the lambda baryon has the quark composition uds. which particle has the same electric charge as the lambda baryon?

Answer:

speed is the rate of change in distance thus it is scalar physical quantity

while velocity is the rate of change in displacement thus it is a vector physical quantity

Explanation:

vector physical quantity: is a quantity that requires both magnitude and direction to identify

scalar quantity: requires only magnitude to identify.

Answer:

\(\frac{n}{t} = 30\ photons/s\)

Explanation:

The radiated power can be given in terms of the wavelength as follows:

\(Rasiated\ Power = \frac{nE}{t} = \frac{nhc}{\lambda t}\)

where,

Radiated Power = 1.2 x 10⁻¹⁷ W

n = no. of photons = ?

h = plank's constant = 6.625 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 500 nm = 5 x 10⁻⁷ m

t = time

Therefore,

\(1.2\ x\ 10^{-17}\ W = \frac{n(6.625\ x\ 10^{-34}\ J.s)(3\ x\ 10^8\ m/s)}{(5\ x\ 10^{-7}\ m)(t) }\\\\\frac{n}{t} = \frac{1.2\ x\ 10^{-17}\ W}{3.975\ x\ 10^{-19}\ J}\\\\\frac{n}{t} = 30\ photons/s\)

Explanation:

physics allow us to understand the electromagnetic radiation we used to transmit data with fibre optics and satellites and to build computers that interpret those signals and transmit data on the internet

Answer:

E_loss  = 690000 [J]

Explanation:

The principle of conservation of Energy tells us that energy is converted from Kinetic to mechanical when the body moves from the highest point to the point where the velocity is maximum. That is, all potential energy is transformed into kinetic energy.

\(E_{pot}=E_{kin}\)

Now we must determine the kinetic energy at the end, when the roller coaster is at the point where the reference point of power energy is zero.

\(E_{kin}=\frac{1}{2} *m*v^{2}\)

where:

m = mass = 800 [kg]

v = 45 [m/s]

Now replacing:

\(E_{kin}=0.5*800*(45)^{2}\\E_{kin}=810000[J]\)

Now by means of the difference of the energy at the beginning minus the final energy, it determines the amount of energy that is lost by the effects of friction and heat.

\(E_{loss}=1500000-810000\\E_{loss}=690000[J]\)

Even though speed doesn't vary, acceleration is a change in motion that slows down, speeds up, or changes direction.

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Answer:

when an object move from its place is called an mation

The value of resistor R that gives the circuit in the figure a time constant of 22 μs is 220 Ω.

The circuit that is in the figure is shown below:Given that time constant (RC) = 22 μs. To find the value of resistor R, we need to use the formula for the time constant:

RC = τ, where R is the resistance and C is the capacitance of the circuit.

Rearranging the above formula, we get:R = τ / C

Where τ is the time constant and C is the capacitance of the circuit.

From the figure, the capacitance is given as 0.1 μF

.Substituting the values of τ and C in the above formula, we get:

R = (22 × 10⁻⁶ s) / (0.1 × 10⁻⁶ F)

R = 220 Ω

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A suitcase's kinetic frictiοn cοefficient with the grοund is 0.272. the required wοrk is 2164 J.

A fοrce acting between surfaces that are mοving is knοwn as kinetic frictiοn. A fοrce is applied in the οppοsite directiοn οf a bοdy's mοvement οn the surface. The kinetic frictiοn cοefficient between the twο materials will determine the fοrce's magnitude.

The relative speed at which the surfaces in cοntact are mοving is nοt a factοr in determining the amοunt οf kinetic frictiοn. Kinetic frictiοn is different frοm οther frictiοn-like phenοmena like air resistance and viscοus fοrce in this way.

Evaluating:

let the distance is d

Using wοrk energy theοrem:

wοrk dοne = u × m × g ×h

642 = 0.272 × 71.5 × 9.8 × d

d = 3.37 m

the suitcase can be pushed a distance οf 3.37 m fοr the wοrk dοne

wοrk needed = change in kinetic energy

wοrk needed = 0.5 × 73 × 7.7²

wοrk needed =  2164 J

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Answer:

A fraction of the wave will be reflected at the junction and the remainder will be transmitted through the junction into the thin rope

Explanation:

When the junction of the thin and the thick rope is reached by the wave, the wave behaves in the following two ways;

1) A fraction of the incident wave from the thick rope to the junction/boundary of the thin rope will be reflected back to in the direction of the thick rope with a lower amplitude and upright

2) A fraction of the incident wave from the thick rope to the junction/boundary of the thin rope will be transmitted through the thin rope such that the transmitted pulse on the thin rope will have an higher amplitude than that of the thick rope due to the available energy in the wave pulse and the lesser weight of the thin rope.

According to the statement, about the design of a radome for the nose of an aircraft, with the objective of protecting an X-band weather radar, we answer the questions:

Explanation to calculate the thickness and the percentage of the incident power of a radome:

(a) The minimum thickness of the foam that will give no reflections at the center frequency of the band can be calculated using the equation:
\(d = λ/4n\)

where d is the thickness of the foam, λ is the wavelength at the center frequency, and n is the refractive index of the foam.

The center frequency of the band is (8.5 + 10.3)/2 = 9.4 GHz. The wavelength at the center frequency is \(λ = c/f = 3 x 10^8 m/s / 9.4 x 10^9 Hz = 0.0319 m.\)

The refractive index of the foam is \(n = √ε_r = √2 = 1.414.\)

Therefore, the minimum thickness of the foam is:
\(d = λ/4n = 0.0319 m / (4 x 1.414) = 0.00565 m = 5.65 mm\)

(b) The percentage of the incident power that is reflected at each end of the operating frequency band can be calculated using the equation:
\(R = |(n - 1)/(n + 1)|^2\)

where R is the reflection coefficient and n is the refractive index of the foam.

At the lower end of the operating frequency band (8.5 GHz), the wavelength is \(λ = c/f = 3 x 10^8 m/s / 8.5 x 10^9 Hz = 0.0353 m\). The refractive index of the foam is \(n = √ε_r = √2 = 1.414.\)

Therefore, the reflection coefficient at the lower end of the operating frequency band is:
\(R = |(1.414 - 1)/(1.414 + 1)|^2 = 0.0174\)

The percentage of the incident power that is reflected at the lower end of the operating frequency band is 0.0174 x 100 = 1.74%.

(c) The percentages of the incident power that is reflected and transmitted through the radome at the center frequency can be calculated using the equation:
\(R = |(n1 - n2)/(n1 + n2)|^2\)
T = 1 – R

where R is the reflection coefficient, T is the transmission coefficient, n1 is the refractive index of the first material, and n2 is the refractive index of the second material.

The refractive index of the first material (foam) is \(n1 = √ε_r1 = √2 =\)1.414. The refractive index of the second material is \(n2 = √ε_r2 = √4.1 = 2.025.\)

Therefore, the reflection coefficient at the center frequency is:
\(R = |(1.414 - 2.025)/(1.414 + 2.025)|^2 = 0.0757\)

The percentage of the incident power that is reflected at the center frequency is 0.0757 × 100 = 7.57%.

The transmission coefficient at the center frequency is:
T = 1 - R = 1 – 0.0757 = 0.9243

The percentage of the incident power that is transmitted through the radome at the center frequency is 0.9243 × 100 = 92.43%.

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(a) To calculate the minimum thickness of the foam required to give no reflections at the center frequency of the band, the formula  t = λ/(4πn) can be used, where t is the minimum thickness, λ is the wavelength of the center frequency of the band, and n is the refractive index of the foam.

Using the given parameters, the minimum thickness of the foam is calculated as t = 9.8 cm/ (4π(-1)) = 1.54 cm.

(b) Using the thickness calculated in part (a), the percentages of incident power reflected at each end of the operating frequency band can be calculated using the formula Reflected power (dB) = 10 log(r^2), where r is the reflection coefficient.

The reflection coefficient is calculated as r = (n2-n1)/(n2+n1), where n1 is the refractive index of air and n2 is the refractive index of the foam.

For the lower frequency of 8.5 GHz, the reflection coefficient is calculated as r = (-1-1)/(-1+1) = -2, and the percentage of incident power reflected is Reflected power (dB) = 10 log(-2^2) = -20 dB, which is equivalent to 1% of the incident power.

For the higher frequency of 10.3 GHz, the reflection coefficient is calculated as r = (-1-1)/(-1+1) = -2, and the percentage of incident power reflected is Reflected power (dB) = 10 log(-2^2) = -20 dB, which is equivalent to 1% of the incident power.

(c) To calculate the percentages of incident power reflected and transmitted through the radome at the center frequency, the Fresnel equation can be used.

Using the given parameters, the reflection coefficient is calculated as r = (4.1-1)/(4.1+1) = 0.75, and the transmission coefficient is calculated as t = 1-r = 0.25.

Therefore, the percentage of incident power reflected is Reflected power (dB) = 10 log(0.75^2) = -3.01 dB, which is equivalent to 19.96% of the incident power, and the percentage of incident power transmitted is Transmitted power (dB) = 10 log(0.25^2) = -12 dB, which is equivalent to 0.25% of the incident power.

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A helium atom (mass = 4m) moving with speed v collides elastically with a deuterium (hydrogen 2) atom (mass = 2m) at rest.  The percentage change in the kinetic energy of the helium atom after the collision is 50%.

Generally, This is because in an elastic collision, kinetic energy is conserved, and the total kinetic energy of the system is the same before and after the collision.

The helium atom has twice the mass of the deuterium atom, so it has half the velocity after the collision.

Since kinetic energy is given by 1/2mv^2, the velocity is squared, so a decrease in velocity by a factor of two results in a decrease in kinetic energy by a factor of four, which is equivalent to a decrease of 50%.

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the average power transferred from the car during the impact is approximately 343 kW.

F = -ΔK/Δx = (1/2)mv^2/Δx

where Δx is the distance that the car is compressed during the impact.

Substituting the given values, we have:

m = 5000 lb / g = 2268 kg

v = 80 mi/h = 35.7632 m/s

Δt = 0.11 s

Converting the units to SI, we get:

m = 2268 kg

v = 35.7632 m/s

Δt = 0.11 s

The distance that the car is compressed can be estimated based on the deformation of the car's structure, but is not provided in the problem statement. Instead, we can assume that the distance is proportional to the initial speed of the car, which gives:

Δx = kv

where k is a proportionality constant. We can estimate k by assuming that the car is deformed by a constant amount during the impact, which gives:

Δx = 0.5 ft = 0.1524 m

Substituting these values, we get:

Δx = kv

0.1524 m = k * 35.7632 m/s

k ≈ 0.00426 s/m

Now we can calculate the force:

F = (1/2)mv^2/Δx

F ≈ 2.47e+5 N

The work done by the barrier is equal to the force multiplied by the distance, which is:

W = FΔx

W ≈ 3.77e+4 J

Finally, the average power transferred during the impact is:

P = W/Δt

P ≈ 3.43e+5 W

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The average power transferred from the car during the impact is approximately 343 kW.

F = -ΔK/Δx = (1/2)mv^2/Δx. where Δx is the distance that the car is compressed during the impact.

Substituting the given values, we have:

m = 5000 lb / g = 2268 kg

v = 80 mi/h = 35.7632 m/s

Δt = 0.11 s

Converting the units to SI, we get:

m = 2268 kg

v = 35.7632 m/s

Δt = 0.11 s

The distance that the car is compressed can be estimated based on the deformation of the car's structure, but is not provided in the problem statement. Instead, we can assume that the distance is proportional to the initial speed of the car, which gives:

Δx = kv

where k is a proportionality constant. We can estimate k by assuming that the car is deformed by a constant amount during the impact, which gives:

Δx = 0.5 ft = 0.1524 m

Substituting these values, we get:

Δx = kv

0.1524 m = k * 35.7632 m/s

k ≈ 0.00426 s/m

Now we can calculate the force:

F = (1/2)mv^2/Δx

F ≈ 2.47e+5 N

The work done by the barrier is equal to the force multiplied by the distance, which is:

W = FΔx

W ≈ 3.77e+4 J

Finally, the average power transferred during the impact is:

P = W/Δt

P ≈ 3.43e+5 W

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If the current in a long straight wire is increasing, the current is induced in the circular loop will be generated in the nearby circular loop.

Faraday's Law states that the electromotive force (EMF) induced in a circuit is proportional to the rate of change of magnetic flux passing through it. As the current in the straight wire increases, the magnetic field around it strengthens, and the change in magnetic flux through the circular loop causes an induced current. The direction of this induced current is determined by Lenz's Law, which states that the induced current will flow in such a way as to oppose the change in magnetic flux causing it.

In this case, as the magnetic field due to the straight wire increases, the induced current in the circular loop will flow in a direction that generates a magnetic field opposing the increase in the straight wire's magnetic field. The magnitude of the induced current depends on the rate of change of the magnetic field, the loop's size, shape, and distance from the straight wire, as well as the resistance of the loop's material. If the current in a long straight wire is increasing, the current is induced in the circular loop will be generated in the nearby circular loop.

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Answer:

The correct answer is "The reactions stop once equilibrum is achieved"

Explanation:

A system is said to have attained equilibrium when the rate of forward reaction is the same/equal to the rate of backward reaction.  The reaction below shows a chemical reaction that is at equilibrium

2N₂O₅ (aq) ⇄ 4NO₂ (aq) + O₂ (aq)

An equilibrium reaction is always a reversible reaction. When the system is at equilibrium (when there is a balance), the concentration of the reactants and products do not change despite the fact that reaction does not stop; what happens however is that the rate at which the reactant(s) are formed (backward reaction) will be equal to the rate at which the product(s) are formed (forward reaction).

From the explanation above, it can be deduced that "The reactions stop once equilibrum is achieved" is a misconception about a system in chemical equilibrium.

The self-inductance of the inductor that will store 20 J of energy when a 3.0-A current flows through it should be 2.22 H.

The formula for calculating the energy stored in an inductor is \(E = \frac{1}{2}  LI^2\), where E is the energy, L is the self-inductance, and I is the current flowing through the inductor.

In this case, we know that the energy to be stored is 20 J and the current is 3.0 A. Therefore, we can rearrange the formula to solve for L as follows:
\(L = \frac{2E}{I^2}\)

Substituting the given values, we get:
\(L = \frac{2 *20 J}{(3.0 A)^2}  = 2.22 H\)
Therefore, the self-inductance of the inductor should be 2.22 H.
The self-inductance of an inductor can be calculated using the formula L = 2E / I^2, where E is the energy to be stored and I is the current flowing through the inductor. In this case, the self-inductance of the inductor that can store 20 J of energy when a 3.0-A current flows through it is 2.22 H.

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The acceleration of the light truck travelling around a flat curve with a radius of 140 m and a speed of 26 m/s is 17.14 m/s².

The acceleration of a light truck travelling around a flat curve with a radius of 140 m and a speed of 26 m/s can be calculated by using the equation: acceleration = (v2/r).

In this equation, ‘v’ is the speed of the truck, and ‘r’ is the radius of the curve.

Therefore, when substituting these values into the equation, we get: acceleration

= (262/140)

= 17.14 m/s².

This means that the acceleration of the light truck travelling around a flat curve with a radius of 140 m and a speed of 26 m/s is 17.14 m/s².

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Answer:

The volume of the gas becomes three times the initial volume.

Explanation:

Given that the pressure is constant, and temperature changes from -173degree C to 27degree C.

So, the initial temperature, \(T_1\) = -173 degree C = -173+273 = 100 K.

The final temperature, \(T_2\)= 27 degree C = 27+273=300 K.

As the pressure is constant, so \(P_1=P_2\).

Let V_1 and V_2 be the initial and final volume respectively.

Assuming that the given gas is ideal gas.

So, applying the ideal gas equation

PV=nRT

where n is the number of moles of the gas and R is the universal gas constant.

For the initial state, \(P_1V_1=n_1RT_1\cdots(i)\)

and for the final state, \(P_2V_2=n_2RT_2 \cdots(ii)\)

Dividing the equation (i) by (ii), we have

\(\frac {P_1V_1}{P_2V_2}=\frac {n_1RT_1}{n_2RT_2} \\\\\frac {P_1V_1}{P_2V_2}=\frac {n_1T_1}{n_2T_2}\)

As the mass of the gas is not changing, so \(n_1=n_2\), then

\(\frac {P_1V_1}{P_2V_2}=\frac {T_1}{T_2}\)

As the pressure is not changing, so \(P_1=P_2\), then

\(\frac {V_1}{V_2}=\frac {100}{300}\)

\(V_2=3V_1\)

So, the volume of the gas becomes three times the initial volume.

The gravitational force would triple between earth and moon attraction.

The gravitational attraction of Earth to the apple are both the same. in step with Newton, every person draws each other frame with a force that, for any  our bodies, is without delay proportional to the prod-cut in their loads and inversely proportional to the square of the distance between their centers.

In line with the regularly occurring regulation of gravitation,  gadgets entice every different with same pressure but in contrary direction. consequently, the earth attracts the moon with the same force because the moon attracts the earth.

The extra an object's mass, the extra gravitational force it exerts. So, to start answering your question, Earth has a more gravitational pull than the moon honestly because the Earth is extra massive.

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The density of air decreases as the altitude increases. The Temperature value decreases as the pressure value decreases with an increase in altitude.

The density of air or atmospheric density, denoted ρ, is the mass per unit volume of Earth's atmosphere. Air density, like air pressure, decreases with increasing altitude. It also changes with variation in atmospheric pressure, temperature and humidity. At 101.325 kPa (abs) and 20 °C (68 °F), air has a density of approximately 1.204 kg/m3 (0.0752 lb/cu ft), according to the International Standard Atmosphere (ISA). At 101.325 kPa (abs) and 15 °C (59 °F), air has a density of approximately 1.225 kg/m3 (0.0765 lb/cu ft), which is about 1⁄800 that of water, according to the International Standard Atmosphere (ISA).[citation needed] Pure liquid water is 1,000 kg/m3 (62 lb/cu ft).

Air density is a property used in many branches of science, engineering, and industry, including aeronautics; gravimetric analysis; the air-conditioning industry; atmospheric research and meteorology; agricultural engineering (modeling and tracking of Soil-Vegetation-Atmosphere-Transfer (SVAT) models); and the engineering community that deals with compressed air.

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Answer:

56.3⁰

Explanation:

∅ = tan⁻¹ (y/x) = tan⁻¹(15/10) = 56.3⁰

The volume of the space would decrease if the piston were to quickly move out, and the process would then reverse itself, taking another hundredth of a second to complete before everything was back to normal. The time it takes for the molecules to balance out is constant, regardless of how quickly or how far the piston is driven.

In mechanical engineering, a piston and cylinder are two sliding cylinders with closed heads that move reciprocally within a somewhat larger cylindrical chamber under pressure from a fluid, as in an engine or pump. The piston rod, which is rigidly coupled to the piston, can pass through one of the end cover plates of a steam engine's cylinder, which is closed by plates at both ends, using a gland and stuffing box (steam-tight joint).

An internal combustion engine's cylinder is open at one end to allow the connecting rod, which connects the piston to the crankshaft, to freely oscillate and is closed at one end by a plate known as the head.

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Answer:

left side

Explanation:

It's smaller in length but still has a pretty close slope to the right side.

C. Series

Consider resistors in a circuit - if all the resistors in the circuit are in series and one of the resistors fails then no current can flow thru the circuit,

If the resistors are in parallel then then each resistor experiences the same voltage drop regardless of whether or not any resistor in particular is carrying current.

If the resistors are connected in series, each one will receive the same pd drop, if withstanding if it is carrying current or not.

Option C is correct  Series

A circuit is simply defined as a  channel that must always be followed in order for electricity to flow and perform something good.

In conclusion, if all of the resistors are connected in series, no current can pass through the circuit.

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Answer: false

Explanation: because if one car is at 2 m/s and the other at 1 m/s one would be farther away from the stating point

Answer:

the answer is C ----- it is the rate of change of velocity per unit time

Explanation:

acceleration

\( = \frac{velocity(v)}{time(t)} \)

The measured initial kinetic energy (J) of the bullet is 158.70 J.

So, the correct answer is A.

We know that the initial kinetic energy (KE) of the bullet is given by:

KE = (1/2) mv²

Where,m = mass of the bullet

v = velocity of the bullet

So, the initial kinetic energy of the bullet can be calculated as:

KE = (1/2) mv²

KE = (1/2) (m) (v)²

To determine the initial kinetic energy of the bullet in Joules, we can use the formula KE = 1/2mv², where m is the mass of the bullet and v is its velocity.

From question 13, we know that the mass of the bullet is 0.023 g, which is equal to 0.023/1000 kg. We also know that its velocity is 330 m/s.

Plugging these values into the formula, we get KE = 1/2 x 0.023/1000 x (330)²= 158.7 J.

Therefore, the measured initial kinetic energy of the bullet is option A, 158.70 J.

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Answer:

A and B

Explanation:

They contain Chlorophyll

They are relatively tall

Answer:

If you mean for exercising  then, The Treadmill. Purists may claim that treadmill running isn't "real" running. ...

The Elliptical Trainer. ...

The Stair-Climber. ...

The Cross-Country Skiing Simulator. are all  machines that help you practice agility

Explanation: