An experiment that produces waves on a string is analyzed, and the experimental value for the wave speed is found to be 66 m/s. If the accepted value for the wave speed is 63 m/s, what is the percent error? 0.045% 0.048% 4.5% 4.8%

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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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).

Answer:

c

Explanation:

just took test on edge but not sure if i was right :) goodluck gs

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

Answer:

cooperating.

Explanation:

:)

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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Forces are vectors, which means they need to be determined by not only their numeric value (amount of force in Newton's) but also their direction.

Some other examples of vector quantities include displacement, acceleration,  velocity, and momentum.  Each of these isn't complete until you know its size and direction.

There are other quantities that have size but no direction.  They include distance, speed, volume, temperature, cost, and loudness.  These are called "scalars".

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.

"V = Frequency x Wavelength.

V = 340;

340 = 20 x Wavelength; divide both sides by 20.

Wavelength = 17"

Explanation:

Answer: apparent weighlessness.

Explanation:

1) Balance of forces on a person falling:

i) To answer this question we will deal with the assumption of non-drag force (abscence of air).

ii) When a person is dropped, and there is not air resistance, the only force acting on the person's body is the Earth's gravitational attraction (downward), which is the responsible for the gravitational acceleration (around 9.8 m/s²).

iii) Under that sceneraio, there is not normal force acting on the person (the normal force is the force that the floor or a chair exerts on a body to balance the gravitational force when the body is on it).

2) This is, the person does not feel a pressure upward, which is he/she does not feel the weight: freefalling is a situation of apparent weigthlessness.

3) True weightlessness is when the object is in a place where there exists not grativational acceleration: for example a point between two planes where the grativational forces are equal in magnitude but opposing in direction and so they cancel each other.

Therefore, you conclude that, assuming no air resistance, a person in this ride experiencing apparent weightlessness.

option (b) is correct.

Overpopulation puts the most demand on water supplies.The reservoirs of the water are fixed but the human population is ever increasing. The needs of water for drinking, washing,industrial use like medicines, construction etc are thus increasing with every passing day.So, we are continuously running out of the water supplies. Because of the overpopulation. the pollution of water is also increasing, thus rendering the water unfit for drinking.

Thus overpopulation puts the most demand on water supplies.

Answer:

option (b) overpopulation is the  correct option for this question

Explanation:

Overpopulation is the reason  in increasing demand on water supplies.Overpopulation is a bothersome condition where the quantity of existing human populace surpasses the conveying limit of Earth. Overpopulation is brought about by number of variables. Decreased death rate, better restorative offices, consumption of valuable assets are few of the causes which results in overpopulation. Because of the overpopulation. the pollution of water is also increasing, thus rendering the water unfit for drinking

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.

Answer:

Given

dq=density*dv

q=lamda*I

Taking double integration

density=lamda/2*pi*r^2

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]

Learn more about work energy theorem : https://brainly.com/question/22236101