A current / ampere flows along an infinitely long straight thin walled tube then the magnetic induction at any point inside the tube is.

The position of the Earth, Sun, and moon during the solar eclipse, a student observes a shadow moving across the sun is Sun, moon, Earth, so option D is correct.

When the Moon passes in front of the Sun, blocking all or part of Earth's view of it, a solar eclipse occurs. The Moon is closest to the plane of the Earth's orbit when there is a new moon and such an alignment. During a total eclipse, the Moon fully obscures the Sun's disk. Partial and annular eclipses only block out a section of the Sun. A solar eclipse occurs naturally. Solar eclipses were sometimes linked to supernatural sources or interpreted as an ill omen in several ancient and modern societies.

During a solar eclipse, the moon comes in between the earth and the sun therefore the shadow moves across the sun during a solar eclipse observed by the student, the position is Sun, moon, and earth.

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Answer:

B

Explanation:

They move at a rate of one to two inches (three to five centimeters) per year.

Answer:

The total displacement 560 m

Explanation:

 Given:

t₀ = 0 s

t = 25 s

V₀ = 10 m/s

V = 35 m/s

___________

D - ?

Acceleration:

a =  (V - V₀) / (t - t₀) = (35 - 10) / (25 - 0) = 25 / 25 = 1.0 m/s²

The total displacement:

D = V₀·t + a·t² / 2

D = 10·25 + 1.0·25² / 2 ≈ 560 m

The momentum flow rate through the pipe carrying liquid ammonia is 1 × \(10^6\) kg·m/s.

The flow rate of momentum (Ṁ) through the pipe can be calculated by multiplying the mass flow rate (ṁ) by the velocity (v). The speed can be determined using the equation v = ṁ / (ρA), where ρ is the density of the liquid ammonia and A is the pipe's cross-sectional area.

Given:

ṁ = 100 kg/s

A = 0.01 m²

Assuming the density (ρ) of liquid ammonia is 700 kg/m³, we can calculate the velocity (v):

v = ṁ / (ρA)

v = 100 kg/s / (700 kg/m³ × 0.01 m²)

v = 10000 m/s

Now, we can calculate the flow rate of momentum (Ṁ):

Ṁ = ṁv

Ṁ = 100 kg/s × 10000 m/s

Ṁ = 1 × \(10^6\) kg·m/s

Therefore, the momentum flow rate through the pipe is 1 × \(10^6\) kg·m/s.

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Answer: The x-component of the acceleration is \(-0.45m/s^{2}\)

There u go (:

The outermost energy shell of the atom, where the electrons are situated in relation to the nucleus. the energy levels are-  1 (only"s"), 2 ("s","p"), 3 ("s","p","d"), 4 ("s", "p", "d", "f").

An atom's energy level is the set distance between its nucleus and any potential location for electrons. Additionally, energy levels might be compared to a staircase's steps.

Most major diseases, including heart disease, many types of cancer, autoimmune diseases like lupus and multiple sclerosis, and anemia, commonly manifest with a lack of energy (too few red blood cells). Another typical indicator of depression and anxiety is fatigue. Additionally, a side effect of several drugs is weariness.

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Answer:

31

Explanation:

Given:

U=3

X=4

Y=7

u + xy

Substitute the given values to the equation:

3 + (4)(7)

3 + 28

31

11.3 g/cm³ equals to 1130  kg/m³.

The mass of a substance per unit of volume is its density. Density is most frequently represented by the symbol ρ, however Latin letter D may also be used. Density is calculated mathematically by dividing mass by volume.

Given that 11.3 g/cm³.

Know that

1000 grams = 1 kg

1 grams = 1/1000 kg.

100 cm = 1 m

1 cm = 1/100 m.

Therefore, 1 cm³ = (1/100)³ m³

= 1/1000000 m³

Hence, 1  g/cm³ = (1/1000) ÷ (1/1000000)  kg/m³

= 1000000/1000 kg/m³

= 1000 kg/m³.

Hence, 11.3 g/cm³ = (11.3 × 1000) kg/m³

= 1130  kg/m³

Hence, 11.3 g/cm³ equals to 1130  kg/m³.

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We presume that the wave function can be written as a product of three one-dimensional wave functions: (x, y, z) = X(x), Y(y), Z.(z). In that case, the three-dimensional harmonic oscillator's Schrödinger equation is as follows:− 2mℏ2(∂x 2∂ 2Ψ+ ∂y 2∂2Ψ+ ∂z 2∂ 2Ψ)+ 21mω 2(x 2+y 2+z 2 )Ψ=EΨ

Each of these equations has the same form as the one-dimensional harmonic oscillator, with a potential proportional to the square of the position and a frequency ω. Therefore, we can use the known energy eigenvalues for the one-dimensional harmonic oscillator to determine the allowed energies for the three-dimensional harmonic oscillator.The energy of a one-dimensional harmonic oscillator is given by En = (n + 1/2)ħω, where n is a non-negative integer. Therefore, the energy of the three-dimensional harmonic oscillator is:E = Ex + Ey + Ez = [(nx + 1/2) + (ny + 1/2) + (nz + 1/2)]ħω

= (nx + ny + nz + 3/2)ħω

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No, law of conservation of energy is not applicable to open systems.

Law of conservation of energy states that energy present anywhere can neither be created nor be destroyed by any means, it can only be converted from one form of energy to other form of energy. It means total energy of an isolated system remains same.

An open system is a system that can exchange energy and matter from the surroundings. The flow of energy takes places in an open system. Therefore, the law of conservation does not apply to an open system as it can lose or gain energy, hence its total energy will not be constant and will violate law of conservation of energy.

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Answer:

Explanation:

Considering that this is parabolic motion, we know that the time the ball is in the air begins the instant it leaves the ground, reaches up to its max height, and then begins falling until it reaches the ground. Duh, right? Some important things happen during this trip. There are a few things we need to know in order to even begin the problem. Parabolic motion has x and y coordinates because it is 2-dimmensional; the acceleration in the x dimension is not the same as the acceleration in the y dimension; the velocity of an object at its max height is always 0; the time it takes to reach its max height (where the max height is half the distance the object travels) is half the time it takes to make the whole trip. Yikes. That's a lot to know and much to remember! Don't you just LOVE physics!?

For a. the hang time is the time the ball was in the air. Some of that stuff we talked about above is pertinent to solving this problem. We know that the velocity of the ball is 0 at its max height, and we also know that if we find the time it takes to reach its max height, we can double that number to find how long it was in the air for the whole trip. Use the one-dimensional equation

\(v=v_0+at\) to find out how long it took to reach the max height. Even though we don't yet know the max height, we DO know that the velocity at that point is 0. BUT before we do that, since we are working in the y-dimension only, it would behoove us (benefit us) to find the velocity particular to this dimension. We are going to answer c. first, then backtrack.

c. wants the initial vertical velocity. That is found in the magnitude of the "blanket" or generic velocity times the sin of the angle, namely:

\(V_y=25sin(45)\) so

\(V_y=\) 18 m/s Now we can use that as the initial upwards velocity in part a:

\(v=v_0+at\) and filling in:

0 = 18 + (-9.8)t and

-18 = -9.8t so

t = 1.8 seconds. But remember, this is only half the time it was in the air. The whole trip, then, takes 2(1.8) which is

t = 3.6 seconds

That's a and c. Now for b:

b. asks for the x component of the velocity:

\(V_x=Vcos\theta\) which works out to be the same as the vertical velocity, since the sin and cos of 45 degrees is the same:

\(V_x=25cos45\) and

\(V_x=\) 18 m/s

Onto d:

d. wants the max height. Remember, it took 1.8 seconds to get to the max height, so using yet another one-dimensional equation:

Δx = v₀t + \(\frac{1}{2}at^2\) where Δx is the displacement, v₀ is the initial upwards velocity, a is the pull of gravity, and t is the time it takes to reach that max height (Δx, our unknown). Filling in:

Δx = \(18(1.8)+\frac{1}{2}(-9.8)(1.8)^2\) and if you do the rounding correctly, you'll end up with this:

Δx = 32 - 16 so

the max height, Δx, is 16 meters.

e. wants the range. That translates to the distance the ball traveled. This is found in a glorified version of d = rt, where d is displacement, r is velocity, and t is...well, time (that doesn't change):

Δx = vt so

Δx = 18(3.6) remember that the ball was in the air for a total of 3.6 seconds, so

Δx = 65 meters.

Phew!!!!! That's a lot! I suggest you learn your physics or this will make you insane by the end of the course!

Answer:

3

Explanation:

The radius of the flywheel and its rotational inertia will be 0.64m and 389kgm² respectively.

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to rotational motion. It is similar to the concept of mass in linear motion. Just as mass is a measure of an object's resistance to linear motion, the moment of inertia is a measure of an object's resistance to rotational motion.

The moment of inertia of an object depends on its shape and mass distribution. Objects with more mass distributed farther from the axis of rotation have a higher moment of inertia than objects with the same mass but a more compact distribution of mass. The moment of inertia is measured in units of kilograms square meters (kg m²) in the SI system.

The radius will be:

= 1000 / 15000(2πrad / 60)

= 0.64m

The inertia will be:

= 2(4.8 × 10^8) / 100 (2π/60)

= 389kgm²

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The maximum height (h) reached by the ball is 5 meters if potential energy of 50 joules (with the potential energy equal to zero at ground level) and is moving upward with a kinetic energy of 50 joules.

To find the maximum height reached by the ball, we need to use the conservation of mechanical energy principle. The total mechanical energy of the ball (E) is the sum of its potential energy (PE) and kinetic energy (KE):
E = PE + KE
Initially, the ball has a potential energy of 50 J and a kinetic energy of 50 J, so the total mechanical energy is:
E = 50 J + 50 J = 100 J
At the maximum height, the ball's kinetic energy will be zero (as it temporarily comes to rest), and its entire mechanical energy will be in the form of potential energy:
\(PE_{max}\) = E = 100 J
We can calculate the maximum height using the formula for potential energy:
\(PE_{max}  = m * g * h_{max}\)
Where m is the mass of the ball, g is the acceleration due to gravity (approximately 9.8 m/s²), and \(h_{max}\) is the maximum height reached by the ball. Rearranging the formula to solve for \(h_{max}\):
\(h_{max}  =\frac{PE_{max}}{(m * g)}\)
Unfortunately, we do not have the mass (m) of the ball provided. However, we can still find the maximum height in terms of the mass:
\(h_{max}  = \frac{ 100 J}{(m * 9.8 m/s^{2} )}\)
This equation shows that the maximum height is directly proportional to the total mechanical energy (100 J) and inversely proportional to the product of the mass and gravitational acceleration.
Assuming the ball has a total mechanical energy of 100 J and considering negligible air friction, the maximum height (h) reached by the ball is 5 meters, which is directly proportional to the total mechanical energy and inversely proportional to the product of the mass and gravitational acceleration.

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The estimated Vmax is always higher than the highest measured velocity, as the estimated Vmax represents the theoretical maximum velocity that the enzyme can achieve.

In enzyme kinetics, Vmax refers to the maximum velocity or rate of an enzyme-catalyzed reaction. It represents the theoretical maximum rate at which the enzyme can convert substrate into product when the enzyme is fully saturated with substrate.

When measuring the velocity of an enzyme-catalyzed reaction, it is possible that the reaction does not reach the maximum velocity due to various factors such as limitations in substrate concentration or other experimental conditions. Therefore, the measured velocity will always be lower than the theoretical maximum velocity or Vmax.

In other words, Vmax represents an asymptotic value that can never be exceeded in practice. It is the maximum velocity that the enzyme can reach under ideal conditions, where the enzyme is fully saturated with substrate. Therefore, it makes sense that the estimated Vmax is always higher than the highest measured velocity, as the estimated Vmax represents the theoretical maximum velocity that the enzyme can achieve.

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7 miles per second.

It would take 5 seconds to stop because 35 divided by 7 = 5.

It would slow 28 miles in 4 seconds because 7 x 4 = 28.

35 - 28 = 7.

The orbital plane of the earth around the Sun is known as the ecliptic or ecliptic plane a path along the ecliptic can see against a background of stars from earth as the Sun moves around the celestial sphere.

Ecliptic in astronomy is the large circle that represents the Sun's apparent movement through the constellations over the course of a year from another angle it is the representation of Earth's orbit around the Sun on the celestial sphere Along the ecliptic the constellations of the zodiac are organized.

The sun's journey is indicated by the ecliptic a fictitious line in the sky Along with the moon planets also follow the ecliptic It designates the plane of the solar system by projecting Earth's orbit onto the cosmic sphere.

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The resultant vector in unit-vector notation is <0.58, 0.98>. Arctan(21.0/11.7) = 67.1° counterclockwise from the +x-axis.

To find the resultant in unit-vector notation, we can calculate the sum of the three displacement vectors. Adding the vector components, we get the resultant vector  = <11.7, 21.0>. Thus, the resultant vector in unit-vector notation is <0.58, 0.98>.

The magnitude of the resultant displacement is the length of the resultant vector, which can be calculated using Pythagorean theorem: Sqrt(11.72 + 21.02) = 24.9 units. The direction of the resultant displacement is the direction of the resultant vector, which can be calculated using inverse tangent: arctan(21.0/11.7) = 67.1° counterclockwise from the +x-axis.

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Answer:

yes/nope

Explanation:

............... :)

Answer:

The lowest energy is for n=4 to n=3 transition.

Statement:

A force is necessary to stretch a spring 0.5 m when the spring constant is 190 N/m.

To find out:

The force required to stretch the spring.

Solution:

Answer:

95 N

Hope you could understand.

If you have any query, feel free to ask.

According to the given data, the angular acceleration of the propeller is approximately 1.49 rad/s².

To find the angular acceleration of the propeller, we can use the following formula:

Δω = α * Δθ

where Δω is the change in angular speed, α is the angular acceleration, and Δθ is the change in angular position (in radians).

First, let's find the change in angular speed (Δω):

Δω = ω_final - ω_initial = 39 rad/s - 11 rad/s = 28 rad/s

Now, let's find the change in angular position (Δθ) for 3.0 revolutions:

Δθ = 3.0 revolutions * 2π radians/revolution = 6π radians

Finally, we can find the angular acceleration (α) using the formula:

we can substitute the values into the formula for angular acceleration,

α = Δω / Δθ = 28 rad/s / 6π radians ≈ 1.49 rad/s²

The angular acceleration of the propeller is approximately 1.49 rad/s².

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The frequency of the sound wave is 1.07 × 10⁹ Hertz

Wavelength is simply the distance over which the shapes of waves are repeated.

From the wavelength, frequency and speed relation,

λ = v ÷ f

Where λ is wavelength, v is velocity/speed and f is frequency.

Given that:

Plug values into the above formula.

λ = v ÷ f

f = v / λ

f = ( 343 m/s ) / ( 3.2 × 10⁻⁷ m )

f = 1.07 × 10⁹ Hz

Therefore, the frequency is 1.07 × 10⁹ Hz.

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If you are finding it difficult to lift an object, it is because of inertial acting on the object.

The term inertial refers to the reluctance to move. We know that according to the Newtons first law of motion, a body will continue in its state of rest unless it is acted upon by an external force.

Given that the more the mass of an object, the greater its inertia, it then follows that if you are finding it difficult to lift an object, it is because of inertial acting on the object.

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Six valence electrons make up sulfur.Since valence electrons are the outermost electrons, they are found on the energy levels with the highest temperatures.

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Answer:

I looked it up but got perihelion so I don't know if that will help at all but try um.... I don't know try L

Explanation:

Answer:

3.485e+6 inches. hope this helps

Answer:

What I got was a big number but it got around 3.485e+6

Explanation: