In the original Bose-Einstein experiment, rubidium atoms were trapped in a small (but macroscopic) device, roughly 0.1 mm on a side. Consider a single Rb atom trapped at t0. If released (from rest), what happens to this particle as time goes by? What are m, a, and Ω here, as numbers? What is the "timescale" in seconds?

Answer:

100j

Explanation:

The position of the image is at infinity, nature is real and the size of the image formed is highly magnified.

A concave mirror has a reflective floor that is curved inward and far from the light source. Concave mirrors mirror mild inward to at least one focal length. not like convex mirrors, the picture shaped by a concave replicate suggests exceptional photo sorts depending on the space between the item and the replicate.

Calculation:-

1/f = 1/v + 1/u

1/10 = 1/v + 1/10

V = infinity

The size of the image is Highly magnified and real.

Concave mirrors form both real and virtual photos. while the concave reflect is positioned very close to the object, a digital and magnified photo is received, and if we grow the gap between the object and the mirror, the size of the picture reduces, and real snapshots are shaped.

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Answer:

A. pushing a car that isn't moving

The average highest height of a professional football kick is 189.728 m if the hang time is 4.4 seconds on average.

A person or an object's total duration in the air after leaving the ground is known as their "hang time." From the time anything leaves the ground until it returns, it is measured.

We know,

y= gt²

Here,

y = Average maximum height

g = acceleration due to gravity

t = Average hang time

Given,

Average hang time (t) = 4.4s

Acceleration due to gravity (g) = 9.8 m/s² (assuming)

Inserting these values in the given equation,

y = gt²

  = 9.8×4.4×4.4

  = 189.728 m.

Hence, the average maximum height of the football is 189.728 m.

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The mass of the planet is  5.98 × 10^24 kg.

To calculate the mass of the planet, we can use Kepler's Third Law of Planetary Motion. This law states that the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit.

First, we need to convert the period of the satellite's orbit to seconds. We know that there are 60 minutes in an hour, so the period can be expressed as (6 × 60 + 12) minutes, which equals 372 minutes. Multiplying this by 60 seconds, we get a period of 22,320 seconds.

Next, we need to find the semi-major axis of the orbit. In a circular orbit, the semi-major axis is equal to the radius of the orbit. Therefore, the semi-major axis is 8.8 × 10^6 m.

Now, we can apply Kepler's Third Law to calculate the mass of the planet. The formula is T^2 = (4π^2/GM) × a^3, where T is the period of revolution, G is the gravitational constant, M is the mass of the planet, and a is the semi-major axis of the orbit.

Rearranging the formula, we can solve for the mass of the planet:
M = (4π^2/G) × a^3 / T^2

Plugging in the values, we get:
M = (4 × π^2 / 6.67430 × 10^-11) × (8.8 × 10^6)^3 / (22,320)^2

Evaluating this expression, we find that the mass of the planet is approximately 5.98 × 10^24 kg.

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Answer:

A. 450 N

B. 3240 J

C. 0.77 KW

Explanation:

From the question given above, the following data were obtained:

Height of one step = 20 cm

Number of steps = 36

Mass of runner = 45 kg

Time taken = 4.2 s

Next, we shall convert 20 cm to metre (m). This can be obtained as follow:

100 cm = 1 m

Therefore,

20 cm = 20 cm × 1 m /100 cm

20 cm = 0.2 m

Next, we shall determine the total height. This can be obtained as follow:

Height of one step = 0.2 m

Number of steps = 36

Total height =?

Total height = 36 × 0.2

Total height = 7.2 m

A. Determination of the runner's weight.

Mass of runner (m) = 45 kg

Acceleration due to gravity (g) = 10 m/s²

Weight (W) =?

W = m × g

W = 45 × 10

W = 450 N

B. Determination of the increase in the potential energy.

At the ground level, the potential energy (PE₁) is 0 J.

Next, we shall determine the potential energy at 7.2 m. This can be obtained as follow:

Mass of runner (m) = 45 kg

Acceleration due to gravity (g) = 10 m/s²

Total height (h) = 7.2 m

Potential energy at height 7.2 m (PE₂) = ?

PE₂ = mgh

PE₂ = 45 × 10 × 7.2

PE₂ = 3240 J

Final, we shall determine the increase in potential energy. This can be obtained as follow:

Potential energy at ground (PE₁) = 0 J

Potential energy at height 7.2 m (PE₂) = 3240 J

Increase in potential energy =?

Increase in potential energy = PE₂ – PE₁

Increase in potential energy = 3240 – 0

Increase in potential energy = 3240 J

C. Determination of the power.

Energy (E) = 3240 J

Time (t) = 4.2 s

Power (P) =?

P = E/t

P = 3240 / 4.2

P = 771.43 W

Finally, we shall convert 771.43 W to kilowatt (KW). This can be obtained as follow:

1000 W = 1 KW

Therefore,

771.43 W = 771.43 W × 1 KW / 1000 W

771.43 W = 0.77 KW

Therefore, her power is 0.77 KW

Answer:

\(E_0=0.173N/C\)

Explanation:

From the question we are told that:

Power \(P=50kw=>50*10^3w\)

Distance \(d=10km=10000m\)

Generally the equation for Intensity is mathematically given by

\(I=\frac{P}{4\pi d^2} w/m^2\)

\(I=\frac{50*10^3}{4 \pi 10000^2} w/m^2\)

\(I=3.98*10^{-5}w/m^2\)

Generally Intensity is also

\(I=\frac{1}{2}cE_0^2e\)

Where

\(e=8.854*10^{-12}Nm^2/c^2\)

Therefore

\(E_0=\sqrt{\frac{2I}{c *e}}\)

\(E_0=\sqrt{\frac{2*3.98*10^{-5}}{3*10^8 *8.854*10^{-12}}}\)

\(E_0=0.173N/C\)

The passage is describing the beauty of Acadie, a place that is home to the speaker. This passage is from Bliss Carman's "LOW TIDES" .

The sun is setting, and the speaker is captivated by the beauty of the landscape, which is enhanced by the unelusive glories of the tide. The speaker also reflects on a beloved face being gone for a long time and is moved by a grievous stream, which meanders through the fields of Acadie as if it is trying to figure out why the beloved face is gone. The speaker almost dreams that the beauty of the place will remain until the beloved face returns. The passage conveys a sense of longing and hope for the beloved to come back to Acadie.

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The radius the star Rigel, a bright blue star in the constellation Orion that radiates energy at a rate of 2.7 × 10³¹ W and has a surface temperature of 11,000 K. (Assume that the star is spherical) = 5,422.17 km

To determine the radius of the star, we can apply the Stefan-Boltzmann law to determine the star output power.

Rigel emits radiation at a rate of 2.7 x 10²³ Watts

The power per unit area, according to the Stefan-equation, Boltzmann's is:

j = σT⁴

Where,

σ = (2\(\pi\)⁵K⁴) / 15c²h³

T = temperature

σ = 5.670 x 10⁻⁸ Wm⁻²K⁻⁴

Hence,

The surface area of the star:

A = P / j

P = power ⇒ 2.7 × 10³¹

A = ( 2.7 × 10³¹) / (5.670 x 10⁻⁸) (11.000⁴)

= 3.52 x 10²³

The area = 4\(\pi\)r²

3.52 x 10²³ = 4\(\pi\)r²

r² = 2.58 x 10²²

r = 5,422.17 km

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The time taken by  the projectile to reach its maximum height is 0.62 second.

Given parameters:

Horizontal component of initial velocity of the projectile: \(u_x\) = 6.1 m/s.

To reach maximum height, the projectile should be throwed in 45 degree. So, vertical component of initial velocity of the projectile: \(u_y\) = 6.1 m/s.

Given,  The acceleration of gravity is 9.8 m/s².

At maximum height the vertical component of velocity of the projectile becomes zero due to acceleration due to gravity acts downwards.

So, from formula \(v_y\) = \(u_y\) - gt; we can write:

0 = 6.1 - 9.8t

⇒ t = 6.1/9.8 = 0.62 s.

Hence, the time taken by  the projectile to reach its maximum height is 0.62 second.

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Answer:

yufjfkfjrruuemgmkfjktjrkttojjfmhtioy ykirouwqdjjdsy

The angle measured counterclockwise from the positive x-axis is θ = 50.4°

Here we know that the vector is:

V = < -177 cm, -214 cm>

To get the correspondent angle for this vector, we can think that this is the hypotenuse of a right triangle, such that the y-component and x-component are the cathetus.

Then, to get the angle (measured counterclockwise from the positive x-axis) is given by:

Tan(θ) = (opposite cathetus)/(adjacent cathetus)

Tan(θ) = (-214cm)/(-177 cm)

Using the inverse tangent function we get:

Atan(Tan(θ)) = Atan((-214cm)/(-177 cm))

θ = 50.4°

So the angle is 50.4°

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Answer:

1 / f = 1 / o + 1 / i = (i + o) / o * 1

f = o * i / (o + i) = 60 * 30 / (60 + 30) = 1800 / 90 = 20 cm

Both the object and image are in positive space for a mirror

Answer: 37.4N

Explanation:

coefficient of friction means how much friction material has, and how difficult would it be to move something on the surface. It's a number between 0 and 1.

e.g. ice has a very low coefficient of friction and it's easy to move and slide on it, while concrete has a higher coefficient of friction and it's going to be difficult to push things on it.

Formula:

coefficient of friction = force of friction/normal (perpendicular) force

on a flat surface normal force is equal to the weight of the object

weight = mass * force of gravity = 4.4 * 10 = 44N

-> normal force = 44N

coefficient of friction = 0.85

force of friction = coefficient of friction * normal force = 0.85 * 44 = 37.4N

-> to just move the object you need at least 37.4N

Answer:

x = A sin (wt + theta)        where w = angular frequency - basic SHM equation

w = 8 pi = 2 pi f

f = 4         basic frequency

N = 8     number of times thru origin

Each cycle the particle will pass thru the origin +x and -x    twice

Answer:

Backyard burning of trash in a barrel, pile or outdoor boiler releases smoke into the air. ... Trash fires in burn barrels can smolder and as a result produce greater amounts of harmful chemicals in the smoke. Harmful chemicals can also be present in the ash.Explanation:

(a) The work done by the 150 N force is 877.5 Joules.

(b) The coefficient of kinetic friction between the crate and the surface is approximately 0.437.

To answer this problem, we must take into account the work done by the applied force as well as the work done by friction.

(a) The applied force's work may be estimated using the following formula:

Work = Force * Distance * cos(theta)

where the force is 150 N and the distance is 5.85 m. Since the force is applied horizontally and the displacement is also horizontal, the angle theta between them is 0 degrees, and the cosine of 0 degrees is 1.

As a result, the applied force's work is:

Work = 150 N * 5.85 m * cos(0) = 877.5 J

So, the work done by the 150 N force is 877.5 Joules.

(b) Frictional work is equal to the force of friction multiplied by the distance. The work done by friction is identical in amount but opposite in direction to the work done by the applied force since the crate travels at a constant speed.

The frictional work may be estimated using the following formula:

Work = Force of Friction * Distance * cos(theta)

The net force applied on the crate is zero since it is travelling at a constant pace. As a result, the friction force must be equal to the applied force, which is 150 N.

Thus, the work done by friction is:

Work = 150 N * 5.85 m * cos(180) = -877.5 J

Since the work done by friction is negative, it indicates that the direction of the frictional force is opposite to the direction of motion.

The coefficient of kinetic friction may be calculated using the following equation:

Friction Force = Kinetic Friction Coefficient * Normal Force

The normal force equals the crate's weight, which may be computed as:

Normal Force = mass * gravity

where the mass is 35.0 kg and the acceleration due to gravity is approximately 9.8 m/s^2.

Normal Force = 35.0 kg * 9.8 m/s^2 = 343 N

Now, we can rearrange the equation for the force of friction to solve for the coefficient of kinetic friction:

Force of Friction = coefficient of kinetic friction * Normal Force

150 N = coefficient of kinetic friction * 343 N

coefficient of kinetic friction = 150 N / 343 N ≈ 0.437

As a result, the kinetic friction coefficient between the container and the surface is roughly 0.437.

In summary, the work done by the 150 N force is 877.5 Joules, and the coefficient of kinetic friction between the crate and the surface is approximately 0.437.

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Certain behaviors and choices during this time can have significant impacts on later-life success are poor diet, lack of exercise, smoking and chronic stress.

Poor diet: Consuming a diet that is high in saturated fats, processed foods, and sugar can lead to weight gain, high blood pressure, and other chronic conditions that can have lasting effects on health.

Lack of exercise: Failing to exercise regularly can increase the risk of chronic diseases such as heart disease, diabetes, and certain cancers, which can impact the quality of life in later years.

Smoking: Smoking is a major risk factor for many chronic diseases, including heart disease, stroke, and lung cancer. Quitting smoking during midlife can significantly reduce the risk of these diseases.

Alcohol abuse: Drinking excessively can increase the risk of liver disease, heart disease, and other chronic health conditions that can negatively impact later-life success.

Chronic stress: Unmanaged stress can contribute to the development of mental health problems, such as anxiety and depression, as well as chronic physical health conditions, such as hypertension and heart disease.

By making healthy choices, managing stress, and maintaining positive perceptions and attitudes, individuals can increase their chances of achieving success in later life.

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A boy pushes a lever down 2 meters with a force of 75 newtons. The box at the other end with a weight of 50 newtons move up 2.5 meters. The efficiency of this machine is 83.33%.

The energy  lost during a machine's operation due to heat and friction is measured as the machine's efficiency. Given that work is the transformation of kinetic energy, a machine's efficiency can be expressed as the ratio of output work divided by input work less work wasted due to friction and heat. It is unitless.

Using the formula:

Efficiency=\(\frac{O}{I}\)×100

where,

O is output work and,

I is the input work.

Work done= Force × displacement

Output work = 50 × 2.5 = 125J

Input work = 75 × 2 = 150J

Substituting the values and solving for efficiency:

Efficiency = 83.33%

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Complete question:

A boy pushes a lever down 2 meters with a force of 75 Newtons. The box on the other end weighs 50 Newtons and moves up 2.5 meters. What is the efficiency of this machine?

Answer:

The giraffe is the tallest of all mammals. It reaches an overall height of 18 ft (5.5 m) or more. The legs and neck are extremely long. The giraffe has a short body, a tufted tail, a short mane, and short skin-covered horns

Answer:

a)  W = 1,639 10⁶ J, b)     v = 38.60 m / s, c)    k = 3.78 10⁵ N / m

Explanation:

a) In this part they ask us the work done by the motor, let's use the work energy theorem

            W = ΔEm = Em_f - Em₀

            W = mh y₂ - mgy₁

            W = mg (y₂ -y₁)

            W = 2200 9.8 (80 -4)

            W = 1,639 10⁶ J

b) for this part let's use conservation of energy

starting point. Highest point

            Em₀ = U = m g y₂

final point . Punoit in contact with the spring

           Em_f = K + U

           Em_f = ½ m v² + m g y₁

           Em₀ = Em_f

           m g y₂ = ½ m v² + mg y₁

           v = √(2g (y₂ - y₁))

           v = √(2 9.8 (80-4))

           v = 38.60 m / s

c) the stiffness of the spring

starting point. Just when it comes into contact with the spring

           Em₀ = K + U

           Em₀ = ½ m v² + mgy₁

Final point. With the spring at maximum compression x = 3 m

           Em_f = Ke + U

           Em_f = ½ k x² + m g y₃

           Em₀ = Em_f

           ½ m v² + mgy₁ = ½ k x² + m g y₃

           ½ k x² = ½ m v² + m g (y₁ - y₃)

let's calculate

            ½ 3² k = ½ 2200 38.60² + 2200 9.8 (4 -1)

            4.5 k = 1.639 10⁶ +6.468 10⁴ = 1.7036 10⁶

            k = 1.7036 10⁶ / 4.5

             k = 3.78 10⁵ N / m

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You need to choose a word that appropriately completes the second pair in order to resolve the analogy. The words in an analogy may appear to be unrelated to one another at first look, but they are always logically connected. The link between the first and second word pairs is comparable.

In the testing analogies, only the first set of words is provided. Therefore, you must first determine how these two words are related. To complete the analogy, you must select a word or words that have an analogous relationship to the pair.

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Small bodies with high thermal conductivity, the medium should be a poor conductor of heat and should be motionless in order to favour lumped system analysis.

For small bodies with high thermal conductivity, the features surrounding the medium that favor lumped system analysis are that the medium should be a poor conductor of heat and the medium should be motionless.

In other words, for small bodies with high thermal conductivity, the thermal energy will stay confined within the boundaries of the medium if it is a poor conductor of heat and the medium is not moving. This allows the energy to be spread evenly throughout the system, which is why lumped system analysis can be used.

Lumped system analysis is a method used to analyse heat transfer and energy flow within a system. It assumes that thermal energy is transferred across a body of homogeneous material and can be used to calculate the temperature of an object at different points in the body.

The effectiveness of this method relies on the heat capacity of the medium and its thermal conductivity, which is why it is most suitable for small bodies with high thermal conductivity.

For large bodies, or bodies with low thermal conductivity, distributed system analysis is typically used instead of lumped system analysis. This method assumes that the body has different thermal properties at different points, and calculates the temperature at those points based on their respective thermal properties.

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Answer: fume hood is used in chemical experiments, which result in toxic fumes, that could potentially have effects on health. fume hood simply acts as an exhaust, to drive away all toxic fumes that result in chemical experiments:

The answer therefore is:

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Answer:

The loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

Given;

mass of the disk, m₁ = 1.64 kg

radius of the disk, r = 0. 61 m

angular velocity of the disk, ω₁ = 17.6 rad/s

mass of the rod, m₂ = 1.51 kg

length of the rod, L = 1.79 m

angular velocity of the rod, ω₂ =  5.12 rad/s (clock-wise)

let the counter-clockwise be the positive direction

let the clock-wise be the negative direction

The common final velocity of the two systems after the collision is calculated by applying principle of conservation of angular momentum ;

m₁ω₁  + m₂ω₂ = ωf(m₁ + m₂)

where;

ωf is the common final angular velocity

1.64 x 17.6    + 1.51(-5.12) = ωf(1.64 + 1.51)

21.1328 = ωf(3.15)

ωf = 21.1328 / 3.15

ωf = 6.709 rad/s

The moment of inertia of the disk is calculated as follows;

\(I_{disk} = \frac{1}{2} mr^2\\\\I_{disk} = \frac{1}{2} (1.64)(0.61)^2\\\\I_{disk} = 0.305 \ kgm^2\)

The moment of inertia of the rod about its center is calculated as follows;

\(I_{rod} = \frac{1}{12} mL^2\\\\I_{rod} = \frac{1}{12} \times 1.51 \times 1.79^2\\\\I _{rod }= 0.4032\ kgm^2\)

The initial rotational kinetic energy of the disk and rod;

\(K.E_i = \frac{1}{2} I_{disk}\omega _1 ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _2 ^2 \\\\K.E_i= \frac{1}{2} (0.305)(17.6) ^2 \ \ + \ \ \frac{1}{2} (0.4032)(-5.12) ^2\\\\K.E_i = 52.523 \ J\)

The final rotational kinetic energy of the disk-rod system is calculated as follows;

\(K.E_f = \frac{1}{2} I_{disk}\omega _f ^2 \ \ + \ \ \frac{1}{2} I_{rod}\omega _f ^2\\\\K.E_f = \frac{1}{2} \omega _f ^2(I_{disk} + I_{rod})\\\\K.E_f = \frac{1}{2} (6.709) ^2(0.305+ 0.4032)\\\\K.E_f = 15.938 \ J\)

The loss in rotational kinetic energy due to the collision is calculated as follows;

\(\Delta K.E = K.E_f \ - \ K.E_i\\\\\Delta K.E = 15.938 J \ - \ 52.523 J\\\\\Delta K.E = - 36.585 \ J\)

Therefore, the loss in rotational kinetic energy due to the collision is 36.585 J.

Explanation:

You want  N/m

 N = 66 * 9.81

 m = 2.3 x 10^-2 m

66* 9.81 / 2.3 x 10^-2  = 28150 =  28 000 N/m    to two S D

Answer:

10 miles

Explanation:

b/c distance all what we go 5+5=10