Review. (b) Is this statement true for all angular displacements, for all displacements less than 180° , or only for small angular displacements? Explain.

's' increases at all times that Rock X falls since Rock X falls with a greater velocity than Rock Y.

The friction force is the opposite force generated when two parts slide over each other if the air is passing over the rocks the frictional force would be the drag force.

Given: Frictional forces are considered to be negligible.

A few seconds after Rock X is released from rest, Rock Y is also released. Consider moving downward to be constructive. Both rocks are launched from rest, thus their initial velocities are equal to zero (u = 0).

We know that the velocity is given by,

v² = u² + 2as

Rock X is released first, followed by Rock Y, and because Rock X travels faster than Rock Y, it travels farther. This is due to the fact that whenever Rock X has already acquired some velocity (t).

Its velocity (v) is higher, increasing the spacing (s). In conclusion, because Rock X falls faster than Rock Y,'s' rises each time Rock X hits the ground.

The complete question is given below:-

Rock X is released from rest at the top of a cliff that is on Earth. A short time later, Rock Y is released from rest from the same location as Rock X. Both rocks fall for several seconds before landing on the ground directly below the cliff. Frictional forces are considered to be negligible. After Rock Y is released from rest several seconds after Rock X is released from rest, what happens to the separation distance S between the rocks as they fall but before they reach the ground, and why? Take the positive direction to be downward.

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6 hours long, give or take a few minutes, you need to wait to see them arranged in a horizontal line if, at 8 pm, you see that the pointer stars of the big dipper and the star Polaris are arranged in a vertical line.

The two prominent stars in Ursa Major known as the Pointer Stars can be utilized to locate Polaris, the North Star. The Big Dipper asterism includes the two stars Dubhe and Merak (Alpha and Beta Ursae Majoris). They trace the bowl of the Dipper's exterior. The direction of the North Star is indicated with a line that runs from Merak via Dubhe. Because it identifies the location of the celestial north pole, Polaris is significant in navigation. While other stars and constellations seem to revolve around it, the star is constantly found in the same location in the sky throughout the year.

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No, we cannot conclude that the photoelectric effect will also occur for different metals under the same conditions.

Mathematically , we can define the photoelectric effect as:

[tex]K.E_m_a_x=E-\phi[/tex]

where  [tex]K.E_m_a_x[/tex] is the kinetic energy of the free electron, E is the energy of the incident photon, and Ф is the work function, the energy needed to release the electron from its parent atom.

For example if we take a light of  2eV that is incident on cesium, which has a work function of 1.26 eV. In this case, since the photon energy is greater than the work function, an electron can be emitted.

but if we hit this light of 2eV on the platinum which has a work function of 3.65eV . Apparently, this time the light doesn't have enough energy to free loosely bound electrons from their parent atoms.

Thus ,the emission of photoelectrons and the kinetic energy of the emitted photoelectrons depend on the frequency of light  incident on the metal surface.

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If the distance d is much greater than the length of the wire, then intuitively, the magnitude of the electric field at point p should be very large. This is because a strong electric field will cause an increase in current flow through any metal conductor. In this case, due to how greater d is compared to l, there would likely be a significant amount of electricity flowing through both wire and ground.

It will happen as a result of a stronger field and more room between charges. In order to determine this magnitude, one method is to apply Coulomb's Law, which stipulates that:

E=q×C where,

E is the electric field.

q is charge per unit area.

C is the amount of charge held by a object.

Coulomb's law can be derived from the principle of electrostatics. This states that an electrical force is equal and opposite to the electric field at any point in space. In other words, if you charge a capacitor up with electricity, the voltages on each terminal will oppose one another and cause a Force between them (Coulomb's Law).

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If a wave has a wavelength of 2.5 m and a frequency of 150 Hz, then the speed of the wave would be

It can be understood in terms of the distance between any two similar successive points across any wave for example wavelength can be calculated by measuring the distance between any two successive crests.

As given in the problem If a wave has a wavelength of 2.5 m and a frequency of 150 Hz, then we have to find the speed of the wave,

The speed of the wave = frequency of the wave ×  wavelength

                                       = 150 × 2.5

                                       =375 meters / seconds

Thus, the speed of the wave would be 375 meters / seconds.

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The new value of the temperature is T₂ = 2T₁ and the new value of the wavelength with the highest intensity is λ₂ = (1/2) λ₁.

Using Stefan's equation, which states that the power of the emitted radiation will be proportional to the fourth power of the absolute temperature, we treat the filament of the evacuated light bulb as a black body.

P = σ AeT⁴

Where Ais the surface area of the emitter, σ is Stefan's constant, and e is the emissivity of the surface.

Now, from Stefan's Law, the radiated power is proportional to the fourth power of temperature.

P =IV

IV ∝ T⁴

Therefore,

I₁ V₁ / (I₂ V₂) = (T₁ / T₂)⁴

T₁ / T₂ = (I₁ V₁ / I₂ V₂)^{1/4}

T₂ = T₂ [ (I₁ V₁ / I₂ V₂)^{1/4} ]

Now we have,

I₂ = 2I₁ and

V₂ = 8V₁

Therefore,

T₂ = T₁ [ (2I₁ )(8V₁)/(I₁ V₁) ]^{1/4}

T₂ = T₁ (16)^{1/4}

T₂ = 2T₁

Let λ₁ be the wavelength emitted with the highest intensity initially and λ₂ be the wavelength emitted with the highest intensity after the change.

Then by Wiens law:

λ₁ T₁ = λ₂ T₂

λ₁ T₁ = λ₂ (2T₁)

λ₂ = (1/2) λ₁

Therefore, the new value of the wavelength is 1/2 λ₁.

The correct option is (d).

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The  force vector that has a magnitude of 12.0 N. and is oriented 60° to the left of the (y) has the followings components:

To solve this exercise the formulas and procedures we will use are:

Where:

Information about the problem:

Applying the formula of the component of the vector in the (x) axis we have:

v(x) = v * cosine (angle).

v(x) = 12.0 N * cosine (60º)

v(x) =6 N

Applying the formula of the component of the vector in the (y) axis we have:

v(y) = v * sine (angle)

v(y) = 12.0 N * sine (60º)

v(y) = 10.39 N

It can be said to be a straight line described by a point (a) and (b) that has direction and sense.

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The magnitude of the force is  295.54 x 10⁻³ N

Attraction betweem nucleus and electron is by electric force. Now, a positive or negative charge generates a field in the void surrounding it. We refer to this void as the electric field.

We know that , force can be found out by given formula

F = q × E where,

F = Force

q = Charge

E = Electric field

Here, the direction of Force is the same as that of the Electric field

Substituting the values, we get

F = 1.4 x 10⁻³ x 211.1

=295.54 x 10⁻³ N

Thus, the magnitude of force is 295.54 x 10⁻³ N

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A physical field called an electric field surrounds electrically charged particles and attracts or repels all other charged particles nearby. The physical field of a system of charged particles can also be referred to as the electric field.

In a uniform electric field E, a particle with electric charge q moves in a straight line. The speed of the particle is u.

The electric force exerted in the charge is qE.

The electric field and the velocity of the particle are both in the x direction.

For a small speed u when compared to the speed of light c, the relativistic equation reduces to the classical acceleration:

a = qE / m

Now, as u approaches c, the acceleration will tend to zero, so that the object can or will never reach the speed of light.

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The study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

The expression nanomachining makes reference to the study of nanometric machines (nanomachines) and related materials, which can be achieved by different approaches including sensor-based strategies related to acoustic auditive phenomena.

In conclusion, the study 'characterizing vibration-assisted atomic force based nanomachining' aims to elucidate nanomachine properties for heterogeneous materials.

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The torque value is 560 Nm, so the result does not depend on the value of Ф.

Given:

Mass=80.0 g

Radius=20.0 cm

Magnetic field=0.35T

In a magnetic field, a loop of current-carrying wire will experience the following torque:

τ=μ×Β

Here, is the μ magnetic moment and Β is the magnetic field.

τ= 1600 × 0.35 T

τ= 560 Nm

The rotating equivalent of force is torque. It is also known as the moment, moment of force, rotating force, or turning effect depending on the field of research. It serves as an example of how a force can alter a body's rotational motion.

The Newton-metre, or kgm2sec-2, is the SI unit for torque. How did we get to this point? Considering the equation Torque = Force X Distance Torque is measured in newton-meters, whereas distance is measured in meters and force is measured in newtons.

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The electric field is 5.39 × 10^3 N/C.

R(radius of the metal rod) = (10cm)( 1m/100cm) = 0.1m

[tex]\lambda\;(\text{charge per unit length})=(30\;\mathrm{nC/m})\left(\dfrac{1\;\mathrm{C}}{10^9\;\mathrm{nC}} \right)=\textcolor{#ff9c8c}{30\times 10^{-9}\;\mathrm{C/m}}[/tex]

Since the rod is long L≫l, we model it as a long filament of infinite length and constant charge per unit length for which the electric field is found from Equation (*):

[tex]E=\dfrac{\lambda}{2\pi \epsilon_0 r} \\[/tex]

Substitute numerical values:

[tex]E&=\dfrac{30\times 10^{-9}}{2\pi (8.8542\times 10^{-12})(0.1)}\\ &={5.39\times 10^{3}\;\mathrm{N/C}}}[/tex]

electric field, an electric property associated with each point in space when the charge is present in any form. The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

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0.125 mm . is the thickness of the sample.

The Hall effect is the creation of a voltage difference (the Hall voltage) across an electrical conductor, which is transverse to an applied magnetic field perpendicular to the current and an electric current in the conductor. Edwin Hall made the discovery in 1879.

We need to know the material's current, magnetic field, length, number of charge carriers, and area in order to calculate the Hall voltage. The Hall voltage is computed using the formula: v=IBlneA=(100A)(1.5T)(1.0102m)(5.91028/m3)(1.61019C)(2.0105m2)=7.9106V.

lof4

First we have to plot those point Then we can use some computer program to fit those point linearly to get slope

of that graph a and interception b. We already know, from theory, that Hall's voltage AVH and magnitude of

magnetic field B are connected as

Δ[tex]V_{H} =\frac{I}{nqt} B[/tex]

where I is current trough probe, n is concentration of charge carriers, q = 1.6 • 10¯19 C is charge of charge

carries and t is thickness of the material. We have put the data from the problem on a graph and fitted linearly and

got

a = 100 μ[tex]\frac{V}{T}[/tex]

b = —0.02  μV.

As we can see, our result are in agreement with theoretical assumptions because interception b is almost O, and a

is asked relation between Hall's voltage A VH and magnitude of magnetic field B. Then we can write

ΔVH =[tex]100X10^{-6}[/tex] V/TB

(4) Then we can use result (4) and numbers from the textbook to calculate the thickness of the sample as

[tex]a=\frac{I}{nqt} \\t=\frac{I}{anq} \\t=\frac{.200A}{100X10^{-6}X 1.6 X10^{-19}X10^{26} } \\t=0.125mm[/tex]

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An object which has a mass of 70 kg is sitting on a cliff 10 m high. the object's potential energy is 7,000 J

Given: m = 70 kg in mass Height h = 10 m and acceleration g = 10 m/s2

Required. Potential Energy of Gravitation, or P.E

Equation: P.E = mgh

= (70 Kg)(10 m/s²) ( 10 m)

P.E. equals 7,000 kg/m2 or

P.E = 7,000 J

Scottish physicist William Rankine initially used the word "potential energy" in the 19th century, and it has since become a cornerstone of the science. Because of their location, which makes them position relative, items that have this energy stored in them have the capacity to unleash it.

Our knowledge of the cosmos and how we fit into it depends on potential energy. In fact, potential energy and the labour necessary to move an item are determined by the position of the object in the universe.

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The amount of heat the cup absorb is 1003200 J.

Given,

Mass of coffee cup = 4.00 × 100 g or 4 kg

Room temperature = 20°C

Hot water temperature = 80°C

We have to find the amount of heat the cup absorbed.

Now, by using the specific heat formula, we have

Q = mCΔT

Where m is the mass of the substance, c is the specific heat, ΔT is the change in temperature, and Q is the amount of heat received or lost.

Putting the values, we get

Q = 4 × 4180 × (80°- 20°)

Q = 1003200 J/kg K

Hence, the amount of heat absorbed by the cup is 1003200 J/kg K.

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Answer:

the fundamental particle that has the least and most massive are the electron and the neutron.

Explaination:

Electrons are the least massive subatomic particles, and they are found in nucleus. neutron is the heaviest sub-atomic Particles among the given subatomic particles with mass of 1.008 amu.

Given the number 0.00210, only the last 210 digits are considered significant. Therefore, the specified number has 3 significant digits.

The significant or significant digits that convey meaning depending on its precision is know as significant digits of a given number .For example, 8.642 has 4 significant digits. These substantive digits give precision to numbers. Also called significant digits.    

All non-zero digits are valid. 136145 has 6 significant digits.

All zeros between two non-zero digits are valid. For example, 885.0085has 7 significant digits.

All zeros to the right  and to the left of non-zero digits of the decimal point are never significant. For example, 0.00897 contained 3 significant digits.

the decimal point are significant, All zeros to the right only if not followed by non-zero digits. For example, 50.00 has 4 significant digits.

The last non-zero and all zeros to the right of decimal place are valid. For example, 0.0045700 has 5 significant digits.

The last non-zero digit , all zeros to the right of are valid if they come from the measurement.

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20855 turns of wire are on the coil.

A substance or thing that creates a magnetic field is called a magnet. The most noteworthy characteristic of a magnet—a force that pulls on other ferromagnetic elements, such as iron, steel, nickel, cobalt, etc.—as well as the ability to attract or repel other magnets—is caused by this invisible magnetic field. An object made of a substance that has been magnetized and produces its own persistent magnetic field is referred to as a permanent magnet. A common example is a refrigerator magnet used to keep notes on the door of the refrigerator. Ferromagnetic materials are those that can be magnetized and are also those that are powerfully drawn to a magnet (or ferrimagnetic).

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Fundamental frequency will be [tex]\frac{1}{\sqrt{2} }[/tex] times that of original frequency on doubling the mass per unit length.

From equation (1) , it can be seen clearly that  fundamental frequency is inversely proportional to the square root of the mass per unit length of the string.

On putting double the mass per unit length in equation (1) keeping the all factor same , we get

               [tex]f'=\frac{1}{2l} \sqrt{\frac{T}{2u} }\\\\f'=\frac{1}{\sqrt{2} } f[/tex]

Where f ' is the new frequency and f is the original frequency .

Thus, on doubling the mass per unit length, with all other factors held constant then the fundamental frequency will be [tex]\frac{1}{\sqrt{2} }[/tex] times that of original frequency.

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The physics problem to find the unknown has applied physics principles will be the following:

The simplified expression for coefficient of kinetic friction is 0.097.

The given parameters:

g = 9.8 m/s²

a = 3.6 m/s²

θ=27.0∘

The simplified expression for coefficient of kinetic friction is calculated as follows;

μkmgcosθ = mgsinθ − ma

m(μkgcosθ) = m(gsinθ − a)

μkgcosθ = gsinθ − a

The practise of using physics as a science to assist people and address their issues is known as applied physics.

Engineering is distinct from it since engineers deal with clearly defined problems. Applying physics to the development of new technologies or the resolution of engineering issues is the work of applied physicists.

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Answer:

Do you have the fomular send please so I can work on it

The rate of increase of the current in the given circuit is determined as 6.67 A/s.

The rate at which the current increases is calculated as follows;

E = L(di/dt)

where;

The rate of increase of the current is calculated as follows;

di/dt = E/L

di/dt = 100 / 15

di/dt = 6.67 A/s

Thus, the rate of increase of the current in the given circuit is determined as 6.67 A/s.

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In the given question, we can infer from the table that the average velocity for the car upto one decimal place is 7.3 m/s.

Velocity is defined as the ratio of Displacement of a body to the time it has take for the displacement. We can find the average velocity but adding up all the velocities and then divide it by number of occurrences. Using the data, after adding up all the average velocities, we get the value as 87.497 m/s. As the number of velocities are 12, we divide the total velocity by 12 and fing the mean velocity. The mean or average velocity is : 87.497 / 12 = 7.291416. Upto one decimal place, we get the value as 7.3 m/s.

Hence the average or mean velocity from the given data is 7.3 m/s

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Achieving a speed of 112.25 km/h is impossible for a human.

The frequency of an event is its repetitions per unit of time. In a contrast to spatial frequency, it is also sometimes referred to as temporal frequency. Hertz (Hz), which equals one (event) per second, is the unit of frequency. The reciprocal of frequency is period, which is defined as the length of time for one cycle of a repeating occurrence.

According to the question:

Emitted frequency, f₀

Frequency heard when athlete approaches horn, f₁

Frequency heard when the athlete passes horn, f₂

Speed of the athlete (observer), v₁

Speed of sound in air = 343m/s

When an athlete approaches the horn, the frequency observed is given by,

f₁=( (v+v₁)/(v-v₂) ) ₓ f₀

f₂=( (v-v₁)/(v-v₂) ) ₓ f₀

From question,

f₂/f₁ = 5/6

5/6 = v-v₁/v+v₁

hence, v₁ = 112.25 km/h

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The Pressure of gas in the syringe when at an initial pressure of 60 kPa and when the volume decreased from 45 cm³ to 40 cm³ is 67.5 kPa.

Pressure can be defined as the force acting perperdicular per unit area to the surface of an object.

To calculate the pressure in the gas, we use the formula below

Formula:

Where:

Make P' the subject of the equation

From the question,

Given:

Substitute these values into equation 2

Hence, the pressure of the gas is 67.5 kPa.

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The scientific experiments are performed under controlled conditions, which can be considered characteristic of science, therefore the correct answer is option C.

The deliberate, empirically supported pursuit and application of knowledge and understanding of the natural and social worlds are what is known as science.

Science is characterized by the controlled environments in which experiments and observations are made.

Thus, option C is the proper response.

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Robert uses a spring scale to measure force. The unit of his measurement is newtons.

In material science, a power is an impact that can change the movement of an item. A power can make an item with mass change its speed, i.e., to speed up. Power can likewise be portrayed naturally as a push or a force. Logicians in classical times involved the idea of power in the investigation of fixed and moving articles and basic machines, yet scholars, for example, Aristotle and Archimedes held crucial mistakes in grasping power. To a limited extent this was because of a deficient comprehension of the occasionally non-clear power of erosion, and a thusly lacking perspective on the idea of normal movement.

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