We will use a video to analyze the dependence of the magnitude of the Coulomb force between two electrically-charged spheres on the distance between the centers of the spheres. The electrical interaction is one of the fundamental forces of nature and acts between any pair of charged objects, therefore it is important to understand how precisely the separation distance affects the corresponding force between them.


1. Study conceptually the nature of Hint 1. Unit conversions Note that 1 nC 109 c. Hint2. The values for the first data point The first suggested data point was 31.9 cm and 0.00016 N electric charge and force

2. Take measurements of the force exerted between two electrically- charged spheres as the distance between them is varied

3. Determine graphically the relationship between electric force and distance nc

Answer:

Perpendicular

Explanation:

write the angle between them as a

cosine theoreme

[tex]0.6^{2} =1.8^{2} +2.4^{2}-2*1.8*2.4*cos(a)[/tex]

cos(a)=(3.24+5.76-0.6)/(2*1.8*2.4)

cos(a)=1

a=90°

To find the arrangement of two vectors with magnitudes of 1.8 m and 2.4 m so that their sum is 0.6 m, we need to consider vector addition. The magnitude of the resultant vector is 4.2 m. One possible arrangement is to align the vectors in opposite directions, cancelling out each other's effects and resulting in a net sum of zero.

To find the arrangement of two vectors with magnitudes of 1.8 m and 2.4 m, so that their sum is 0.6 m, we need to understand vector addition. The sum of two vectors can be found by aligning the vectors head to tail and drawing a line connecting the head of the first vector to the tail of the second vector. The resulting vector, called the resultant vector, is the sum of the two vectors.

In this case, let's call the first vector A with magnitude 1.8 m and the second vector B with magnitude 2.4 m. To find the arrangement where their sum is 0.6 m, we can set up the equation: A + B = 0.6. Since the magnitudes are given, we can rearrange the equation to solve for the magnitude of the resultant vector: |A| + |B| = |0.6|. Substituting the given values, we get: 1.8 + 2.4 = 0.6. Simplifying the equation, we find that the magnitude of the resultant vector is 4.2 m.

Now, to find the arrangement, we need to consider the direction of the vectors. Since the sum of the two vectors is not zero, their directions cannot be directly opposite. Therefore, we need to arrange them in a way that their individual directions cancel out each other to result in a net sum of zero. One possible arrangement is to align the vectors such that they form a straight line in opposite directions. In this case, A and B would have the same magnitude but opposite directions, cancelling out each other's effects and resulting in a net sum of zero.

Answer:

Explanation:

Let the charge particle have charge equal to +q .

force due to electric field will be along the field that is along y - axis . To balance it force by magnetic force must be along   - y axis. ( negative of y axis )

force due to magnetic field = q ( v x B ) , v is velocity and B is magnetic field.

F = q ( v i x B k )  , ( velocity is along x direction and magnetic field is along z axis. )

= (Bqv) - j  

= - Bqv j

The force will be along - ve y - direction .

If we take charge as negative or - q

force due to electric field will be along - y axis .

magnetic force = F = -q ( v i x B k )

= +  Bqv j

magnetic force will be along  + y axis

So it is difficult to find out the nature of charge on the  particle from this experiment.

Answer:

The solution to the question above is explained below:

Explanation:

(a) The work function of the surface is:

The work function of a metal, Φ, it's the minimum amount of energy required to remove electron from the conduction band and remove it to outside the metal. It is typically exhibited in units of eV (electron volts) or J (Joules). Work Function, is the minimum thermodynamic work.

hf = hc /λ = φ + Kmax = φ + 1 /2 mev ²max

where,   φ=work function of a metal

               eV=electron volts

              J= Joules

             λ=the Plank constant 6.63 x 10∧-34 J s

              f = the frequency of the incident light in hertz

              ∧= signifies raised to the power in the solution

Kmax= the maximum kinetic energy of the emitted electrons in joules

φ= ( hc /λ -1/2mev ²max= 6.63 × 10∧-34Js × 3.00 × 10∧8 m/s ÷ 625 × 10∧-9) -

 (1/2 × 9.11 × 10∧-31 kg × 4.60 × 10∧5 m/s) = 2.21 × 10∧-19 J

ans= 1.38 eV

(b) The cutoff frequency for this surface is:

The cutoff frequency is the minimum frequency that is required for the emission of electrons from a metallic surface at which energy flowing through the metallic surface begins to be reduced  rather than passing through.

Light at the cutoff frequency only barely supplies enough energy to overcome the work function. The value of the cutoff frequency is in unit hertz (Hz)

hfcut = φ

fcut = φ /h

ans= 334 THz

Explanation:

Given that,

Length of the spring, l = 50 cm

Mass, m = 330 g = 0.33 kg

(A) The mass is released and falls, stretching the spring by 28 cm before coming to rest at its lowest point. On applying second law of Newton at 14 cm below the lowest point we get :

[tex]kx=mg\\\\k=\dfrac{mg}{x}\\\\k=\dfrac{0.33\times 9.8}{0.14}\\\\k=23.1\ N/m[/tex]

(B) The amplitude of the oscillation is half of the total distance covered. So, amplitude is 14 cm.

(C) The frequency of the oscillation is given by :

[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f=\dfrac{1}{2\pi}\sqrt{\dfrac{23.1}{0.33}} \\\\f=1.33\ Hz[/tex]

Answer:

the wavelength λ of the light when it is traveling in air = 560 nm

the smallest thickness t of the air film = 140 nm

Explanation:

From the question; the path difference is Δx = 2t  (since the condition of the phase difference in the maxima and minima gets interchanged)

Now for constructive interference;

Δx= [tex](m+ \frac{1}{2} \lambda)[/tex]

replacing ;

Δx = 2t   ; we have:

2t = [tex](m+ \frac{1}{2} \lambda)[/tex]

Given that thickness t = 700 nm

Then

2× 700 = [tex](m+ \frac{1}{2} \lambda)[/tex]     --- equation (1)

For thickness t = 980 nm that is next to constructive interference

2× 980 = [tex](m+ \frac{1}{2} \lambda)[/tex]     ----- equation (2)

Equating the difference of equation (2) and equation (1); we have:'

λ = (2 × 980) - ( 2× 700 )

λ = 1960 - 1400

λ = 560 nm

Thus;  the wavelength λ of the light when it is traveling in air = 560 nm

b)  

For the smallest thickness [tex]t_{min} ; \ \ \ m =0[/tex]

∴ [tex]2t_{min} =\frac{\lambda}{2}[/tex]

[tex]t_{min} =\frac{\lambda}{4}[/tex]

[tex]t_{min} =\frac{560}{4}[/tex]

[tex]t_{min} =140 \ \ nm[/tex]

Thus, the smallest thickness t of the air film = 140 nm

Answer:

1.4x10^7m & 98nm

Explanation:

Pls the calculation is in the attached file

The instantaneous rate of change of the force between two masses with respect to distance is given by the derivative of Newton's Law of Universal Gravitation, resulting in [tex]dF/dd = -2G(m1*m2)/d^3.[/tex]

The question is asking for the instantaneous rate of change of the force between the two masses with respect to the distance. This can be found by taking the derivative of Newton's Law of Universal Gravitation.

The gravitational force, F, is defined by the equation[tex]F = G(m1*m2)/d^2.[/tex]Taking the derivative of this function with respect to d, gives us [tex]dF/dd = -2G(m1*m2)/d^3.[/tex] This equation represents how the force changes with a small change in distance.

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Final answer:

To find the total length of the wire in a 65-turn square coil subjected to a changing magnetic field, apply Faraday's law of induction to calculate the magnetic flux change and then determine the side length of the coil. Multiply the coil's perimeter by the number of turns to obtain the total length.

Explanation:

The student's question pertains to electromagnetic induction in a coil exposed to a changing magnetic field. To solve for the total length of the wire, we can use Faraday's law of induction, which states that the induced emf (electromotive force) in a coil is proportional to the rate of change of magnetic flux through the coil. Given an induced emf of 80.0 mV and a change in magnetic field from 200 µT to 600 µT over 0.400 s, the change in magnetic flux φ can be calculated.

Since the coil is square and positioned at a 30.0° angle to the magnetic field, the effective area A for inducing emf can be determined using the cosine of the angle and the side length of the square, assuming all sides are equal. With the number of turns (N) being 65, we can apply Faraday's law to find the magnetic flux and subsequently the side length of the square. The total length of the wire is simply the perimeter of the square (4 times the side length) multiplied by the number of turns (N).

Answer:

[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]

Explanation:

The standard form of the 2nd order differential equation governing the motion of mass-spring system is given by

[tex]my'' + \zeta y' + ky=0[/tex]

Where m is the mass, ζ is the damping constant, and k is the spring constant.

The spring constant k can be found by

[tex]w - kL=0[/tex]

[tex]mg - kL=0[/tex]

[tex]4 - k\frac{1}{6}=0[/tex]

[tex]k = 4\times 6 =24[/tex]

The damping constant can be found by

[tex]F = -\zeta y'[/tex]

[tex]6 = 3\zeta[/tex]

[tex]\zeta = \frac{6}{3} = 2[/tex]

Finally, the mass m can be found by

[tex]w = 4[/tex]

[tex]mg=4[/tex]

[tex]m = \frac{4}{g}[/tex]

Where g is approximately 32 ft/s²

[tex]m = \frac{4}{32} = \frac{1}{8}[/tex]

Therefore, the required differential equation is

[tex]my'' + \zeta y' + ky=0[/tex]

[tex]\frac{1}{8} y'' + 2y' + 24y=0[/tex]

The initial position is

[tex]y(0) = \frac{1}{2}[/tex]

The initial velocity is

[tex]y'(0) = 0[/tex]

We formulated the initial value problem for a damped spring-mass system:

1/8 * d²x/dt² + 2 * dx/dt + 24 * x = 0

with initial conditions x(0) = 0.5 ft and dx/dt(0) = 0 ft/s.

Let's break down the problem with the given data:

First, find the spring constant k:

The weight of the mass stretches the spring by 2 inches (0.1667 feet).

The force exerted by the weight = 4 lb = mg

The spring force F = kx

So,

The general form of the second-order differential equation governing the motion of the spring-mass-damper system is:

The viscous resistance given is 6 lb at 3 ft/s. Therefore, the damping coefficient c:

The initial conditions are displacement 6 inches (0.5 feet) and initial velocity 0:

Combining these elements, the initial value problem is:

Answer:

Power emitted will be 0.314 mW

So option (A) will be correct option

Explanation:

We have given threshold hearing [tex]I=10^{-12}W/m^2[/tex]

Distance is given r = 5 km =5000 m

We have to find the power emitted

Power emitted is equal to

[tex]P=I\times A[/tex]

[tex]=10^{-12}\times 4\pi r^2[/tex]

[tex]=10^{-12}\times 4\times 3.14\times (5000)^2[/tex]

=[tex]314\times 10^{-6}watt=0.314mW[/tex]

So power emitted will be 0.314 mW

So option (A) will be correct option.

Answer: 3.60*10^-4 s

Explanation:

Given

Mass of bullet, m = 26 g = 0.026 kg

Initial speed of bullet, u = 1000 m/s

Length of target, s = 35 cm = 0.35 m

Mass of target, M = 400 kg

Final speed of bullet, v = 950 m/s

Using equation of motion

v² = u² + 2as, making a subject of formula, we have

a = (v² - u²) / (2*s)

a = 950² - 1000² / 2 * 0.35

a = 902500 - 1000000 / 0.7

a = -97500 / 0.7

The acceleration = - 1.39*10^5 m/s² ( it is worthy of note that the acceleration is negative)

now, using another equation of motion, we have

v = u + a*t

we know our a, so we make t subject of formula

time t = (v-u) / a

t = (950 - 1000) / -1.39*10^5

t = -50 / -1.39*10^5

t = 3.60*10^-4 s

Answer:

the penny loses contact at the piston's highest point.

f = 2.5 Hz

Explanation:

Concepts and Principles  

1- Newton's Second Law: The net force F on a body with mass m is related to the body's acceleration a by  

∑F = ma                                                          (1)  

2- The maximum transverse acceleration of a particle in simple harmonic motion is found in terms of the angular speed w and the amplitude A as follows:  

a_max = -w^2A                                                (2)  

3- The angular frequency w of a wave is related to the frequency f by:  

w = 2π f                                                              (3)  

Given Data  

- The amplitude of the piston is: A = (4.0 cm) ( 1/ 100 cm)=  0.04 m.  

- The frequency of oscillation of the piston is steadily increased.

Required Data

In part (a), we are asked to determine the point at which the penny first loses contact with the piston.  

In part (b), we are asked to determine the maximum frequency for which the penny just barely remains in place for a full cycle.  

Solution  

(a)  

The free-body diagram in Figure 1 shows the forces acting on the penny; mi is the gravitational force exerted by the Earth on the penny andrt is the normal contact force exerted by the piston on the penny.  

figure 1 is attached

Apply Newton's second law from Equation (1) in the vertical direction to the penny:  

∑F_y -mg= ma        

Solve for n=m(g+a) The penny loses contact with the surface of the oscillating piston when the normal force n exerted by the piston is zero. So  

0 = m(g + a)

a = —g  

Therefore, the penny loses contact with the piston when the piston starts accelerating downwards. The piston first acceleratesdownward at its highest point and hence the penny loses contact at the piston's highest point.

(b)  

The maximum acceleration of the penny at the highest point of the piston is found from Equation (2):  

a = —w^2A  

where a = —g at the highest point. So  

g = w^2A  

Solve for w:  

w =√g/A

Substitute for w from Equation (3):

2πf =  √g/A

Solve for f :  

f = 1/2π√g/A

Substitute numerical values:  

f = 1/2π√9.8 m/s^2/0.04

f = 2.5 Hz

The question is based on Simple Harmonic Motion: The penny first loses contact with the piston at its highest point (option c) in the motion. The maximum frequency for which the penny just barely remains in place for the full cycle is approximately 12.5 Hz.

The penny will first lose contact with the piston at its highest point, which is option c) highest point in the cycle.

To determine the maximum frequency for which the penny just barely remains in place for the full cycle, we need to consider the condition for the penny to stay in contact with the piston. At the highest point of the motion, the penny experiences an upward gravitational force and a downward centripetal force due to the circular motion.

When the centripetal force becomes greater than or equal to the gravitational force, the penny will lose contact with the piston. This occurs when

mvmax2/r = mg

Solving for the maximum velocity using vmax = 2πfA, where A is the amplitude, and substituting into the equation above, we find that

fmax = g / (4π2A)

Substituting the given values, with a=4.0 cm = 0.04 m and g=9.8 m/s2:

fmax = 9.8 / (4π2(0.04)) ≈ 12.5 Hz

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Answer: 0.5 m/s

Explanation:

Given

Speed of the sled, v = 0.55 m/s

Total mass, m = 96.5 kg

Mass of the rock, m1 = 0.3 kg

Speed of the rock, v1 = 17.5 m/s

To solve this, we would use the law of conservation of momentum

Momentum before throwing the rock: m*V = 96.5 kg * 0.550 m/s = 53.08 Ns

When the man throws the rock forward

rock:

m1 = 0.300 kg

V1 = 17.5 m/s, in the same direction of the sled with the man

m2 = 96.5 kg - 0.300 kg = 96.2 kg

v2 = ?

Law of conservation of momentum states that the momentum is equal before and after the throw.

momentum before throw = momentum after throw

53.08 = 0.300 * 17.5 + 96.2 * v2

53.08 = 5.25 + 96.2 * v2

v2 = [53.08 - 5.25 ] / 96.2

v2 = 47.83 / 96.2

v2 = 0.497 ~= 0.50 m/s

Answer:

It made landfall.

Explanation:

On land there is more friction than in open water, so it slowed down the hurricane.

Answer:

The wavelength of incident light is [tex]1.38x10^{-6}m[/tex]

Explanation:

The physicist Thomas Young established, through his double slit experiment, a relation between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

[tex]\Lambda x = L\frac{\lambda}{d} [/tex]  (1)

Where [tex]\Lambda x[/tex] is the distance between two adjacent maxima, L is the distance of the screen from the slits, [tex]\lambda[/tex] is the wavelength and d is the separation between the slits.  

The values for this particular case are:

[tex]L = 2.60m[/tex]

[tex]d = 0.230mm[/tex]

[tex]\Lambda x = 1.57cm[/tex]

Then, [tex]\lambda[/tex] can be isolated from equation 1

[tex]\lambda = \frac{d \Lambda x}{L}[/tex]  (2)

However, before equation 2 can be used, it is necessary to express [tex]\Lambda x[/tex] and d in units of meters.

[tex]\Lambda x= 1.57cm \cdot \frac{1m}{100cm}[/tex] ⇒ [tex]0.0157m[/tex]

[tex]d = 0.230mm \cdot \frac{1m}{1000mm}[/tex] ⇒ [tex]2.3x10^{-4}m[/tex]

Finally, equation 2 can be used.

[tex]\lambda = \frac{(2.3x10^{-4}m)(0.0157m}{(2.60m)}[/tex]

[tex]\lambda = 1.38x10^{-6}m[/tex]

Hence, the wavelength of incident light is [tex]1.38x10^{-6}m[/tex]

Answer:

[tex]h = 83.093\,m[/tex]

Explanation:

The speed of the firework shell just before the explosion is:

[tex]v = \sqrt{(44\,\frac{m}{s})^{2}-2\cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (65\,m)}[/tex]

[tex]v \approx 25.712\,\frac{m}{s}[/tex]

After the explosion, the initial speed of one of the mass halves is:

[tex]v_{f}^{2} = v_{o}^{2} -2\cdot g \cdot s[/tex]

[tex]v_{o}^{2} = v_{f}^{2} + 2\cdot g \cdot s[/tex]

[tex]v_{o} = \sqrt{v_{f}^{2}+2\cdot g \cdot s}[/tex]

[tex]v_{o} = \sqrt{\left(0\,\frac{m}{s}\right)^{2}+ 2 \cdot \left(9.807\,\frac{m}{s^{2}}\right)\cdot (120\,m-65\,m)}[/tex]

[tex]v_{o} \approx 32.845\,\frac{m}{s}[/tex]

The initial speed of the other mass half is determined from the Principle of Momentum Conservation:

[tex]m \cdot (25.712\,\frac{m}{s} ) = 0.5\cdot m \cdot (32.845\,\frac{m}{s} ) + 0.5\cdot m \cdot v[/tex]

[tex]25.842\,\frac{m}{s} = 16.423\,\frac{m}{s} + 0.5\cdot v[/tex]

[tex]v = 18.838\,\frac{m}{s}[/tex]

The height reached by this half is:

[tex]h = h_{o} -\frac{v_{f}^{2}-v_{o}^{2}}{2\cdot g}[/tex]

[tex]h = 65\,m - \frac{(0\,\frac{m}{s} )^{2}- (18.838\,\frac{m}{s} )^{2}}{2\cdot (9.807\,\frac{m}{s^{2}} )}[/tex]

[tex]h = 83.093\,m[/tex]

Answer:

The other half goes 17.4m high

Explanation:

Pls see calculation in the attached file

Answer:

Nuclear reaction is the disintegration or integration of nucleus or nuclei of an atom, to produce a different nuclei or nucleus, thereby given out a large amount of heat energy. The energy been extracted from this reaction are very much.

At a successful fusion reaction, the world will have more better opportunity to preserve it's natural resources, as they will be enough energy to produce electricity, cooking, industrial power, and many more. It will also reduce spending, as energy from fusion reaction are not too expensive in obtaining, when compared to energy from natural resources.

ADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR BASED STATION.

1) Central power stations are reliable, because of the abundance of natural resources been used to produce the electricity. Solar based station are not reliable because during winter, when the sun is hardly seen, they will not be enough resource to produce electricity.

2) Central power stations are very efficient because they produce a stable electricity power, while the smaller solar stations are not efficient, because the amount of power produced is dependent on the amount of solar energy it has stored.

DISADVANTAGES OF USING CENTRAL POWER STATION THAN SMALLER SOLAR-BASED STATION.

1) Central power stations causes pollution, and green house effect, because the natural gas been used to produce the electricity causes pollution and green house effect during venting. While smaller solar-based station are pollution free and does not cause green house effect, because it only need sun to store it's charge.

2) Central power station are more dangerous for people to work, as risk of being electrocuted is high. Smaller-based solar stations has less risk, as the power been stored from the sun requires less complex wiring.

3) Central power stations are more expensive to produce and requires much land space. While smaller solar-based station are less expensive, and requires little land space.

Final answer:

Successful fusion reactors could dramatically reshape global politics and economies by reducing reliance on fossil fuels. Centralized fusion power offers consistency, whereas decentralized solar power fosters resilience and local control but is intermittent. Fusion faces technological challenges, while solar technology is readily expanding.

Explanation:

Worldwide Changes with Successful Fusion Reactors

The advent of successful fusion reactors would be a transformative event worldwide. Fusion promises a cleaner, nearly inexhaustible energy source compared to fossil fuels, which could significantly alter the current geopolitical landscape where oil and natural gas are dominant factors in international politics and economics. A shift toward fusion energy could lessen global dependence on fossil fuels, potentially reducing the geopolitical power of oil-rich nations and creating a new paradigm in the world economy. With hydrogen from water as fuel, countries may realign alliances and shift focus towards technological advancements and harnessing fusion energy efficiently.

Central vs. Distributed Power Generation

Electricity from a large central power station, such as a fusion reactor, offers economies of scale and a consistent power supply but requires a significant up-front investment and complex infrastructure. Meanwhile, smaller solar-based stations, which are often owned and operated by individuals, offer distributed generation, empowering local communities and enhancing energy resilience. However, solar energy faces challenges with intermittency and requires storage solutions to provide a reliable power supply.

The disadavantages of fusion, despite its potential, include the complexities and high costs associated with maintaining such a power source, as well as addressing issues related to radioactivity. On the other hand, the technology for solar power is already available today, with deployment expanding rapidly thanks to decreasing costs and improving battery storage technologies.

Challenges in Fusion Technology

Developing fusion reactors for electricity generation requires overcoming significant technical hurdles, such as achieving the necessary high temperatures and creating materials that can contain the fusion reaction without melting. Scientists are optimistic, yet the timeline for fusion reactors to become a commercial reality remains uncertain.

Answer:

θ₁ = 3.35 10⁻⁴ rad ,  θ₂ = 8.39 10⁻⁵ rad

Explanation:

This is a diffraction problem for a slit that is described by the expression

       sin θ = m λ

the resolution is obtained from the angle between the central maximum and the first minimum corresponding to m = 1

      sin θ = λ / a

as in these experiments the angle is very small we can approximate the sine to its angle

        θ = λ / a

In this case, the circular openings are explicit, so the system must be solved in polar coordinates, which introduces a numerical constant.

       θ = 1.22 λ / D

where D is the diameter of the opening

 let's apply this expression to our case

indicates that the wavelength is λ = 550 nm = 550 10⁻⁹ m

the case of a lot of light D = 2 mm = 2 10⁻³ m

       θ₁ = 1.22 550 10-9 / 2 10⁻³

       θ₁ = 3.35 10⁻⁴ rad

For the low light case D = 8 mm = 8 10⁻³

      θ₂ = 1.22 550 10-9 / 8 10⁻³

      θ₂ = 8.39 10⁻⁵ rad

Answer:

b. The reflection of light from a smooth surface is called specular reflection.

c. The reflection of light from a rough surface is called diffuse reflection.

Explanation:

a. The angle of incidence is equal to the angle of reflection only when a ray of light strikes a plane mirror.

This is wrong: Based on law of reflection "The angle of incidence is equal to the angle of reflection when light strikes any plane surface" examples plane mirrors, still waters, plane tables, etc

b. The reflection of light from a smooth surface is called specular reflection.

This is correct

c. The reflection of light from a rough surface is called diffuse reflection.

This is correct

d. For diffuse reflection, the angle of incidence is greater than the angle of reflection.

This is wrong: the angle of incident is equal to angle of reflection. The only difference between this type of reflection and specular reflection, is that the normal for diffuse reflection is not parallel to each due to the rough surface in which the light incidents.

For specular reflection, the angle of incidence is less than the angle of reflection.

This is wrong: the angle of incident is equal to angle of reflection

The correct statements are statement 2 and statement 3.The true statements are: the reflection of light from a smooth surface is called specular reflection, and the reflection of light from a rough surface is called diffuse reflection. The others are incorrect as the angle of incidence always equals the angle of reflection.

To answer the question about the reflection of light, let's analyze each statement:

In summary, the true statements concerning the reflection of light are:

Answer:

Approximately [tex]1.44\times 10^3 \; \rm N \cdot m^{-1}[/tex] assuming that the spring has zero mass.

Explanation:

Without any external force, a piece of mass connected to an ideal spring (like the chair in this question) will undergo simple harmonic oscillation.

On the other hand, the force constant of a spring (i.e., its stiffness) can be found using Hooke's Law. If the spring exerts a restoring force [tex]\mathbf{F}[/tex] when its displacement is [tex]\mathbf{x}[/tex], then its force constant would be:

[tex]\displaystyle k = -\frac{\mathbf{F}}{\mathbf{x}}[/tex].  

The goal here is to find the expressions for [tex]F[/tex] and for [tex]x[/tex]. By Hooke's Law, the spring constant would be ratio of these two expressions.

Let [tex]T[/tex] represent the time period of this oscillation. With the chair alone, the period of oscillation is [tex]T = 1.00\; \rm s[/tex].

For a simple harmonic oscillation, the angular frequency [tex]\omega[/tex] can be found from the period:

[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].

Let [tex]A[/tex] stands for the amplitude of this oscillation. In a simple harmonic oscillation, both [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex] are proportional to [tex]A[/tex]. Keep in mind that the spring constant [tex]k[/tex] is simply the opposite of the ratio between [tex]\mathbf{F}[/tex] and [tex]\mathbf{x}[/tex]. Therefore, the exact value of [tex]A[/tex] shouldn't really affect the value of the spring constant.

In a simple harmonic motion (one that starts with maximum displacement and zero velocity,) the displacement (from equilibrium position) at time [tex]t[/tex] would be:

[tex]\displaystyle \mathbf{x}(t) = A \cos(\omega \cdot t)[/tex].

The restoring velocity at time [tex]t[/tex] would be:

[tex]\displaystyle \mathbf{v}(t) = \mathbf{x}^\prime(t) = -A\, \omega \sin(\omega\cdot t)[/tex].

The restoring acceleration at time [tex]t[/tex] would be:

[tex]\displaystyle \mathbf{a}(t) = \mathbf{v}^\prime(t) = -A\, \omega^2 \cos(\omega\cdot t)[/tex].

Assume that the spring has zero mass. By Newton's Second Law of motion, the restoring force at time [tex]t[/tex] would be:

[tex]\begin{aligned}& \mathbf{F}(t) \\ &= m(\text{chair}) \cdot \mathbf{a}(t) \\&= -m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)\end{aligned}[/tex].

Apply Hooke's Law to find the spring constant, [tex]k[/tex]:

[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= -\left(\frac{-m(\text{chair}) \, A\, \omega^2 \cos(\omega \cdot t)}{A\cos(\omega \cdot t)}\right) \\ &= \omega^2 \cdot m(\text{chair}) \end{aligned}[/tex].

Again, [tex]\omega[/tex] stands for the angular frequency of this oscillation, where

[tex]\displaystyle \omega = \frac{2\pi}{T}[/tex].

Before proceeding, note how [tex]A[/tex] was eliminated from the ratio (as expected.) Additionally, [tex]t[/tex] is also eliminated from the ratio. In other words, the spring constant is "constant" at all time. That agrees with the assumption that this spring is indeed ideal. Back to [tex]k[/tex]:

[tex]\begin{aligned} k & = -\frac{\mathbf{F}}{\mathbf{x}} \\ &= \cdots \\ &= \omega^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{T}\right)^2 \cdot m(\text{chair}) \\ &= \left(\frac{2\pi}{1.00\; \rm s}\right)^2 \times 36.4\; \rm kg\end{aligned}[/tex].

Side note on the unit of [tex]k[/tex]:

[tex]\begin{aligned} & 1\; \rm kg \cdot s^{-2} \\ &= 1\rm \; \left(kg \cdot m \cdot s^{-2}\right) \cdot m^{-1} \\ &= 1\; \rm N \cdot m^{-1}\end{aligned}[/tex].

Answer:

Statements 1, 2 and 3 are all correct.

Complete Question

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,000 km above the surface of the earth, and we assume it has an isotropic power output of 1 kW (although, in practice, satellite antennas transmit signals that are less powerful but more directional).

Imagine that the satellite described in the problem introduction is used to transmit television signals. You have a satellite TV reciever consisting of a circular dish of radius R which focuses the electromagnetic energy incident from the satellite onto a receiver which has a surface area of 5 cm2.

How large does the radius R of the dish have to be to achieve an electric field vector amplitude of 0.1 mV/m at the receiver?

For simplicity, assume that your house is located directly beneath the satellite (i.e. the situation you calculated in the first part), that the dish reflects all of the incident signal onto the receiver, and that there are no losses associated with the reception process. The dish has a curvature, but the radius R refers to the projection of the dish into the plane perpendicular to the direction of the incoming signal.

Give your answer in centimeters, to two significant figures.

Answer:

 The radius  of  the dish is [tex]R = 18cm[/tex]

Explanation:

  From the question we are told that

     The radius of the orbit is  = [tex]R = 35,000km = 35,000 *10^3 m[/tex]

    The power output of the power is  [tex]P = 1 kW = 1000W[/tex]

   The electric vector amplitude is given as [tex]E = 0.1 mV/m = 0.1 *10^{-3}V/m[/tex]

    The area of thereciever  is   [tex]A_R = 5cm^2[/tex]

Generally the intensity of the dish is mathematically represented as

         [tex]I = \frac{P}{A}[/tex]

Where A is the area orbit which is a sphere so this is obtained as

          [tex]A = 4 \pi r^2[/tex]

              [tex]= (4 * 3.142 * (35,000 *10^3)^2)[/tex]

              [tex]=1.5395*10^{16} m^2[/tex]

  Then substituting into the equation for intensity

          [tex]I_s = \frac{1000}{1.5395*10^{16}}[/tex]

            [tex]= 6.5*10^ {-14}W/m2[/tex]

 Now the intensity received by the dish can be mathematically evaluated as

              [tex]I_d = \frac{1}{2} * c \epsilon_o E_D ^2[/tex]

  Where c is thesped of light with a constant value  [tex]c = 3.0*10^8 m/s[/tex]

              [tex]\epsilon_o[/tex] is the permitivity of free space  with a value  [tex]8.85*10^{-12} N/m[/tex]

              [tex]E_D[/tex] is the electric filed on the dish

So  since we are to assume to loss then the intensity of the satellite is equal to the intensity incident on the receiver dish

      Now making the eletric field intensity the subject of the formula

                  [tex]E_D = \sqrt{\frac{2 * I_d}{c * \epsilon_o} }[/tex]

substituting values

                 [tex]E_D = \sqrt{\frac{2 * 6.5*10^{-14}}{3.0*10^{8} * 8.85*10^{-12}} }[/tex]

                       [tex]= 7*10^{-6} V/m[/tex]

The incident power on the dish is what is been reflected to the receiver

                [tex]P_D = P_R[/tex]

Where [tex]P_D[/tex] is the power incident on the dish which is mathematically represented as

              [tex]P_D = I_d A_d[/tex]

                   [tex]= \frac{1}{2} c \epsilon_o E_D^2 (\pi R^2)[/tex]

And  [tex]P_R[/tex] is the power incident on the dish which is mathematically represented as

                 [tex]P_R = I_R A_R[/tex]

                       [tex]= \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]

Now equating the two

                [tex]\frac{1}{2} c \epsilon_o E_D^2 (\pi R^2) = \frac{1}{2} c \epsilon_o E_R^2 A_R[/tex]

   Making R the subject we have

                   [tex]R = \sqrt{\frac{E_R^2 A_R}{\pi E_D^2} }[/tex]

Substituting values

                   [tex]R = \sqrt{\frac{(0.1 *10^{-3})^2 * 5}{\pi (7*10^{-6})^ 2} }[/tex]

                     [tex]R = 18cm[/tex]

Answer:

A. Up-down

B. East-west & north -south

C. North or south

Explanation:

See attached handwritten document for more details

Answer:

56.5 m/s²

Explanation:

From the law of conservation of momentum,

mu+m'u' = mv+m'v'........................ Equation 1

Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.

From the question,

The golf ball is at rest, Hence u' = 0 m/s

mu = mv+m'v'

Make v' the subject of the equation

v' = (mu-mv)/m'........................... Equation 2

Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.

Substitute into equation 2

v' = (0.152×44.8+0.152×27.7)/0.046

v' = (6.8096-4.2104)/0.046

v' = 2.5992/0.046

v' = 56.5 m/s²

Find the given attachments

Answer:

0.366kgm²

Explanation:

F = 1.8*10³N

r = 2.85cm = 0.0285m

α = 140rad/s²

Torque = applied force * distance

τ = r * F

τ = 0.0285 * 1.8*10³

τ = 51.3N.m

but τ = I * α

I = τ / α

I = 51.3 / 140

I = 0.366kgm²

Answer:

B = (2.80 × 10⁻¹²) T

Explanation:

First of, we convert the 19.5 MeV to Joules

19.5 MeV = 19.5 × 10⁶ × 1.602 × 10⁻¹⁹

= (3.124 × 10⁻¹²) J

The velocity of the cosmic-ray proton can be calculated from the kinetic energy formula

E = (1/2) mv²

m = mass of a proton = (1.67 × 10⁻²⁷) kg

(3.124 × 10⁻¹²) = (1/2)(1.67 × 10⁻²⁷)v²

v = (6.117 × 10⁷) m/s

And since the magnetic force keeps the cosmic ray proton in uniform circular motion,

Magnetic force = Centripetal force keeping the proton in circular motion.

qvB = (mv²/r)

q = charge on a proton = (1.602 × 10⁻¹⁹) C

v = velocity of the proton = (6.117 × 10⁷) m/s

B = magnetic field = ?

r = radius of circular motion = (2.28 × 10¹¹) m

B = (mv/qr)

B = (1.67 × 10⁻²⁷ × 6.117 × 10⁷) ÷ (1.602 × 10⁻¹⁹ × 2.28 × 10¹¹)

B = (2.797 × 10⁻¹²) T

Hope this Helps!!!

Answer:

The magnetic field strength in that region of space is 2.7983 x 10⁻¹² T

Explanation:

Given;

Energy of the cosmic-ray proton, E = 19.5 MeV = 19.5 x 10⁶ x 1.6 x 10⁻¹⁹

                                                       E = 3.12 x 10⁻¹² J

Radius of the circular orbit, r = 2.28 x 10¹¹ m

Step 1:

determine the speed of cosmic-ray proton

Kinetic energy of the cosmic-ray proton;

K.E = ¹/₂mv²

[tex]v = \sqrt{\frac{2K.E}{m} }[/tex]

where;

m is mass of proton, m = 1.67 x 10⁻²⁷ kg

[tex]v = \sqrt{\frac{2*3.12*10^{-12}}{1.67*10^{-27}} } \\\\v = 6.1127*10^7 \ m/s[/tex]

Step 2:

determine the magnetic field strength in that region of space

magnetic force = centripetal force

[tex]qvB = \frac{mv^2}{r} \\\\B = \frac{mv}{rq}\\\\[/tex]

Where;

q is charge of proton, q = 1.6 x 10⁻¹⁹ C

[tex]B = \frac{mv}{rq} = \frac{(1.67*10^{-27})(6.1127*10^7)}{(2.28*10^{11})(1.6*10^{-19})}\\\\B =2.7983 *10^{-12} \ T[/tex]

Therefore, the magnetic field strength in that region of space is 2.7983 x 10⁻¹² T

Answer:

[tex]f = 0.806\,hz[/tex], [tex]T = 1.241\,s[/tex]

Explanation:

The problem can be modelled as a vertical mass-spring system exhibiting a simple harmonic motion. The spring constant is:

[tex]k = \frac{970\,N}{0.03\,m}[/tex]

[tex]k = 32333.333\,\frac{N}{m}[/tex]

The angular frequency is:

[tex]\omega = \sqrt{\frac{32333.333\,\frac{N}{m} }{1258.879\,kg} }[/tex]

[tex]\omega = 5.068\,\frac{rad}{s}[/tex]

The frequency and period of oscillations are, respectively:

[tex]f = \frac{5.068\,\frac{rad}{s} }{2\pi}[/tex]

[tex]f = 0.806\,hz[/tex]

[tex]T = \frac{1}{0.806\,hz}[/tex]

[tex]T = 1.241\,s[/tex]

Answer:

a). 675J

b). 337.5J

c). 1012.5J

Explanation:

M = 6.0kg

V = 15.0m/s

a). Translational energy

E = ½ *mv²

E = ½ * 6 * 15²

E = 675J

b). Rotational kinetic energy K.E(rot) = Iw²

But moment of inertia of a cylinder (I) = ½Mr²

I = ½mr²

V = wr, r = v / w

K.E(rot) = ¼ mv²

K.E(rot) = ¼* 6 * 15²

K.E(rot) = 337.5J

Total energy of the system = K.E(rot) + Translational energy = 337.5 + 675

T.E = 1012.5J

Answer:

Remember that Kinetic energy is a scalar quantity and it only depends on the speed and not necessary not the angle

Thus,Since the masses and the speed are same for both A and B, the initial kinetic energy of A and B are same.

b]

The difference or variation in gravitational potential energy is again a scalar quantity and so as long as the initial speed is same, the change in gravitational potential energy will also be the same [though they may not occur at the same horizontal position].

therefore, from the ground to the highest point of both A and B, both will have same potential energy.

Also The energy in part (a) differs from part (b),

In part (a) energy mention is kinetic energy that depends on mass and velocity of particle whereas in part (b) energy is potential energy that depends on mass and the position with reference of ground. Potential energy is a state function but kinetic energy is not.

Answer: they are the same

Explanation:

Answer:

2.837% less than actual value.

Explanation:

Based on given information let's calculate our value.

S = Vxt = 331m/s x 5s = 1655m, that is the total distance that sound would travel in 5 seconds.

1mile = 1609.34meters.

percentage error is.

[tex]\frac{actual-calculated}{actual} *100 = \frac{1609.34-1655}{1609.34} *100 = -2.83%[/tex]

negative indicates less than actual value.

The rule of thumb that states the lightning bolt was one mile away for every five seconds between seeing the flash and hearing the thunder is not very accurate. The actual distance the lightning bolt would be approximately 3.19 miles away, which results in a percent error of approximately 219%. The rule tends to underestimate the distance to the lightning bolt.

The "rule of thumb that many people follow" during a thunder and lightning storm is based on the fact that light travels much faster than sound. When you see a flash of lightning, the sound wave created by the thunder associated with the lightning bolt takes more time to reach the observer than the light from the flash.

According to the rule of thumb, we estimate one mile per five seconds because sound travels at a speed of approximately 331 meters per second. However, to calculate the actual distance in miles, you would multiply the time (in seconds) by the speed of sound and convert to miles (1 meter = 0.00062137 miles). This gives an actual distance of about 0.00063741 miles/second. Therefore, for every second delay between the lightning and the thunder, the lightning bolt would be about 0.00063741 miles away.

So, if we see a flash of lightning and hear the thunder 5 seconds later, according to the accurate calculation, the lightning bolt was just over 3.19 miles away (5 seconds * 0.00063741 miles/second). That's a sizeable difference from the one-mile estimate given by the rule of thumb. To find the percent error, we subtract the accepted value from the experimental value, divide by the accepted value, and multiply by 100. That gives us a percent error of approximately 219%. So the rule of thumb is not particularly accurate.

Learn more about Speed of Sound here:

https://brainly.com/question/35989321

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