A sound source A and a reflecting surface B move directly toward each other. Relative to the air, the speed of source A is 28.7 m/s, the speed of surface B is 62.2 m/s, and the speed of sound is 334 m/s. The source emits waves at frequency 1110 Hz as measured in the source frame. In the reflector frame, what are (a) the frequency and (b) the wavelength of the arriving sound waves? In the source frame, what are (c) the frequency and (d) the wavelength of the sound waves reflected back to the source?

Answer:

Quarks

Explanation:

Quarks are the smallest constituents of protons and neutrons. In particular, the proton and the neutron are a combination of two types of quarks:

- Quark up: it has a charge of [tex]+\frac{2}{3}e[/tex]

- Quark down: it has a charge of [tex]-\frac{1}{3}e[/tex]

Both the proton and the neutron consists of 3 quarks:

- The proton consists of 2 quarks up and 1 quark down, so that its charge is

[tex]q_p = +\frac{2}{3}e+\frac{2}{3}e-\frac{1}{3}e = +e[/tex]

- The neutron consists of 1 quark up and 2 quarks down, so that its charge is

[tex]q_n = +\frac{2}{3}e-\frac{1}{3}e-\frac{1}{3}e = 0[/tex]

Protons and neutrons, particles of matter, are composed of smaller particles known as quarks. Two types (up and down) of the six kinds of quarks form protons and neutrons. These are held together by strong forces carried by particles called gluons.

The particles of matter that make up protons and neutrons are known as quarks. There are six types of quarks, but protons and neutrons are made from only two types: up quarks and down quarks. A proton is composed of two up quarks and one down quark while a neutron is composed of one up quark and two down quarks. These quarks are held together by strong forces through an exchange of particles called gluons.

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a) The differential  equation for its volume, V be - dV/dt = kA.

b) If the initial volume is s, the value of volume  be  (4/3)π(  (3s/4π)^(1/3)  - kt)³.

The space that any three-dimensional solid occupies is known as its volume. These solids can take the form of a cube, cuboid, cone, cylinder, or sphere.

According to the question:

the spherical snowball melts at a rate proportional to its surface area.

Let V be its volume and S be its  surface area. So, the differential equation of it can be represented as:

- dV/dt ∝ A

⇒ - dV/dt = k A

Where, negative sign represents that volume of the show ball is decreasing during melting.

⇒  - dV = k A dt

for spherical snowball

dV = 4πr²dr

A = 4πr²

So, from this equation,

-dr = kdt

Integrating, we get:

-r = kt +c

At t=0, V = s, so: r = (3s/4π)^(1/3) and c = - (3s/4π)^(1/3)

Therefore:

- r  = kt +  - (3s/4π)^(1/3)

r =   (3s/4π)^(1/3)  - kt

Hence, the  value of volume at time t be (4/3)π(  (3s/4π)^(1/3)  - kt)³

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Final answer:

The differential equation for the volume, V, of a spherical snowball as it melts at a rate proportional to its surface area is dv/dt = -k * dS/dt. To solve the differential equation, we can integrate both sides to obtain V = -k * S + C, where C is the constant of integration. By substituting the initial volume s into the equation, we find that V = s.

Explanation:

(a) The differential equation for the volume, V, of the spherical snowball can be expressed as:

dV/dt = -k * dS/dt

where k is the constant of proportionality and dS/dt is the rate of change of the surface area.

(b) To solve the differential equation, we need to integrate both sides with respect to time:

dV = -k * dS

Assuming the initial volume is s, we can integrate:

∫dV = -k * ∫dS

The integral of dV is V and the integral of dS is S, so we get:

V = -k * S + C

where C is the constant of integration. Since we know that the initial volume is s, we can substitute the values:

s = -k * S + C

To solve for C:

C = s + k * S

Substituting back into the equation:

V = -k * S + s + k * S

V = s

Noble gases have what is known as a stable octet of electrons - the completeness of this outer shell of electrons legislates the reactivity of the compound. For example, Sodium, in its elemental form, has only one electron in the outermost shell - this makes it extremely reactive with other elements. As you move to the right of the periodic table, the shell begins to fill and the relative reactivity decreases. Once the shell is full, the need to react no longer exists. Forming any bonds would actually result in an unfavorable coupling, so naturally occurring compounds are practically nonexistent.

Hope this helps!

Noble gases do not form compounds readily due to their stable electronic configuration with a full valence shell, which makes them unreactive. They have high ionization energies and do not favor sharing or transferring electrons except under high pressure and temperature.

The noble gases such as helium, neon, argon, and others do not form compounds readily because they possess a complete valence shell, often referred to as a full octet. This stable electronic configuration means they have no tendency to gain, lose, or share electrons in chemical reactions, making them very unreactive. These gases have high ionization energies, which means it requires a considerable amount of energy to remove an electron from them. Additionally, if they were to accept an extra electron, it would be placed in a higher and less stable energy level, making the compound formed less stable.

Noble gases like xenon, krypton, and radon are large enough that their outer electrons are further from the nucleus, hence they can form compounds under specific conditions such as high pressure and temperature. Chemical reactions that involve sharing or transferring electrons are generally not favored by noble gases because of their stable valence shells, except under extreme conditions.

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Answer:

red shift, indicating that the universe is expanding

Explanation:

Doppler effect occurs when a source of a wave (e.g. light, or sound waves) moves relative to an observer; as a result of this relative motion, the wavelength of the wave appears lengthened/shortened to the observer. Two situations can occur:

- The source of the wave is moving towards the observer - in this case, the wavelength of the wave becomes shorter. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the blue-end of the spectrum (blue-shift)

- The source of the wave is moving away from the observer - in this case, the wavelength of the wave becomes longer. If the wave is visible light, such as the light emitted by distant galaxies, this means that the wavelength of the light shifts towards the red-end of the spectrum (red-shift)

In our universe, we observe a red-shift for all the distant galaxies: this means that these galaxies are moving away from us, so this is an indication that the universe is expanding.

Light from distant galaxies most likely shows a: A. red shift, indicating that the universe is expanding.

A redshift can be defined as a displacement (shift) of the spectral lines of celestial or astronomical objects toward longer wavelengths (the red end of an electromagnetic spectrum), as a result of the Doppler effect.

Basically, a redshift occur when observing a star from planet Earth because the star is moving away from planet Earth.

Hence, a redshift is considered to be a subtle change in the color of visible electromagnetic radiation from stars (starlight), as observed from planet Earth.

In Astronomy, a red shift is used by astronomers in the following ways:

I. For tracking the expansion (increase in size) of the universe.

II. To find a planet.

III. For measuring the speeds of galaxies.

In conclusion, light from distant galaxies most likely shows a red shift, which indicates that the universe is expanding.

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Answer:

d = 84 m

Explanation:

As we know that when an object moves with uniform acceleration or deceleration then we can use equation of kinematics to find the distance moved by the object

here we know that

initial speed [tex]v_i = 42 m/s[/tex]

final speed [tex]v_f = 0[/tex]

time taken by the car to stop

[tex]t = 4s[/tex]

now the distance moved by the car before it stop is given as

[tex]d = \frac{v_f + v_i}{2} \times t [/tex]

now we have

[tex]d = \frac{42 + 0}{2} \times 4[/tex]

[tex]d = 84 m[/tex]

The car that travels to the right with a speed of 42 m/s, skids 84 meters for 4 seconds before it comes to a stop.

The distance traveled by car before coming to a stop can be calculated with the following equation:

[tex] v_{f}^{2} = v_{i}^{2} + 2ad [/tex]    (1)

Where:

[tex] v_{f}[/tex]: is the final speed = 0 (it stops)

[tex] v_{i}[/tex]: is the initial speed = 42 m/s

a: is the acceleration

d: is the distance =?  

We need to find the acceleration. We can use the next equation:

[tex] v_{f} = v_{i} + at [/tex]    (2)  

Where:

t: is the time = 4.0 s

Hence, the acceleration is:

[tex]a = \frac{v_{f} - v_{i}}{t} = \frac{0 - 42 m/s}{4.0 s} = -10.5 m/s^{2}[/tex]

Now, the car skid the following meters before coming to a stop (eq 1).

[tex]d = \frac{v_{f}^{2} - v_{i}^{2}}{2a} = \frac{-(42 m/s)^{2}}{2(-10.5 m/s^{2})} = 84 m[/tex]  

Therefore, the car skids 84 meters before coming to a stop.

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Answer:

[tex]2.25\cdot 10^5 J[/tex]

Explanation:

First of all, we have to find the power used by the hair dryer, which is given by

[tex]P=\frac{V^2}{R}[/tex]

where

V = 120 V is the voltage

[tex]R=9.6 \Omega[/tex] is the resistance of the hair dryer

Substituting,

[tex]P=\frac{(120 V)^2}{9.6 \Omega}=1500 W[/tex]

Now we can find the total electrical energy used, given by

[tex]E=Pt[/tex]

where P is the power and

t = 2.5 min = 150 s is the time

Substituting,

[tex]E=(1500 W)(150 s)=2.25\cdot 10^5 J[/tex]

Using the given equation:

di = 20.0 * 10.0 / 20.0 - 10.0

di = 200/10

di = 20.0 cm

The answer is A.

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[tex]\boxed{Explained\:Answer}[/tex]

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A is the answer.

The decibel rating of sound source B is 80 dB. So, the decibel rating of sound source B would be 50 dB + 10 * log10(1000) = 50 dB + 10 * 3 = 50 dB + 30 dB = 80 dB

The decibel rating of sound source B can be calculated by taking the decibel rating of sound source A and adding 10 times the logarithm base 10 of the intensity ratio between sound source B and A.

Since sound source B is 1000 times more intense than sound source A, the intensity ratio is 1000.

So, the decibel rating of sound source B would be 50 dB + 10 * log10(1000) = 50 dB + 10 * 3 = 50 dB + 30 dB = 80 dB.

Answer:

F. The mesosphere contains fewer oxygen molecules than the stratosphere.

Explanation:

The layers of the atmosphere are divided into:

1. Troposphere

2. Stratosphere

3. Mesosphere

4. Thermosphere

5. Exosphere

The troposphere extends from the earth surface to about 10km upwards. This is the region of the greatest atmospheric pressure and where all weather conditions arises. In the troposphere, the higher one goes the cooler it becomes.

The stratosphere lies on the troposphere and it is about 50km from the top of the troposphere. The stratosphere is the region where ozone, an oxygen molecule, forms a layer. The higher you go in the stratosphere, the warmer it becomes.

The mesosphere is about 90km thick and it has less gas density. It extends from the top of the stratosphere upwards. The gases here are sparse and atmospheric pressure is lesser than that of the surface. Here, the higher you go in the mesosphere, the cooler it becomes.

Only option F is correct: the mesosphere contains fewer oxygen molecules than the stratosphere.

Answer:

F: The mososphere contains fewer oxygen molecules than the Stratosphere.

Explanation:

Answer:

1/2

Explanation:

The energy stored in a capacitor is given by:

[tex]U=\frac{1}{2}CV^2[/tex]

where

C is the capacitance

V is the potential difference

For capacitor 1, we have

[tex]U_1=\frac{1}{2}C_1V_1^2[/tex]

Capacitor 2 has

[tex]C_2 = \frac{C_1}{2}[/tex] (half the capacitance of capacitor 1)

[tex]V_2 = 2 V_1[/tex] (twice the potential difference of capacitor 1)

So the energy of capacitor 2 is

[tex]U_2=\frac{1}{2}C_2V_2^2=\frac{1}{2}(\frac{C_1}{2})(2V_1)^2=C_1 V_1^2[/tex]

So, the ratio between the two energies is

[tex]\frac{U_1}{U_2}=\frac{\frac{1}{2}C_1 V_1^2}{C_1 V_1^2}=\frac{1}{2}[/tex]

What happens when you put them together is you will make a stronger magnet that will pick up alot of metal

To calculate the index of refraction of the glass block in Part 2, we use Snell's law. The calculated value is 1.638, with a percentage error of 7.05% compared to the accepted value of 1.53. Experiments like this one can help identify materials based on their index of refraction, although similar indexes can make it difficult to differentiate between materials.

To calculate the index of refraction of the glass block in Part 2, we can use Snell's law which states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the velocities of light in the two mediums. Using the values of the angles of incidence and refraction provided in the question, we can calculate the index of refraction for each trial.

For Trial 1, the index of refraction is calculated as sin(20°) / sin(14°) = 1.554. For Trial 2, the index of refraction is sin(40°) / sin(25°) = 1.695. And for Trial 3, the index of refraction is sin(60°) / sin(35°) = 1.666. We can calculate the average index of refraction by taking the average of these three values, which is (1.554 + 1.695 + 1.666) / 3 = 1.638.

The accepted value of the index of refraction for glass is 1.53. To calculate the percentage error, we can use the formula (|experimental value - accepted value| / accepted value) x 100%. So, the percentage error is (|1.638 - 1.53| / 1.53) x 100% = 7.05%.

Experiments like this one can be used to identify materials based on their index of refraction. Each material has a unique index of refraction, so by measuring the index of refraction of an unknown material and comparing it to known values, we can determine the material. However, one difficulty in identifying materials based on their indexes of refraction is that different materials can have similar indexes. This can cause confusion and make it challenging to differentiate between materials solely based on their index of refraction.

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Answer:

754.6 m

Explanation:

The GPE (Gravitational potential energy) of an object with respect to the ground is given by

[tex]GPE = mgh[/tex]

where

m is the mass of the object

g = 9.8 m/s^2 is the acceleration due to gravity

h is the heigth above the ground

Here we have

m = 12,400.05 kg is the mass

GPE = 91,700,000.00J is the GPE

Solving the formula for h, we find the heigth:

[tex]h=\frac{GPE}{mg}=\frac{91,700,000.00J}{(12,400.05 kg)(9.8 m/s^2)}=754.6 m[/tex]

Final answer:

The 12,400.05kg generator with a gravitational potential energy of 91,700,000J was approximately 750 meters above the ground, calculated using the formula for GPE which is GPE = mgh.

Explanation:

The student asked how far above the ground a 12,400.05kg generator was when it had a gravitational potential energy (GPE) of 91,700,000.00J. To find the height, we use the formula for gravitational potential energy: GPE = mgh, where m is mass in kilograms, g is the acceleration due to gravity (9.8 m/s2), and h is the height in meters. Solving for h, we rearrange the formula to h = GPE / (mg).

By inserting the values we have:

h = 91,700,000J / (12,400.05kg × 9.8m/s2)

After calculating:

h ≈ 750 meters

Therefore, the generator was approximately 750 meters above the ground.

Answer:

1. The current will drop to half of its original value.

Explanation:

The problem can be solved by using Ohm's law:

[tex]V=RI[/tex]

where

V is the voltage across the circuit

R is the resistance of the circuit

I is the current

We can rewrite it as

[tex]I=\frac{V}{R}[/tex]

In this problem, we have:

- the resistance of the circuit remains the same: R' = R

- the voltage is decreased to half of its original value: [tex]V'=\frac{V}{2}[/tex]

So, the new current will be

[tex]I'=\frac{V'}{R'}=\frac{V/2}{R}=\frac{1}{2}\frac{V}{R}=\frac{I}{2}[/tex]

so, the current will drop to half of its original value.

Final answer:

If the voltage across a circuit decreases to half its original value while the resistance remains constant, the current will drop to half of its original value.

Explanation:

According to Ohm's law, the current in a circuit is directly proportional to the voltage and inversely proportional to the resistance. If the voltage across a circuit decreases to half its original value while the resistance remains constant, the current will also decrease proportionally. Therefore, the correct answer is option 1: The current will drop to half of its original value.

Answer:

Option c

Explanation:

A concave lens always produces a virtual upright and small image of an object irrespective to the location of object.

Option d is also a concave lens it is called plan concave lens.

A lens that has at least one curved surface, analogous to the inner surface of a spherical. A concave lens is an optical device that can only produce pictures of objects that are smaller than the original.

A lens that has at least one curved surface, analogous to the inner surface of a spherical. A concave lens bends a straight light beam away from the source, creating a smaller, more upright virtual image.

They're used to treat myopia since they reduce the size of distant objects.

A concave lens is an optical instrument that can only generate pictures smaller than the original object. The concave lenses are B and D . As a result, option c is correct.

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Answer:

its temperature must increase.

Explanation:

An isobaric expansion is a transformation in which the pressure of the gas does not change (isobaric), while the volume increases (expansion).

Since the pressure does not change,

"its pressure must increase."

is a false statement.

The 1st law of thermodynamics is

[tex]\Delta U = Q - W[/tex]

where

[tex]\Delta U[/tex] is the change in internal energy of the gas, which is proportional to the change in temperature: [tex]\Delta U \propto \Delta T[/tex]

Q is the heat supplied to the gas

W is the work done by the gas, which is given by

[tex]W=p\Delta V[/tex]

where p is the pressure and [tex]\Delta V[/tex] is the change in volume. Since the gas is expanding, we can say that [tex]\Delta V>0[/tex], so the gas does positive work:

[tex]W>0[/tex]

This means that the option

"the gas does no work."

is false.

Moreover, from the ideal gas law

[tex]pV=nRT[/tex] (2)

we also know that the temperature of the gas is increasing (because p, the pressure, n the number of moles, and R, the gas constant, are all constant in this process, and since the volume V is increasing, than the temperature T must be increasing also)

So, we know that the option

"its internal (thermal) energy does not change. "

is false.

Finally, in an isobaric expansion, in order to keep the pressure constant heat should be supplied to the system, so

"no heat enters or leaves the gas."

is also wrong

We also said from (2) that the temperature of the gas is increasing, therefore the statement

"its temperature must increase."

is the only correct one.

Answer:

16 times

Explanation:

The international space station orbits the earth every 90 minutes. That means the international space station will orbit the earth 16 times within a 24 hour period.

Final answer:

The International Space Station orbits Earth once every 90 minutes in a Low Earth Orbit, circling the planet around 16 times per day, whereas geostationary satellites orbit once every 24 hours at a fixed point above the Earth's surface.

Explanation:

The International Space Station (ISS) orbits Earth approximately once every 90 minutes. It is in a type of orbit known as Low Earth Orbit (LEO), which ranges in altitude from 160 km to 2,000 km above Earth's surface, with the ISS specifically orbiting at around 370 km in the thermosphere. This rapid orbit allows the ISS to circumnavigate Earth around 16 times per day. In contrast, satellites in a geostationary orbit, such as weather and communications satellites, orbit at a much higher altitude of 36,000 km and take exactly 24 hours to complete one orbit. This synchronized period with Earth's rotation keeps them stationary over one specific location on the planet's surface.

A) [tex]C_3 < C_1 < C_2[/tex]

The capacitance of a parallel-plate capacitor is given by

[tex]C=\epsilon_0 \frac{A}{d}[/tex]

where

A is the plate area

d is the plate separation

Here we have:

- Capacitor 1: plate area A, plate separation d

capacitance: [tex]C_1=\epsilon_0 \frac{A}{d}[/tex]

- Capacitor 2: plate area 2A, plate separation d

capacitance: [tex]C_2=\epsilon_0 \frac{2A}{d} = 2C_1[/tex]

- Capacitor 3: plate area A, plate separation 2d

capacitance: [tex]C_3=\epsilon_0 \frac{A}{2d}=\frac{C_1}{2}[/tex]

So ranking the three capacitor from least to greatest capacitance we have:

[tex]C_3 < C_1 < C_2[/tex]

2. [tex]V_2 < V_1 < V_3[/tex]

The three capacitors have same amont of charge, Q.

The potential difference between the plates on each capacitor is given by

[tex] V = \frac{Q}{C}[/tex]

so here we have

- Capacitor 1: [tex]C = C_1[/tex]

Potential difference: [tex] V_1 = \frac{Q}{C_1}[/tex]

- Capacitor 2: [tex]C = 2C_1[/tex]

Potential difference: [tex] V_2= \frac{Q}{2C_1}=\frac{ V_1}{2}[/tex]

- Capacitor 3: [tex]C = \frac{C_1}{2}[/tex]

Potential difference: [tex] V_3 = \frac{Q}{C_1/2}=2 V_1 [/tex]

So ranking the three capacitor from least to greatest potential difference we have:

[tex]V_2 < V_1 < V_3[/tex]

C. [tex]E_2 < E_1 = E_3[/tex]

The electric field magnitude between the plates of a capacitor is given by

[tex]E=\frac{V}{d}[/tex]

where V is the potential difference between the plates and d is the plate separation

So here we have

- Capacitor 1: potential difference [tex]V_1[/tex], plate separation d

electric field: [tex]E_1 = \frac{V_1}{d}[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex], plate separation d

electric field: [tex]E_2=\frac{V_1/2}{d} =\frac{V_1}{2d}= \frac{E_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex], plate separation 2d

electric field: [tex]E_3=\frac{2 V_1}{2d} =\frac{V_1}{d}= E_1[/tex]

So ranking the three capacitor from least to greatest electric field we have:

[tex]E_2 < E_1 = E_3[/tex]

D. [tex]U_2 < U_1 < U_3[/tex]

The energy stored in a capacitor is

[tex]U=\frac{1}{2}QV[/tex]

where Q is the same for the three capacitors

Here we have

- Capacitor 1: potential difference [tex]V_1[/tex]

energy: [tex]U_1 = \frac{1}{2}QV_1[/tex]

- Capacitor 2: potential difference [tex]\frac{V_1}{2}[/tex]

energy: [tex]U_2 = \frac{1}{2}Q\frac{V_1}{2}=\frac{U_1}{2}[/tex]

- Capacitor 3: potential difference [tex]2V_1[/tex]

energy: [tex]U_3 = \frac{1}{2}Q(2 V_1)=2 U_1[/tex]

So ranking the three capacitor from least to greatest energy we have:

[tex]U_2 < U_1 < U_3[/tex]

E. [tex]u_2 < u_1 = u_3[/tex]

The energy density in a capacitor is given by

[tex]u=\frac{1}{2}\epsilon_0 E^2[/tex]

where E is the electric field strength

Here we have

- Capacitor 1: electric field [tex]E_1[/tex]

Energy density: [tex]u_1=\frac{1}{2}\epsilon_0 E_1^2[/tex]

- Capacitor 2: electric field [tex]\frac{E_1}{2}[/tex]

energy density: [tex]u_2=\frac{1}{2}\epsilon_0 (\frac{E_1}{2})^2=\frac{E_1}{4}[/tex]

- Capacitor 3: electric field [tex]E_1[/tex]

Energy density: [tex]u_3=\frac{1}{2}\epsilon_0 E_1^2[/tex]

So ranking the three capacitor from least to greatest energy density we have:

[tex]u_2 < u_1 = u_3[/tex]

Capacitor 2 has the highest capacitance, potential difference, energy stored, and energy density, followed by Capacitor 1, and then Capacitor 3.

Part A: The formula for capacitance is C = εA/d, where C is the capacitance, ε is the permittivity of the material between the plates, A is the plate area, and d is the plate separation. Comparing the three capacitors, Capacitor 1 has a capacitance of C1 = εA/d, Capacitor 2 has a capacitance of C2 = ε(2A)/d, and Capacitor 3 has a capacitance of C3 = εA/(2d). Since C2 = 2C1 and C3 = 1/2C1, Capacitor 2 has the highest capacitance, followed by Capacitor 1, and then Capacitor 3.

Part B: The potential difference between the plates of a capacitor is given by V = Q/C, where V is the potential difference, Q is the charge stored on the plates, and C is the capacitance. Since all three capacitors store the same amount of charge, their potential differences are directly proportional to their capacitances. Therefore, Capacitor 2 has the highest potential difference, followed by Capacitor 1, and then Capacitor 3.

Part C: The electric field magnitude between the plates of a capacitor is given by E = V/d, where E is the electric field magnitude and d is the plate separation. Comparing the three capacitors, Capacitor 1 has an electric field magnitude of E1 = V/d, Capacitor 2 has an electric field magnitude of E2 = (2V)/d, and Capacitor 3 has an electric field magnitude of E3 = V/(2d). Since E2 = 2E1 and E3 = 1/2E1, Capacitor 2 has the highest electric field magnitude, followed by Capacitor 1, and then Capacitor 3.

Part D: The energy stored in a capacitor is given by U = (1/2)CV^2, where U is the energy stored, C is the capacitance, and V is the potential difference. Comparing the three capacitors, Capacitor 1 has an energy stored of U1 = (1/2)C1V^2, Capacitor 2 has an energy stored of U2 = (1/2)C2V^2, and Capacitor 3 has an energy stored of U3 = (1/2)C3V^2. Since C2 > C1 > C3, Capacitor 2 has the highest energy stored, followed by Capacitor 1, and then Capacitor 3.

Part E: The energy density of a capacitor is given by u = U/Vd, where u is the energy density, U is the energy stored, V is the volume between the plates, and d is the plate separation. Comparing the three capacitors, Capacitor 1 has an energy density of u1 = U1/(Vd), Capacitor 2 has an energy density of u2 = U2/(2Vd), and Capacitor 3 has an energy density of u3 = U3/(2Vd). Since u2 = 2u1 and u3 = 1/2u1, Capacitor 2 has the highest energy density, followed by Capacitor 1, and then Capacitor 3.

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Answer:

0.004 m

Explanation:

For light passing through a single slit, the position of the nth-minimum in the diffraction pattern is given by

[tex]y=\frac{n\lambda D}{d}[/tex]

where

[tex]\lambda[/tex] is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

Therefore, the width of the central maximum is equal to twice the value of y for n=1 (first minimum):

[tex]w=2\frac{\lambda D}{d}[/tex]

where we have

[tex]\lambda=500 nm = 5\cdot 10^{-7}m[/tex] is the wavelength

D = 2.0 m is the distance of the screen

[tex]d=0.50 mm=5\cdot 10^{-4}m[/tex] is the width of the slit

Substituting, we find

[tex]w=2\frac{(5\cdot 10^{-7} m)(2.0 m)}{5\cdot 10^{-4} m}=0.004 m[/tex]

Explanation:

The Compton Effect is related to an experiment the American physicist Arthur H. Compton made, consisting in the scattering of photons from electrons, proving photons have momentum.

In additon, Compton proved that when a photon collides with a free electron, the photon loses part of its energy [tex]E[/tex], which is given by the following equation:

[tex]E=h.\nu[/tex]   (1)

Where:

[tex]h[/tex] is the Planck constant

[tex]\nu[/tex] is the frequency of the photon

So, the energy of the photon is directly proportional to its frequency, therefore, if energy decreases in the Compton effect, the frequency decreases as well.

This also means, the photon wavelength increases (because there is an inverse relation between frequency and wavelength)

It is important to note the frequency (or wavelength) of the scattered radiation depends only on the scattering angle [tex]\theta[/tex], which is an important element in the Compton Shift [tex]\Delta \lambda[/tex] equation:

[tex]\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta)[/tex]  

Where:  

[tex]\lambda_{c}[/tex] is a constant whose value is given by [tex]\frac{h}{m_{e}.c}[/tex], being [tex]m_{e}[/tex] the mass of the electron and [tex]c[/tex] the speed of light in vacuum.  

Hence, the correct option is B.

Answer:

Because on different surfaces there's more/less friction. Smooth surfaces will allow the duster to accelerate while rough surfaces will decrease the acceleration.

Answer:

Because of the difference in the magnitude of friction.

Explanation:

Friction is defined as the force which is produced between the two surfaces when these two surfaces are in contact with each other.

When we observing the different surfaces there will be different value of friction it is less or high with respect to different surfaces.

Smooth surfaces which posses less friction allow duster to accelerate while on the other hand rough surface will decrease the acceleration of the duster due to high friction.

Answer:-10 miles per second

Explanation:

Acceleration = (change in speed) / (time for the change)

Change in speed = (speed at the end) - (speed at the beginning)

Change in speed = (10 mph) - (50 mph) = -40 mph

Acceleration = (-40 mi/hr) / (4 sec)

Acceleration = -10 mi/hr-sec

To get a more comfortable, familiar unit for the answer . . .

Acceleration =

(-10 mile/hour-second) x (1,609.3 meter/mile) x (1 hour / 3,600 sec)

Acceleration = (-10 x 1,609.3 / 3,600) (mile-meter-hour / hour-sec-mile-sec)

Acceleration = -4.47 m/s²

The Ecliptic refers to the orbit of the Earth around the Sun. Therefore, for an observer on Earth it will be the apparent path of the Sun in the sky during the year, with respect to the "immobile background" of the other stars.

It should be noted that the ecliptic plane (which is the same orbital plane of the Earth in its translation movement) is tilted with respect to the equator of the planet about [tex]23\°[/tex] approximately. This is due to the inclination of the Earth's axis.

Hence, the correct option is Earth's orbital path around the Sun.

The ecliptic plane is defined by Earth's orbit around the Sun. Therefore, Earth's orbital path around the Sun lies in the ecliptic plane, while lines connecting Earth and Polaris, Earth's north and south poles, or Earth's equator do not.

The ecliptic plane is an imaginary flat surface defined by the Earth's orbit around the sun. It is on this plane that the Earth travels in its path around the Sun. Thus, out of the choices given, the option that lies in the ecliptic plane is Earth's orbital path around the Sun.

Polaris, the North Star, is not in alignment with this plane, and neither is a line connecting Earth’s north and south poles. The Earth's equator doesn't lie in the ecliptic plane because Earth’s rotational axis is tilted relative to the ecliptic by about 23.5 degrees.

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Answer:

C

Explanation:

The units of velocity are m/s, not Hz.  Make sure you wrote it correctly.

The speed of sound in air at sea level is approximately:

v ≈ 331.4 + 0.6 T

where v is velocity in m/s and T is the temperature in Celsius.

At 56°C:

v ≈ 331.4 + 0.6 (56)

v ≈ 365 m/s

The closest is C.

The velocity of the sound wave at the room temperature of 56° C is

364.6 m/s.

To find the velocity of sound wave, the given temperature is 56° C.

Here, we have to find the velocity not the frequency. The unit is given as Hz but we have to find in m/s.

Velocity:

           The velocity of an object can be defined as the rate of change in the object's position corresponding to a frame of reference and time.

                             [tex]\overline{v}={\frac{\Delta x}{\Delta t}}\\\overline{v} = average velocity\\{\Delta x} = displacement\\{\Delta t} = change in time[/tex]

The speed of sound in air at sea level is approximately:

                             v ≈ 331.4 + 0.6 T

where,

      v - velocity in m/s

      T - temperature in Celsius.

Substituting, at 56°C:

      v ≈ 331.4 + 0.6 (56)

      v ≈ 365 m/s.

Thus, Option C is the correct answer.

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Answer:

c would be your answer just did this on usatest prer

Explanation:

Answer:

Option (c) is correct.

Explanation:

When a ray of light passes from one optical medium to another optical medium, the path of ray of light is deviated, this phenomena is called refraction.

It occurs due to the change in speed of light, as it passes from medium to another.

So, when a laser beam passes through a glass block, the refraction takes place.

Answer:

b. AG, work function=4.74eV

Explanation:

Ultraviolet light starts at the end of the visible light spectrum, where violet light ends:

[tex]\lambda=380 nm =3.8\cdot 10^{-7}m[/tex] (wavelength of lowest-energy ultraviolet light)

So, the lowest energy of ultraviolet light can be found by using the formula

[tex]E=\frac{hc}{\lambda}[/tex]

where

h is the Planck constant

c is the speed of light

Substituting,

[tex]E=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{3.8\cdot 10^{-7} m}=5.23\cdot 10^{-19}J[/tex]

And keeping in mind that

[tex]1 eV = 1.6\cdot 10^{-19}J[/tex]

This energy converted into electronvolts is

[tex]E=\frac{5.23\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=3.27 eV[/tex]

The work function of a metal is the minimum energy needed to extract a photoelectron from the surface of the metal. Therefore, the metals that exhibit photoelectric effect are the ones whose work function is larger than the energy we found previously, so:

b. AG, work function=4.74eV

Because for all the other metals, visible light will be enough to extract photoelectrons.

There are many different types of precipitation —rain, snow, hail, and sleet for example—yet they all have a few things in common. They all come from clouds. They are all forms of water that fall from the sky.

(a)

For this part of the problem, we can ignore the horizontal motion of the engine and consider only the vertical motion.

The vertical position of the engine at time t is given by

[tex]y(t) = h - \frac{1}{2}gt^2[/tex]

where

h is the initial altitude of the airplane

g = 9.81 m/s^2 is the acceleration due to gravity

t is the time

Since the engine takes 30 seconds to hit the ground, t = 30 s when y(t) = 0 (the ground). Substituting into the equation, we find h:

[tex]0 = h - \frac{1}{2}gt^2\\h= \frac{1}{2}gt^2 = \frac{1}{2}(9.81 m/s^2)(30 s)^2=4,415 m \sim 4.5 km[/tex]

(b)

For this part of the problem, we can ignore the vertical motion and consider the horizontal motion only.

Since the engine travels at constant speed along the horizontal direction:

v = 280 m/s

its horizontal position after time t is given by

[tex]x(t) = v t[/tex]

If we substitute

t = 30 s

which is the total duration of the fall, we can find the horizontal distance covered by the airplane during this time:

[tex]x(t) = (280 m/s)(30 s)=8,400 m[/tex]

(c) The engine will be exactly 4.5 km under the plane

Here we have:

- the airplane is moving horizontally, at 4.5 km of altitude, at constant velocity of 280 m/s

- The engine moves both horizontally, also with a horizontal velocity of 280 m/s, and vertically, with acceleration 9.81 m/s^2 towards the ground

Both the plane and the engine moves with same horizontal velocity, so they cover the same horizontal distance (8400 m) during the 30 seconds. The only difference is that the engine falls down approx. 4.5 km, so the engine will be 4.5 km under the plane, when it hits the ground.

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

To find the height if the plane from ground, we use:

H = ut + (1/2)gt²

Where u = initial velocity = 0 m/s, t = time = 30 s, g = acceleration due to gravity = 10 m/s²

H = 0(30) + (1/2)(10)(30)²

H = 4500 m = 4.5 km

The horizontal distance (R) is given by:

R = ut = 280(30) = 8400 m

Let d represent the distance from the engine to the air plane at the ground, hence:

d² = 8400² + 4500²

d = 9529 m

The airplane is 4.5 km high and the horizontal distance that the aircraft engine moves during its fall is 8400 m.

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Answer:

No

Explanation:

The period of a pendulum is given by

[tex]T=2\pi \sqrt{\frac{L}{g}}[/tex]

where

L is the length of the pendulum

g is the acceleration due to gravity

We see that the period of the pendulum depends on the value of g. However, the value of the gravitational acceleration is different at different locations on Earth. In particular, at the top of the mountain the value of g is slightly lower than the value of g at the base of the mountain; in fact, g is given by

[tex]g=\frac{GM}{r^2}[/tex]

where

G is the gravitational constant

M is the Earth's mass

r is the distance from the Earth's center

so since r is greater at the top of the mountain, g is lower, and therefore the period of the pendulum will be slightly longer.

A pendulum clock will not keep perfect time when moved to the top of a mountain due to the decrease in gravity, which causes the pendulum's swing period to lengthen and the clock to run slower.

No, a pendulum clock will not keep perfect time when moved from the base to the top of a mountain. The reason lies in the nature of pendulums and how they keep time. A pendulum on a clock maintains a constant period, which is influenced by the length of the pendulum and the acceleration due to gravity. Simple harmonic oscillation, which is observed when the displacement of the pendulum is small, ensures that the resting force is directly proportional to its displacement.  

As a result, as we ascend to higher altitudes, the acceleration due to gravity reduces, producing a slightly longer period in the pendulum's swing. This change is small but would affect the timekeeping of a pendulum clock, making it run a little slower at higher altitudes. Therefore, a pendulum clock that keeps perfect time at the base of a mountain would lose time if taken to the top of the mountain due to the decrease in gravity.

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-- A transmitter has 50 mW output power.  

50 mW is equivalent to +17 dBm.

(+17 is a magic number.  It tells us that the transmitter could very well be based on a single modulated Gunn diode oscillator, which, after resonating and filtering to remove the unwanted puree, hash, and garbage, typically delivers right around +17 dBm at the output.  

-- The power passes through a piece of lossy cable, where it loses 3 dB.

+17 dBm went into the cable.   +14 dBm came out of the other end.  

(The lost 3 dBm warmed the cable.)  

-- The power was then coupled (losslessly) to an antenna with +16 dB "gain".

+14 dBm went into the antenna.  It was shaped and focused so that coming out of the antenna in a certain direction, it sounded as loud as a source that's radiating (+14 + 16) = +30 dBm = 1 watt .

This is NOT 1 watt of real power output.  The antenna has no batteries, it isn't plugged into a wall outlet, and it has no actual 'gain'.  

That 1 watt is "eirp" . . . "Effective Isotropic Radiated Power".  The antenna focuses most of its power in one certain narrow direction, and then, in that direction, it sounds as loud as an antenna would that took 1 watt and spread it equally in all directions.

The EIRP power output of the system in question is 26 dBm.

EIRP (Effective Isotropic Radiated Power) is a measure of the power output of a transmitter, taking into account the gain of the antenna.

To calculate the EIRP, you can use the following formula:

EIRP = transmit power + antenna gain - cable loss

In your scenario, the transmitter has a power output of 50 mW and the cable has a loss of 3 dB. Antenna has 16 dbi of gain.

First convert the power to dBm

50 mW = 10log10(5010^-3) = 13 dBm

Then the EIRP is calculated as:

EIRP = 13 dBm + 16 dBi - 3 dB = 26 dBm

So the EIRP power output of the system in question is 26 dBm.

It's important to note that the max allowed EIRP power in a specific area could be different depending on the standards of the area and the device, Thus this output should be compared to the allowed EIRP in that area.

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