What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth

Answers

Answer 1

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

g' = 1.09 m/s²

Answer 2

Answer:

The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]

Explanation:

Surface of earth,

[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]

free fall acceleration at height h,

[tex]g_h = \frac{GM}{(R+h)^2}[/tex]

where

G = gravitational constant

R = Radius of earth

M = mass of earth

therefore,

[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]

Where height h = 2R

[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]

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Related Questions

ametal of mass 0.6kg is heated by an electric heater connected to 15v batter when the ammeter reading is 3A its tempeeature rises feom 20c to 85c in 10 minutes calculate the s.h.c of metal cylinder​

Answers

Answer:

692 J/kg/°C

Explanation:

Electric energy added = amount of heat

Power × time = mass × SHC × increase in temperature

Pt = mCΔT

(15 V × 3 A) (10 min × 60 s/min) = (0.6 kg) C (85°C − 20°C)

C = 692 J/kg/°C

The sun generates both mechanical and electromagnetic waves. Which statement about those waves is true?
OA. The mechanical waves reach Earth, while the electromagnetic waves do not.
OB. The electromagnetic waves reach Earth, while the mechanical waves do not.
OC. Both the mechanical waves and the electromagnetic waves reach Earth.
OD. Neither the mechanical waves nor the electromagnetic waves reach Earth.

Answers

Answer: The correct answer for this question is letter (B) The electromagnetic waves reach Earth, while the mechanical waves do not. The sun generates both mechanical and electromagnetic waves. Space, between the sun and the earth is a nearly vacuum. So mechanical wave can not spread out in the vacuum.

Hope this helps!

Answer:

The electromagnetic waves reach Earth, while the mechanical waves do not

A major strike-slip earthquake on the San Andreas fault in California will cause a catastrophic tsunami affecting residents of San Francisco.
a) true
b) false

Answers

Answer:

a) true

Explanation:

The san andres is a transform fault that forms boundary between the Pacific and the North Atlantic plate and this slip strike is characterized by the latex motion the fault runs in the length of the California state. This plate is widely estimated for the high magnitude of earthquakes and varies from 7.7 to 8.3 magnitude. They are capable of producing a deadly tsunami that can devastate the pacific northwest.

The ability of a water strider to move along the surface of water without breaking the surface is due to:

Answers

Answer:

The ability of a water strider to move along the surface of water without breaking the surface is due to:

SURFACE TENSION

Explanation:

this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface

One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.

Answers

Answer:

ΔP.E = 6.48 x 10⁸ J

Explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

ΔP.E = 6.48 x 10⁸ J

A parallel plate air-filled capacitor is being charged.The circular plates have a radius of 4.00cm, and at a particular instant the conduction current in the wiresis 0.280A.


a. What is the displacement current density (jD) in theair space between the plates?

b. What is the rate at which the electric field between the platesis changing?

c. What is the induced magnetic field between the plates at adistance of 2.00 cm from the axis?

d. What is the induced magnetic field between the plates at adistance of 1.00 cm form the axis?

Answers

Answer:

Given that

capacitor being charged by a current i c has a displacementcurrent equal to i c between the plates∴

displacement current iD =i c=0.280 Aradius of the circular plate r = 4 cm=0.04 m

( A ) . displacement current density j D = iD / ( π r 2 )=0.28 / ( 3.14 * 0.04 2 )=55.73 A / m 2

( B ) . displacement current density j

D = ε( dE / dt )the rate at which the electric field between the plates is changing(

dE / dt ) = jD/ εdE/ dt ) = 55.73/ 8.85 * 10 -12=6.3*10 12 N / C - s( C ) . the induced magnetic field between theplates B = ( μ * r / 2π R 2) * i c ----( 1 )whereR= 2 cm=0.02 mr= 4 cm=0.04 mμ=permeability of free space=4π* 10 -7 H( D )substitute R = 1 cm = 0.01 m inequation ( 1 ),wegetanswer

In a certain process a gas ends in its original thermodynamic state. Of the following, which is possible as the net result of the process?
A. It is adiabatic and the gas does 50 J of work.
B. The gas does no work but absorbs so J of energy as heat.
C. The gas does no work but rejects 50 J of energy as heat.
D. The gas rejects 50 J of energy as heat and does 501 of work.
E. The gas absorbs 50 of energy as heat and does 50」ot work.

Answers

Answer:

E. The gas absorbs 50 of energy as heat and does 50」ot work

Explanation:

This is following the law of thermodynamics that energy is neither created nor destroyed

g suppose he used an alpha particle with an energy of 8.3 MeV, what would be the speed of this alpha particle

Answers

Answer:

speed of the alpha particle is 2 x 10^7 m/s.

Explanation:

energy of alpha particle = 8.3 Mev

1 Mev = 1.602 x 10^-13 J

8.3 Mev = [tex]x[/tex]

solving, [tex]x[/tex] = 8.3 x 1.602 x 10^-13 = 1.329 x 10^-12 J

mass of a alpha particle = 6.645 x 10^−27 kg

The energy of the alpha particle is the kinetic energy KE of the alpha particle

KE = [tex]\frac{1}{2}mv^{2}[/tex]

where m is the mass of the alpha particle

v is  the velocity of the alpha particle

substituting values, we have

1.329 x 10^-12 = [tex]\frac{1}{2}*6.645*10^{-27}*v^{2}[/tex]

[tex]v^{2}[/tex] = 4 x 10^14

[tex]v = \sqrt{4*10^{14} }[/tex] = 2 x 10^7 m/s

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye. Part A If you hold the book 28.0 cmcm from your eye, what is your nearpoint distance

Answers

Answer:

The near point is  [tex]n =44.8 \ cm[/tex]

Explanation:

From the question we are told that

   The power is  [tex]P = 1.50[/tex]

   The  distance from the eye is  [tex]k = 1.8 \ cm[/tex]

    The distance of the book from the eye is [tex]z = -28 \ cm[/tex]

Generally the focal length of the glasses is  

       [tex]f = \frac{1}{P}[/tex]

=>   [tex]f = \frac{1}{1.50 }[/tex]

=>   [tex]f = 0.667 \ m[/tex]

=>   [tex]f = 66.7 \ cm[/tex]

The object distance is evaluated as

     [tex]u = z + k[/tex]

=>   [tex]u = -28 + 1.8[/tex]

=>  [tex]u = -26.2 \ cm[/tex]

The image distance is evaluated from lens formula as

       [tex]\frac{1}{v} = \frac{1}{f} + \frac{1}{u}[/tex]

=>   [tex]\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}[/tex]

=>   [tex]v=- \frac{1}{0.0232}[/tex]

=>    [tex]v=- 43 \ cm[/tex]

The  near point is evaluated as

      [tex]n = -v + k[/tex]

=>    [tex]n =-(-43) + 1.8[/tex]

=>    [tex]n =44.8 \ cm[/tex]

A 1300-turn coil of wire 2.40 cm in diameter is in a magnetic field that increases from 0 T to 0.120 T in 9.00 ms . The axis of the coil is parallel to the field. What is the emf of the coil?

Answers

Answer:

The induced emf in the coil is 7.843 V

Explanation:

Given;

number of turns of the coil, N = 1300 turn

diameter of the coil, d = 2.4 cm = 0.024 m

initial magnetic field, B₁ = 0 T

final magnetic field, B₂ = 0.12 T

change in time, dt = 9.0 ms = 9 x 10⁻³ s

Area of the coil is given by;

A = πr²

radius of the coil, r = 0.024 / 2

radius of the coil, r = 0.012 m

A = π(0.012)²

A = 4.525 x 10⁻⁴ m²

The induced emf in the coil is given by;

E = NA(dB/dt)

E =  NA [(B₂ - B₁) /dt]

E = 1300 x 4.525 x 10⁻⁴ (0.12 - 0) / (9 x 10⁻³)

E = 7.843 V

Therefore, the induced emf in the coil is 7.843 V

A car moving at 36 m/s passes a stationary police car whose siren has a frequency of 500 Hz. What is the change in the frequency (in Hz) heard by an observer in the moving car as he passes the police car? (The speed of sound in air is 343 m/s.)

Answers

Answer:

Change in the frequency (in Hz) = 104.96 Hz

Explanation:

Given:

Speed of sound in air (v) = 343 m/s

Speed of car (v1) 36 m/s

Frequency(f) = 500 Hz

Find:

Change in the frequency (in Hz)

Computation:

Frequency hear by the observer(before)(f1) = [f(v+v1)] / v

Frequency hear by the observer(f1) = [500(343+36)] / 343

Frequency hear by the observer(f1) = 552.48 Hz

Frequency hear by the observer(after)(f2) = [f(v-v1)] / v

Frequency hear by the observer(f2) = [500(343-36)] / 343

Frequency hear by the observer(f2) = 447.52 Hz

Change in the frequency (in Hz) = f1 - f2

Change in the frequency (in Hz) = 552.48 Hz - 447.52 Hz

Change in the frequency (in Hz) = 104.96 Hz

15.Restore the battery setting to 10 V. Now change the number of loops from 4 to 3. Explain what happens to the magnitude and direction of the magnetic field. Now change to 2 loops, then to 1 loop. What do you observe the relationship to be between the magnitude of the magnetic field and the number of loops for the same current

Answers

Answer:

we see it is a linear relationship.

Explanation:

The magnetic flux is u solenoid is

      B = μ₀ N/L   I

where N is the number of loops, L the length and I the current

By applying this expression to our case we have that the current is the same in all cases and we can assume the constant length. Consequently we see that the magnitude of the magnetic field decreases with the number of loops

      B = (μ₀ I / L)  N

the amount between paracentesis constant, in the case of 4 loop the field is worth

      B = cte 4

N       B

4       4 cte

3       3 cte

2       2 cte

1        1 cte

as we see it is a linear relationship.

In addition, this effect for such a small number of turns the direction of the field that is parallel to the normal of the lines will oscillate,

What physical feature of a wave is related to the depth of the wave base? What is the difference between the wave base and still water level?

Answers

Answer:

physical feature of a wave is related to the depth of the wave base is The circular orbital motion

B. The wave base is the depth, and the still water level is the horizontal level

A group of students conducted several trials of an experiment to study Newton’s second law of motion. They concluded that tripling the mass required tripling the net force applied. What quantity were the students holding constant?

Answers

Answer:

1) Mass and acceleration

2) 4.0

3)When the net force applied to an object changes, the acceleration changes by the same factor.

4)acceleration

5)The acceleration is half of its original value

Explanation:

A wise man once said if you cheat together you succeed together! lol i am jk  

A mechanic wants to unscrew some bolts. She has two wrenches available: one is 35 cm long, and one is 50 cm long. Which wrench makes her job easier and why?

Answers

Answer:

50 cm long

When 35cm long wrench is compared to 50cm long wrench, we find that the 50cm long wrench produces more turning effect of force because it has longer distance between fulcrum and line of action of force. At conclusion, the more the turning effect of force the more it is easy to unscrew bolts.

An electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Which of the following are also the same for the two particles?
(A) speed
(B) kinetic energy
(C) frequency
(D) momentum

Answers

Explanation:

The De-Broglie wavelength is given by :

[tex]\lambda=\dfrac{h}{p}[/tex]

h is Planck's constant

p is momentum

In this case, an electron and a proton both moving at nonrelativistic speeds have the same de Broglie wavelength. Mass of electron and proton is different. It means their velocity and energy are different.

Only momentum is the factor that remains same for both particles i.e. momentum.

Parallel light rays with a wavelength of 563 nm fall on a single slit. On a screen 3.30 m away, the distance between the first dark fringes on either side of the central maximum is 4.70 mm . Part A What is the width of the slit

Answers

Answer:

The width of the slit is 0.4 mm (0.00040 m).

Explanation:

From the Young's interference expression, we have;

(λ ÷ d) = (Δy ÷ D)

where λ is the wavelength of the light, D is the distance of the slit to the screen, d is the width of slit and Δy is the fringe separation.

Thus,

d = (Dλ) ÷ Δy

D = 3.30 m, Δy = 4.7 mm (0.0047 m) and λ = 563 nm (563 ×[tex]10^{-9}[/tex] m)

d = (3.30 × 563 ×[tex]10^{-9}[/tex] ) ÷ (0.0047)

  = 1.8579 × [tex]10^{-6}[/tex] ÷ 0.0047

  = 0.0003951 m

d = 0.00040 m

The width of the slit is 0.4 mm (0.00040 m).

At the B end of the recessed horizontal bar, forces F1 and F2 of magnitudes 3 KN and 2KN respectively are applied and oriented as shown in the figure. Determine the magnitude of the resulting force and its orientation with respect to the horizontal.

Answers

Answer:

2.98 kN, 69.1° below the horizontal

Explanation:

Sum of forces in the x direction:

Fₓ = 3 kN cos 30° − 2 kN cos 40°

Fₓ ≈ 1.07 kN

Sum of forces in the y direction:

Fᵧ = -3 kN sin 30° − 2 kN sin 40°

Fᵧ = -2.79 kN

Magnitude of the resultant force is:

F² = Fₓ² + Fᵧ²

F = 2.98 kN

The direction of the resultant force is:

tan θ = Fᵧ / Fₓ

θ = -69.1°

Electrons are accelerated through a voltage difference of 270 kV inside a high voltage accelerator tube. What is the final kinetic energy of the electrons?

Answers

Each electron winds up with kinetic energy of

(270 keV)

plus

(whatever KE it had when it started accelerating).

A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastward. What are the magnitude (in T) and direction of the magnetic field at this instant?

Answers

Answer:

The values is  [tex]B = 3.2 *10^{-8} \ T[/tex]

The  direction is out of the plane

Explanation:

From the question we are told that

  The  magnitude of the electric field is  [tex]E = 9.6 \ V/m[/tex]

 

The  magnitude of the magnetic field is mathematically represented as

       [tex]B = \frac{E}{c}[/tex]

where c is the speed of light with value

      [tex]B = \frac{ 9.6}{3.0 *10^{8}}[/tex]

     [tex]B = 3.2 *10^{-8} \ T[/tex]

Given that the direction off the electromagnetic wave( c ) is  northward(y-plane ) and  the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page  (z-plane  )

     

The temperature of the hot spots caused by the impact of transferred matter onto the surface of a pulsar can be 108 K. What is the peak wavelength in the blackbody spectrum of such a spot, and in what range of the electromagnetic spectrum does it occur

Answers

Given that,

Temperature = 10⁸ K

We need to calculate the peak wavelength in the blackbody spectrum

Using formula of peak wavelength

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{T}[/tex]

Where, T= temperature

Put the value into the formula

[tex]peak\ wavelength = \dfrac{2.898\times10^{-3}}{10^{8}}[/tex]

[tex]peak\ wavelength = 2.90\times10^{-11}\ m[/tex]

[tex]peak\ wavelength = 290\ nm[/tex]

This range of wavelength is ultraviolet.

Hence, The peak wavelength in the blackbody spectrum is 290 nm and the range of wavelength is ultraviolet electromagnetic spectrum .

In the direction perpendicular to the drift velocity, there is a magnetic force on the electrons that must be cancelled out by an electric force. What is the magnitude of the electric field that produces this force

Answers

Answer:

E = VdB

Explanation:

This is because canceling the electric and magnetic force means

q.vd. B= we

E= Vd. B

Suppose a 500 mb chart valid today at 12 Z indicates a large trough over the eastern US and a large ridge over the western US. An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will _____ altitude if the altimeter is not corrected. Group of answer choices

Answers

Answer:

An aircraft, flying in the vicinity of 18,000 ft altitude from west to east over the US at 12 Z today, will __LOSE___ altitude if the altimeter is not corrected

Can a car moving with a negative velocity moves faster than a car moving with a positive velocity? explain.

Answers

Answer:

Yes.

Explanation:

This is because "negative velocity" just means it is in the negative in relation to the point of 0. Negative velocity doesn't equal a decrease in velocity. For example lets say you were parked next to a cone (this cone represents zero) if you accelerate forwards then that would be positive acceleration. If you were to accelerate backwards, this would be in the negative direction, aka negative velocity.

SUMMARY:

A negative velocity means that the object which has the negative velocity is moving in the opposite direction of an object moving at a positive velocity. This is a question of frame of reference. The possibility for the velocity is what makes it different to the speed. Speed is only positive.

Which examination technique is the visualization of body parts in motion by projecting x-ray images on a luminous fluorescent screen?

Answers

Answer:

Fluoroscopy

Explanation:

A Fluoroscopy is an imaging technique that uses X-rays to obtain real-time moving images of the interior of an object. In its primary application of medical imaging, a fluoroscope allows a physician to see the internal structure and function of a patient, so that the pumping action of the heart or the motion of swallowing, for example, can be watched.

An electron has an initial velocity to the south but is observed to curve upward as the result of a magnetic field. This magnetic field must have a component:___________
a) north
b) upwards
c) downwards
d) east
e) west

Answers

Answer:

e) west

Explanation:

According to Lorentz left hand rule, the left hand is used to represent the motion of an electron in a magnetic field. Hold out the left hand with the fingers held out parallel to the palm, and the thumb held at right angle to the other fingers. If the thumb represents the motion of the electron though the field, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the particle.

In this case, if we point the thumb to the south (towards your body), with the palm facing up, then the fingers will point west.

Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.490 mm . In the resulting interference pattern on a screen 2.12 m away, adjacent bright fringes are separated by 2.86 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Determining wavelength. Part A What is the wavelength of the light that falls on the slits

Answers

Answer:

λ = 6.61 x 10⁻⁷ m = 661 nm

Explanation:

From the Young's Double Slit experiment, the the spacing between adjacent bright or dark fringes is given by the following formula:

Δx = λL/d

where,

Δx = fringe spacing = 2.86 mm = 2.86 x ⁻³ m

L = Distance between slits and screen = 2.12 m

d = slit separation = 0.49 mm = 0.49 x 10⁻³ m

λ = wavelength of light = ?

Therefore,

2.86 x 10⁻³ m = λ(2.12 m)/(0.49 x 10⁻³ m)

(2.86 x 10⁻³ m)(0.49 x 10⁻³ m)/(2.12 m) = λ

λ = 6.61 x 10⁻⁷ m = 661 nm

A scientist is carrying out an experiment to determine the index of refraction for a partially reflective material. To do this, he aims a narrow beam of light at a sample of this material, which has a smooth surface. He then varies the angle of incidence. (The incident beam is traveling through air.)
The light that gets reflected by the sample is completely polarized when the angle of incidence is 46.5°.
(a) What index of refraction describes the material?
n =
(b) If some of the incident light (at θi = 46.5°) enters the material and travels below the surface, what is the angle of refraction (in degrees)?

Answers

Answer:

a) 1.05

b) 43.6°

Explanation:

a) The index refraction that describes the material can be found using Brewster's law:

[tex] \theta_{1} = arctan(\frac{n_{2}}{n_{1}}) [/tex]

where:

n₁ is the refractive index of the initial medium through which the light propagates (air) = 1

n₂ is the index of the material=?

θ₁ = 46.5°    

[tex] n_{2} = n_{1}tan(\theta_{1}) = tan(46.5) = 1.05 [/tex]

Hence, the material's index refraction is 1.05.

b) The angle of refraction can be found as follows:

[tex] n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2}) [/tex]

[tex]sin(\theta_{2}) = \frac{n_{1}sin(\theta_{1})}{n_{2}} = \frac{sin(46.5)}{1.05} = 0.69[/tex]

[tex] \theta_{2} = arcsin(0.69) = 43.6^{\circ} [/tex]

Therefore, the angle of refraction is 43.6°.

I hope it helps you!

UV radiaGon having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum kineGc energy of the ejected photoelectrons

Answers

Answer:

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

Explanation:

First we calculate the energy of photon:

E = hc/λ

where,

E = Energy of Photon = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength = 120 nm = 1.2 x 10⁻⁷ m

Therefore,

E = (6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(1.2 x 10⁻⁷ m)

E = (16.565 x 10⁻¹⁹ J)(1 eV/1.6 x 10⁻¹⁹ J)

E = 10.35 eV

Now, from Einstein's Photoelectric equation we know that:

Energy of Photon = Work Function + K.E of Electron

10.35 eV = 4.82 eV + K.E

K.E = 10.35 eV - 4.82 eV

K.E = 5.53 eV = 8.85 x 10⁻¹⁹ J

The maximum kinetic energy of the ejected photoelectrons will be "8.85 × 10⁻¹⁹ J".

Kinetic energy

According to the question,

Speed of light, c = 3 × 10⁸ m/s

Wavelength, λ = 120 nm or,

                        = 1.2 × 10⁻⁷ m

Plank's Constant, h = 6.626 × 10⁻³⁴ J.s

Now,

The energy of photon will be:

→ E = [tex]\frac{hc}{\lambda}[/tex]

By substituting the values,

     = [tex]\frac{6.626\times 10^{-34}\times 3\times 20^8}{1.2\times 10^{-7}}[/tex]

     = [tex]\frac{16.565\times 10^{-19}}{\frac{1 \ eV}{1.6\times 10^{-19}} }[/tex]

     = 10.35 eV

By using Einstein's Photoelectric equation,  

Energy of Photon = Work function + K.E

                   10.35 = 4.82 + K.E

                       K.E = 10.35 - 4.82

                             = 5.53 eV or,

                             = 8.85 × 10⁻¹⁹ J

Thus the response above is correct.

Find out more information about Kinetic energy here:

https://brainly.com/question/25959744

Two separate disks are connected by a belt traveling at 5m/s. Disk 1 has a mass of 10kg and radius of 35cm. Disk 2 has a mass of 3kg and radius of 7cm.
a. What is the angular velocity of disk 1?
b. What is the angular velocity of disk 2?
c. What is the moment of inertia for the two disk system?

Answers

Explanation:

Given that,

Linear speed of both disks is 5 m/s

Mass of disk 1 is 10 kg

Radius of disk 1 is 35 cm or 0.35 m

Mass of disk 2 is 3 kg

Radius of disk 2 is 7 cm or 0.07 m

(a) The angular velocity of disk 1 is :

[tex]v=r_1\omega_1\\\\\omega_1=\dfrac{v}{r_1}\\\\\omega_1=\dfrac{5}{0.35}\\\\\omega_1=14.28\ rad/s[/tex]

(b) The angular velocity of disk 2 is :

[tex]v=r_2\omega_2\\\\\omega_2=\dfrac{v}{r_2}\\\\\omega_2=\dfrac{5}{0.07}\\\\\omega_2=71.42\ rad/s[/tex]

(c) The moment of inertia for the two disk system is given by :

[tex]I=I_1+I_2\\\\I=\dfrac{1}{2}m_1r_1^2+\dfrac{1}{2}m_2r_2^2\\\\I=\dfrac{1}{2}(m_1r_1^2+m_2r_2^2)\\\\I=\dfrac{1}{2}\times (10\times (0.35)^2+3\times (0.07)^2)\\\\I=0.619\ kg-m^2[/tex]

Hence, this is the required solution.

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