Calculate the mean free path of air molecules at a pressure of 3.00×10−13 atm and a temperature of 304 K . (This pressure is readily attainable in the laboratory.) Model the air molecules as spheres with a radius of 2.00×10−10 m .

The concept required to develop this problem is Hook's Law and potential elastic energy.

By definition the force by Hooke's law is defined as

[tex]F = kx[/tex]

Where,

k = Spring Constant

x = Displacement

On the other hand, the elastic potential energy is defined as

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

With the given values we can find the value of the spring constant, that is,

[tex]F = kx[/tex]

[tex]k=\frac{F}{x}[/tex]

[tex]k= \frac{83}{0.02}[/tex]

[tex]k = 4120N/m[/tex]

Applying the concepts of energy conservation then we can find the position of the block, that is,

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

[tex]E = \frac{1}{2} k\Delta x^2[/tex]

[tex]4.8 = \frac{1}{2} (4120)(x^2-(-0.02)^2)[/tex]

[tex]x = \sqrt{\frac{2*4.8}{4120}+(-0.02)^2}[/tex]

[tex]x = \pm 0.05225m[/tex]

Therefore the position of the block can be then,

[tex]x_1 = 5.225cm[/tex]

[tex]x_2 = -5.225cm[/tex]

Answer:

17.7 cm^3

Explanation:

depth, h = 120 m

density of water, d = 1000 kg/m^3

V1 = 1.4 cm^3

P1 = P0 + h x d x g

P2 = P0

where, P0 be the atmospheric pressure

Let V2 be the volume of the bubble at the surface of water.

P0 = 1.01 x 10^5 Pa

P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa

Use

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Thus, the volume of bubble at the surface of water is 17.7 cm^3.

The volume (in cm3) of the bubble just before it pops at the surface of the lake is mathematically given as

V2 = 17.7 cm^3

Question Parameter(s):

An air bubble released by a remotely operated underwater vehicle, 120 m

below the surface of a lake, has a volume of 1.40 cm3.

Generally, the equation for the initial Pressure  is mathematically given as

P1 = P0 + h x d x g

Where,atmospheric pressure

P0 = 1.01 x 10^5 Pa

Therefore

P1 = 1.01 * 10^5 + 120 * 1000 &* 9.8

P1= 12.77 * 10^5 Pa

In conclusion

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Read more about volume

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Answer:

Power, P = 924.15 watts

Explanation:

Given that,

Length of the ramp, l = 12 m

Mass of the person, m = 55.8 kg

Angle between the inclined plane and the horizontal, [tex]\theta=25^{\circ}[/tex]

Time, t = 3 s

Let h is the height of the hill from the horizontal,

[tex]h=l\ sin\theta[/tex]

[tex]h=12\times \ sin(25)[/tex]

h = 5.07 m

Let P is the power output necessary for a person to run up long hill side as :

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

[tex]P=\dfrac{mgh}{t}[/tex]

[tex]P=\dfrac{55.8\times 9.8\times 5.07}{3}[/tex]

P = 924.15 watts

So, the minimum average power output necessary for a person to run up is 924.15 watts. Hence, this is the required solution.

We find the height of the 25° incline and then calculate the potential energy per kg, which is then turned into power by dividing by time. The result is 16.6 W or 0.022 hp. Therefore, the minimum average power output necessary for a person to run up a 12.0 m, 25° long hillside in 3.00 s is approximately 2.2% of a horsepower.

This problem is solved in several steps. First, we need to determine the potential energy gain of the person running up the hillside. The formula for potential energy (PE) is PE = m*g*h, where m is mass, g is gravity (approximately 9.8 m/s²), and h is the height of the hill. Since we're not given the person's mass, we'll imagine the person has a mass of 1 kg just to calculate the potential energy gain per kg, but it isn't necessary to know the exact weight for calculating the minimum average power.

The height of the incline is given by 12.0m*sin(25°) = 5.09 m. So, the potential energy gain is PE = 1 kg * 9.8 m/s² * 5.09 m = 49.88 J. Converting this to power (P) by dividing energy by time, P = 49.88 J / 3.0 s = 16.6 W.

Since one horsepower (hp) is approximately 746 watts (W), the power in horsepower is 16.6 W / 746 W/hp = 0.022 hp, or 2.2% of a horsepower. This is the minimum average power output required for a person to climb this hill in 3 seconds.

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Answer:

The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]

Explanation:

Given that,

Distance = 25000 light year

Speed of light [tex]c=3\times10^{8}\ m/s[/tex]

Speed of astronomers = 230 m/s

Suppose the orbit is circular, we need to find the time period of the solar system's orbit.

We need to calculate the time period

Using formula of time period

[tex]t = \dfrac{d}{v}[/tex]

[tex]t = \dfrac{2\pi r}{v}[/tex]

r = distance

t = time

v = speed

Put the value into the formula

[tex]t=\dfrac{2\pi\times25000\times9.46\times10^{15}}{230\times10^{3}}[/tex]

[tex]t=6.460\times10^{15}\ sec[/tex]

Time in years

[tex]t=\dfrac{6.460\times10^{15}}{3.15\times10^{7}}[/tex]

[tex]t=2.05\times10^{8}\ year[/tex]

Hence, The time period of the solar system's orbit is [tex]2.05\times10^{8}\ year[/tex]

The solar system is about 25,000 light years away from the centre of the Milky Way galaxy. It orbits the galaxy at a speed of 230 km/s, which translates to a galactic year of approximately 225 million Earth years. The galaxy's mass, including dark matter, influences the orbital velocities of all celestial bodies.

The solar system, orbiting at a speed of 230 km/s, is located approximately 25,000 light years from the Milky Way galaxy's center. A light year is defined as the distance light travels in one year with a speed of 3.0×108m/s. Understanding this concept helps us comprehend the vast size of our galaxy. Additionally, the solar system's movement around the Milky Way is referred to as a galactic year, which lasts approximately 225 million years in Earth's time.

The sun, and all stars in the galaxy, orbit the galactic center in a nearly circular path lying in the galaxy's disk. Moreover, the total mass of our Galaxy can be determined by measuring the orbital velocities of stars and interstellar matter, including the Sun. This vast mass, around 2 × 1012 Msun, includes a significant part composed of dark matter that emits no electromagnetic radiation.

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Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

[tex]I=I_{0}Cos^{2}\theta[/tex]

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².

Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.

Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².

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Answer:

Explanation:

If the acceleration due to gravity on the Moon's surface is [tex]a_M[/tex] and the acceleration due to gravity on the Earth is [tex]a_E[/tex], we can write that:

[tex]a_M=\frac{1}{6}a_E[/tex]

[tex]\frac{a_M}{a_E}=\frac{1}{6}[/tex]

If the radius of the Moon is [tex]r_M[/tex] and the radius of the Earth is [tex]r_E[/tex], we can write that:

[tex]r_M=\frac{1}{4}r_E[/tex]

[tex]\frac{r_M}{r_E}=\frac{1}{4}[/tex]

By Newton's 2nd Law we know that F=ma and using Newton's law of universal gravitation  we can calculate the gravitational force an object with mass m experiments from a planet with mass M being at a distance r from it. We will assume our object is on the surface so this distance will be the radius of the planet.

Since the force the object experiments is the force of gravitation we can write, for Earth:

[tex]F=ma_E=\frac{GM_Em}{r_E^2}[/tex]

which means:

[tex]a_E=\frac{GM_E}{r_E^2}[/tex]

[tex]M_E=\frac{a_Er_E^2}{G}[/tex]

And for the Moon:

[tex]F=ma_M=\frac{GM_Mm}{r_M^2}[/tex]

which means:

[tex]a_M=\frac{GM_M}{r_M^2}[/tex]

[tex]M_M=\frac{a_Mr_M^2}{G}[/tex]

We can then write the fraction:

[tex]\frac{M_M}{M_E}=\frac{a_Mr_M^2}{G}\frac{G}{a_Er_E^2}=\frac{a_M}{a_E}(\frac{r_M}{r_E})^2=\frac{1}{6}(\frac{1}{4})^2=0.01[/tex]

Which means:

[tex]M_M=0.01M_E[/tex]

Final answer:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration and the given ratios between the Moon and Earth's acceleration and radii. Using these values, we can solve for the mass of the Moon in terms of the mass of the Earth.

Explanation:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration: g = GM/r². Given that the acceleration due to gravity on the Moon is one-sixth that of Earth and the radius of the Moon is one-quarter that of Earth, we can set up the following equation:

(1/6) * (9.8 m/s²) = GM/(1/4 * R)²

Simplifying, we get: 1/6 * 9.8 = GM/(1/16)

Now, we can solve for the mass of the Moon (M) in terms of the mass of the Earth (m):

M = (1/6 * 9.8 * R²)/(1/16 * G)

Substituting the values for R and G, we get:

M = (1/6 * 9.8 * (1/4)²)/(1/16 * 6.67 × 10⁻¹¹)

M ≈ 0.123m

To determine the maximum amplitude of the piston's oscillation without the bolt losing contact, we need to set the gravitational force equal to the spring force, and solve for the amplitude. Using the given values, including the mass of the bolt, the acceleration due to gravity, and the frequency of oscillation, we can calculate the maximum amplitude.

In order for the bolt to stay in contact with the piston's surface, the maximum amplitude of the piston's oscillation needs to be determined.

For simple harmonic motion, the restoring force is proportional to the displacement and acts opposite to the direction of motion. In this case, the restoring force is provided by the weight of the bolt (mg) and the force exerted by the piston (kx), where x is the displacement from equilibrium and k is the spring constant of the piston.

At maximum amplitude, the net force acting on the bolt is zero, so we can set the gravitational force equal to the spring force: mg = kA. Rearranging this equation gives us the maximum amplitude A = mg/k.

Now we can calculate the maximum amplitude using the given values: mass of bolt = 33.0 g = 0.033 kg, acceleration due to gravity g = 9.81 m/s^2, and frequency of oscillation f = 3.05 Hz. The period of oscillation is T = 1/f.

Using the equation T = 2*pi*sqrt(m/k) for the period of a mass-spring system, we can solve for the spring constant k: k = (2*pi/T)^2*m = (2*pi/1/f)^2*0.033 = (2*pi*f)^2*0.033. Plugging in the given value for f, we find k = (2*pi*3.05)^2*0.033.

Finally, we can calculate the maximum amplitude A = mg/k = 0.033*9.81/(2*pi*3.05)^2*0.033. Evaluating this expression gives us the maximum amplitude of the piston's oscillation.

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Answer:

Part A:

Unknown variables:

velocity of the astronaut after throwing the tank.

maximum distance the astronaut can be away from the spacecraft to make it back before she runs out of oxygen.

Known variables:

velocity and mass of the tank.

mass of the astronaut after and before throwing the tank.

maximum time it can take the astronaut to return to the spacecraft.

Part B:

To obtain the velocity of the astronaut we use this equation:

-(momentum of the oxygen tank) = momentum of the astronaut

-mt · vt = ma · vt

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

To obtain the maximum distance the astronaut can be away from the spacecraft we use this equation:

x = x0 + v · t

Where:

x = position of the astronaut at time t.

x0 = initial position.

v = velocity.

t = time.

Part C:

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Explanation:

Hi there!

Due to conservation of momentum, the momentum of the oxygen tank when it is thrown away must be equal to the momentum of the astronaut but in opposite direction. In other words, the momentum of the system astronaut-oxygen tank is the same before and after throwing the tank.

The momentum of the system before throwing the tank is zero because the astronaut is at rest:

Initial momentum = m · v

Where m is the mass of the astronaut plus the equipment (100 kg) and v is its velocity (0 m/s).

Then:

initial momentum = 0

After throwing the tank, the momentum of the system is the sum of the momentums of the astronaut plus the momentum of the tank.

final momentum = mt · vt + ma · va

Where:

mt = mass of the tank

vt = velocity of the tank

ma = mass of the astronaut

va = velocity of the astronaut

Since the initial momentum is equal to final momentum:

initial momentum = final momentum

0 = mt · vt + ma · va

- mt · vt = ma · va

Now, we have proved that the momentum of the tank must be equal to the momentum of the astronaut but in opposite direction.

Solving that equation for the velocity of the astronaut (va):

- (mt · vt)/ma = va

mt = 15 kg

vt = 10 m/s

ma = 100 kg - 15 kg = 85 kg

-(15 kg · 10 m/s)/ 85 kg = -1.8 m/s

The velocity of the astronaut is 1.8 m/s in direction to the spacecraft.

Let´s place the origin of the frame of reference at the spacecraft. The equation of position for an object moving in a straight line at constant velocity is the following:

x = x0 + v · t

where:

x = position of the object at time t.

x0 = initial position.

v = velocity.

t = time.

Initially, the astronaut is at a distance x away from the spacecraft so that

the initial position of the astronaut, x0, is equal to x.

Since the origin of the frame of reference is located at the spacecraft, the position of the spacecraft will be 0 m.

The velocity of the astronaut is directed towards the spacecraft (the origin of the frame of reference), then, v = -1.8 m/s

The maximum time it can take the astronaut to reach the position of the spacecraft is 1.5 min = 90 s.

Then:

x = x0 + v · t

0 m = x - 1.8 m/s · 90 s

Solving for x:

1.8 m/s · 90 s = x

x = 162 m

The maximum distance the astronaut can be away from the spacecraft is 162 m.

Answer:

N₂/N₁ = 45

Explanation:

The relation of the number of turns with the voltage is given by the EMF equation of transformer:  

[tex] \frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}} [/tex] (1)

where V₁: voltage of the primary winding, V₂: voltage of the secondary winding, N₁: number of turns in the primary winding, and N₂: number of turns in the secondary winding

Assuming that the primary winding is connected to the input voltage supply, and using equation (1), we can calculate the ratio of N₂ to N₁:

[tex] \frac{N_{2}}{N_{1}} = \frac{V_{2}}{V_{1}} = \frac{5.0 \cdot 10^{3}V}{110V} = 45 [/tex]       

So, the ratio N₂/N₁ is 45.

Have a nice day!

Answer:

Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz

Explanation:

Since, the two are in out of phase,

their path difference is

d= nλ

[tex]d_2-d_1= n\lambda[/tex]

Given d1= 2.75 m

D= 3.90 m

[tex]d_2= \sqrt{D^2- d_1^2}[/tex]

[tex]d_2= \sqrt{3.90^2- 2.75^2}[/tex]

d_2= 2.76 m

2.76-2.75= 1×λ

λ= 0.01 m

0.01= 1*λ

λ =0.01

frequency ν = v/λ = 340/0.01

f= 3400 Hz

Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz

Answer:

Option b

Solution:

As per the question:

Speed of the mouse, v = 1.3 m/s

Speed of the cat, v' = 2.5 m/s

Angle, [tex]\theta = 38^{\circ}[/tex]

Now,

To calculate the distance between the mouse and the cat:

The distance that the cat moved is given by:

[tex]x = v'cos\theta t[/tex]

[tex]x = 2.5cos38^{\circ}\times t = 1.97t[/tex]

The position of the cat and the mouse can be given by:

[tex]x = x' + vt[/tex]

[tex]1.97t = x' + 1.3t[/tex]

x' = 0.67 t           (1)

The initial speed of the cat ahead of the mouse:

u = [tex]v'sin\theta = 2.5sin38^{\circ} = 1.539\ m/s[/tex]

When the time is 0.5t, the speed of the cat is 0, thus:

[tex]0 = u - 0.5tg[/tex]

[tex]t = \frac{1.539}{0.5\times 9.8} = 0.314\ s[/tex]

Substituting the value of t in eqn (1):

x' = 0.67(0.314) = 0.210 m

Thus the distance comes out to be 0.210 m

Answer: It should be one that matches the symmetry of the charge distribution

Explanation: Gaussian surfaces are usually carefully chosen to exploit symmetries of a situation to simplify the calculation of the surface integral. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constants which arise can be taken out of the integral.It is defined as the closed surface in three dimensional space by which the flux of vector field be calculated.

Final answer:

The best choice for a Gaussian surface is one that matches the symmetry of the charge distribution, for example, a spherical Gaussian surface for a point charge and a cylindrical Gaussian surface for a cylindrical charge distribution, as it simplifies the integral for the flux.

Explanation:

The best choice for the shape of a Gaussian surface is E) It should be one that matches the symmetry of the charge distribution. When dealing with point charges that are spherically symmetric, selecting a spherical Gaussian surface makes sense since the electric field is radial. Likewise, for a cylindrical charge distribution, a cylindrical Gaussian surface is most appropriate. The goal is to make the flux integral easy to evaluate and to find a surface over which the electric field is constant in magnitude and makes the same angle with every element of the surface.

Using a Gaussian surface that matches the symmetry of the charge distribution, such as a spherical shell for a point charge or spherical charge distribution, ensures the electric field (E) is constant in magnitude across the surface, simplifying the calculations. Conversely, using a less ideal surface, like a cylindrical surface to envelop a spherical charge, leads to a more complex integral due to the variations in electric field strength (E) across the surface.

Explanation:

Part 1.

The force of gravity on the apple, [tex]F_1=9.39\ N[/tex] (downward)

The force of the wind on the apple, [tex]F_2=1.5\ N[/tex] (right)

Let F is the magnitude of the net force acting on it. The resultant of two vectors is given by :

[tex]F=\sqrt{F_1^2+F_2^2}[/tex]

[tex]F=\sqrt{9.39^2+1.5^2}[/tex]

F = 9.509 N

Part 2.

[tex]tan\theta=\dfrac{F_2}{F_1}[/tex]

[tex]tan\theta=\dfrac{1.5}{9.39}[/tex]

[tex]\theta=9.07^{\circ}[/tex]

So, the direction of the net external force is 9.07 degrees wrt vertical. Hence, this is the required solution.

Answer:

e=2356.125 V

Explanation:

Given that

D= 10 cm  = 0.1 m

R= 0.05

B= 10,000

Angular speed ,ω = 1800 rev/min

Speed in rad/s given as

[tex]\omega =\dfrac{2\pi N}{60}[/tex]

[tex]\omega =\dfrac{2\pi \times 1800}{60}[/tex]

ω= 188.49 rad/s

The potential difference given as

e= B R v

v=average speed

[tex]v=\dfrac{\omega}{2}R[/tex]

[tex]e=\dfrac{B}{2}R^2 {\omega}[/tex]

By putting the values

[tex]e=\dfrac{10000}{2}0.05^2 \times {188.49}[/tex]

e=2356.125 V

We can find the distance the sled travels before it comes to rest by equating the work done by the frictional force with the initial kinetic energy of the sled. We first calculate the initial kinetic energy with (1/2)mv². The force of friction is given by μmg. The work done by friction is equal to this force times the distance traveled, and we set this equal to the negative initial kinetic energy to account for the sled stopping.

The subject of this question is about the concept of work and energy in a physics problem involving a sled on a frozen pond. To solve this, we would use the principle of work and energy where the work done on the sled is equal to the change in its kinetic energy according to the work-energy theorem. Since the sled comes to rest, the final kinetic energy is zero.

First, we calculate the initial kinetic energy of the sled using the formula (1/2)mv², where m is the mass (8.54 kg), and v is the velocity (1.87 m/s). Next, we know that the work done by the friction on the sled is equal to the force of friction times the distance traveled which is equal to the change in kinetic energy. The force of friction can be calculated by multiplying the mass of the sled, the acceleration due to gravity (9.8 m/s²), and the coefficient of kinetic friction (0.087). Set this work equal to the negative initial kinetic energy (as it acts to bring the sled to a stop) and solve for the distance.

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To find the distance the sled moves before coming to rest, we can use the work and energy concepts. Work done by the friction force is equal to the change in kinetic energy. By setting the work done by friction equal to the change in kinetic energy and using the given values, we can solve for the distance the sled moves.

To find the distance the sled moves before coming to rest, we will use the work and energy concepts. The work done on an object is equal to the change in its kinetic energy, so we can use the work-energy principle to solve the problem. The work done by the friction force is equal to the force of friction multiplied by the distance the sled moves. Setting the work done by friction equal to the change in kinetic energy, we can solve for the distance the sled moves.

The work done by the friction force is given by the equation:
W = F * d * cos(180°)

where W is the work done, F is the force of friction, d is the distance, and cos(180°) is the angle between the force of friction and the displacement. The work done by the friction force is equal to the negative change in kinetic energy of the sled, which can be calculated as:
ΔKE = -(1/2) * m * v²

where ΔKE is the change in kinetic energy, m is the mass of the sled, and v is the final velocity of the sled.

Setting the work done by friction equal to the negative change in kinetic energy and solving for the distance d:

W = -(1/2) * m * v²
F * d * cos(180°) = -(1/2) * m * v²

Using the given values for the mass of the sled (m = 8.54 kg), the initial velocity (v = 1.87 m/s), and the coefficient of kinetic friction (µ = 0.087), we can solve for the distance d.

First, we need to calculate the magnitude of the friction force using the equation:
F = µ * N

where µ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the sled, which can be calculated as:
N = m * g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the equations:

F = 0.087 * (8.54 kg * 9.8 m/s²) = 7.94 N

Now, we can substitute the values for the force of friction and the change in kinetic energy into the equation:

F * d * cos(180°) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Using the cosine of 180° as -1:

7.94 N * d * (-1) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Simplifying the equation:

-7.94 N * d = -(1/2) * (8.54 kg) * (1.87 m/s)²

Dividing both sides of the equation by -7.94 N:

d = [(1/2) * (8.54 kg) * (1.87 m/s)²] / 7.94 N

Calculating the distance d:

d = 2.008 m

Therefore, the sled moves a distance of 2.008 meters before coming to rest.

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Answer:

The wavelength of the light is 530 nm.

Explanation:

Given that,

Distance D= 1.0 m

Distance between slits d= 0.30 mm

Number of fringe = 9

Width = 1.6 cm

We need to calculate the angle

Using formula of angle

[tex]\tan\theta=\dfrac{y}{D}[/tex]

[tex]tan\theta=\dfrac{1.6\times10^{-2}}{1.0}[/tex]

[tex]\theta=\tan^{-1}(\dfrac{1.6\times10^{-2}}{1.0})[/tex]

[tex]\theta=0.91^{\circ}[/tex]

We need to calculate the wavelength of the light

Using formula of wavelength

[tex]d\sin\theta=m\lambda[/tex]

[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]

Put the value into the formula

[tex]\lambda= \dfrac{0.30\times10^{-3}\times\sin0.91}{9}[/tex]

[tex]\lambda=5.29\times10^{-7}\ m[/tex]

[tex]\lambda=530\ nm[/tex]

Hence, The wavelength of the light is 530 nm.

Answer:

530 nm

Explanation:

Screen distance, D = 1 m

slit distance, d = 0.3 mm

n = 9 th bright

y = 1.6 cm

Let λ be the wavelength of light used.

y = n x D x λ / d

1.6 x 10^-2 = 9 x 1 x λ / (0.3 x 10^-3)

λ = 5.3333 x 10^-7 m

λ = 533.33 nm

λ = 530 nm ( by rounding off)

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

[tex]Q_1 = Q_2[/tex]

[tex]V_1A_1=V_2A_2[/tex]

Our values are given as,

[tex]A_1=\frac{1}{2}^2*\pi=0.785 in^2[/tex]

[tex]A_2=\frac{5}{8}^2*\pi=1.227 in^2[/tex]

Re-arrange the equation to find the first ratio of rates we have:

[tex]\frac{V_1}{V_2}=\frac{A_2}{A_1}[/tex]

[tex]\frac{V_1}{V_2}=\frac{1.227}{0.785}[/tex]

[tex]\frac{V_1}{V_2}=1.56[/tex]

The second ratio of rates is

[tex]\frac{V2}{V1}=\frac{A_1}{A2}[/tex]

[tex]\frac{V2}{V1}=\frac{0.785}{1.227}[/tex]

[tex]\frac{V2}{V1}=0.640[/tex]

The ratio of the flow rate of a ½ inch hose to that of a 5/8 inch hose is 16:25. This means that under equal pressures, for every 16 gallons of water that pass through the ½ inch hose, 25 gallons could flow through the 5/8 inch hose.

To compare the flow rates within the two hoses, we'll utilize a principle of fluid dynamics which states that the flow rate of an incompressible fluid (like water) is proportional to the cross-sectional area of the pipe. The area can be calculated using the formula for the area of a circle: A=πr², where r is the radius of the pipe.

To transform the diameters into radii, we divide them by 2: the radii are therefore 1/4 inch and 5/16 inch. As we're primarily interested in the ratio of areas (and consequently flow rates), we can ignore the π in the equation, leaving us with r² as directly representing 'area' for this comparison.

(1/4)² = 1/16 and (5/16)² = 25/256. Hence, the flow rate of the ½-inch hose to the 5/8 inch hose is 1/16: 25/256.

To simplify, we can multiply both portions of the ratio by 256 to achieve 16:25. This means for every 16 gallons per minute through the ½ inch hose, 25 gallons per minute could flow through the 5/8-inch hose, assuming the pressure in both systems is equal.

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Answer:

Width of central bright fringe will be 0.00228 m

Explanation:

We have given wavelength of light [tex]\lambda =632.8nm=632.8\times 10^{-9}m[/tex]

Distance between the slits [tex]d=0.5mm=0.5\times 10^{-3}m[/tex]

Distance between slit and screen D = 1.8 m

We have to find the width

We know that width is given by

[tex]width=\frac{\lambda D}{d}=\frac{632.8\times 10^{-9}\times 1.8}{0.5\times 10^{-3}}=0.00228m[/tex]

So width of central bright fringe will be 0.00228 m

Applying the given values to the formula for the width of the central bright fringe in a double-slit interference pattern yields a result of approximately 4.556 × 10^(-3) m.

The width of the central bright fringe in a double-slit interference pattern can be calculated using the formula:

w = λL / d

where:

w is the width of the central bright fringe,

λ is the wavelength of the light,

L is the distance from the slits to the screen, and

d is the distance between the slits.

In this case:

λ = 632.8 nm = 6.328 × 10^(-7) m,

L = 1.8 m, and

d = 0.5 mm = 5 × 10^(-4) m.

Plug these values into the formula:

w = (6.328 × 10^(-7) m × 1.8 m) / (5 × 10^(-4) m)

Calculating this expression gives:

w ≈ 4.556 × 10^(-3) m

So, the width of the central bright fringe is indeed approximately 4.556 × 10^(-3) m

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Answer:

Zero

Explanation:

Total energy of the EARTH  is the sum total of the all the kinetic and the gravitation potential energy of the earth.

Which can be written as

[tex]E_{total} = KE+U[/tex]

[tex]= \frac{1}{2}mv^2+ \frac{GMm}{R}[/tex]

At very large distances the kinetic as well as the gravitational potential energies become zero.

therefore, E_total= 0

When an object is launched at its escape velocity, its total mechanical energy at infinite distance from the planet is zero. The object loses all its kinetic energy, and its potential energy also becomes zero as it reaches infinity.

Escape velocity is the minimum initial velocity required for an object to escape the gravitational pull of a planet or other large body and reach an infinite distance where gravitational force becomes zero. When an object is launched at this escape velocity, its total mechanical energy (Etotal) at an infinite distance is zero. This is because the object gives up all its kinetic energy and the potential energy approaches zero as the distance (r) approaches infinity.

Using conservation of energy, we can illustrate this with the following steps:

Answer:

[tex]\frac{d\phi}{dt}=\pi r^2\frac{dB}{dt}[/tex]

Explanation:

the rate of the magnetic flux is given by,

[tex]\frac{d\phi}{dt}= \frac{d(BA)}{dt}[/tex]

where B= magnetic strength

A= area of cross section

r= radius of circle

\phi= flux

=[tex]A\frac{dB}{dt}[/tex]

=[tex]\pi r^2\frac{dB}{dt}[/tex]

Answer:

a. T = 119.68 N.m

b. r = 140 rev

Explanation:

first we know that:

∑T = Iα

where ∑T is the sumatory of the torques, I is the moment of inertia and α is the angular aceleration.

so, if the prop goes from rest to 400 rpm in 14 seconds we can find the α of the system:

1. change the 400 rpm to radians as:

                W = 400*2π/60

                W = 41.888 rad /s

2. Then, using the next equation, we find the α as:

                w = αt

solving for α

               α = [tex]\frac{w}{t}[/tex]

note: t is the time, so:

               α = [tex]\frac{41.888}{14}[/tex]

               α = 2.992 rad/s^2

Now using the first equation, we get:

T = Iα

T = 40(2.992)

T = 119.68 N.m

On the other hand, for know the the number of revolutions turned as the prop achieves operating speed, we use the following equation:

θ = wt +[tex]\frac{1}{2}[/tex]α[tex]t^2[/tex]

Where w =  41.888 rad /s, α = 2.992 rad/s^2, t is the time and θ give as the number of radians that the prop made in the fisrt 14 seconds, so:

θ = (41.888)(14)+[tex]\frac{1}{2}[/tex](2.992)(14[tex])^2[/tex]

θ = 879.648 rad

and that divided by 2π give us the number of revolutions r, so:

r = 140 rev

Answer:

They have the same momentum, but the car has more kinetic energy than the truck.

Explanation:

Let [tex]m_c[/tex] is the mass of the car and [tex]m_t[/tex] is the mass of truck such that,

[tex]m_t=2m_c[/tex]

You push both the car and the truck for the same amount of time with the same force. Momentum of an object is given by :

[tex]p=mv=F\times t[/tex]

Since, force and time are same, so they have same momentum. The kinetic energy of an object is given by :

[tex]K=\dfrac{1}{2}mv^2[/tex]

Since, the mass of truck is more, it will have maximum kinetic energy. So, the correct option is (a) "They have the same momentum, but the car has more kinetic energy than the truck".

The car has more momentum and more KE than the truck because car mass is lower so it moves fast as compared to truck.

Momentum is the product of its mass and velocity. When we compare car with truck , car has a lower mass so when the mass is lower the velocity will be higher so car has more momentum.

We also know that kinetic energy also depends on mass so less mass have more kinetic energy.

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Answer:

[tex]\lambda=3.20*10^{-7}m[/tex]

Explanation:

The wavelength is inversely proportional to the frequency. The wavelength is equal to the speed of the wave, divided by the frequency. In the case of electromagnetic waves like ultraviolet radiation, the speed of propagation is the speed of light.

[tex]\lambda=\frac{c}{f}\\\lambda=\frac{3*10^8\frac{m}{s}}{9.38*10^{14}Hz}\\\lambda=3.20*10^{-7}m[/tex]

Answer :  The wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]

Explanation : Given,

Frequency = [tex]9.38\times 10^{14}Hz=9.38\times 10^{14}s^{-1}[/tex]

Formula used :

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency

[tex]\lambda[/tex] = wavelength

c = speed of light = [tex]3\times 10^8m/s[/tex]

Now put all the given values in the above formula, we get:

[tex]9.38\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}[/tex]

[tex]\lambda=3.20\times 10^{-7}m[/tex]

Therefore, the wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]

Answer:

Speed of second block will be 20 m /sec in opposite direction

Explanation:

We have given the mass of two blocks [tex]m_1=40kg[/tex] and [tex]m_2=30kg[/tex]

As both the blocks are initially at rest so [tex]u_1=0m/sec\ and\ u_2=0m/sec[/tex]

Speed of first block [tex]v_1=15m/sec[/tex]

We have to find the speed of second block, that is [tex]v_2[/tex]

From momentum conservation we know that

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]40\times 0+30\times 0=40\times 15+30\times v_2[/tex]

[tex]v_2=-20m/sec[/tex]

So speed of second block will be 20 m /sec in opposite direction.

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

[tex]v^2=u^2+2gs[/tex]

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

[tex]0=u^2-2\times9.8\times11.0[/tex]

[tex]u=\sqrt{2\times9.8\times11.0}[/tex]

[tex]u=14.68\ m/s[/tex]

Hence, The speed of the water is 14.68 m/s.

Answer:

The angular acceleration increases.

Explanation:

The relationship between the torque and angular acceleration is :

[tex]\tau=I\times \alpha[/tex]

Where

I is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

We can see that the torque is directly proportional to the angular acceleration. So, when we have more torque it means angular acceleration increases. Hence, the correct option is (A).

Final answer:

When there is more torque applied to an object with a constant rotational inertia, the angular acceleration of the object increases proportionally. So the correct option is A.

Explanation:

According to the relationship between torque and angular acceleration, when you have more torque (given a constant rotational inertia), the angular acceleration increases. This is because torque (τ) is directly related to angular acceleration (α) through the equation τ = Iα, where I represent the moment of inertia. If the moment of inertia remains constant, an increase in torque results in a proportional increase in angular acceleration, similar to how pushing a merry-go-round harder makes it accelerate faster.

The final Celsius temperature is C. greater by a factor between 1 and 2.

Explanation:

We can solve this problem by applying the pressure law, which states that for an ideal gas kept at constant volume, the ratio between the pressure and the absolute temperature of the gas is constant:

[tex]\frac{p}{T}=constant[/tex]

or

[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]

where in this problem:

[tex]p_1 = p_0[/tex] is the initial pressure of the gas

[tex]T_1[/tex] is the initial absolute temperature (in Kelvin)

[tex]p_2 = 2 p_0[/tex] is the final pressure

[tex]T_2[/tex] is the final temperature in Kelvin

Re-arranging the equation,

[tex]\frac{p_0}{T_1}=\frac{2p_0}{T_2}\\T_2 = 2T_1[/tex]

So, the temperature in Kelvin has doubled. We can now rewrite this relationship by rewriting the Kelvin temperature in Celsius degrees:

[tex]T_{C2}-273 = 2(T_{C1}-273)\\T_{C2}=2T_{C1}-273[/tex]

where [tex]T_{C1}, T_{C2}[/tex] are the initial and final temperatures in Celsius degrees.

This means that the temperature in Celsius increases by a factor between 1 and 2, so the correct answer is C.

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The final Celsius temperature of the gas after being warmed at constant volume to reach a pressure of 2p0 is greater by a factor between 1 and 2, according to Gay-Lussac's law. So the correct option is C.

The question deals with the behavior of gas under the influence of temperature changes at constant volume, which is described by Gay-Lussac's law. This law states that for a given mass of gas at constant volume, the pressure of the gas is directly proportional to its Kelvin (absolute) temperature. Since the initial pressure of the gas is p0 and the final pressure is 2p0, and we know that the volume remains constant, the final Kelvin temperature must also double. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. If the initial temperature in Celsius T1 corresponds to the Kelvin temperature K1, and after the change, the final Kelvin temperature is 2K1, then when converted back to Celsius, the final temperature T2 will not be twice T1—because the relationship between Kelvin and Celsius is linear and not proportional. Therefore, the final Celsius temperature will be higher by a certain amount but not double.

Given this, the correct answer is that the final Celsius temperature of the gas is greater by a factor between 1 and 2.

To solve this problem it is necessary to apply the concepts presented in Kepler's third law in which the period is described, as well as the theorems developed for acceleration based on gravity.

Acceleration in gravitational terms can be expressed as

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

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Distance

At the same time the Period by Kepler's law the Period is described as

[tex]T^2 = \frac{4\pi^2r^3}{GM}[/tex]

That is equal to

[tex]T^2 = \frac{4\pi^2r}{\frac{GM}{r^2}}[/tex]

Using the equation of acceleration,

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

Re-arrange to find a,

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

Our values are given as

[tex]r = 1.5*10^{11}m[/tex]

[tex]T = 365days(\frac{86400s}{1days}) = 31536000s[/tex]

Replacing we have,

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

[tex]a = \frac{4\pi^2(1.5*10^{11})}{(31536000)^2}[/tex]

[tex]a = 0.005954m/s^2[/tex]

Therefore the magnitude of the acceleration of the earth is [tex]0.005954m/s^2[/tex]

The magnitude of the acceleration of the Earth in its orbit around the sun, assuming a circular orbit, is approximately 0.0059 m/s².

The concept at play here is centripetal acceleration, which is the rate of change of tangential velocity and points toward the center of the circle around which Earth, or any object, orbits. For Earth orbiting the sun, we can use the formula ac = v²/r (where v is velocity, r is radius), but given that v isn't directly stated, one must compute it first by v = 2πr/T (where T is the period of the orbit in seconds).

So, the velocity of the Earth in its orbit is approximated as 2π(1.50x10¹¹ m)/(365.25x24x60x60 s) = 29,785.6 m/s. To get the acceleration, plug this computed v value into the ac equation: ac = (29,785.6 m/s)²/(1.50 x 10¹¹ m) = 0.0059 m/s².

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Answer:

The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

Explanation:

Given that,

Energy density = 0.1 J/m³

Separation = 0.2 mm

We need to calculate the potential difference

Using formula of energy density

[tex]J=\dfrac{1}{2}\epsilon_{0}E^2[/tex]

[tex]J=\dfrac{1}{2}\epsilon_{0}\dfrac{V^2}{d^2}[/tex]

[tex]V^2=\dfrac{0.1\times(0.2\times10^{-3})^2\times2}{8.85\times10^{-12}}[/tex]

[tex]V^2=\sqrt{903.95}[/tex]

[tex]V=30.06\ V[/tex]

We need to calculate the speed of electron

Using energy conservation

[tex]U=eV=\dfrac{1}{2}mv^2[/tex]

Put the value into the formula

[tex]1.6\times10^{-19}\times30.06=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2[/tex]

[tex]v^2=\dfrac{1.6\times10^{-19}\times30.06\times2}{9.1\times10^{-31}}[/tex]

[tex]v=\sqrt{1.057\times10^{13}}[/tex]

[tex]v=3.2\times10^{6}\ m/s[/tex]

Hence, The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

Answer:

Explanation:

Diameter of the circular coil d = 17.7 cm = 0.177m

Current in the coil, I = 7.4 A

Number of loops in the coil, N = 18

Magnetic field, B = 5.5 x 10-5 T

Angle below line, (θ) = 66°

Write the expression for the torque in the coil

Torque = N x I x B x A x sin (θ)

Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.

The cross- sectional area of the coil = (pie)(d/2)²

A = (π)(0.177/2)² = 0.0246 m²

The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north

Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°

Theta = 90 – 66 = 24°

Substitute the value to find out torque

Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m

The torque on the coil is 7.33 x 10⁻⁵ N-m