What marine life zones are warm well lit and have lots of marine life

Air resistance is the pushing of air against an object that is moving. Both the air and the object rub together and this slows the down the object and making it use more energy to reach a required speed. If an object moves faster, the greater the air resistance it encounters.

Sample Response: Digital signals do not pass noise along, so the sound heard is likely very close to the signal that was sent. Digital signals also can be stored more easily than analog signals. Recorded analog signals degrade over time, while recorded digital signals do not.

The change in kinetic energy during that time is : [tex]\frac{26}{3} cl^{3}[/tex]

Given data :

Direction of particle = x

Net force = F(x) = cx²

Initial position = l

Final position = 3l

we will apply the work energy theorem which is : ΔK = W

To determine change in kinetic energy we have to determine the work done on the particle.

Work done on particle ( W ) =  [tex]\int\limits^3_l F({x}) \, dx[/tex] =  [tex]\int\limits^3_l c{x^{2} } \, dx =[/tex] [tex]\frac{26}{3} cl^{3}[/tex]

Hence the change in kinetic energy during that time is : [tex]\frac{26}{3} cl^{3}[/tex]

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Answer:

cooperating.

Explanation:

:)

The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds from rest, convert the revolutions to radians and use the rotational kinematic equation, resulting in an angular acceleration of approximately 3.93 radians/s².

The angular acceleration of a grindstone that makes 20 revolutions in 8 seconds, starting from rest.

To find the angular acceleration, we can use the kinematic equation for rotational motion:

θ = ω₀t + 1/2αt²

where:

θ is the angular displacement in radians,

ω₀ is the initial angular velocity (which is 0 because it starts from rest),

t is the time in seconds,

and α is the angular acceleration in radians per second squared.

First, we convert the revolutions to radians:

20 revolutions * 2π radians/revolution = 40π radians

We then have:

40π radians = 0 + 1/2α(8²)

Solving for α gives:

α = (40π radians) / (1/2 * 64 s²)

α = 80π / 64 radians/s²

α ≈ 3.93 radians/s²

Therefore, the angular acceleration of the grindstone is approximately 3.93 radians/s².

Answer:

Given

dq=density*dv

q=lamda*I

Taking double integration

density=lamda/2*pi*r^2

Answer:

Density of the wooden block is [tex]7805.5\ kg/m^3[/tex]          

Explanation:

It is given that,

Mass of the wooden block, m = 562 g = 0.562 kg

Volume of the block, [tex]V=72\ cm^3=7.2\times 10^{-5}\ m^3[/tex]

We need to find the density of the block. Mass per unit volume of an object is called its density. It is given by :

[tex]d=\dfrac{m}{V}[/tex]

[tex]d=\dfrac{0.562\ kg}{7.2\times 10^{-5}\ m^3}[/tex]

[tex]d=7805.5\ kg/m^3[/tex]

So, the density of the wooden block is [tex]7805.5\ kg/m^3[/tex]. Hence, this is the required solution.

Density of a Material

The density of a material is defined as the mass per unit volume

it formula is given as

Density = Mass/Volume

and the unit is g/cm^3

Explanation:

Given data

Mass = 562 g

Volume = 72 cm3

Hence the density is expressed as

Density =  562 /72

Density = 7.80 g/cm^3

therefore the density is 7.80 g/cm^3

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The potential difference required to accelerate the helium ion to a speed of   [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] is [tex]\boxed{21\,{\text{kV}}}[/tex] or [tex]\boxed{20875\,{\text{V}}}[/tex].

Further Explanation:

The potential difference, through which the helium ion passes, provides the potential energy and this potential energy provided by the potential difference to the Helium ion leads to the increase in the kinetic energy of the ion.

The charge on the [tex]H{e^ + }[/tex] ion is equal to the charge of one electron and the mass of the helium atom is equal to the mass of [tex]4\,{\text{amu}}[/tex].

The expression for the conservation of the energy of the Helium ion in the potential difference is:

[tex]\Delta PE = \Delta KE[/tex]  

Substitute [tex]q\Delta V[/tex] for [tex]\Delta PE[/tex] and [tex]\dfrac{1}{2}m{v^2}[/tex] for [tex]\Delta KE[/tex] in above expression.

[tex]\begin{aligned}q\Delta V &= \frac{1}{2}m{v^2} \hfill \\\Delta V &= \frac{{m{v^2}}}{{2q}} \hfill \\\end{aligned}[/tex]  

Here, [tex]\Delta V[/tex] is the potential difference, [tex]m[/tex] is the mass of the ion, [tex]q[/tex] is the charge over the ion.

Substitute [tex]1.6 \times {10^{ - 19}}\,{\text{C}}[/tex] for [tex]q[/tex], [tex]\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right)[/tex] for [tex]m[/tex] and [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]v[/tex] in above expression.

[tex]\begin{aligned}\Delta V &= \frac{{\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right){{\left( {1.0 \times {{10}^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} \right)}^2}}}{{2\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)}} \\&= \frac{{6.68 \times {{10}^{ - 15}}}}{{3.2 \times {{10}^{ - 19}}}}\,{\text{V}} \\&= 2{\text{0875}}\,{\text{V}} \\&\approx {\text{21}}\,{\text{kV}}\\\end{aligned}[/tex]  

Thus, the potential difference required to accelerate the helium ion to a speed of [tex]1.0 \times {10^6}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}[/tex] is [tex]\boxed{21\,{\text{kV}}}[/tex] or [tex]\boxed{20875\,{\text{V}}}[/tex].

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords:  Potential energy, potential difference, accelerate, kinetic energy, speed, helium ion, charge, significant figures, rest to a speed.

"The height of the tower is given by the equation:

[tex]\[ h = \frac{1}{2} g (t + \hat{t})^2 - \frac{1}{2} g t^2 \][/tex]

 This equation represents the difference in height between the maximum height the rock reaches and the height it reaches at time \( t \) when it passes the top of the tower. Here, \( g \) is the acceleration due to gravity, \( t \) is the time it takes for the rock to reach the top of the tower, and \( \hat{t} \) is the additional time it takes for the rock to reach the maximum height after passing the top of the tower.

To find the height of the tower, we can simplify the equation:

[tex]\[ h = \frac{1}{2} g ((t + \hat{t})^2 - t^2) \][/tex]

[tex]\[ h = \frac{1}{2} g (t^2 + 2t\hat{t} + \hat{t}^2 - t^2) \][/tex]

[tex]\[ h = \frac{1}{2} g (2t\hat{t} + \hat{t}^2) \][/tex]

[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]

 Since [tex]\[ u = g \hat{t} \][/tex]is the time it takes for the rock to reach the maximum height from the top of the tower, and at the maximum height the velocity of the rock is zero, we can use the kinematic equation for the final velocity \( v \) at the maximum height:

[tex]\[ v = u + g t \][/tex]

Here, \( u \) is the initial velocity (which is the velocity of the rock at the top of the tower), \( g \) is the acceleration due to gravity, and \( t \) is the time taken to reach the maximum height, which is \( \hat{t} \). At the maximum height, \( v = 0 \), so:

[tex]\[ 0 = u - g \hat{t} \][/tex]

[tex]\[ u = g \hat{t} \][/tex]

 Now, we can substitute \( u \) back into the height equation:

[tex]\[ h = g t\hat{t} + \frac{1}{2} g \hat{t}^2 \][/tex]

[tex]\[ h = g t\left(\frac{u}{g}\right) + \frac{1}{2} g \left(\frac{u}{g}\right)^2 \][/tex]

[tex]\[ h = u t + \frac{1}{2} \frac{u^2}{g} \][/tex]

Since \( u t \) represents the distance the rock would have traveled in time \( t \) if it continued at a constant velocity \( u \), and \( \frac{1}{2} \frac{u^2}{g} \) represents the total height the rock would reach if it continued to move upward with initial velocity \( u \) and was only acted upon by gravity, the term \( u t \) is actually the height of the tower. Thus, the height of the tower is:

[tex]\[ h = u t \][/tex]

 This is the final answer, and it shows that the height of the tower is the product of the initial velocity of the rock at the top of the tower and the time it takes to reach the top of the tower. The term[tex]\( \frac{1}{2} \frac{u^2}{g} \)[/tex]is not needed for the height of the tower, as it represents the additional height the rock reaches after passing the top of the tower until it reaches the maximum height."

When Trinity pulls on the rope with her weight, Newton's Third Law of Motion tells us that the rope will "pull back".

Newton's third law of motion expresses that, at whatever point a first question applies a power on a second object, the first object encounters a power meet in extent however inverse in heading to the power that it applies.  

Newton's third law of movement reveals to us that powers dependably happen in sets, and one question can't apply a power on another without encountering a similar quality power consequently. We once in a while allude to these power matches as "action-reaction" sets, where the power applied is the activity, and the power experienced in kind is the response (despite the fact that which will be which relies upon your perspective).

The pressure difference between the inner and outer ear at the top of the mountain is given by option a. 1.2 × 10³ Pa.

The pressure difference between the inner and outer ear at the top of the mountain can be calculated by evaluating the difference between atmospheric pressure at top and bottom of the lift. Let this difference be denoted as ΔP, then:

Δ P = P₁ - P₂

P₁ = pressure at the top of the lift = 1.010 × 10⁵ Pa

P₂ = pressure at the bottom of the lift = 0.998 × 10⁵ Pa

or, Δ P = (1.010 × 10⁵ Pa - 0.998 × 10⁵ Pa)

or, Δ P = 0.012 × 10⁵ Pa

or, Δ P = 1.2 × 10³ Pa

The ratio of the magnitudes of the force of air resistance to the force of gravity acting on the ball is approximately 1:24.

The student is asking about the ratio of the force of air resistance to the force of gravity acting on a blue ball, when the ball has an acceleration of 9.40 m/s2. To calculate this ratio, we can use Newton's second law, F = ma, where F is the force, m is the mass of the ball, and a is the acceleration of the ball due to the net force acting on it. If we denote the force of air resistance as Fair and the force of gravity as Fgrav, we can write Fnet = Fgrav - Fair, since the air resistance works in the opposite direction of gravity.

Given that the acceleration due to gravity is approximately 9.81 m/s2, the force due to gravity (weight) can be expressed as Fgrav = mg. Here, the acceleration is less than g due to air resistance, thus we can write Fnet = ma = mg - Fair. Since the acceleration of the ball is given as 9.40 m/s2, we can now express Fair as mg - ma. Taking the ratio of Fair to mg (the weight), we get:

(mg - ma) / mg = (g - a) / g = (9.81 - 9.40) / 9.81.

Therefore, the ratio of the magnitudes of the force of air resistance to the force of gravity is equal to 0.0418, which simplifies to approximately 1:24.

The transition of an electron from n = 2 to n = 6 in a hydrogen atom in the Bohr model involves the absorption of a photon. The wavelength of this photon can be calculated using the Rydberg formula, but a specific value cannot be provided without additional information.

The student's question relates to the Bohr model of the hydrogen atom and specifically to the transition of an electron from the n = 2 energy level to the n = 6 energy level. According to Bohr's model, when an electron in a hydrogen atom transitions between two energy levels, it either absorbs or emits a photon whose energy equals the difference in energy between the two levels.

To find the wavelength of the photon associated with the transition, you can use the Rydberg formula:

The exact calculations require substituting the values of h, c, and E (with E determined by the specific energy levels involved in the transition - here from n = 2 to n = 6).

Because this question does not provide all the numerical values needed for the calculation, such as Planck's constant and the speed of light, and does not specify whether the transition represents absorption or emission, we cannot provide a specific value for the wavelength. However, it's important to note that transitions to higher energy levels (n > 2) correspond typically to absorption, and the wavelength would fall into a specific region of the electromagnetic spectrum depending on its energy.

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When 12.8 eV energy electrons excite hydrogen atoms, the atoms are elevated to an energy level just below ionization. To determine the specific energy state, the energy formula for hydrogen atoms is used, then the energies of emitted photons are found by calculating the energy difference between the excited and final states.

If electrons with an energy of 12.8 eV are incident on hydrogen atoms in their ground state, we first need to determine the energy level to which the hydrogen atoms are excited. The ground state energy of hydrogen is -13.6 eV. If 12.8 eV is provided, the electron reaches an energy level just below 0 eV (ionization energy), since 12.8 eV is not sufficient to completely ionize the atom (13.6 eV needed).

To find out which energy level the electron will jump to, we use the formula for the total energy of an electron in a hydrogen atom, En = (- 13.6 eV/n²), and solve for n.

Once the electron is excited, it will eventually drop back to a lower energy state, emitting a photon in the process.

The energy of the emitted photon can be calculated using the difference in energy levels, ΔE = E1 - Ef.