What is EMF?

When are we exposed to EMF?

Does EMF increase or decrease with distance?

What are the primary differences in direct current and alternating current?

17.

There are three different methods for charging objects:

- Friction: in friction, two objects are rubbed against each other. As a result, electrons can be passed from one object to the other, so one object will gain a net negative charge while the other object will gain a net positive charge due to the lack of electrons.

- Conduction: this occurs when two conductive objects are put in contact with each other, and charges (electrons, usually) are transferred from one object to the other one.

- Induction: this occurs when two objects are brought closer to each other, but not in contact. If one of the two objects has a net charge (different from zero) on its surface, then it will induce a movement of charges in the second object: in particular, in the second object, charges of the opposite polarity will be attracted towards the first object, while charges of same polarity will be repelled further away.

18.

Charged objects produce around themselves an electric field. The strenght of the electric field is given by (assuming the charged objects are spherical)

[tex]E=k\frac{q}{r^2}[/tex]

where k is the Coulomb's constant, q is the magnitude of the charge and r the distance from the centre of the charge. As we see, the strength of the field is inversely proportional to the square of the distance.

Also, the direction of the field is determined by the sign of the charge:

- if the charge is positive, the electric field points away from the charge (this means that other positive charges in the field will be repelled away)

- if the charge is negative, the electric field points towards the charge (this means that other positive charges in the field will be attracted towards it)

19.

Electrical force is given by:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where k is the Coulomb's constant, q1 and q2 are the two charges, and r their separation.

Gravitational force is given by:

[tex]F=G\frac{m_1 m_2}{r^2}[/tex]

where G is the gravitational constant, m1 and m2 are the masses of the two objects, and r their separation.

Similarities between the two forces:

- Both are inversely proportional to the square of the distance between the two objects, r

- Both are non-contact forces (the two objects can experience the forces even if they are not in contact)

- Both forces have infinite range

Differencies between the two forces:

- The electric force can be either attractive or repulsive, while the gravitational force is attractive only

- The electric force is much stronger than the gravitational force, due to the much larger value of the Coulomb's constant k compared to the gravitational constant G

Answer:

4.2°

Explanation:

Snell’s law gives:-

n₁ sin θ₁ = n₂ sin θ₂

Given, θ₁ = 54.0°

For the refraction of violet light ( n = 1.66 ) from air to glass:

( 1.00 ) sin 54.0° = ( 1.66 ) sin θ₂₁

sin θ₂₁ = ( 1.00 / 1.66 ) sin 54.0° = 29.2

For the refraction of violet light from glass to air …  

n₂ sin θ₂ = n₃ sin θ₃₁

θ₂ = 60 -29.2 = 30.8

n₂ = 1.66

n₃ = 1

sin θ₃₁ = ( n₂ / n₃ ) sin θ₂ = n₂ sin θ₂₂ = ( 1.66 ) sin 30.8°  

θ₃₁ = sin ⁻ ¹ [ ( 1.66 ) sin 30.8° ] = 58.2°

For the refraction of red light ( n = 1.62 ) from air to glass

( 1.000 ) sin 54° = ( 1.62 ) sin θ₂

sin θ₂ = ( 1.000 / 1.62 ) sin 54°  

θ₂ = sin ⁻ ¹ [ ( 1.000 / 1.62 ) sin 54° ] = 29.95986° = 30.0°

For the refraction of red light from glass to air

n₂ sin θ₂ = n₃ sin θ₃₂

n₂ = 1.62 , θ₂ = 30° , n₃ = 1.000 …  

sin θ₃₂ = ( n₂ / n₃ ) sin θ₂ = n₂ sin θ₂ = ( 1.62 ) sin 30°

θ₃₂ = sin ⁻ ¹ [ ( 1.62 ) sin 30° ] = 54°

Angular spread =   γ = θ₃₁ - θ₃₂ = 4.2°

The angular spread in degrees of visible light passing through a prism of ap ex angle 60.0° if the angle of incidence is 54.0° is; 4.1°

For incoming rays, sin sin θ₂ = (sin θ₁)/n

Thus;

(θ₂)_violet = sin⁻¹ ((sin 54)/1.66)

(θ₂)_violet = 29.17°

(θ₂)_red = sin⁻¹ ((sin 54)/1.62)

(θ₂)_red = 29.96°

For the outgoing ray;

(90 - θ₂) + (90 - θ₃) + 60° = 180°

Also, θ₃ = 60 - θ₂ and sin θ₄ = n sin θ₃

(θ₄)_violet = sin⁻¹ (1.66 * sin 30.83))

(θ₄)_violet = 58.29°

θ₄)_red = sin⁻¹ (1.62 * sin 30.04))

(θ₄)_red = 54.19°

The angular dispersion is the difference and so;

Angular dispersion = (θ₄)_violet - (θ₄)_red

Angular dispersion = 58.29° - 54.19°

Angular Dispersion = 4.1°

Read more about angular dispersion at; https://brainly.com/question/13972198

Answer:

[tex]9.8\cdot 10^{-6}m[/tex]

Explanation:

For light passing through a single slit, the position of the nth-minimum from the central bright fringe in the diffraction pattern is given by

[tex]y=\frac{n \lambda D}{d}[/tex]

where

[tex]\lambda[/tex] is the wavelength

D is the distance of the screen from the slit

d is the width of the slit

In this problem, we have

[tex]\lambda=700 nm = 7.00\cdot 10^{-7}m[/tex] is the wavelength of the red light

D = 14 m is the distance of the screen from the doorway

d = 1.0 m is the width of the doorway

Substituting n=1 into the equation, we find the distance between the central bright fringe and the first-order dark fringe (the first minimum):

[tex]y=\frac{(1)(7.00\cdot 10^{-7} m)(14 m)}{1.0 m}=9.8\cdot 10^{-6}m[/tex]

(a) [tex]7.32\cdot 10^7 Hz[/tex]

The frequency of an electromagnetic waves is given by:

[tex]f=\frac{c}{\lambda}[/tex]

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

[tex]\lambda=4.1 m[/tex] is the wavelength of the wave in the problem

Substituting into the equation, we find

[tex]f=\frac{3.0\cdot 10^8 m/s}{4.1 m}=7.32\cdot 10^7 Hz[/tex]

(b) [tex]4.60\cdot 10^8 rad/s[/tex]

The angular frequency of a wave is given by

[tex]\omega = 2\pi f[/tex]

where

f is the frequency

For this wave,

[tex]f=7.32\cdot 10^7 Hz[/tex]

So the angular frequency is

[tex]\omega=2\pi(7.32\cdot 10^7 Hz)=4.60\cdot 10^8 rad/s[/tex]

(c) [tex]1.53 m^{-1}[/tex]

The angular wave number of a wave is given by

[tex]k=\frac{2\pi}{\lambda}[/tex]

where

[tex]\lambda[/tex] is the wavelength of the wave

For this wave, we have

[tex]\lambda=4.1 m[/tex]

so the angular wave number is

[tex]k=\frac{2\pi}{4.1 m}=1.53 m^{-1}[/tex]

(d) [tex]1.03\cdot 10^{-6}T[/tex]

For an electromagnetic wave,

[tex]E=cB[/tex]

where

E is the magnitude of the electric field component

c is the speed of light

B is the magnitude of the magnetic field component

For this wave,

E = 310 V/m

So we can re-arrange the equation to find B:

[tex]B=\frac{E}{c}=\frac{310 V/m}{3\cdot 10^8 m/s}=1.03\cdot 10^{-6}T[/tex]

(e) z-axis

In an electromagnetic wave, the electric field and the magnetic field oscillate perpendicular to each other, and they both oscillate perpendicular to the direction of propagation of the wave. Therefore, we have:

- direction of propagation of the wave --> positive x axis

- direction of oscillation of electric field --> y axis

- direction of oscillation of magnetic field --> perpendicular to both, so it must be z-axis

(f) 127.5 W/m^2

The time-averaged rate of energy flow of an electromagnetic wave is given by:

[tex]I=\frac{E^2}{2\mu_0 c}[/tex]

where we have

E = 310 V/m is the amplitude of the electric field

[tex]\mu_0[/tex] is the vacuum permeability

c is the speed of light

Substituting into the formula,

[tex]I=\frac{(310 V/m)^2}{2(4\pi\cdot 10^{-7} H/m) (3\cdot 10^8 m/s)}=127.5 W/m^2[/tex]

(g) [tex]1.53\cdot 10^{-8} kg m/s[/tex]

For a surface that totally absorbs the wave, the rate at which momentum is transferred to the surface given by

[tex]\frac{dp}{dt}=\frac{<S>A}{c}[/tex]

where the <S> is the magnitude of the Poynting vector, given by

[tex]<S>=\frac{EB}{\mu_0}=\frac{(310 V/m)(1.03\cdot 10^{-6} T)}{4\pi \cdot 10^{-7}H/m}=254.2 W/m^2[/tex]

and where the surface is

A = 1.8 m^2

Substituting, we find

[tex]\frac{dp}{dt}=\frac{(254.2 W/m^2)(1.8 m^2)}{3\cdot 10^8 m/s}=1.53\cdot 10^{-8} kg m/s[/tex]

(h) [tex]8.47\cdot 10^{-7} N/m^2[/tex]

For a surface that totally absorbs the wave, the radiation pressure is given by

[tex]p=\frac{<S>}{c}[/tex]

where we have

[tex]<S>=254.2 W/m^2[/tex]

[tex]c=3\cdot 10^8 m/s[/tex]

Substituting, we find

[tex]p=\frac{254.2 W/m^2}{3\cdot 10^8 m/s}=8.47\cdot 10^{-7} N/m^2[/tex]

Answer:

The correct answer is the third option: The kinetic energy of the water molecules decreases.

Explanation:

Temperature is, in depth, a statistical value; kind of an average of the particles movement in any physical system (such as a glass filled with water). Kinetic energy, for sure, is the energy resulting from movement (technically depending on mass and velocity of a system; in other words, the faster something moves, the greater its kinetic energy.

Since temperature is related to the total average random movement in a system, and so is the kinetic energy (related to movement through velocity), as the thermometer measures less temperature, that would mean that the particles (in this case: water particles) are moving slowly, so that: the slower something moves, the lower its kinetic energy.

In summary: temperature tells about how fast are moving and colliding the particles within a system, and since it is directly proportional to the amount of movement, it can be related (also directly proportional) to the kinectic energy.

Answer:

1.171

Explanation:

if n₁sinΘ₁=n₂sinΘ₂, then n₂=n₁sinΘ₁ / sinΘ₂;

[tex]n_2=\frac{1.5*sin45}{sin65}=\frac{1.5*0.707}{0.906} =1.1705[/tex]

Answer:

463.4 m/s

Explanation:

The escape velocity on the surface of a planet/asteroid is given by

[tex]v=\sqrt{\frac{2GM}{R}}[/tex] (1)

where

G is the gravitational constant

M is the mass of the planet/asteroid

R is the radius of the planet/asteroid

For the asteroid in this problem, we know

[tex]\rho=4.49\cdot 10^6 g/m^3[/tex] is the density

[tex]V=3.32\cdot 10^{12} m^3[/tex] is the volume

So we can find its mass:

[tex]M=\frac{\rho}{V}=(4.49\cdot 10^6 g/m^3)(3.32\cdot 10^{12}m^3)=1.49\cdot 10^{19} kg[/tex]

Also, the asteroid is approximately spherical, so its volume is given by

[tex]V=\frac{4}{3}\pi R^3[/tex]

where R is the radius. Solving the formula for R, we find its radius:

[tex]R=\sqrt[3]{\frac{3V}{4\pi}}=\sqrt[3]{\frac{3(3.32\cdot 10^{12}m^3)}{4\pi}}=9256 m[/tex]

So now we can use eq.(1) to find the escape velocity:

[tex]v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.49\cdot 10^{19}kg)}{9256 m}}=463.4 m/s[/tex]

Answer:

it must be possible to prove it wrong

Explanation:

Answer:

it must be possible to prove it wrong

Explanation:

Falsibiable is the word used to make the refernce that it is possible to prove something wrong or to prove that somthing is wrong with a statement ot hypothesis, since the hypothsesis is falsifiable it must be possible to prove it wrong, that is the correct answer.

(a) 3.33

For a capacitor with dielectric disconnected from the battery, the relationship between the voltage across the capacitor without the dielectric (V) and with the dielectric (V) is given by

[tex]V' = \frac{V}{k}[/tex]

where

k is the dielectric constant of the material

In this problem, we have

[tex]V' = 3.6 V[/tex]

[tex]V=12 V[/tex]

So we can re-arrange the formula to find the dielectric constant:

[tex]k=\frac{V}{V'}=\frac{12 V}{3.6 V}=3.33[/tex]

(b) The energy stored reduces by a factor 3.33

The energy stored in a capacitor is

[tex]U=\frac{1}{2}QV[/tex]

where

Q is the charge stored on the capacitor

V is the voltage across the capacitor

Here we can write the initial energy stored in the capacitor (without dielectric) as

[tex]U=\frac{1}{2}QV[/tex]

while after inserting the dielectric is

[tex]U'=\frac{1}{2}QV' = \frac{1}{2}Q\frac{V}{k}[/tex]

since Q, the charge, has not changed (the capacitor is disconnected, so the charge cannot flow away from the capacitor).

So the ratio between the two energies is

[tex]\frac{U'}{U}=\frac{\frac{1}{2}Q\frac{V}{k}}{\frac{1}{2}QV}=\frac{1}{k}[/tex]

which means

[tex]U' = \frac{U}{k}=\frac{U}{3.33}[/tex]

So, the energy stored has decreased by a factor 3.33.

(c) 5.5 V

Pulling the dielectric only partway so that it fills half of the space between the plates is equivalent to a system of 2 capacitors in parallel, each of them with area A/2 (where A is the original area of the plates of the capacitor), of which one of the two is filled with dielectric while the other one is not.

Calling

[tex]C=\frac{\epsilon_0 A}{d}[/tex] the initial capacitance of the capacitor without dielectric

The capacitance of the part of the capacitor of area A/2 without dielectric is

[tex]C_1 = \frac{\epsilon_0 \frac{A}{2}}{d}= \frac{C}{2}[/tex]

while the capacitance of the part of the capacitor with dielectric is

[tex]C_2 = \frac{k \epsilon_0 \frac{A}{2}}{d}= \frac{kC}{2}[/tex]

The two are in parallel, so their total capacitance is

[tex]C' = C_1 + C_2 = \frac{C}{2}+\frac{kC}{2}=(1+k)\frac{C}{2}=(1+3.33)\frac{C}{2}=2.17 C[/tex]

We also have that

[tex]V=\frac{Q}{C}=12 V[/tex] this is the initial voltage

So the final voltage will be

[tex]V' = \frac{Q}{C'}=\frac{Q}{2.17 C}=\frac{1}{2.17}V=\frac{12 V}{2.17}=5.5 V[/tex]

Final answer:

The dielectric constant of the material is 3.3 based on the voltage drop after insertion between the capacitor's plates from 12 V to 3.6 V. The insertion of the dielectric changes the stored energy to 9% of the original. With half the space filled by the dielectric, the voltmeter would read approximately 5.6 V.

Explanation:

The question concerns the concept of capacitors in physics, specifically the effects of inserting a dielectric material between the plates of a parallel-plate capacitor. Assuming that the battery has been disconnected and there is no charge loss during the process, let's address the parts of the question step by step.

(a) Dielectric constant of the material

The dielectric constant (K) of a material can be determined by the ratio of the initial voltage (Vo) to the voltage (V) after the dielectric has been inserted. Given that the initial voltage is 12 V and it drops to 3.6 V, the dielectric constant is K = Vo / V = 12 / 3.6 = 3.3. Therefore, the dielectric constant of this material is 3.3.

(b) Fraction of the stored energy changed

The stored energy in a capacitor is proportional to the square of the voltage across it. When the voltage drops due to the insertion of the dielectric, the energy stored changes by a factor of (V / Vo)2. Hence, the fraction of the energy change is (3.6 / 12)2 = 0.09, or 9% of the original stored energy.

(c) Voltmeter reading with half space filled by the dielectric

When the dielectric only fills half of the space between the plates, the effective dielectric constant becomes a weighted average of the dielectric and air (K=1). Assuming the dielectric constant of the inserted material is 3.3, and it fills half the space, the effective dielectric constant is (3.3 + 1) / 2 = 2.15. Thus, the new voltage is Vnew = Vo / Keff = 12 / 2.15 = 5.58 V. Therefore, the voltmeter will read approximately 5.6 V when the dielectric fills only half of the space between the plates.

Answer:

1440 W

Explanation:

First of all, we can find the total displacement of the sled, which is given by

[tex]d=\frac{1}{2}at^2[/tex]

where

a = 0.08 m/s^2 is the acceleration

t = 1 min = 60 s is the time

Substituting,

[tex]d=\frac{1}{2}(0.08 m/s^2)(60 s)^2=144 m[/tex]

Now we can find the wotk done on the sled, equal to the product between force and displacement:

[tex]W=Fd=(600 N)(144 m)=86,400 J[/tex]

And finally we can fidn the average power, which is the ratio between the work done and the time taken:

[tex]P=\frac{W}{t}=\frac{86,400 J}{60 s}=1440 W[/tex]

Answer:

The answer would be D 56,320

Explanation:

A. Lithium

The equation for the photoelectric effect is:

[tex]E=\phi + K[/tex]

where

[tex]E=\frac{hc}{\lambda}[/tex] is the energy of the incident light, with h being the Planck constant, c being the speed of light, and [tex]\lambda[/tex] being the wavelength

[tex]\phi[/tex] is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

[tex]\lambda=190 nm=1.9\cdot 10^{-7}m[/tex], so the energy of the incident light is

[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J[/tex]

Converting in electronvolts,

[tex]E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV[/tex]

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

[tex]\phi = E-K=6.5 eV-4.0 eV=2.5 eV[/tex]

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

[tex]\lambda=510 nm=5.1\cdot 10^{-7} m[/tex]

So its energy is

[tex]E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J[/tex]

Converting in electronvolts,

[tex]E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV[/tex]

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: [tex]\phi = 4.5 eV[/tex]

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

[tex]E=\phi+K=4.5 eV+2.7 eV=7.2 eV[/tex]

Then the copper is replaced with sodium, which has work function of

[tex]\phi = 2.3 eV[/tex]

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

[tex]K=E-\phi = 7.2 eV-2.3 eV=4.9 eV[/tex]

The surface is most likely to be made of Lithium. Four metals (Cesium, Potassium, Sodium, and Lithium) will eject electrons when a green laser is shone on them. The maximum kinetic energy of the electrons emitted from Sodium will be approximately 4.9 eV.

Part A: Let's calculate the energy of the incident photon using Planck's equation E = h * c / λ , where h is Planck's constant (6.626 * 10^-34 J.s), c is the speed of light (3 * 10^8 m/s) and λ is the wavelength of the incident light (convert 190 nm to 1.9 * 10^-7 m). The energy of the photon (Ep) is approximately 6.6 eV. The given kinetic energy (Ek) of the most energetic electrons is 4.0 eV. According to the photoelectric effect, Ep = Work Function + Ek. So, the work function of the metal surface (Wf) is Ep - Ek, which gives us a work function of 2.6 eV. The closest value to this in the table is that of Lithium with 2.5 eV, so the surface is likely made of Lithium.

Part B: To find out the number of metals that will eject electrons when a green laser of wavelength 510 nm is shone on them, we first need to convert this wavelength into energy using the same formula as in Part A. The energy hence calculated is approximately 2.4 eV. As per the photoelectric effect, a metal can only eject electrons when its work function is less than or equal to the energy of the incident light. Therefore, only metals with work functions less than or equal to 2.4 eV will eject electrons. Looking at the table, we see that this covers Cesium, Potassium, Sodium, and Lithium. So, 4 metals will eject electrons.

Part C: Given the kinetic energy of electrons emitted from copper is 2.7 eV, and the work function of copper is 4.5 eV. According to the photoelectric effect, the energy of the incident radiation should be the sum of these two, which yields 7.2 eV. Now, if we replace the copper sample with sodium (having a work function of 2.3 eV), then the maximum kinetic energy of the emitted electrons from the sodium sample (Ek) will be the energy of the incident light minus the work function of Sodium i.e., 7.2 eV - 2.3 eV which is approximately 4.9 eV.

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(a) [tex]4.29\cdot 10^{14}Hz[/tex]

The frequency of an electromagnetic wave is given by

[tex]f=\frac{c}{\lambda}[/tex]

where

c is the speed of light

[tex]\lambda[/tex] is the wavelength

We notice from the formula that the frequency is inversely proportional to the wavelength, so the minimum frequency corresponds to the maximum wavelength, and viceversa

The maximum value of the wavelength of the visible light waves is

[tex]\lambda_{max} = 700 nm = 7.0\cdot 10^{-7} m[/tex] (red light)

so the minimum frequency of visible light is

[tex]f_{min} = \frac{c}{\lambda_{max}}=\frac{3.0\cdot 10^8 m/s}{7.00\cdot 10^{-7}m}=4.29\cdot 10^{14}Hz[/tex]

(b) [tex]7.50\cdot 10^{14}Hz[/tex]

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

[tex]\lambda_{min} = 400 nm = 4.0\cdot 10^{-7} m[/tex] (violet light)

so the maximum frequency of visible light is

[tex]f_{max} = \frac{c}{\lambda_{min}}=\frac{3.0\cdot 10^8 m/s}{4.00\cdot 10^{-7}m}=7.50\cdot 10^{14}Hz[/tex]

(c) 1 m

The wavelength of an electromagnetic wave is given by

[tex]\lambda=\frac{c}{f}[/tex]

as before, we notice that the minimum wavelength corresponds to the maximum frequency, and viceversa.

The maximum frequency of shortwave radio waves is

[tex]f_{max}=300 MHz = 3.0\cdot 10^8 Hz[/tex]

so the minimum wavelength of these waves is

[tex]\lambda_{min} = \frac{c}{f_{max}}=\frac{3.0\cdot 10^8 m/s}{3.0\cdot 10^8 Hz}=1 m[/tex]

(d) 200 m

The minimum frequency of shortwave radio waves is

[tex]f_{min}=1.5 MHz = 1.5\cdot 10^6 Hz[/tex]

so the maximum wavelength of these waves is

[tex]\lambda_{max} = \frac{c}{f_{min}}=\frac{3.0\cdot 10^8 m/s}{1.5\cdot 10^6 Hz}=200 m[/tex]

(e) [tex]6.0\cdot 10^{16}Hz[/tex]

As in part (a) and (b), we can find the frequency of the X-rays by using the formula

[tex]f=\frac{c}{\lambda}[/tex]

The maximum wavelength of the x-rays is

[tex]\lambda_{max} = 5.0 nm = 5.0\cdot 10^{-9} m[/tex]

so the minimum frequency is

[tex]f_{min} = \frac{c}{\lambda_{max}}=\frac{3.0\cdot 10^8 m/s}{5.0\cdot 10^{-9}m}=6.0\cdot 10^{16}Hz[/tex]

(f) [tex]3.0\cdot 10^{19}Hz[/tex]

The maximum frequency corresponds to the minimum wavelength;

The minimum wavelength is

[tex]\lambda_{min} = 1.0\cdot 10^{-2} nm = 1.0\cdot 10^{-11} m[/tex]

so the maximum frequency of the x-rays is

[tex]f_{max} = \frac{c}{\lambda_{min}}=\frac{3.0\cdot 10^8 m/s}{1.0\cdot 10^{-11}m}=3.0\cdot 10^{19}Hz[/tex]

Final answer:

The frequency of electromagnetic waves varies inversely with the wavelength. For visible light, the frequencies range between 4.29 x 10⁹⁴ Hz and 7.5 x 10⁹⁴ Hz; for shortwave radio, the wavelengths range from 1 to 200 meters; for x-rays, the frequencies vary between 6 x 10⁹⁶ Hz and 3 x 10⁹⁹ Hz.

Explanation:

Calculating Frequency and Wavelength of Electromagnetic Waves

To calculate the frequency (ν) and wavelength (λ) of electromagnetic waves, we use the relation c = λν, where c is the speed of light in a vacuum (approximately 3.0 x 10⁸ m/s).

(a) The minimum frequency of visible light (700 nm) is obtained by c/λ, yielding approximately 4.29 x 10¹⁴ Hz.
(b) The maximum frequency for visible light (400 nm) is reached with c/λ, giving around 7.5 x 10¹⁴ Hz.

(c) The minimum wavelength for shortwave radio at 300 MHz is found by c/ν, resulting in 1 meter (m).
(d) The maximum wavelength at 1.5 MHz of shortwave radio is calculated with c/ν, equating to 200 meters (m).

(e) The minimum frequency of x-rays (1.0 x 10⁻² nm) is computed using c/λ, and is approximately 3 x 10¹⁹ Hz.
(f) The maximum frequency for x-rays (5.0 nm) is also obtained with c/λ, yielding around 6 x 10¹⁶ Hz.

A) 1153 N/m

We can find the spring constant by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force applied to the spring

k is the spring constant

x is the displacement of the spring

In this problem, a fish of mass m = 4.0 kg is hanging on the spring, so the force applied is the weight of the fish:

[tex]F=mg=(4.0 kg)(9.8 m/s^2)=39.2 N[/tex]

and the displacement of the spring is:

[tex]x = 13.4 cm - 10.0 cm = 3.4 cm = 0.034 m[/tex]

so, the spring constant is

[tex]k=\frac{F}{x}=\frac{39.2 N}{0.034 m}=1153 N/m[/tex]

B) 16.8 cm

In this case, a fish of mass

m = 8.0 kg

is hanging on the spring. Therefore, the force applied to the spring is

[tex]F=mg=(8.0 kg)(9.8 m/s^2)=78.4 N[/tex]

So we can find the displacement of the spring:

[tex]x=\frac{F}{k}=\frac{78.4 N}{1153 N/m}=0.068 m = 6.8 cm[/tex]

And since the equilibrium length of the spring is

[tex]x_0 = 10.0 cm[/tex]

the new length of the spring will be

[tex]x' = 10.0 cm + 6.8 cm = 16.8 cm[/tex]

The spring constant of the spring is 1,152.94 N/m.

The new length of the spring when 8kg fish is suspended on it is 16.8 cm.

The given parameters;

The extension of the spring is calculated as follows;

x = L₂ - L₁

x = 13.4 cm - 10.0 cm

x = 3.4 cm

The spring constant of the spring is calculated as follows;

F = kx

mg = kx

[tex]k = \frac{mg}{x} \\\\k = \frac{4 \times 9.8}{0.034} \\\\k = 1,152.94 \ N/m[/tex]

The new extension of the spring when 8 kg fish is suspended on it;

[tex]x = \frac{mg}{k} \\\\x = \frac{8 \times 9.8}{1152.94} \\\\ x= 0.068 \ m\\\\x = 6.8 \ cm[/tex]

The new length of the spring = 6.8 cm + 10 cm = 16.8 cm

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The answer is:

The answer is the first option, 100 ft-lbs.

To calculate the work done by an object over a distance, we need to multiply the applied force by the amount of movement (distance).

So, to calculate the work that Gary needs to do to move the chair, we need to use the following equation:

[tex]Work=Force*Distance[/tex]

We are given,

[tex]Force=20lb\\Distance=5ft[/tex]

[tex]Work=F*D\\Work=(20lb)*(5ft)=100ft-lbs[/tex]

Hence, the answer is the first option, 100 ft-lbs.

Have a nice day!

answer- 100 ft-lbs

Hope I helped

Answer:

50 Hz

Explanation:

The frequency of a wave is given by

[tex]f=\frac{v}{\lambda}[/tex]

where

v is the speed of the wave

[tex]\lambda[/tex] is the wavelength

In this problem,

v = 60 m/s

[tex]\lambda=1.2 m[/tex]

So the frequency is

[tex]f=\frac{60 m/s}{1.2 m}=50 Hz[/tex]

Answer:

50 hertz

Explanation:

(a) -4667 N

First of all, we can calculate the acceleration of the arm and the glove, using the following equation:

[tex]v^2 - u^2 = 2ad[/tex]

where

v = 0 is the final speed

u = 10 m/s is the initial speed

a is the acceleration

d = 7.50 cm = 0.075 m is the distance through which the arm and the glove move before coming to a stop

Solving for a,

[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.075 m)}=-666.7 m/s^2[/tex]

And since the know the mass of the arm+glove:

m = 7.00 kg

We can now calculate the force exerted:

[tex]F=ma=(7.00 kg)(-666.7 N)=-4,667 N[/tex]

(b) -17500 N

We can repeat the problem, but this time the stopping distance is different:

d = 2.00 cm = 0.02 m

So the acceleration is

[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(10.0 m/s)^2}{2(0.02 m)}=-2500 m/s^2[/tex]

and so the force is

[tex]F=ma=(7.00 kg)(-2500 m/s^2)=-17,500 N[/tex]

(c) Yes

The force exerted when the glove is used is

F = 4667 N

We see that this force corresponds approximately to the weight of an object of mass m=476 kg, in fact:

[tex]W=mg=(476 kg)(9.81 m/s^2)=4670 N[/tex]

Which is quite a lot. Therefore, the force even when gloves are used seems enough to cause damage.

To calculate the force exerted by a boxing glove on an opponent's face, we can use the equation for impulse.

In order to calculate the force exerted by a boxing glove on an opponent's face, we need to use the equation for impulse:

Impulse = change in momentum = force x time

We know the initial speed of the arm and glove (10.0 m/s), the final speed (0 m/s), and the time it takes for the glove and face to compress (which can be calculated using the distance and the initial speed).

By substituting these values into the equation, we can find the average force exerted by the boxing glove on the opponent's face.

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Answer:

Binding energy

Explanation:

The energy required to bind the nucleus is called binding energy.

Binding energy is equal to the product of mass defect and square of velocity of light.

Mass defect is the difference of mass of neucleons and the mass of atom.

Answer:

Binding Energy

Explanation:

Answer:

[tex]2.84\cdot 10^{-8} m[/tex]

Explanation:

Due to the law of conservation of energy, the energy of the emitted X-ray photon is equal to the energy lost by the electron.

The initial kinetic energy of the electron is:

[tex]K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(9.11\cdot 10^{-31}kg)(5.90\cdot 10^6 m/s)^2=1.59\cdot 10^{-17}J[/tex]

The electrons decelerates to 3/4 of its speed, so the new speed is

[tex]v_f = \frac{3}{4}v_i = \frac{3}{4}(5.90\cdot 10^6 m/s)=4.425\cdot 10^6 m/s[/tex]

So the final kinetic energy is

[tex]K_f = \frac{1}{2}mv_f^2=\frac{1}{2}(9.11 \cdot 10^{-31} kg)(4.425\cdot 10^6 m/s)^2=8.9\cdot 10^{-18} J[/tex]

So, the energy lost by the electron, which is equal to the energy of the emitted photon, is

[tex]E=K_i - K_f =1.59\cdot 10^{-17} J-8.9\cdot 10^{-18} J=7\cdot 10^{-18} J[/tex]

The wavelength of the photon is related to its energy by

[tex]\lambda=\frac{hc}{E}[/tex]

where h is the Planck constant and c the speed of light. Substituting E, we find

[tex]\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{7\cdot 10^{-18} J}=2.84\cdot 10^{-8} m[/tex]

To determine the wavelength of the photon emitted when an electron decelerates in an X-ray tube, we can use Planck's equation and the information given about the electron's speed.

When an electron decelerates, it emits electromagnetic waves known as X-rays. The wavelength of the X-ray photon can be determined using Planck's equation, E = hv, where E is energy, h is Planck's constant, and v is frequency. Given that the electron's velocity decreases to three-quarters of its original speed and the initial speed is 5.90x10^6 m/s, we can calculate the velocity of the electron after deceleration. Using this value, we can then calculate the wavelength of the photon emitted. Therefore, using the provided information, we can find the wavelength of the photon.

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Answer:

radio waves, microwaves, infrared, visible light, ultraviolet light, X-rays, gamma rays

Explanation:

Electromagnetic waves are oscillations of electric and magnetic fields in a direction perpendicular to the direction of motion of the wave (transverse waves). They are classified into 7 different types, according to their frequencies.

From lowest to highest frequency, we have:

Radio waves [tex]<10^9 Hz[/tex]

Microwaves [tex]10^9 Hz - 4\cdot 10^{13}Hz[/tex]

Infrared [tex]4\cdot 10^{13} - 4\cdot 10^{14} Hz[/tex]

Visible light [tex]4\cdot 10^{14} - 8\cdot 10^{14}Hz[/tex]

Ultraviolet [tex]8\cdot 10^{14} - 2.4\cdot 10^{16} Hz[/tex]

X-rays [tex]2.4\cdot 10^{16} -5 \cdot 10^{19}Hz[/tex]

Gamma rays [tex]>5\cdot 10^{19} Hz[/tex]

(a) 0.0026 Hz

The relationship between frequency and wavelength for an electromagnetic wave is

[tex]f=\frac{c}{\lambda}[/tex] (1)

where

[tex]c=3.0\cdot 10^8 m/s[/tex] is the speed of light

f is the frequency

[tex]\lambda[/tex] is the wavelength

For the wave in the problem, the wavelength is [tex]1.8\cdot 10^4[/tex] Earth radii. The Earth radius is

[tex]R=6370 km = 6.37\cdot 10^6 m[/tex]

so the wavelength would be

[tex]\lambda = (1.8\cdot 10^4 )(6.37\cdot 10^6 m)=1.14\cdot 10^{11}m[/tex]

So by using eq.(1) we find the frequency:

[tex]f=\frac{3\cdot 10^8 m/s}{1.14\cdot 10^{11}m}=0.0026 Hz[/tex]

(b) 384.6 s

The period of a wave is given by:

[tex]T=\frac{1}{f}[/tex]

where

T is the period

f is the frequency

For the wave in the problem,

f = 0.0026 Hz

so the period is

[tex]T=\frac{1}{0.0026 Hz}=384.6 s[/tex]

To find the frequency and period of the radiation emitted by Project Seafarer, use the formula: Frequency = Speed of Light / Wavelength and Period = 1 / Frequency. Convert the effective wavelength given to kilometers and then use the formulas to calculate the frequency and period.

To find the frequency and period of the radiation emitted by the Project Seafarer antenna, we need to use the formula:

Frequency = Speed of Light / Wavelength

Period = 1 / Frequency

The effective wavelength given in the question is 1.8 × 104 Earth radii. We can convert this to kilometers by multiplying it by the Earth's radius of 6370 km. So the wavelength is 1.8 × 104 * 6370 = 1.1486 × 108 km.

Now we can calculate the frequency using the formula: Frequency = 3 × 108 m/s / 1.1486 × 108 km.

Finally, we can calculate the period by taking the reciprocal of the frequency.

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Answer:

general internet site

Explanation:

Answer:

General internet site

Explanation:

The TIL will support your answer I believe if you just think about of the provide the inslutoin which that mean money is gong fast to the enconmomy

Answer:

0.05 V/m

Explanation:

For a uniform electric field, electric field strength and potential difference are related by

[tex]E=\frac{\Delta V}{d}[/tex]

where

E is the electric field strength

[tex]\Delta V[/tex] is the potential difference

d is the distance between the two points

Here we have

[tex]\Delta V= 0.50 mV=5\cdot 10^{-4}V[/tex]

[tex]d=1.0 cm=0.01 m[/tex]

So, the electric field strength is

[tex]E=\frac{5\cdot 10^{-4} V}{0.01 m}=0.05 V/m[/tex]

Answer:

it is a.health record documentation

Explanation:hope this helps

Answer: Reimbursement and research

Explanation: I got it right on the PF exam

(c) the fraction of energy lost during the impact

I really don’t know maybe energy from the poles just attaches and the soil comes and energy lifts the matter of soil and the magnet follows and lifts and goes up to the soil and the poles attach the magnet to the soil and the energy lifts the magnet and the soil follows and it is attracted to the magnet

The North Pole of the bar magnet lines up magnetic domains in the paper clip, in such a fashion that the domain’s South Pole are on the closer side. In the same way, the South Pole induces North Pole in the paper clip. This causes the attraction

Answer:

The Sun Crosses the Equator. on March 19, 20, or 21 every year. The September equinox occurs the moment the Sun crosses the celestial equator, this happens either on September 22, 23, or 24 every year.

Explanation:

The Sun Crosses the Equator. on March 19, 20, or 21 every year. The September equinox occurs the moment the Sun crosses the celestial equator, this happens either on September 22, 23, or 24 every year.

Answer:

The sun is directly above the equator on the equinoxes, the spring equinox being on March 20, and the fall equinox being on September 22.

Explanation:

(a) 80 N/m

The spring constant can be found by using Hooke's law:

[tex]F=kx[/tex]

where

F is the force on the spring

k is the spring constant

x is the displacement of the spring relative to the equilibrium position

At the beginning, we have

F = 16.0 N is the force applied

x = 0.200 m is the displacement from the equilibrium position

Solving the formula for k, we find

[tex]k=\frac{F}{m}=\frac{16.0 N}{0.200 m}=80 N/m[/tex]

(b) 0.84 Hz

The frequency of oscillation of the system is given by

[tex]f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}[/tex]

where

k = 80 N/m is the spring constant

m = 2.90 kg is the mass attached to the spring

Substituting the numbers into the formula, we find

[tex]f=\frac{1}{2\pi}\sqrt{\frac{80 N/m}{2.90 kg}}=0.84 Hz[/tex]

(c) 1.05 m/s

The maximum speed of a spring-mass system is given by

[tex]v=\omega A[/tex]

where

[tex]\omega[/tex] is the angular frequency

A is the amplitude of the motion

For this system, we have

[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]

[tex]A=0.200 m[/tex] (the amplitude corresponds to the maximum displacement, so it is equal to the initial displacement)

Substituting into the formula, we find the maximum speed:

[tex]v=(5.25 rad/s)(0.200 m)=1.05 m/s[/tex]

(d) x = 0

The maximum speed in a simple harmonic motion occurs at the equilibrium position. In fact, the total mechanical energy of the system is equal to the sum of the elastic potential energy (U) and the kinetic energy (K):

[tex]E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2[/tex]

where

k is the spring constant

x is the displacement

m is the mass

v is the speed

The mechanical energy E is constant: this means that when U increases, K decreases, and viceversa. Therefore, the maximum kinetic energy (and so the maximum speed) will occur when the elastic potential energy is minimum (zero), and this occurs when x=0.

(e) 5.51 m/s^2

In a simple harmonic motion, the maximum acceleration is given by

[tex]a=\omega^2 A[/tex]

Using the numbers we calculated in part c):

[tex]\omega=2\pi f=2\pi (0.84 Hz)=5.25 rad/s[/tex]

[tex]A=0.200 m[/tex]

we find immediately the maximum acceleration:

[tex]a=(5.25 rad/s)^2(0.200 m)=5.51 m/s^2[/tex]

(f) At the position of maximum displacement: [tex]x=\pm 0.200 m[/tex]

According to Newton's second law, the acceleration is directly proportional to the force on the mass:

[tex]a=\frac{F}{m}[/tex]

this means that the acceleration will be maximum when the force is maximum.

However, the force is given by Hooke's law:

[tex]F=kx[/tex]

so, the force is maximum when the displacement x is maximum: so, the maximum acceleration occurs at the position of maximum displacement.

(g) 1.60 J

The total mechanical energy of the system can be found by calculating the kinetic energy of the system at the equilibrium position, where x=0 and so the elastic potential energy U is zero. So we have

[tex]E=K=\frac{1}{2}mv_{max}^2[/tex]

where

m = 2.90 kg is the mass

[tex]v_{max}=1.05 m/s[/tex] is the maximum speed

Solving for E, we find

[tex]E=\frac{1}{2}(2.90 kg)(1.05 m/s)^2=1.60 J[/tex]

(h) 0.99 m/s

When the position is equal to 1/3 of the maximum displacement, we have

[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]

so the elastic potential energy is

[tex]U=\frac{1}{2}kx^2=\frac{1}{2}(80 N/m)(0.0667 m)^2=0.18 J[/tex]

and since the total energy E = 1.60 J is conserved, the kinetic energy is

[tex]K=E-U=1.60 J-0.18 J=1.42 J[/tex]

And from the relationship between kinetic energy and speed, we can find the speed of the system:

[tex]v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(1.42 J)}{2.90 kg}}=0.99 m/s[/tex]

(i) 1.84 m/s^2

When the position is equal to 1/3 of the maximum displacement, we have

[tex]x=\frac{1}{3}(0.200 m)=0.0667 m[/tex]

So the restoring force exerted by the spring on the mass is

[tex]F=kx=(80 N/m)(0.0667 m)=5.34 N[/tex]

And so, we can calculate the acceleration by using Newton's second law:

[tex]a=\frac{F}{m}=\frac{5.34 N}{2.90 kg}=1.84 m/s^2[/tex]

The spring constant of the dense object will be 80N/m.

The spring constant will be calculated thus:

k = f/m = 16/0.2 = 80N/m.

The frequency oscillation will be:

= 1/2π × ✓k/✓m

= 1/2π × ✓80/✓2.9

= 0.84 Hz

The maximum speed of the spring mass system will be:

v = [2π(0.84)] × 0.2

= 1.05m/s

The position on the x-axis where the maximum speed occur is at x = 0.

The maximum acceleration will be:

a = w²A

a = 5.25² × 0.2

= 5.51m/s²

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(a) 9.8 m/s^2, downward

There is only one force acting on the ball while it is in flight: the force of gravity, which is

F = mg

where

m is the mass of the ball

g is the gravitational acceleration

According to Newton's second law, the force acting on the ball is equal to the product between the mass of the ball and its acceleration, so

F = mg = ma

which means

a = g

So, the acceleration of the ball during the whole flight is equal to the acceleration of gravity:

g = -9.8 m/s^2

where the negative sign means the direction is downward.

(b) v = 0

Any object thrown upward reaches its maximum height when its velocity is zero:

v = 0

In fact, at that moment, the object's velocity is turning from upward to downward: that means that at that instant, the velocity must be zero.

(c) 8.72 m/s, upward

The initial velocity of the ball can be found by using the equation:

v = u + at

Where

v = 0 is the velocity at the maximum height

u is the initial velocity

a = g = -9.8 m/s^2 is the acceleration

t is the time at which the ball reaches the maximum height: this is half of the time it takes for the ball to reach again the starting point of the motion, so

[tex]t=\frac{1.77 s}{2}=0.89 s[/tex]

So we can now solve the equation for u, and we find:

[tex]u=v-at=0-(-9.8 m/s^2)(0.89 s)=8.72 m/s[/tex]

(d) 3.88 m

The maximum height reached by the ball can be found by using the equation:

[tex]v^2 - u^2 = 2ad[/tex]

where

v = 0 is the velocity at the maximum height

u = 8.72 m/s is the initial velocity

a = g = -9.8 m/s^2 is the gravitational acceleration

d is the maximum height reached

Solving the equation for d, we find

[tex]d=\frac{v^2-u^2}{2a}=\frac{0^2-(8.72 m/s)^2}{2(-9.8 m/s^2)}=3.88 m[/tex]