An 7.80-mF capacitor is connected across an AC source with an rms voltage of 71.0 V oscillating with a frequency of 30.0 Hz. What are the following for this circult? (a) capacitive reactance (b) rms current (c) maximum current

The Gravitational potential energy depends on mass , acceleration and path and potential also as G=V/m.

a).  As you can imagine, gravitational energy is the energy associated with gravity. This is the potential energy stored due to an object being in a higher position compared to a lower position. (e.g. far from the ground or close to the ground).

Gravitational energy is the energy associated with gravity.

b). Systems can increase their gravitational energy as their mass moves away from the center of the Earth and away from other bodies (such as the sun, planets, and stars) large enough to generate a large amount of gravity. there is.

Potential energy can be defined as the amount of work expended to lift an object from the ground to its current height. In another sense, this is the amount of work the body does before returning to ground level (the position of maximum stability) when translated from its current level. Therefore, potential energy can be related to work.

c). A negative sign indicates that gravity is doing positive work as the mass approaches. A negative potential indicates a bonding state. That is, when mass is placed under its influence, it becomes trapped in its potential, and if something can provide the energy to escape the clutches of gravity, it must not escape.

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The acceleration versus time graphs are the mathematical representation of the variation of acceleration with time.

There are several methods of representing the variation of two quantities one such method is graphs.

When the variation of the acceleration of the object and the time taken by the object to reach at acceleration are plotted on a graph the provides us with a few certain quantities.

For an example the slope of the acceleration-time graph tell us about the sudden change in acceleration at that particular instant.

Also the area under the acceleration vs. time graph represents the change in velocity of the object.

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Answer: When the light hits the water, it causes the light to slow down, and when it does this, the light changes directions.

Explanation:

More massive stars emit more energy and run out of hydrogen fuel in their cores faster.

In order to create helium and energy, hydrogen nuclei must fuse. The procedure uses hydrogen as its fuel. As the hydrogen is depleted, the star's core condenses and warms up further. As a result, all the elements up to iron are formed through the fusing of heavier and heavier elements.

Nuclear fusion is how stars generate their energy. This process, known as the "proton-proton chain," which involves a series of events that turns four hydrogen atoms into one helium atom, dominates for the majority of stars.

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= [ 4 (C-H) + 1 (C=C) + 1 (X2) ] - [ 2 (C-X) + 1 + (C-C) + 4 (C-H)

= [ (C=C) +(X2) ] - [ 2 (C-X) + 1 (C-C) ]

= [ 604 +151.9 ] - [ 2 (276.3) + (338)]

= -134.7 KJ

= -135 KJ

Bond energies depend on the number of bonds between atoms. Even though π bonds are weaker than σ bonds, a double bond, which consists of a σ and π is bond, is stronger than a single bond because there are  bonds.

A quantity of strength, equal to the distinction among the energies of the Bond energies depend on the number of bonds between atoms. Atoms and the energies of the separated atoms, is released, usually as warmness. that is, the bonded atoms have a lower strength than the character atoms do.

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The wavelength in a vacuum is 5.08 x 10-7  m and the wavelength of the light in the glass is 3.367 x 10-7  m

As we know that

 Velocity = wavelength x frequency

So,    wavelength = Velocity / frequency

If the light travels in a vacuum,

Then,    wavelength = speed of light/frequency

Here  the frequency is 5.9 x1014 Hz  

And the speed of light = 3 x 10 8 m/s

So, wavelength in vacuum = 3 x 10 8 / 5.9 x1014

    = 5.08 x 10-7  m

Again,  the wavelength of the light in the glass

 λ=c/μn            where μ is a block of glass that has an index of refraction

Then  λ=(3 x 10 8 )/(1.51x5.9 x1014 )            

 = 3.367 x 10-7  m

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(a) Refer to part (a)

(b) Refer to part (b)

(c) [tex]nr^{n-1}\bold{\hat{r}}[/tex]

To answer the questions, I will show the derivations of each part by applying vector calculus methods such as the gradient operator, denoted by '▽', to the given quantities. I will use the definition of the gradient operator in Cartesian coordinates to solve each part.

Given:

[tex]\bold{r} = (x-x')\hat{\imath}+ (y-y')\hat{\jmath} + (z-z')\hat{k}; \ r=\sqrt{ (x-x')^2+(y-y')^2+(z-z')^2}[/tex]

Show that:

(a) [tex]\nabla (r^2)=2 \bold{r}[/tex]

(b)[tex]\nabla (1/r)=- \bold{\hat{r}}/r^2[/tex]

(c) What is the general formula for [tex]\nabla (r^n)[/tex]?

[tex]\hrulefill[/tex]

The gradient operator in Cartesian coordinates is:

[tex]\nabla = \hat{\imath}\dfrac{\partial}{\partial x}+ \hat{k}\dfrac{\partial}{\partial y}+ \hat{\jmath}\dfrac{\partial}{\partial z}[/tex]

The square of the vector length is given by:

[tex]\Longrightarrow r^2 = [(x-x')\hat{\imath}+ (y-y')\hat{\jmath} + (z-z')\hat{k}]^2\\\\\\\\\Longrightarrow r^2 = (x-x')^2\hat{\imath}+ (y-y')^2\hat{\jmath} + (z-z')^2\hat{k}[/tex]

Now, applying the gradient operator to 'r²':

[tex]\Longrightarrow \nabla r^2 = \\\\\bullet \dfrac{\partial}{\partial x}[(x-x')^2\hat{\imath}+ (y-y')^2\hat{\jmath} + (z-z')^2\hat{k}]=2(x-x')\\\\\bullet \dfrac{\partial}{\partial y}[(x-x')^2\hat{\imath}+ (y-y')^2\hat{\jmath} + (z-z')^2\hat{k}] = 2(y-y')\\\\\bullet\dfrac{\partial}{\partial z}[(x-x')^2\hat{\imath}+ (y-y')^2\hat{\jmath} + (z-z')^2\hat{k}] =2(z-z')\\\\\\\\\therefore \nabla r^2=2(x-x')\hat{\imath}+ 2(y-y')\hat{\jmath} + 2(z-z')\hat{k}\\\\\\\\[/tex]

Factoring out a 2:

[tex]\Longrightarrow \nabla r^2=2[(x-x')\hat{\imath}+ (y-y')\hat{\jmath} + (z-z')\hat{k}]\\\\\\\\\therefore \boxed{\nabla r^2=2 \bold{r}}[/tex]

[tex]\hrulefill[/tex]

The length 'r' is given by:

[tex]r=\sqrt{ (x-x')^2+(y-y')^2+(z-z')^2}[/tex]

Now applying the gradient operator to 1/r:

[tex]\dfrac{1}{r}=((x-x')^2+(y-y')^2+(z-z')^2)^{-1/2}[/tex]

[tex]\nabla \dfrac{1}{r} =\\\\\bullet \dfrac{\partial}{\partial x}[((x-x')^2+(y-y')^2+(z-z')^2)^{-1/2}]=-(x-x')[(x-x')^2+(y-y')^2+(z-z')^2]^{-3/2}\\\\\bullet\dfrac{\partial}{\partial y}[((x-x')^2+(y-y')^2+(z-z')^2)^{-1/2}]=-(y-y')[(x-x')^2+(y-y')^2+(z-z')^2]^{-3/2} \\\\\bullet\dfrac{\partial}{\partial z}[((x-x')^2+(y-y')^2+(z-z')^2)^{-1/2}]=-(z-z')[(x-x')^2+(y-y')^2+(z-z')^2]^{-3/2}[/tex]

[tex]\Longrightarrow \nabla \dfrac{1}{r} = - \dfrac{(x-x')\hat{\imath}+ (y-y')\hat{\jmath} + (z-z')\hat{k}}{[(x-x')^2+ (y-y')^2 + (z-z')^2]^{3/2}}[/tex]

[tex]\Longrightarrow \nabla \dfrac{1}{r} = -\dfrac{\bold{r}}{r^3}; \ \text{Where:} \ \bold{\hat{r}} =\dfrac{\bold{r}}{r} \rightarrow \bold{r}=\bold{\hat{r}}\cdot r \\\\\\\\\Longrightarrow \nabla \dfrac{1}{r} = -\dfrac{ \bold{\hat{r}} \cdot r}{r^3}\\\\\\\\\therefore \boxed{\nabla \dfrac{1}{r} = \dfrac{\bold{-\hat{r}}}{r^2}}[/tex]

[tex]\hrulefill[/tex]

The general formula is as follows:

[tex]\Longrightarrow \nabla(r^n)=\dfrac{\partial}{\partial r}[r^n] \bold{\hat{r}} \\\\\\\\\therefore \boxed{\nabla(r^n)=nr^{n-1}\bold{\hat{r}}}[/tex]

The two planets are similar in size, inclination, structure, composition, and the presence of water on their surfaces. They also have many key differences.

While Earth has 23h 56m and 4 s period of rotation, Mars's period of rotation is about 24 hours and 40 minutes. Mars is inclined on its axis, similar to Earth., It is inclined at 25.19° to its orbital plane.

They are both terrestrial planets. This means that both have a dense metallic core and also an overlying mantle and crust, which is composed of less dense materials (like silicate rock).

Mars also has a surface covered by a mountain range. Ice water exists beneath much of the Martian surface. Radar data and soil samples have confirmed the presence of shallow subsurface water at the middle latitudes as well.

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The angle between the two vectors is 59 degrees.

The angle between the two vectors is calculated by applying the following equation as shown below.

Mathematically, the formula for angle between two vectors is given as;

tan θ = Vy / Vx

where;

The resultant vertical vector is calculated as follows;

∑ Vy = ( 1 + 4 ) = 5 y

The resultant horizontal vector is calculated as follows;

∑ Vx = ( -4 + 1 ) = -3x

The angle between the two vectors is calculated as follows;

tan θ = Vy / Vx

tan θ = 5 / 3

tan θ = 1.667

θ = arc tan ( 1.667 )

θ = 59 degrees

Thus, the angle between the two vectors is a function of the resultant vertical and horizontal vector.

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The velocity with which the jumper leaves the floor is 5.1 m/s.

The initial velocity of the jumper or the velocity with which the jumper leaves the floor is calculated by applying the principle of conservation of energy as shown below.

Kinetic energy of the jumper at the floor = Potential energy of the jumper at the maximum height

¹/₂mv² = mgh

v² = 2gh

v = √2gh

where;

v = √(2 x 9.8 x 1.3)

v = 5.1 m/s

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When something bounces off a wall or other object, reflection happens. Following the rule of reflection, waves are reflected off of straight boundaries.

Reflection

When light reflects off of something, it does so. The light will reflect at the same angle it struck the surface if the surface is smooth and shining, such as glass, water, or polished metal.

Echoes. A sound that is repeated as a result of sound waves being reflected back is known as an echo. The way a rubber ball bounces off the ground is the same way that sound waves can bounce off smooth, hard things.

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The boundary layer thickness is 7.92 x 10^(-3) m.

A) Formula for Reynolds number to reflect this question is given as;

R_e = Vx/v

Where;

V is velocity

x is polar coordinate where the boundary layer will become turbulent.

R_e is reynolds number.

Reynolds number for turbulent flow of flat plates has a value of 5 x 10^(5)

v is kinematic velocity

From water tables at standard conditions, kinematic viscosity is 1.12 x 10^(-6) m²/s

Now, R_e = Vx/v

We are looking for x, so let's make x the subject.

x = v•R_e/V

Thus, plugging in the relevant values to obtain;

x = [1.12 x 10^(-6) x 5 x 10^(5)]/0.5

x = 1.12m

The boundary layer thickness at the point where the flow becomes turbulent is given by;

δ = 5√(vx/V)

Where;

V is velocity

x is polar coordinate where the boundary layer will become turbulent.

R_e is reynolds number.

Reynolds number for turbulent flow of flat plates has a value of 5 x 10^(5)

v is kinematic velocity

Plugging in the relevant values to get ;

δ = 5√(1.12 x 10^(-6)•1.12/0.5)

δ = 5√(2.5088 x 10^(-6)

δ = 7.92 x 10^(-3) m

Therefore, the boundary layer thickness is 7.92 x 10^(-3) m.

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The wave is traveling in the negative x direction.The wave speed is 12.0m/s.The frequency is 5HzThe wavelength is 2.4m.The displacement is 0.029m.

The general wave equation is given by;

y(x, t) = y cos (2(x/λ) - 2ft)    --------------------------------(i)

y(x, t) is the displacement of the wave at position x and a given time t

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]   ------------------(ii)

Which can be re-written as;

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m)) + 2π(t/(0.20 s))]  -------------(iii)

Comparing equations (i) and (iii) we have that;

=> 2π(x/(2.4 m) = 2π(x/λ)

=> λ = 2.4m

Therefore the wavelength of the wave is 2.4m

Also, still comparing the two equations;

=> 2π(t/(0.20 s) = 2πft

=> f = 1 / 0.20

=> f = 5Hz

Therefore the frequency of the wave is 5Hz

To get the wave speed (v), it is given by;

v = f x λ

Where f = 5Hz and λ = 2.4m

=> v = 5 x 2.4

=> v = 12.0m/s

Therefore, the speed of the wave is 12.0m/s

At t = 0.50s and x = 0.20m;

The displacement, y(x,t) of the string wave is given by

y(x, t) = (3.0 cm) × cos[2π(x/(2.4 m) + t/(0.20 s))]

Convert the amplitude of 3.0cm to m

=> 3.0cm = 0.03m

Carefully solve the equation

=> y(x, t) = (0.03m) × cos[2π((0.20)/(2.4 m)) + 2π(0.50/(0.20 s))]

=> y(x, t) = (0.03m) × -0.9684

=> y(x, t) = 0.029m

Therefore the displacement at those points is -0.029m

Also, the sign of the displacement shows that the direction of the wave is in the negative x direction.

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The magnitude and direction of its momentum are 23259.5 kg*m/s. and eastward direction respectively.

Its velocity is 10.66 m/s

To find the magnitude and direction of the momentum of the compact car, we can use the formula for momentum, which is defined as the product of an object's mass and velocity:

momentum = mass * velocity

In this case, the mass of the compact car is 725 kg and its velocity is 115 km/h. Converting the velocity to meters per second (m/s) by dividing by 3.6 gives us a velocity of 31.94 m/s.

Plugging these values into the formula above gives us a momentum of 23259.5 kg*m/s.

The direction of the momentum is the same as the direction of the velocity, which in this case is toward the east. Therefore, we can represent the momentum of the compact car as an arrow pointing eastward.

To find the velocity of the second car, we can use the formula for momentum and solve for velocity:

velocity = momentum / mass

In this case, the mass of the second car is 2175 kg and its momentum is the same as that of the compact car, which we calculated to be 23259.5 kg*m/s. Plugging these values into the formula above gives us a velocity of 10.66 m/s.

Note that the velocity of the second car is much lower than that of the compact car, despite having the same momentum. This is because the second car has a much larger mass, so it takes a larger force to change its momentum by a given amount.

Therefore, the correct answer is as given above

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Speed of the spacecraft right before the collision:[tex]\sqrt{} \frac{2G.Me}{Re}[/tex] .

Assumption: the earth is exactly spherical with a uniform density.

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and

the (gravitational) potential energy of this spacecraft.

[tex]GPE=\frac{G.Me.M}{R}[/tex]

[tex]\sqrt{} \frac{2KE}{M}[/tex]=[tex]\sqrt{\frac{2G.Me}{Re} }[/tex]

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.

The emf induced in the loop at t = 0 is

Part a)

[tex]$$\text { EMF }=14 \times 10^{-3} \mathrm{~V}$$[/tex]

The emf induced in the loop at t = 1.00 s

Part b)

[tex]$$E M F=15.67 \times 10^{-3} \mathrm{~V}$$[/tex]

What is emf induced ?

As we know that magnetic flux through the loop is given as [tex]$\phi=B . A$[/tex]

now we have

[tex]$$\phi=B \pi r^2$$[/tex]

now rate of change in flux is given as

[tex]\frac{d \phi}{d t}=B(2 \pi r) \frac{d r}{d t}$$[/tex]

now we know that

[tex]$$\begin{aligned}& A=\pi r^2 \\& 0.285=\pi r^2 \\& r=0.30 m\end{aligned}$$[/tex]

Now plug in all data

[tex]$$\begin{aligned}& E M F=(0.20) \times 2 \pi \times(0.30) \times(0.037) \\& E M F=14 \times 10^{-3} V\end{aligned}$$[/tex]

Now the radius of the loop after [tex]$t=1 \mathrm{~s}$[/tex]

[tex]$$\begin{aligned}& r_1=r_0+\frac{d r}{d t} \\& r_1=0.30+0.037 \\& r_1=0.337 m\end{aligned}$$[/tex]

Now plug in data in above equation

[tex]$$\begin{aligned}& E M F=(0.20) \times 2 \pi \times(0.337) \times(0.037) \\& E M F=15.67 \times 10^{-3} V\end{aligned}$$[/tex]

Complete question:The loop is in a magnetic field 0.40 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A = 0.285 square m.Suppose the radius of the elastic loop increases at a constant rate, dr/dt = 3.70 cm/s .

A) Determine the emf induced in the loop at t = 0.

B)Determine the emf induced in the loop at t = 1.00 s .

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The metals on the basis of the maximum kinetic energy of the emitted electrons is Cesium > potasium > Beryllium = aluminum = Megnesium = Platinum.

From Einstein photo electic effect equation

E = 3.1 eV

E = Wo+ Kmax

Kmax = E - Wo

For Be:  E< Wo, Kmax =  0

There is no emission of photo electron

For K : Kmax = 3.1 -2.3 = 0.8 eV

For Al : E< Wo, Kmax =  0

There is no emission of photo electron

For  Mg: E< Wo, Kmax =  0

There is no emission of photo electron

For Cs: Kmax = 3.1 -2.1 = 1.0 eV

For Pt : E< Wo, Kmax =  0

There is no emission of photo electron

Cesium > potasium > Beryllium = aluminum = Megnesium = Platinum

Einstein photo electic effect:

Einstein said that energy is a beam of particles related to frequency according to Planck's formula. When this beam is directed at a metal, photons collide with atoms. If the photon frequency is high enough to knock off an electron, the collision will produce the photoelectric effect.

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The wavelength λ of the laser is 632.8 nm and the index of refraction for this mystery fluid is 1.329.

Refraction can change the wavelength of light or sound as it travels through a medium with a density greater than that of air. The wavelength reduces as a result of the wave's decreasing speed as it moves through the denser medium. If the wave travels faster when in a rarer medium, the wavelength rises. Because the speed and, consequently, the wavelength of the wave are lowered while traveling through the denser media, the index of refraction is greater for the denser medium than the rarer one.

An interferometer is a device that employs the interference phenomena to calculate the wavelength of light in terms of a standard length or the distance in terms of the wavelength of light that is already known. Through partial reflection and transmission, light from an extended source is split into two halves. The two directions in which these two beams are transmitted are at right angles to one another. They are noticed and studied as interference fringes that are created when they are reflected off the mirror.The angle between mirrors M1 and M2 determines how the fringes in the Michelson interferometer are shaped. Let S' be the replica of the source S, and let M2 be the virtual replica of mirror M1.Let D represent the distance between the mirror M1 and its virtual counterpart, M2'. As a result, the air film that is enclosed between M1 and M2' is what will cause the interference pattern to appear. Let S1' and S2' represent the respective Saras M1 and M2's photos.As a result of the amplitude division process, we are able to produce two coherent sources, S1' and S2'.

A)

2D = nλ

2 x 0.03164 x 10^-3 = 100 x λ

Wavelength, λ = 632.8 nm

B)

2D = n λ

2 x 0.02381 x 10^-3 = 100 λ

Wavelength,λ = 476.2 nm

Refractive index =λ₁/λ₂

Thus,

Refractive index = 632.8/476.2 = 1.329

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The force applied due to friction is 32.2 N.

Friction is the force resisting the relative motion of solid surfaces, fluid layers, and material elements sliding against each other. There are several types of friction: Dry friction is a force that opposes the relative lateral motion of two solid surfaces in contact.

Coefficient of friction, ratio of the frictional force resisting the motion of two surfaces in contact to the normal force pressing the two surfaces together. It is usually symbolized by the Greek letter mu (μ). Mathematically, μ = F/N, where F is the frictional force and N is the normal force.

To solve for the minimum force, let us assume it to be F (N)

So

F=mg sin A

But

=>>>> coefficient of static friction x (F + mg cos A

=>3 x 9.8 x sin35 = 0.3 x (F + 3 x 9.8 x cos35)

So making F subject of formula

F + 24.0 = 56.2

F = 32.2N

Therefore, the force applied due to friction is 32.2 N.

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The qualitative sketch of the process on the PV plane is attached below:

The ideal gas law (also called the perfect gas law), the relationship between pressure P, volume V, and temperature T of a gas in the boundary region between low pressure and high temperature. Gas molecules move almost independently. each other. In such cases, all gases obey an equation of state known as the ideal gas law.

PV = nRT, where n is the number of moles of gas and R is the universal (or complete) gas constant, 8.31446261815324 joules/kelvin/mol (the universal gas constant is defined as Avogadro's number NA multiplied by Boltzmann's constant k) . ) In the International System of Units, energy is measured in joules, volume in cubic meters (m3), force in newtons (N), and pressure in pascals (Pa). where 1 Pa = 1 N/m2. A force of 1 Newton moving a distance of 1 meter does 1 Joule of work. Therefore, the dimension of both products PV and nRT is work (energy).

The qualitative sketch of the process is drawn on the PV plane.

At A,

P₁ = 100 kPa

V₁ = 0.50 m³
T₁ = 375 K

At B,

T₂ = 375 K

V₂ = 0.25 m³
P₁V₁ = P₂V₂

So, P₂ = (100× 10³ ×0.50)/ 0.25

= 200 kPa

At C,

P₃ = 200 kPa

V₃ = 0.75 m³

We know, V₂/ T₂ = V₃/ T₃

So,

T₃ = V₃T₂/ V₂

= (0.75 × 375)/ 0.25

= 1125 K

At D,

P₄ = 100 kPa

T₄ = 1125 K

V₄ = P₃V₃/ P₄

= (200 × 0.75)/ 100

= 1.5 m³

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The complete question is:

The physical quantity that does the slope of the v2/2 v/s h graph represent is the slope-g. The slope is representing the change in v over the change in time.

By definition, that is the acceleration. Also extra than only a gimmick, it is geometry's manner of displaying us how variables now no longer most effective relate to 1 another, however how they are able to constitute the fee of alteration among the variables.

The slope of the displacement time graph offers us the bodily amount called pace. The slope of a pace graph represents an object's acceleration. As a result, the cost of the slope at a given time represents the object's acceleration at that time. The fee of changing of an object's pace with admire to time is described as acceleration.

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Answer:

The acceleration of the luggage is:

a = (33.3 N - 12.3 N) / 50.0 kg = 0.466 m/s²

Explanation:

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An express magnitude is a positive number and angle between. ±180∘.  v1 = ∠ ∘ V.

Nodal stress analysis complements the previous mesh analysis in that it is equally powerful and based on the same concepts as matrix analysis. As the name suggests, node voltage analysis uses Kirchhoff's first law node equation to find the voltage potential around a circuit.

When we use the term node voltage, we are referring to the potential difference between two nodes in a circuit. Select one of the nodes in your circuit as the reference node. All other node voltages are measured with respect to this one reference node.

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Gravitational Potential Energy is demonstrated by a stationary ball put at the top of a ramp. The energy that a body has or acquires is known as gravitational potential energy.

because when something is in a gravitational field, it changes position. It is, to put it simply, energy that results from gravitational force or gravity. Taken two pencils is the most typical illustration. If one pencil is set down on a table and the other is raised above it, the pencil raised will have more gravitational potential energy than the pencil set down on the table. The equation U = mgh is used to compute it. The mass, m, is expressed in kilograms.

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The following are some properties of images formed by a plane mirror are; virtual, upright, left-right reversed, the same distance from the mirror as the object's distance, and the same size as the object.

When a ray of light hits a surface, then it can undergo one of the following reflection, refraction, or absorption. So mirrors are polished surfaces coated with mercury such that they reflect most of the light hitting them.

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The graph that best represents the relationship between the gravitational potential energy of a freely falling object and the object's height above the ground is the last graph.

What is Gravitational potential energy ?

Gravitational potential energy is the energy possessed by a body by virtue of its position. The formula is expressed as:

E = mgh

m is the mass of the object

g is the acceleration due to gravity

h is the height of the object

Note that the formula shows that the gravitational potential energy is directly proportional to the height of the object

This direct proportionality shows that the required graph will have a positive slope.

This shows that the graph that best represents the relationship between the gravitational potential energy of a freely falling object and the object's height above the ground is the last graph.

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The angular acceleration is 0.19 and the number of turns is 41.07 revolutions.

m = 87 kg be the mass of the grindstone,

r = 0.35 m be the radius of the grindstone,

I = mr²/2 be the grindstone's moment of inertia,

ω = 95 revolution/min = 95 2π/60s = 9.9 s-1 be initial angular velocity,

F = 15 N be the force pressing radially,

f = 0.2 be the coefficient of friction,

α (unknown) be the grindstones angular acceleration.

a)The tangential force slowing the grindstone is

fF

Its torque is

rfF

By Newton's Second Law

rfF = Iα

Therefore

α = rfF/I = 2rfF/mr² = 2fF/mr

Substituting actual numbers

α = 2×0.2×15/(87×0.35) = 0.19 s-2

Note that the above is the absolute value of the acceleration.

Depending on which direction is considered positive or negative,

the answer may need to be negated.

b)

Let

t be the time it takes the grindstone to come to rest,

θ be the total angle by which the grindstone will turn before coming to rest.We have the following identities, which follow from the definition of acceleration:

ω = αt

θ = αt²/2

From the first

t = ω/α

Plug into the second

θ = α(ω/α)²/2 = ω²/2α

Substitute actual numbers

θ = 9.9²/2×0.19 = 257.92

The above result is in radians.

Since the result is supposed to be given in revolutions,

we need to divide by the number of radians per revolutions:

257.92/2π = 41.07 revolutions

Therefore, the angular acceleration is 0.19 and the number of turns is 41.07 revolutions.

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The wave bounces off an item when there is a reflection kind of wave interaction.

When waves strike a surface and rebound, reflection occurs. When a wave encounters a surface, it might reflect in all directions. For instance, visible light bounces off the surfaces it encounters. Your eyes receive the reflected light, which makes it possible for you to see the objects.

When two or more waves collide, interaction occurs. Wave interference is the term used to describe how waves interact with one another. Wave interference may occur when two waves that are traveling in opposite directions meet. The two waves pass through each other, and this affects their amplitude.

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In a two-slit interference and diffraction experiment, the fifth interference maximum and the first diffraction minimum can occur at the same angle relative to the center of the screen under certain conditions.

In a diffraction experiment, light waves passing through the two slits can also diffract and spread out, producing a pattern of bright and dark fringes on the screen. The position of the dark fringes, also known as diffraction minima, is determined by the wavelength of the light and the size of the slits.

If the fifth interference maximum and the first diffraction minimum occur at the same angle relative to the center of the screen, it means that the path difference between the waves passing through the two slits is equal to the wavelength of the light. This can be achieved by adjusting the separation between the slits, the distance between the slits and the screen, and the wavelength of the light.

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