Enter the answer as two numbers (either fraction or decimal), separated by a comma, with C first.

Answers

Answer:

D)     D = [tex]\frac{5}{4} - \frac{3}{4} \ C[/tex], E)  (C, D) = ( [tex]\frac{17}{7}, \ \frac{-4}{7}[/tex]

Explanation:

Part D) two expressions are indicated

          3C + 4D = 5

          2C +5 D = 2

let's simplify each expression

         3C + 4D = 5

         4D = 5 - 3C

we divide by 4

            D = [tex]\frac{5}{4} - \frac{3}{4} \ C[/tex]

The other expression

       2C +5 D = 2

       2C = 2 - 5D

        C = [tex]1 - \frac{5}{2} \ D[/tex]

we can see that the correct result is 1

Part E.

It is asked to solve the problem by the substitution method, we already have

          D =  [tex]\frac{5}{4} - \frac{3}{4} \ C[/tex]

we substitute in the other equation

            2C +5 D = 2

             2C +5 (5/4 - ¾ C) = 2

we solve

            C (2 - 15/4) + 25/4 = 2

             -7 / 4 C = 2 - 25/4

             -7 / 4 C = -17/4

              7C = 17

               C = [tex]\frac{17}{7}[/tex]

now we calculate D

               D = [tex]\frac{5}{4} - \frac{3}{4} \ \frac{17}{7}[/tex]

               D = 5/4 - 51/28

               D =[tex]\frac{35-51}{28}[/tex]

               D = - 16/28

               D = [tex]- \frac{4}{7}[/tex]

the result is (C, D) = ( [tex]\frac{17}{7}, \ \frac{-4}{7}[/tex] )

Two scientists setup an investigation in which a radio transmitter and a sound source are both placed inside a sealed Eontainer, The scientists then pump all of the air out of the container, creating a vacuum. The scientists then notice that the radio waves can still be detected outside of the container but the sound waves cannot be detected outside of the container. Which of the following explanations fit the results of their investigation?
A. Radio waves are electromagnetic but sound waves are mechanical because the radio waves could travel without a medium but the sound waves could not
B. Radio waves are mechanical and sound waves are electromagnetic because the radio waves could travel without a medium but the radio waves could not
C. Radio waves are electromagnetic but sound waves are mechanical because radio waves cannot travel through air but sound waves can travel through air.
D. Radio waves are mechanical but sound waves are electromagnetic because radio waves cannot travel through air but sound waves can travel through air.

Answers

Answer:

A. Radio waves are electromagnetic but sound waves are mechanical because the radio waves could travel without a medium but the sound waves could not.

Explanation:

Electromagnetic waves are waves that do not require a medium for traveling. Therefore, they can travel in a vacuum as well. For example, the sunlight travels from the Sun to the Earth in a vacuum.

Mechanical waves are those waves that require a material medium to travel. Therefore, they can not travel in a vacuum. For example, sound waves can not travel in space.

Therefore, the correct option is:

A. Radio waves are electromagnetic but sound waves are mechanical because the radio waves could travel without a medium but the sound waves could not.

What is the period of a wave that completes 3 waves in 2 seconds?

Answers

P = 3÷2

= 1.5

Explanation:

hope it helps

Greta is on the chairlift going up the mountain when one of her skis falls off.
The ski has a mass of 5 kg and hits the snow below with a speed of 72 km/h.
How high is the chairlift above the snow when the ski falls off?
(Ignore air resistance.)

Give your answer to the nearest 0.1 m.

Use g = 9.8 m/s2 for the acceleration due to gravity.

Answers

Answer:

[tex]20.4\:\text{m}[/tex]

Explanation:

We can use the following kinematics equation to find the height of the chairlift above the snow:

[tex]v_f^2=v_i^2+2a\Delta y[/tex].

First, let's convert 72 km/h to m/s:

[tex]72\: \text{km/h}=20\:\text{m/s}[/tex]

Since the ski starts with an initial vertical velocity of zero, we have [tex]v_i=0[/tex]. We can now substitute [tex]v_f=20,v_i=0, a=9.8[/tex] and solve for how high the chairlift is above the snow:

[tex]20^2=0^2+2\cdot9.8\cdot \Delta y,\\400=19.6\cdot \Delta y,\\\Delta y =\frac{400}{19.6}=\boxed{20.4\:\text{m}}[/tex].

What is hydrometer? Describe its use and any one type of hydrometer.​

Answers

Answer:

An instrument for measuring the density of liquids.

Explanation:

A hydrometer is an instrument used to determine specific gravity. It operates based on the Archimedes principle that a solid body displaces its own weight within a liquid in which it floats. Hydrometers can be divided into two general classes: liquids heavier than water and liquids lighter than water.

Thermohydrometers is one type of hydrometer.

ANSWER ASAP: Keiyah and Sam canoe down the Guadalupe. As Keiyah pulls the oar through the water, the water against the oar is pushing the canoe forward. This is an example of Newton's:


First Law

Second Law

Third Law

Law of Conservation of Mass

Answers

- Newton's first law states that, if a body is at rest or moving at a constant speed in a straight line, it will remain at rest or keep moving in a straight line at constant speed unless it is acted upon by a force.

- His second law defines a force to be equal to change in momentum (mass times velocity) per change in time. Momentum is defined to be the mass m of an object times its velocity V.

- His third law states that for every action (force) in nature there is an equal and opposite reaction. If object A exerts a force on object B, object B also exerts an equal and opposite force on object A.

Now which law do you think is happening in the problem?

The answer will be Newton's third law.

why the weight of the body varries from place to place on the earth's surface ​

Answers

Answer: Gravity

Explanation: the weight of the body varries because of varitations on earths gravity. Thank you!

Which formula is used to find an object's acceleration?

Answers

The answer is:

C.  a = △t/ △v

The formula used to find an object's acceleration is a = Δv / Δt.

What is acceleration?

Acceleration is the rate of change of the velocity of an object with respect to time. It is a vector quantity.

The equation for acceleration is  

a = Δv / Δt

where a is acceleration,

Δv is the change in velocity,

Δt is the amount of time it took for that change to occur.

The unit is [tex]\frac{m}{s^{2} }[/tex].

There are two ways by which an object can accelerate. One is to change the speed and other one is to change the direction. Even you can change both at same time.

If not changing any of them then object can not accelerate.

Some of examples of acceleration are

Turning a corner at high speed as turning changes the direction and giving rise to acceleration.Slowing or speeding a car is acceleration as speed is changing. So acceleration happens.Falling from the bridge. Here acceleration equals to gravitational acceleration.

 

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You attach a speaker to an air track cart and then attach the cart to one end of the air track by means of a spring and start the system oscillating by stretching the spring 0.797 m from its equilibrium position and then releasing it. The force constant for the spring is k = 49.8 N/m, the total mass of the speaker and air track cart are 4.00 kg, and the speaker emits sound with a frequency of 405 Hz. If your lab partner Hal stands at the end of the air track, determine the highest and lowest frequencies he hears. The speed of sound in air is 343 m/s. (Enter answers to at least the nearest Hz.)
Lowest Frequency =
Highest Freqency =

Answers

☞ANSWER☜“LOWEST FREQUENCY”

The lowest possible frequency at which a string could vibrate to form a standing wave pattern is known as the fundamental frequency or the first harmonic. The second lowest frequency at which a string could vibrate is known as the second harmonic; the third lowest frequency is known as the third harmonic; and so on.

HIGHEST FREQUENCY

Gamma rays have the highest energies, the shortest wavelengths, and the highest frequencies. Radio waves, on the other hand, have the lowest energies, longest wavelengths, and lowest frequencies of any type of EM radiation.

A microbe is ejected from its home planet by a violent event such as a volcanic eruption or a meteorite impact. The subsequent fate of the microbe depends on the balance of the two main forces that act on it, namely, the gravitational attraction of the star around which the planet orbits and radiation pressure due to the luminosity of the star.

Required:
Model the microbe as a dark sphere of radius a at a distance r from the star. Assume that the total power radiated by the star P = 1.0 x 1031 J/s (greater than that of the sun by a factor of 30,000). Derive an expression for the radiation force on the microbe. Give your answer in terms of p, r,c and a.

Answers

Answer:

  [tex]F = P_r (\frac{a}{4r})^2[/tex]

Explanation:

The radiation pressure for a total absorption of the incident light is

         P = S / c

Where S is the pointing vector, which is defined as the energy flux per unit area

         S = I = P_r / A

where P_r  powre irradiate by star, A is the area over which the energy emitted by the star is distributed, which is the surface of a sphere

         A = 4π r²

substituting

         S =[tex]\frac{P_r}{4 \pi r^2 }[/tex]

Pressure is defined by

         P = F / A₂

Where A₂ is the area over which the force is applied, in this case the area of ​​the microbe

          A₂ = π a²

         P = F / π a²

 

we substitute

      [tex]\frac{F}{\pi a^2} = \frac{P_r}{ 4 \pi r^2}[/tex]

       [tex]F = P_r (\frac{a}{4r})^2[/tex]

An ideal parallel plate capacitor is charged by connecting it to a battery. The battery is then removed so that charges may no longer enter onto or exit from the plates of the capacitor. There is nothing between the capacitor plates. As the plates are moved closer together, the magnitude of the electric field between them:_________
a. decreases
b. increases
c. remains the same

Answers

Answer:

c. remains the same

Explanation:

The capacitance of a capacitor is can be represented as;

C = Q/V

There will be no charge flow from the battery to the capacitor or vice versa if the battery is disconnected.

Due to energy conservation, the energy contained and stored in a capacitor will continue to retain and be the same, since an increase in capacitance reduces the voltage, resulting in the same final energy stored.

A team at a university in Pennsylvania has concerns about the health of streams in a watershed because of the various land uses in the area. There has been an increase in road repair work, and the area has many dairy farms. Which of the following would be most effective at preventing sediment runoff into the streams of the watershed? *
a) Have construction crews steepen the slopes of the valleys of streams near roadwork zones.
b) Maintain zones of grass at least wide on the sides of streams in the area.
c) Construct at least one dam on a stream in the watershed to improve flood control.
d) Shift from dairy farming to corn planted in agricultural fields located within of streams.

Answers

The ecological methods allows to find the correct answer to the question of how to prevent river pollution is:

   b)  Maintain zones of grass at least wide on the sides of streams in the area.

Stream pollution from human activities can occur in a number of ways:

The rubble boat in the river. The dragging of products from the bargains, fertilizers by the rains.   Dragging of land by rain and machinery.

Let's analyze the different claims.

a) False. It is too expensive and the slopes cannot be changed due to possible flooding problems.

b) True. Maintaining a protective zone with grass on the sides of the rivers, prevents the waste of the frarmer from reaching the river, and the problems of the debris reaching the river bank. This un  ecological methods.

c) False. The dam controls flooding but does not eliminate the problems of river pollution.

d) False. Many lands are not acts for agriculture, but for livestock.

In conclusion we can find the correct answer to the question of how to prevent river pollution is:

   b) Maintain zones of grass at least wide on the sides of streams in the area.

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An organ pipe is 116 cm long. Determine the fundamental and first three audible overtones if the pipe is (a) closed at one end, and (b) open at both ends.

Answers

Answer:

(a) when the pipe is closed at one end;

F₀ = 73.92 Hz

F₁ = 221.76 Hz

F₂ = 369.6 Hz

F₃ = 517.44

(b) when the pipe is open at both ends;

F₀ = 147.85 Hz

F₁ =  295.7 Hz

F₂ = 443.55 Hz

F₃ = 591.4 Hz

Explanation:

Given;

length of the pipe, L = 116 cm = 1.16 m

speed of sound in air, v = 343 m/s

The formula below will be used to determine the different frequencies of the pipe at different wavelengths

[tex]F = \frac{V}{\lambda}[/tex]

where;

F is frequency

λ is wavelength

(a) when the pipe is closed at one end;

Wavelength for fundamental frequency;  

L = Node -------> Antinode

L = λ/4

λ = 4L

the fundamental mental frequency;

[tex]F_0 = \frac{V}{4L} = \frac{343}{4 \times 1.16} = 73.92 \ Hz[/tex]

Wavelength for first overtone or audible frequency;

L = Node ----->  Node  +  Node ----> Antinode

L =          λ/2                 +        λ/4

L = 3λ/4

λ = 4L/3

[tex]F_1 = \frac{V}{4L/3} = 3(\frac{V}{4L} ) = 3F_0 = 3 (73.92) = 221.76 \ Hz[/tex]

Thus, the next two audible frequencies will be multiple of next consecutive odd numbers after 3 (i.e  5 and 7);

F₂ = 5F₀ = 5(73.92) = 369.6 Hz

F₃ = 7F₀ = 7(73.92) = 517.44 Hz

(b) when the pipe is open at both ends;

Wavelength for fundamental frequency;

L = Antinode ----> Node   +  Node -----> Antinode

L =    λ/4  + λ/4

L = λ/2

λ = 2L

the fundamental frequency;

[tex]F_0 = \frac{V}{2L} = \frac{343}{2\times 1.16} = 147.85 \ Hz[/tex]

The first three audible overtones will be multiples of the 3 consecutive positive integers after 1 (i.e 2, 3, 4)

F₁ = 2F₀  = 2(147.85) = 295.7 Hz

F₂ = 3F₀ = 3(147.85) = 443.55 Hz

F₃ = 4F₀ = 4(147.85) = 591.4 Hz

E._____________________

Answers

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A block of mass 3m is placed on a frictionless horizontal surface, and a second block of mass m is placed on top of the first block. The surfaces of the blocks are rough. A constant force of magnitude F is applied to the first block as shown in the figure. (a) Construct free-body diagrams for each block. (b) Identify the horizontal force that causes the block of mass m to accelerate. (c) Assume that the upper block does not slip on the lower block, and find the acceleration of each block in terms of m and F.

Answers

By Newton's second law, assuming F is horizontal,

• the net horizontal force on the larger block is

F - µmg = 3mA

where µmg is the magnitude of friction felt by the larger block due to rubbing with the smaller one, µ is the coefficient of static friction between the two blocks, and A is the block's acceleration;

• the net vertical force on the larger block is

4mg - 3mg - mg = 0

where 4mg is the mag. of the normal force of the surface pushing up on the combined mass of the two blocks, 3mg is the weight of the larger block, and mg is the weight of the smaller block;

• the net horizontal force on the smaller block is

µmg = ma

where µmg is again the friction between the two blocks, but notice that this points in the same direction as F. It is the only force acting on the smaller block in the horizontal direction, so (b) static friction is causing the smaller block to accelerate;

• the net vertical force on the smaller block is

mg - mg = 0

where mg is the magnitude of both the normal force of the larger block pushing up on the smaller one, and the weight of the smaller block.

(You should be able to draw your own FBD's based on the forces mentioned above.)

(c) Solve the equations above for A and a :

A = (F - µmg) / (3m)

a = µg

1. A 10.-ohm resistor and a 20. -ohm resistor are connected in series
to a voltage source. When the current through the 10.- ohm resistor
is 2.0 amperes, what is the current through the 20.-ohm resistor?
(5F)
a. 1.0 A
b. 2.0 A
C. 0.50 A
d. 4.0 A

Answers

A 10.-ohm resistor and a 20. -ohm resistor are connected in series

to a voltage source. When the current through the 10.- ohm resistor

is 2.0 amperes, what is the current through the 20.-ohm resistor?

Answer:

B --> 2.0 A

the current going through the 10.-ohm resistor is what goes through 20.-ohm resistor

The current through the 20Ω resistor is 2.0 A. Hence option B is correct.

What is Current ?

Current is a flow of charges. it is denoted by i and expressed in ampere A. Mathematically it is expressed as i = q/t, where q is the amount of charge and t is time. Current is nothing but amount of charges flown in the unit time in the electric wire. Charge is expressed in coulomb C and time in second s. hence coulomb per second (C/s) is ampere A. Charge on electron is 1.60217663 × 10⁻¹⁹C which is called as elementary charge.

There are two types of the current, Convectional current and non-conventional current. Convectional current is the current flows from positive to negative. Non convectional current flows from negative to positive. Note that flow of electrons is from negative to positive. Hence direction of flow of conventional current is from positive to negative.

In this problem, two resistors are connected in series. In the series combination of resistor, same current flows through all resistor.

Hence current through 20Ω Resistor is 2.0A.

Hence option B is correct.

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what is conductivity?​

Answers

Answer:

Conductivity is the measure of the ease at which an electric charge or heat can pass through a material. A conductor is a material that gives very little resistance to the flow of an electric current or thermal energy.

Explanation:

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By how many times would the gravitational force between two objects decrease if the distance between the two objects was tripled?

Answers

Answer:

If the separation distance between any two objects is tripled (increased by a factor of 3), then the force of gravitational attraction is decreased by a factor of 9 (3 raised to the second power).

distance travelled/time taken gives?

Answers

Answer:

speed

Explanation:

Speed = distance travelled/time taken

Answer: Speed

Explanation:

Speed is defined as the rate of change of the position of an object in a direction. The speed of an object is calculated as the distance travelled divided by the time that is taken.

Speed is regarded as a scalar quantity since it has only direction but doesn't have magnitude. The answer to the above question is speed.

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to -13.60 eV.

Answers

Answer:

(a) The energy of the photon is 1.632 x [tex]10^{-8}[/tex] J.

(b) The wavelength of the photon is 1.2 x [tex]10^{-17}[/tex] m.

(c) The frequency of the photon is 2.47 x [tex]10^{25}[/tex] Hz.

Explanation:

Let;

[tex]E_{1}[/tex] = -13.60 ev

[tex]E_{2}[/tex] = -3.40 ev

(a) Energy of the emitted photon can be determined as;

[tex]E_{2}[/tex] - [tex]E_{1}[/tex] = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x [tex]10^{-9}[/tex])

[tex]E_{2}[/tex] - [tex]E_{1}[/tex] = 1.632 x [tex]10^{-8}[/tex] Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x [tex]10^{-8}[/tex] Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x [tex]10^{-34}[/tex] Js), c is the speed of light (3 x [tex]10^{8}[/tex] m/s) and λ is the wavelength.

10.20(1.6 x [tex]10^{-9}[/tex]) = (6.6 x [tex]10^{-34}[/tex] * 3 x [tex]10^{8}[/tex])/ λ

λ = [tex]\frac{1.98*10^{-25} }{1.632*10^{-8} }[/tex]

  = 1.213 x [tex]10^{-17}[/tex]

Wavelength of the photon is 1.2 x [tex]10^{-17}[/tex] m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x [tex]10^{-8}[/tex]  = 6.6 x [tex]10^{-34}[/tex] x f

f = [tex]\frac{1.632*10^{-8} }{6.6*10^{-34} }[/tex]

 = 2.47 x [tex]10^{25}[/tex] Hz

Frequency of the emitted photon is 2.47 x [tex]10^{25}[/tex] Hz.

A bar magnet is held in place while another bar magnet is placed near it. The second bar magnet spins around and
attaches to the first magnet on one end. Which statement is correct about the energy stored in the magnetic field?
(1 point)
O The increase in the energy stored in the system is proportional to the increase in kinetic energy.
O The decrease in the energy stored in the system is proportional to the decrease in kinetic energy.
O The increase in the energy stored in the system is proportional to the decrease in kinetic energy.
O The decrease in the energy stored in the system is proportional to the increase in kinetic energy.

Answers

Answer:

o

Explanation:

the increase energy stored in thw system is proportional to the decrease in kinetic energy

The increase in the energy stored in the system is proportional to the decrease in kinetic energy.

What is energy stored in the system ?

"The energy stored in the system means the sum of energies of both magnets when second magnet stop spin and stick with first magnet. It is clear that the rotational kinetic energy of second magnet decrease and becomes part of the system. Hence the increase in the energy stored in the system is proportional to the decrease in kinetic energy."

What is kinetic energy ?

"Kinetic energy is a form of energy that an object or a particle has by reason of its motion. If work, which transfers energy, is done on an object by applying a net force, the object speeds up and thereby gains kinetic energy."

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Wegener’s theories of plate movement became the basis for the development of the theory of

Answers

Answer:

Wegener proposed his theory of "continental drift." Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines 

Plate tectonic theory had its beginnings in 1915 when Alfred Wegener proposed his theory of "continental drift." Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines of many coastlines (like South America and Africa) look like they fit together like a puzzle.

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