In an X-ray photoelectron experiment, a photon of wavelength 121 pm ejects an electron from the inner shell of an atom and it emerges with a speed of 56.9 Mm s−1 . Calculate the binding energy of the electron.

Answer:

Volume.

Explanation:

Boyle’s law experiment :

This is also known as Mariotte's law or in the other words it is also known as Boyle–Mariotte law.

This law tell us about the variation of gas pressure when the volume of the gas changes at the constant temperature.According to this law the abslute pressure is inversely proportional to the volume .

We can say that

[tex]P\alpha \dfrac{1}{V}[/tex]

Where P=Pressure

V=Volume

PV = K

K=Constant

When volume decrease then the pressure of the gas will increase.

That is why the answer is "Volume".

To solve this problem we will apply the geometric concepts of the Volume based on the consideration made of the radius measurement. The Volume must be written in differential terms of the radius and from the formula of the margin of error the respective response will be obtained.

The error in radius of sphere is not exceeding 2%

[tex]\frac{dr}{r} = \pm 0.02[/tex]

The objective is to find the percentage error in the volume.

The volume can be defined as

[tex]V = \frac{4}{3} \pi r^3[/tex]

Differentiate with respect the radius we have,

[tex]\frac{dV}{dr} = 4\pi r^2[/tex]

[tex]dV = 4\pi r^2 \times dr[/tex]

[tex]dV = 4\pi r^2 (\pm 0.02r)[/tex]

[tex]dV = \pm 4\times 0.02 \times \pi r^3[/tex]

The percentage change in the volume is as follows

[tex]\% change = \frac{dV}{V} \times 100[/tex]

[tex]\% change = \frac{\pm 4 \times 0.02 \times \pi r^3 \times 3}{4\pi r^3}\times 100[/tex]

[tex]\% change = \pm 6\%[/tex]

Therefore the percentage change in volume is [tex]\pm 6\%[/tex]

The question pertains to physics, specifically the concept of density. The mass of the gold in the ring is computed to be 10.615 g based on the provided volume displacement and density of gold. To determine the cost of the ring, additional information such as the current market price for gold and labor costs would be necessary.

The subject matter of this question relates to the physical property of density and its application in determining the mass of gold in a wedding ring. Given that the density of gold is 19.3 g/cm³ and the ring displaced 0.55 mL of liquid, one can calculate the mass of the ring using the rule that 1 mL equals 1 cm³. Hence, if the ring displaces 0.55 mL of the liquid, it means the volume of the ring is 0.55 cm³.

To find mass, we use Density = Mass/Volume, so Mass = Density * Volume. Therefore, the mass of the ring will be 19.3 g/cm³ * 0.55 cm³ = 10.615 g.

This calculation gives us the mass of the gold in the ring.

However, to determine the cost of the ring, we need more information such as the market price of gold per gram and the cost of workmanship and other potential elements in the ring, which are not provided in the question.

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Answer:

a) [tex]\lambda=1\ m[/tex]

b) [tex]f=122.47\ Hz[/tex]

c) [tex]\lambda_s=2.8\ m[/tex]

Explanation:

Given:

distance between the fixed end of strings, [tex]l=1.5\ m[/tex]

mass of string, [tex]m=5\ g=0.005\ kg[/tex]

tension in the string, [tex]F_T=50\ N[/tex]

a)

Since the wave vibrating in the string is in third harmonic:

Therefore wavelength λ of the string:

[tex]l=1.5\lambda[/tex]

[tex]\lambda=\frac{1.5}{1.5}[/tex]

[tex]\lambda=1\ m[/tex]

b)

We know that the velocity of the wave in this case is given by:

[tex]v=\sqrt{\frac{F_T}{\mu} }[/tex]

where:

[tex]\mu=[/tex] linear mass density

[tex]v=\sqrt{\frac{50}{(\frac{m}{l}) } }[/tex]

[tex]v=\sqrt{\frac{50}{(\frac{0.005}{1.5}) } }[/tex]

[tex]v=122.47\ m.s^{-1}[/tex]

Now, frequency:

[tex]f=\frac{v}{\lambda}[/tex]

[tex]f=\frac{122.47}{1}[/tex]

[tex]f=122.47\ Hz[/tex]

c)

When the vibrations produce the sound of the same frequency:

[tex]f_s=122.47\ Hz[/tex]

Velocity of sound in air:

[tex]v_s=343\ m.s^{-1}[/tex]

Wavelength of the sound waves in air:

[tex]\lambda_s=\frac{v_s}{f_s}[/tex]

[tex]\lambda_s=2.8\ m[/tex]

Answer:

[tex]\mathbf{1\ L=61.02384\ in^3}[/tex]

[tex]\mathbf{0.135\ kW=99.5709185406\ ft-lb/s}[/tex]

[tex]\mathbf{304\ kPa=44.091435811\ psi}[/tex]

[tex]\mathbf{122\ ft^3=3.45465495258\ m^3}[/tex]

[tex]\mathbf{100\ hp=74.6\ kW}[/tex]

[tex]\mathbf{1000\ lbm=453.592\ kg}[/tex]

Explanation:

[tex]1\ L=10^{-3}\ m^3[/tex]

[tex]1\ m=39.3701\ in[/tex]

[tex]10^{-3}\ m^3=10^{-3}\times 39.3701^3=61.02384\ in^3[/tex]

[tex]\mathbf{1\ L=61.02384\ in^3}[/tex]

[tex]0.135\ kW=135\ W[/tex]

[tex]135\ W=135\ Nm/s[/tex]

[tex]1\ N=0.224809\ lbf[/tex]

[tex]1\ m=3.28084\ ft[/tex]

[tex]135\times 0.224809\times 3.28084=99.5709185406\ ft-lb/s[/tex]

[tex]\mathbf{0.135\ kW=99.5709185406\ ft-lb/s}[/tex]

[tex]304\ kPa=304000\ N/m^2[/tex]

[tex]1\ m=39.3701\ in[/tex]

[tex]304000\ N/m^2=304000\times 0.224809\times \dfrac{1}{39.3701^2}=44.091435811\ psi[/tex]

[tex]\mathbf{304\ kPa=44.091435811\ psi}[/tex]

[tex]122\ ft^3=\dfrac{122}{3.28084^3}=3.45465495258\ m^3[/tex]

[tex]\mathbf{122\ ft^3=3.45465495258\ m^3}[/tex]

[tex]1\ hp=746\ W[/tex]

[tex]100\ hp=100\times 746\ W=74.6\ kW[/tex]

[tex]\mathbf{100\ hp=74.6\ kW}[/tex]

[tex]1\ lbm=0.224809\ kg[/tex]

[tex]1000\ lbm=1000\times 0.453592=453.592\ kg[/tex]

[tex]\mathbf{1000\ lbm=453.592\ kg}[/tex]

Answer: E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

Explanation:

The efficiency (e) of a Carnot engine is defined as the ratio of the work (W) done by the engine to the input heat QH

E = W/QH.

W=QH – QC,

Where Qc is the output heat.

That is,

E=1 - Qc/QH

E =1 - Tc/TH

where Tc for a temperature of the cold reservoir and TH for a temperature of the hot reservoir.

Note: The unit of temperature must be in Kelvin.

Tc = 300K

TH = 2000K

Substituting the values of E, we have;

E = 1 - 300K/2000K

E = 1 - 0.15

E = 0.85

Therefore the efficiency is: E = 0.85 or 85%

To calculate the ideal efficiency of an engine with specified temperatures, convert to Kelvin and use the Carnot efficiency formula.

The Carnot Engine Efficiency Calculation:

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Answer:

A

Explanation:

i took the test pimp

Answer:

Van will stop after a distance of 183.216 ft

Explanation:

We have given initial speed of the van = 36 mi/hr

As the van finally stops so final velocity v = 0 mi/hr

Distance after which van stop = 50 ft

As 1 ft = 0.000189 mi

So 50 ft [tex]=50\times 0.000189=0.00945mi[/tex]

From third equation of motion [tex]v^2=u^2+2as[/tex]

[tex]0^2=36^2+2\times a\times 0.00945[/tex]

[tex]a=-68571.42mi/hr^2[/tex]

In second case u = 69 mi/hr

And acceleration is same

So [tex]0^2=69^2-2\times 68571.42\times s[/tex]

[tex]s=0.03471mi[/tex]

As 1 mi = 5280 ft

So [tex]0.0347mi=0.0347\times 5280=183.216ft[/tex]

Answer:

No, they are not able to make this calculation

Explanation:

Acceleration Equation of satellite:

                                g = (G • Mcentral)/R2 ..............(1)

Acceleration equation of a satellite in circular motion:

                                a=(G • Mcentral)/R2...................(2)

Newton's form of Kepler's third law .

The period of a satellite (T) and the mean distance from the central body (R) are related by the following equation:

                                     [tex]\frac{T^{2} }{R^{3} }[/tex]=4×[tex]\pi ^{2}[/tex]/G×[tex]M_{central}[/tex].............(3)

T= the period of the satellite

R= The average radius of orbit for the satellite

G=6.673 x 10-11 N•m2/kg2.

According to all these three equations(1)(2)(3)

The period, speed and the acceleration of an orbiting satellite are independent upon the mass of the satellite.

Answer:

The conversion factor used here will be 1000 (kg/m^3)/(g/cm^3).

Which is a combination of two conversion factors:

1 kg = 1000 g

1 x 10^6 cm^3 = 1 m^3

Explanation:

We will use unitary method to convert g/cm^3 into kg/m^3. This is shown below:

Since, 1 kilogram is equivalent to 1000 gram and 1 meter is equivalent to 100 centimeter. Therefore:

1 g/cm^3 = (1 g/ cm^3)(1 kg/ 1000 g)(100 cm / 1 m)^3

1 g/cm^3 = 1000 kg/m^3

Hence, the conversion factor that will be multiplied is found to be 1000.

Using this in our case, we get:

Density of silicon = (2.33)(1000) kg/m^3

Density of Silicon = 2330 kg/m^3

Answer:

1000

Explanation:

Conversion from 'g' to 'kg': divide by 1000g/kg

Conversion from 'cm^3' to 'm^3': divide by 1000000cm^3/m^3

2.33g/cm^3 = [tex]\frac{2.33*1000000}{100}[/tex]

                    = 2330 kg/m^3

we simply multiply by 1000 to get the units converted to kg/m^3

Answer:

a)time t = 47.4s

b)speed v = 38.7cm/s

Explanation:

Given:

Total volume of blood = 4.93L × 1000cm^3/L = 4930cm^3

Volumetric rate of flow = 104.1cm^3/s

a) Time taken for all the blood to pass through the aorta is:

Time t = total volume/ volumetric rate

t = 4930/104.1

t = 47.4s

b) Given that the diameter of the aorta is 1.85cm.

V = Av

Where V = Volumetric rate

A = area of aorta

v = speed of blood

v = V/A ...1

Area of a circular aorta = πr^2 = (πd^2)/4

d = 1.85cm

A = (π×1.85^2)/4

A = 2.69cm^2

From equation 1.

v = V/A = 104.1/2.69

v = 38.7 cm/s

Answer:

1. True

2. True

3. False

4. False

5. True

6. False

Explanation:

Acceleration: It refers to the change in velocity/speed of an object with respect to time. When the speed increases with time we call it acceleration and when its decreases it is called as deceleration.  Let us analyze each instance individually:

1. When roller coaster starts to roll down the track its speed will increase with time. That means it is accelerating.

2. When the ball reaches at the peak of its trajectory, it comes to a stop for a fraction of a second that means it decelerates.

3. Since the velocity remains constant there is no acceleration.

4. Since the speed is no changing with time, there is no acceleration.

5. Since the moving plane comes to a stop, it is a case of deceleration.

6. Since the truck is moving at a constant speed so the acceleration is zero.

A roller coaster as it starts to roll down a track and an airplane skidding to a stop are examples of acceleration, while a ball tossed straight up, a planet tracing a circular orbit, a block sliding down a ramp, and a dump truck carrying a load at a constant speed are not examples of acceleration.

1) True. When a roller coaster starts to roll down the track, it experiences a change in velocity, which means it is accelerating.

2) False. At the peak of its trajectory, a ball tossed straight up has zero velocity and is momentarily at rest, so it is not accelerating.

3) False. A planet tracing out a circular orbit at a constant speed is not experiencing a change in velocity, so it is not accelerating.

4) False. The block sliding down a ramp at a constant speed is not experiencing a change in velocity, so it is not accelerating.

5) True. An airplane skidding to a stop on a runway experiences a change in velocity, so it is accelerating.

6) False. A dump truck carrying a load straight forward at a constant speed is not experiencing a change in velocity, so it is not accelerating.

Answer:

[tex]F_G=G. \frac{m_1.m_2}{R^2}[/tex] gravitational force

[tex]F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}[/tex] electrostatic force

Explanation:

The forces that balloons may exert on each other can be gravitational pull due to the mass of the balloon membrane and the mass of the gas contained in each. This force is inversely proportional to the square of the radial distance between their center of masses.

The Mutual force of gravitational pull that they exert on each other can be given as:

[tex]F_G=G. \frac{m_1.m_2}{R^2}[/tex]

where:

[tex]G=[/tex] gravitational constant  [tex]=6.67\times 10^{-11} m^3.kg^{-1}.s^{-2}[/tex]

[tex]m_1\ \&\ m_2[/tex] are the masses of individual balloons

[tex]R=[/tex] the radial distance between the  center of masses of the balloons.

But when  there are charges on the balloons, the electrostatic force comes into act which is governed by Coulomb's law.

Given as:

[tex]F=\frac{1}{4\pi \epsilon_0} \times \frac{q_1.q_2}{R^2}[/tex]

where:

[tex]\rm \epsilon_0= permittivity\ of\ free\ space[/tex]

[tex]q_1\ \&\ q_2[/tex] are the charges on the individual balloons

R = radial distance between the charges.

Answer:

[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.

Explanation:

We have a system of Earth & Moon:

Now we assume the origin of the system to be at the center of the earth.

Now for the center of mass of this system:

[tex]\bar x=\frac{m_e.x_e+m_m.x_m}{m_e+m_m}[/tex]

here:

[tex]x_e\ \&\ x_m[/tex] are the distance of the centers (center of masses) of the Earth and the Moon from the origin of the system.

[tex]x_e=0[/tex] ∵ since we have taken the point as the origin of the system.

[tex]x_m=d[/tex]

now putting the values in the above equation:

[tex]\bar x=\frac{(5.972\times 10^{24}\times 0)+(7.348\times 10^{22}\times 385000\times 1000)}{5.972\times 10^{24}+7.348\times 10^{22}}[/tex]

[tex]\bar x=4679496.086\ m=4679.496086\ km[/tex] from the center of the earth.

We will make the comparison between each of the sizes against the known wavelengths.

In the case of the hydrogen atom, we know that this is equivalent to [tex]10^{-10}[/tex] m on average, which corresponds to the wavelength corresponding to X-rays.

In the case of the Virus we know that it is oscillating in a size of 30nm to 200 nm, so the size of the virus is equivalent to the range of the wavelength of an ultraviolet ray.

In the case of height, it fluctuates in a person around [tex]10 ^ 0[/tex] to [tex]10 ^ 1[/tex] m, which falls to the wavelength of a radio wave.

The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region. The visible light spectrum corresponds to objects around the same size as the wavelength. The size of a virus is much smaller than the wavelength of visible light

The region of the spectrum that best corresponds to light with a wavelength equal to the diameter of a hydrogen atom is the X-ray region of the electromagnetic spectrum.

The wavelength of visible light corresponds to the size of objects that are around the same size as the wavelength. For example, if you want to use light to see a human, you would need to use a wavelength at or below 1 meter, since humans are about 1 meter in size.

The size of a virus is much smaller than the wavelength of visible light, so the wavelength is very small compared to the object's size.

Answer: Maximum distance

= {s²/g} * sine(2*theta)unit

Explanation: This is a projectile motion problem. The horizontal distance between the tennis player and where the tennis reaches over the net is given by the horizontal Range.

Range = {s² * sine2*theta}/g

(s)is the initial speed of projection

Theta is the angle of projection

g is acceleration due to gravity 10m/s².

Answer:

option B

Explanation:

given,

Q = energy input = 70 J

U = increase in internal energy

W is the work done = 20 J

using first law of thermodynamics

The change in internal energy of a system is equal to the heat added to the system minus the work done.

U = Q - W

U = 70 - 20

U = 50 J

The change in internal energy is equal to 50 J

Hence, the correct answer is option B

a) Total power output: [tex]3.845\cdot 10^{26} W[/tex]

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is [tex]540 W/m^2[/tex]

Explanation:

a)

The intensity of electromagnetic radiation is given by

[tex]I=\frac{P}{A}[/tex]

where

P is the power output

A is the surface area considered

In this problem, we have

[tex]I=1360 W/m^2[/tex] is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

[tex]r=1.5\cdot 10^{11} m[/tex] (distance Earth-Sun)

Therefore

[tex]A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2[/tex]

And now, using the first equation, we can find the total power output of the Sun:

[tex]P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W[/tex]

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

[tex]E=hf[/tex]

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

[tex]P=\frac{E}{t}[/tex]

where t is the time.

This means that the power output is proportional to the frequency:

[tex]P\propto f[/tex]

Here the frequency increases by 1 MHz: the original frequency was

[tex]f_0 = 60 MHz[/tex]

so the relative percentage change in frequency is

[tex]\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%[/tex]

And therefore, the power also increases by 1.67 %.

c)

In this second  case, we have to calculate the new power output of the Sun:

[tex]P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W[/tex]

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

[tex]r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m[/tex]

Now we can find the radiation intensity with the equation

[tex]I=\frac{P}{A}[/tex]

Where the area is

[tex]A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2[/tex]

And substituting,

[tex]I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2[/tex]

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The total power output of the Sun is 3.8 x 10^26 W and does not change with frequency variations. Therefore, increasing frequency by 1MHz won't change the power output. Using a distance that takes into account Mars being farther from the Sun, the radiation intensity at Mars can be calculated.

The total power output of the sun can be calculated using the equation for the intensity of radiant power (P = I * 4πd^2). So, the total power output of the sun is P = 1360 W/m^2 * 4π * (1.5 x 10^11 m)^2 ≈ 3.8 x 10^26 W.

As for the relative change in power output with frequency change, it is important to note that the power output of the Sun is independent of frequency. Therefore, an increase in frequency by 1 MHz would not result in any relative change in power output. The power output remains 3.8 x 10^26 W regardless of frequency.

Lastly, to find the intensity at Mars which is 60% farther from the Sun than Earth, we use the same intensity equation (I = P / 4πd²) but adjust the distance accordingly. If r is the distance to Earth, then the distance to Mars is 1.6r. Substituting these values, we get I = 3.8 x 10^26 W / 4π * (1.6 * 1.5 x 10^11 m)². From this, you can calculate the radiation intensity at Mars.

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Final answer:

The wavelength of the 6th harmonic of a stretched string fixed at both ends with a length of 160 cm is 53.33 cm.

Explanation:

To calculate the wavelength of the 6th harmonic of a stretched string fixed at both ends, we need to use the formula for the wavelength of standing waves on a string. The formula for the nth harmonic on a string of length L is given by:

\[\lambda = \frac{2L}{n}\]

In this case, the length L is 160 cm, and the harmonic number n is 6.

So, the wavelength for the 6th harmonic is:

\[\lambda_6 = \frac{2 \times 160}{6}\]

\[\lambda_6 = \frac{320}{6}\]

\[\lambda_6 = 53.33 \text{ cm}\]

Therefore, the wavelength of the 6th harmonic is 53.33 cm.

Answer:

a) [tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]

b) [tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]

Explanation:

Part a

For this case we have the situation illustrated on Figure 1.  We will have two forces involved in equilibrium the weight [tex] W_B[/tex] and the Bouyance force[tex] F_B[/tex], and since the system is on equilibrium we have:

[tex] \sum F_{vertical}=0[/tex]

So then we have:

[tex] W_B = F_B = \gamma_{w} V_s[/tex]

Where [tex] V_s[/tex] represent the submerged volume. [tex]\gamma_w[/tex] represent the specific weight for the fluid. So we can replace and we have:

[tex] W_B = F_B = 62.4 \frac{lb}{ft^3} (6ft*28ft*93ft)= 974937.6 lb[/tex]

Part b

As we can see on figure 2 attached we have the illustration for this case. We add the weight for the grain and now the depth is 9ft.

W can do the balance of forces in the vertical and we got again:

[tex] W_B +W_g = F_B[/tex]

Where [tex] W_g[/tex] represent the weight for the grain.

And if we solve for [tex] W_g[/tex] we got:

[tex] W_g = F_B -W_B[/tex]

[tex] W_g =\gamma_w V_S -W_B[/tex]

Where [tex] \gamma_w[/tex] represent the specific weight of rthe water and [tex] V_s[/tex] the submerged volume. If we replace we got:

[tex] W_g= 62.4 \frac{lb}{ft^3} * (9ft*28ft*93ft) -974937.6 lb =487468.8 lb[/tex]

The weight of the unloaded barge is 975,769.6 lbs, and the weight of the grain is 486,572.8 lbs. These calculations were made based on Archimedes' principle, which states that the buoyant force (the weight of water displaced) is equal to the weight of the object.

This question is about calculating the weight of a river barge and its load based on the principles of fluid mechanics. Here, we are examining the fact that the weight of the water displaced by the barge equals the weight of the barge according to Archimedes' principle.

Firstly, we calculate the unloaded weight of the barge. The volume of water displaced by the barge when it is empty is the volume of a rectangular prism with dimensions 28ft x 93ft x 6ft, which gives us 15,624 cubic feet. Given that the density of water is 62.4 lbs/ft³, the weight of this water, which is equal to the weight of the empty barge, would be (Volume x Density) = 15,624ft³ x 62.4 lbs/ft³ = 975,769.6 lbs.

Secondly, let's calculate the weight of the grain. The volume of water displaced when the barge is loaded is the volume of a rectangular prism with dimensions 28ft x 93ft x 9ft, which equals 23,436 cubic feet. The weight of this water, which is equal to the weight of the loaded barge, is (Volume x Density) = 23,436ft³ x 62.4 lbs/ft³ = 1,462,342.4 lbs. Hence, the weight of the grain is the weight of the loaded barge minus the weight of the barge itself = 1,462,342.4 lbs - 975,769.6 lbs = 486,572.8 lbs.

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Answer:

option a

Explanation:

Size of an atom (diameter) = 10⁻¹⁰ m

There are approximately 10²² atoms in a single drop of water. If they are put in  a straight line, the length would be

l = diameter of an atom × number of atoms

l = 10²²×  10⁻¹⁰ m = 10¹² m

Distance between the Sun and the Earth is 1.47 × 10¹¹ m. The calculated length is greater than the distance between the Sun and the Earth.

Thus, option a is correct.

Answer:

Biological system is one of the major causes of oscillation due to sensitive negative feedback loops. For instance, imagine a father teaching his son how to drive, the teen is trying to keep the car in the centre lane and his father tell him to go right or go left as the case may be. This is a example of a negative feedback loop of a biological system. If the father's sensitivity to the car's position on the road is reasonable, the car will travel in a fairly straight line down the centre of the road. On the other hand, what happens if the father raise his voice at the son "go right" or when the car drifts a bit to the left? The startled the son will over correct, taking the car too far to the right. The father will then starts yelling "go left" then the boy will over correct again and the car will definitely oscillate back and forth. A scenario that indicates the behavior of a car driver under a very steep feedback control mechanism. Since the driver over corrects in each direction. Therefore causes oscillations.

Explanation:

In physics, a system's equilibrium point is considered stable if the system returns to that point after being slightly disturbed. For an n value less than 8, the system remains in stable equilibrium; once n exceeds 8, stable oscillations indicate an unstable equilibrium. This concept can be visualized by the marble in a bowl analogy, where the bowl's orientation determines the stability of the marble's equilibrium.

The stability of an equilibrium point is determined by the response of a system to a disturbance. If an object at a stable equilibrium point is slightly disturbed, it will oscillate around that point. The stable equilibrium point is characterized by forces that are directed toward it on either side. On the contrary, an unstable equilibrium point will not allow the object to return to its initial position after a slight disturbance, since the forces are directed away from that point. For values of n less than 8, the system finds a stable equilibrium. However, when n surpasses 8, the system exhibits unstable equilibrium, leading to stable oscillations.

As an example, consider a marble in a bowl. When the bowl is right-side up, the marble represents a stable equilibrium; when disturbed, it returns to the center. If the bowl were inverted, the marble on top would be at an unstable equilibrium point; any disturbance would cause it to roll off, as the forces on either side would be directed outwards. Extending this concept to potential energy, n in a potential energy function acts as an adjustable parameter. For n=<8, the system remains in stable equilibrium, as with NaCl, which has an n value close to 8. Beyond this value, the equilibrium becomes unstable, giving rise to oscillatory behavior.

Stability also depends on the nature of damping in the system. An overdamped system moves slowly towards equilibrium without oscillations, an underdamped system quickly returns but oscillates, and a critically damped system reaches equilibrium as quickly as possible without any oscillations.

Answer: according to the Avagadro's law, volume is directly propotional to no of moles: VXn

according to the Charles law, volume is directly propotional to  temperatue: VXT

according to the Boyle's law, volume is inversely propotional to P: VX1/P

when we combine them we get:

VXnT1/P

V=knT/P

k= R(universal gas constant)

V=RnT/P

PV=nRT  

Answer:

Explanation:

q = - 4.25 nC = - 4.5 x 10^-9 C

(A) d = 0.250 m

The formula for the electric field is given by

[tex]E = \frac{1}{4\pi \epsilon _{0}}\frac{q}{d^{2}}[/tex]

By substituting the values

[tex]E = \frac{9\times 10^{9}\times 4.5\times10^{-9}}{0.25\times 0.25}[/tex]

E = 648 N/C

(B) As the charge is negative in nature so the direction of electric field is towards the charge and downwards.

(a) The magnitude of the electric field is 612 N/C.

(b) The direction of the electric field will be up, away from the particle.

(c) The distance from the particle is 1.71 m.

The magnitude of the electric field is calculated as follows;

E = (kq)/r²

where;

k is Coulomb's constant

q is the charge

r is distance

E = ( 9 x 10⁹ x 4.25 x 10⁻⁹)/(0.25 x 0.25)

E =  612 N/C

The direction of the electric field is always opposite to the direction of the negative charge.

Thus, the direction of the electric field will be up, away from the particle.

The distance from the particle is determined using the following formula;

E = (kq)/r²

r² = kq/E

r² = (9 x 10⁹ x 4.25 x 10⁻⁹) / 13

r² = 2.94

r = √2.94

r = 1.71 m

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Answer:

The diameters of the disk is 1.23 cm

Explanation:

Given information:

the space of two disks, d =0.5 mm = 0.0005 m =  5 x [tex]10^{-4}[/tex] m

the transferred electron, n = 1.70 x [tex]10^{9}[/tex]

electric field strength, E = 2.6 x [tex]10^{5}[/tex] N/C

to find the diameter of the disk we can use the following equation

A = πD²/4 ...........................................(1)

where

D = the distance of the disk

A = the area of the disk

first, we have to find the are of the disk using the capacitance equation

C = ε₀A/d...........................................(2)

A = Cd/ε₀ where C = Q/V (Q is total charge and V is potential difference)

thus

A = Qd/Vε₀.........................................(3)

now substitute V = Ed and Q = ne, so

A = (ned)/(Edε₀)

   = ne/Eε₀..........................................(4)

e = 1.6 x [tex]10^{-19}[/tex] C

now can substitute equation (4) to the first equation

A = πD²/4

D² = 4A/π

D  = √4A/π

    = √(4ne)/(πEε₀) , ε₀ = 9.85 x [tex]10^{-12}[/tex] C²/Nm²

    = √4(1.70 x [tex]10^{9}[/tex])(1.6 x [tex]10^{-19}[/tex])/π(2.6 x [tex]10^{5}[/tex] )(8.85 x [tex]10^{-12}[/tex] )

    = 0.0123 m

    = 1.23 cm

To solve this problem we will apply the physical equations of the angular kinematic movement, for which it defines the square of the final angular velocity as the sum between the square of the initial angular velocity and the product between 2 times the angular acceleration and angular displacement. We will clear said angular displacement to find the correct response

Using,

[tex]\omega^2 = \omega_0^2 +2\alpha \theta[/tex]

Here,

[tex]\omega[/tex] = Final angular velocity

[tex]\omega_0[/tex] = Initial angular velocity

[tex]\alpha =[/tex] Angular acceleration

[tex]\theta =[/tex] Angular displacement

Replacing,

[tex]59^2 = 0+2*58\theta[/tex]

[tex]\theta = 30rad[/tex]

Therefore the correct answer is C.

The angle at which the wheel turns while accelerating is 30 radians and this can be determined by using the kinematics equation.

Given :

A wheel accelerates from rest to 59 rad/[tex]\rm s^2[/tex] at a uniform rate of 58 rad/[tex]\rm s^2[/tex].

The equation of kinematics is used in order to determine the angle at which the wheel turn while accelerating.

[tex]\omega^2 = \omega^2_0+2\alpha \theta[/tex]

where [tex]\omega[/tex] is the final angular velocity, [tex]\omega_0[/tex] is the initial angular velocity, [tex]\alpha[/tex] is the angular acceleration, and [tex]\theta[/tex] is the angular displacement.

Now, substitute the values of the known terms in the above formula.

[tex]59^2 =0+2\times 58 \times \theta[/tex]

Simplify the above expression.

[tex]\rm \theta = 30\; rad[/tex]

Therefore, the correct option is C).

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Answer:

[tex]2.537\times 10^{6}\ years[/tex]

Explanation:

Distance to Andromeda galaxy

[tex]2.537\times 10^{6}\ ly=2.537\times 10^{6}\times c\times y[/tex]

Speed of light is

[tex]c=3\times 10^8[/tex]

Time is given by

[tex]t=\dfrac{Distance}{Speed}\\\Rightarrow t=\dfrac{2.537\times 10^{6}\times c\times y}{c}\\\Rightarrow t=2.537\times 10^{6}\ y[/tex]

Hence, the light from Andromeda left [tex]2.537\times 10^{6}\ years[/tex] ago

The light observed from the Andromeda galaxy by the astronomer left Andromeda around 2 million years ago, showcasing the vast distances in space and providing insights into the universe's evolution.

The light from Andromeda galaxy that the astronomer observes left Andromeda approximately 2 million years ago. This is because the distance in light years is the same as the time it takes for the light to reach us.

This phenomenon is due to the vast distances in space. The Andromeda galaxy is the closest large galaxy to the Milky Way, located 2 million light years away from us.

Understanding the age of the light we observe helps us gain insights into the history of distant galaxies and the evolution of the universe.

Answer:

Explanation:

Given

speed of locomotive [tex]v=80\ km/h[/tex]

distance [tex]d=9\ km[/tex]

Relative velocity of two locomotive [tex]v_r=80-(-80)=160\ km/h[/tex]

Time taken[tex]=\frac{distance}{speed}[/tex]

[tex]t=\frac{9}{160} \hr[/tex]

[tex]t=\frac{9}{160}\times 60\ min[/tex]

[tex]t=3.375\ min[/tex]

Answer:

[tex]<3068.2352, 800, 0>\ m/s[/tex]

Explanation:

F = Force = [tex]<-1.12\times 10^{-11}, 0, 0>[/tex]

m = Mass of proton = [tex]1.7\times 10^{-27\ kg[/tex]

t = Time taken = [tex]2\times 10^{-14}\ s[/tex]

Acceleration is given by

[tex]a=\dfrac{F}{m}\\\Rightarrow a=\dfrac{<-1.12\times 10^{-11}, 0, 0>}{1.7\times 10^{-27}}\\\Rightarrow a=<-6.58824\times 10^{15}, 0, 0>\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-6.58824\times 10^{15}, 0, 0>\times 2\times 10^{-14}\\\Rightarrow v=<3200, 800, 0>+<-131.7648, 0, 0>\\\Rightarrow v=<3068.2352, 800, 0>\ m/s[/tex]

The velocity of the proton is [tex]<3068.2352, 800, 0>\ m/s[/tex]

Answer:

a) See attachment

b) The pattern converges towards central order

c) Very bright spot at central order and rest is dark and contrast of pattern increases.

Explanation:

b) According to Young's Double slit experiment the following relationship is given:

[tex]wavelength = \frac{x*d}{nL}[/tex]

where,

λ = wavelength of light used (m)

x = distance from central fringe (m)

d = distance between the slits (m)

n = the order of the fringe

L = length from the screen with slits to the viewing screen (m)

Using the formula if we increase the (d) i.e distance between slits we see that (x) distance between fringes decreases and the patterns of bright and dark spots is alternating more frequently.

c) When d is arbitrarily large the x is arbitrarily small.

Hence, the entire pattern converges on to the film in a small space with millions spots of bright and dark spots alternating together to forma big bright spot and contrast of pattern increases.

Adding an identical slit close to the first results in an interference pattern of light and dark fridges due to light interference. Increasing the slits' distance results in narrower fringes, while in the limit that this distance becomes arbitrarily large, interference effects vanish and we see two independent single-slit diffraction patterns.

The question discusses an experiment in Physics related to wave interference, specifically light interference in a double-slit experiment. When you add an identical slit a distance of 0.25mm away from the first, an interference pattern of light and dark fringes (bright and dark spots) is created on the screen due to constructive (light) and destructive (dark) interference of the light waves passing through the slits.

Now, as we increase the distance d between the slits, the interference fringes on the screen will become narrower, and their separation will increase. This occurs because the path difference between the light waves emerging from the two slits and reaching the screen changes.

In the limit that the distance d between the slits becomes arbitrarily large, interference effects would vanish, and one would observe two independent single-slit diffraction patterns.

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