Estimate the radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb. Assume that light is completely absorbed.

Answers

Answer 1

..........................................................

Estimate The Radiation Pressure Due To A Bulb That Emits 25 W Of EM Radiation At A Distance Of 6.5 Cm
Answer 2

The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².

What is the Radiation pressure?

Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.

Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.

E.g: Black-body radiation

Given the values are,

Wattage of bulb = W = 25 W

distance = d = 6.5 cm = 0.065 m

To know the Radiation Pressure,

It can be given by

P = I/c

Where, c = 299792458 m/s is the speed of light,

I is the intensity of radiation and given by

I = W/4πd²

Where W is the Wattage of bulb and d is the distance

I = 25/4π*0.065²

I = 470.872 w/m²

so, the radiation pressure becomes

P = 470.872/299792458

P = 1.5707x10⁻⁶ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²

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Related Questions

Solve 3* +5-220t = 0​

Answers

Answer:

t = 27.5

Explanation:

[tex]3 + 5 -220t = 0[/tex]

Well to solve for t we need to combine like terms and seperate t.

So 3+5= 8

8 - 220t = 0

We do +220 to both sides

8 = 220t

And now we divide 220 by 8 which is 27.5

Hence, t = 27.5

. If you live in a region that has a particular TV station, you can sometimes pick up some of its audio portion on your FM radio receiver. Explain how this is possible. Does it imply that TV audio is broadcast as FM

Answers

Answer:

Please see below as the answer is self-explanatory.

Explanation:

The low band of the VHF TV Spectrum, spans channels 2-6, from 54 to 88 Mhz.

In the analog TV, in the Americas, the total bandwidth of any channel is 6 Mhz, with the visual carrier modulated in VSS (Vestigial Side Band) at 1.25 Mhz from the lowest frequency of the channel.

The aural carrier is located at 4.5 Mhz from the visual carrier, and is FM modulated.

For Channel 6, which spans between 82 and  88 Mhz, the visual carrier is at 83.25 Mhz, so the aural carrier is at 87.75 Mhz, which falls within the FM Band, so it is possible to listen the audio part of this channel in a FM radio receiver, even at a lower volume, due to the FM radio has a greater deviation than TV aural carrier.

The reason why it is possible for TV station to sometimes pick up some of the audio portion on your FM radio receiver is because; TV waves can sometimes deviate into the FM radio frequency range.

Let us start with explaining the waves of TV and radio.

The frequency range utilized by TV stations is either the range 54 MHz to 88 MHz or 174 MHz to 222 MHz. In contrast, the frequency range utilized by FM Radio band is between 88 MHz and 174 MHz.

Now, in some cases, it is possible that the TV signal may deviate into the range of the FM Radio and as such in that case, the TV signal will pick the audio portion of an FM Radio. These TV waves are very high frequency waves.

Finally, it does not imply that the TV wave is broadcasting as an FM because it only deviated a bit from the TV range and not like that is where it is made to operate.

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A person bends over to grab a 20 kg object. The back muscle responsible for supporting his upper body weight and the object is located 2/3 of the way up his back (where it attaches to the spine) and makes an angle of 12 degrees with the spine. His upper body weighs 36 kg. What is the tension in the back muscle

Answers

Answer:

T = 2689.6N

Explanation:

Considering the situation, one can say that torque due to tension in the spine is counter balanced by the torque due to weight of upper part of the body and the weight of the object. Hence, the tension force is acting at an angle of 12 degree

while both weight are acting perpendicular to the length. Hence we have :

Torque ( clockwise) = Torque ( anticlockwise)

m1g (L/2)+ m2g(L) = Tsin 12(2L/3)........1

Where m1 = 36kg

m2 = 20kg

g = 9.81m/s^2

Theta = 12

Substituting into equation 1

36(9.81) * (L/2)+20(9.81)(L) = Tsin12(2L/3)

353.16L/2+196.2L = T ×0.2079(2L/3)

176.58L+196.2L = T × 0.1386L

372.78L = 0.1386LT

T = 372.78L/0.1386L

T = 2689.6N

Light emitted by element X passes through a diffraction grating that has 1200 slits/mm. The interference pattern is observed on a screen 77.0 cm behind the grating. First-order maxima are observed at distances of 58.0 cm , 65.4 cm , and 94.5 cm from the central maximum. What are the wavelengths of light emitted by element X?

Answers

Answer:

500 nm, 530 nm, 650 nm

Explanation:

Let's say that there is diffraction grating observed with a slit spacing of s. Respectively we must determine the angle θ which will help us determine the 3 wavelengths ( λ ) of the light emitted by element X. This can be done applying the following formulas,

s( sin θ ) = m [tex]*[/tex] λ, such that y = L( tan θ ) - where y = positioning, or the distance of the first - order maxima, and L = constant, of 77 cm

Now the grating has a slit spacing of -

s = 1 / N = 1 / 1200 = 0.833 [tex]*[/tex] 10⁻³ mm

The diffraction angles of the " positionings " should thus be -

θ = tan⁻¹ [tex]*[/tex] ( 0.58 / 0.77 ) = 37°,

θ = tan⁻¹ [tex]*[/tex] ( 0.654 / 0.77 ) = 40°,

θ = tan⁻¹ [tex]*[/tex] ( 0.945 / 0.77 ) = 51°

The wavelengths of these three bright fringes should thus be calculated through the formula : λ = s( sin θ ) -

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 37° ) = ( 500 [tex]*[/tex] 10⁻⁹ m )

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 40° ) = ( 530 [tex]*[/tex] 10⁻⁹ m )

λ = 0.833 [tex]*[/tex] 10⁻³ [tex]*[/tex] sin( 51° ) = ( 650 [tex]*[/tex] 10⁻⁹ m )

Wavelengths : 500 nm, 530 nm, 650 nm

This question will be solved using the "grating equation".

The wavelengths of the light emitted by element X are:

"1. 6.654 x 10⁻⁷ m = 665.4 nm

2. 6.349 x 10⁻⁷ m = 634.9 nm

3. 5.262 x 10⁻⁷ m = 526.2 nm"

The diffraction grating equation is given as follows:

[tex]m\lambda = d Sin\ \theta[/tex]

where,

m = order of maxima = 1

λ = wavelength of light = ?

d = grating element = [tex]\frac{1}{no.\ of\ slits\ per\ unit\ length} = \frac{1}{1200\ slits/mm}[/tex]

d =  (8.33 x 10⁻⁴ mm/slit)(1 m/ 1000 mm) = 8.33 x 10⁻⁷ m/slit

θ = angle of diffraction = [tex]tan^{-1}(\frac{L}{y})[/tex]

where,

L = distance of grating from the screen = 77 cm

y = distance of maxima from central maxima

Hence, the general equation after substituting constant values becomes:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{y}))[/tex]

FOR y = 58 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{58\ cm}))[/tex]

λ = 6.654 x 10⁻⁷ m = 665.4 nm

FOR y = 65.4 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{65.4\ cm}))[/tex]

λ = 6.349 x 10⁻⁷ m = 634.9 nm

FOR y = 94.5 cm:

[tex]\lambda = (8.33\ x\ 10^{-7}\ m/slits)\ Sin(tan^{-1}(\frac{77\ cm}{94.5\ cm}))[/tex]

λ = 5.262 x 10⁻⁷ m = 526.2 nm

The attached picture shows the arrangement of the light rays in a diffraction grating.

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An optical fiber uses one glass clad with another glass. What is the critical angle? (Assume the glass in the fiber has an index of refraction of 1.69, and the glass in the cladding has an index of refraction of 1.50.)

Answers

Answer:

The critical angle is [tex]\theta _c = 62.57^o[/tex]

Explanation:

From the question we are told that

    The  index of refraction of the glass in the fiber is [tex]n_f = 1.69[/tex]

     The index of refraction of the glass in the cladding is [tex]n_c = 1.50[/tex]

The critical angle is mathematically evaluated as

       [tex]\theta_c = sin^{-1}[\frac{n_c }{n_f } ][/tex]

substituting values

       [tex]\theta_c = sin^{-1}[\frac{1.50 }{1.69 } ][/tex]

       [tex]\theta _c = 62.57^o[/tex]

1. Notice that the voltmeter moves in response to the coil entering or leaving the magnetic gap.
2. Let's apply Faraday's Law to this situation. Faraday's Law says that the induced voltage (or emf )in a loop of wire caused by a changing magnetic field is:
€ = 1
Where is the magnetic flux which is
Q = BA
In this case, the flux density B is not changing. Instead, the changing flux is due to the motion of the coil as it enters or leaves the magnetic gap:
do = BdA
Given that the area immersed in the gap is changing as the coil enters the gap, what is the correct expression of Faraday's Law for this situation?

Answers

Answer:

Explanation:

let the coil of length l and breathe b entering the magnetic field B with speed v.

So, the magnetic flux through the coil is

Ф = B(l×b)

length × breathe = area

Ф = BA

dФ = BdA

therefore induced emf is given as

ε = [tex]Bl(\frac{db}{dt})[/tex]

note: [tex]\frac{db}{dt} = v[/tex]

ε[tex]= Blv[/tex]

attached is the diagram for the solution


I need help with this question?

Answers

Answer:

You got it right, didn't you?

c) vector C

Explanation:

opposite to vector V

Answer:

A, vector B

Explanation:

A negative vector is a vector which points in the opposite direction, even though it’s in a different quadranot it’s still the opposite direction.

Recent technological developments like high-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have refined and extended the camera’s capacity to provide information. Which passage assertion does this information support most strongly?

Answers

Answer:

D) Photography can be used to both control and benefit society.

Explanation:

High-resolution satellite imagery and diagnostic positron emission tomography (PET scans) have been used to both control and benefits the society in the sense that it has helped to take records of information of crime, traffic offenders such drunk drivers and over speeding drivers, e.t.c. it helps control by given their information and automatically penalizing them or ensuring the agency penalized them and also benefit the society by preventing people from committing crime thereby, protecting them from offenders.

Your professor is conducting a chemical isotope analysis of diet for her latest paleoanthropology project. Which of the following is true regarding isotopic analysis of diet?
a. Carbon analysis of bones and teeth will give insight to the types of plant material the hominin consumed.
b. Strontium analysis of bones and teeth will provide information regarding whether the hominin was a meat-eater.
c. Carbon analysis of bones and teeth will enable your professor to determine whether the meat consumed was from land or water sources.
d. Nitrogen analysis of bones and teeth will allow your professor to determine whether the hominin relied on C3 or C4 plants.

Answers

Answer:

Carbon analysis of bones and teeth will give insight to the types of plant material the hominin consumed.

Explanation:

This is because carbon being the principal composition of plants, analysis of the teeth's and bones for carbon will give an insight of what the homini n consumed

Estimate the volume of a human heart (in mL) using the following measurements/assumptions:_______.
1. Blood flow through the aorta is approximately 11.2 cm/s
2. The diameter of the aorta is approximately 3.0 cm
3. Assume the heart pumps its own volume with each beat
4. Assume a pulse rate of 67 beats per minute.

Answers

Answer:

Explanation:

radius of aorta = 1.5 cm

cross sectional area = π r²

= 3.14 x 1.5²

= 7.065 cm²

volume of blood flowing out per second out of heart

= a x v , a is cross sectional area , v is velocity of flow

= 7.065 x 11.2

= 79.128 cm³

heart beat per second = 67 / 60

= 1.116666

If V be the volume of heart

1.116666 V = 79.128

V = 70.86 cm³.

Tuning a guitar string, you play a pure 330 Hz note using a tuning device, and pluck the string. The combined notes produce a beat frequency of 5 Hz. You then play a pure 350 Hz note and pluck the string, finding a beat frequency of 25 Hz. What is the frequency of the string note?

Answers

Answer:

The  frequency is  [tex]F = 325 Hz[/tex]

Explanation:

From the question we are told that

    The frequency for the first note is  [tex]F_1 = 330 Hz[/tex]

     The  beat frequency of the first note is  [tex]f_b = 5 \ Hz[/tex]

     The  frequency for the second note is  [tex]F_2 = 350 \ H_z[/tex]

      The  beat frequency of the first note is [tex]f_a = 25 \ Hz[/tex]

Generally beat frequency is mathematically represented as

        [tex]F_{beat} = | F_a - F_b |[/tex]

Where [tex]F_a \ and \ F_b[/tex] are frequencies of two sound source

  Now in the case of this question

For the first note

     [tex]f_b = F_1 - F \ \ \ \ \ ...(1)[/tex]

Where  F is the frequency of the string note

For the second note  

      [tex]f_a = F_2 - F \ \ \ \ \ ...(2)[/tex]

Adding  equation 1 from 2

      [tex]f_b + f_a = F_1 + F_2 + ( - F) + (-F) )[/tex]

      [tex]f_b + f_a = F_1 + F_2 -2F[/tex]

substituting values

       [tex]5 +25 = 330 + 350 -2F[/tex]

=>     [tex]F = 325 Hz[/tex]

       

A converging lens of focal length 7.40 cm is 18.0 cm to the left of a diverging lens of focal length -7.00 cm . A coin is placed 12.0 cm to the left of the converging lens.
A) Find the location of the coin's final image relative to the diverging lens.
B) Find the magnification of the coin's final image.

Answers

Answer:

Explanation:

The set up is a compound microscope. The converging lens is the objective lens while the diverging lens is the eyepiece lens.

In compound microscopes, the distance between the two lenses is expressed as L = v0+ue

v0 is the image distance of the objective lens and ue is the object distance of the eye piece lens.

Befre we can get the location of the coin's final image relative to the diverging lens (ve), we need to get ue first.

Given L = 18.0cm

Using the lens formula to get v0 where u0 = 12.0cm and f0 = 7.40cm

1/f0 = 1/u0+1/v0

f0 and u0 are the focal length and object distance of the converging lens (objective lens)respectively.

1/v0 = 1/7.4-1/12

1/v0 = 0.1351-0.0833

1/v0 = 0.0518

v0 = 1/0.2184

v0 = 19.31cm

Note that v0 = ue = 19.31cm

To get ve, we will use the lens formula 1/fe = 1/ue+1/ve

1/ve = 1/fe-1/ue

Given ue = 19.31cm and fe = -7.00cm

1/ve = -1/7.0-1/19.31

1/ve = -0.1429-0.0518

1/ve = -0.1947

ve = 1/-0.1947

ve = -5.14cm

Hence, the location of the coin's final image relative to the diverging lens is 5.14cm to the lens

b) Magnification of the final image M = ve/ue

M = 5.14/19.31

M = 0.27

Magnification of the final image is 0.27

greater than: The electric potential energy of a proton at point A is _____ the electric potential energy of an proton at point B.

Answers

Answer:

[similar to]

Explanation:

it is the missing word

Asteroid A has 3.5 times the mass and 2.0 times the velocity of Asteroid B. If
Asteroid B has a kinetic energy of 2,300,000 J then what is the kinetic energy of
Asteroid A?

Answers

Answer:

   K_A = 32.2 10⁶ J

Explanation:

In this exercise we must relate the quantities given to find the kinetic energy

   

Asteroid A data

              m_A = 3.5 m_B

               v_A = 2.0 v

they also give the value of the kinetic energy of asteroid A

              K_B = 2.3 10⁶ J

the expression for scientific energy is

               K = ½ m v²

let's replace

              K_A = ½ m_a V_a2

               K_A = ½ 3.5 m_B (2.0 v_B)^2

                K_A = 3.5 2² (½ m_B v_B²)

                K_A = 14 K_B

                   

               K_A = 32.2 10⁶ J

The radius of curvature of the path of a charged particle in a uniform magnetic field is directly proportional toA) the particle's charge.B) the particle's momentum.C) the particle's energy.D) the flux density of the field.E)All of these are correct

Answers

Answer:

B) the particle's momentum.

Explanation:

We know that

The centripetal force  on the particle when its moving in the radius R and velocity V

[tex]F_c=\dfrac{m\times V^2}{R}[/tex]

The magnetic force on the particle when the its moving with velocity V in the magnetic filed B and having charge q

[tex]F_m=q\times V\times B[/tex]

At the equilibrium condition

[tex]F_m=F_c[/tex]

[tex]q\times V\times B=\dfrac{m\times V^2}{R}[/tex]

[tex]R=\dfrac{m\times V}{q\times B}[/tex]

Momentum = m V

Therefore we can say that the radius of curvature is directly proportional to the particle momentum.

B) the particle's momentum.

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?

Answers

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

which example describes a nonrenewable resource?
A. everyone in our neighborhood uses solar panels to generate electricity to run their pool pumps.
B. once up and running, the power plant will convert the energy from tides and waves into electricity.
C. there is a long stretch of land in the desert with many windmills that are able to generate enough electricity to run the town.
D. there are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity.

Answers

Answer:

D. There are drilling platforms all along the coast that are used to drill for natural gas that can be used to generate electricity

Explanation:

Solar panels are a renewable resource because the sun will not run out. The power plant uses water, so it is also a renewable resource. Windmills use wind, and wind will not run out so it is a renewable resource. However, natural gas and oil are not renewable resources because they will run out one day.

An 1300-turn coil of wire that is 2.2 cmcm in diameter is in a magnetic field that drops from 0.14 TT to 0 TT in 9.0 msms . The axis of the coil is parallel to the field.
What is the emf of the coil? (in V)

Answers

Answer:

The induced  emf is  [tex]\epsilon =7.68 \ V[/tex]

Explanation:

From the question we are told that

     The number of turns is  [tex]N = 1300 \ turns[/tex]

    The diameter is  [tex]d = 2.2 \ cm = 2.2*10^{-2}[/tex]

     The initial magnetic field is  [tex]B_i = 0.14 \ T[/tex]

      The final magnetic field is  [tex]B_f = 0 \ T[/tex]

      The  time taken is  [tex]dt = 9.0ms = 9.0*10^{-3} \ s[/tex]

 

The radius is mathematically evaluated as

      [tex]r = \frac{d}{2 }[/tex]

substituting values

     [tex]r = \frac{2.2 *10^{-2}}{2 }[/tex]

     [tex]r = 1.1*10^{-2} \ m[/tex]

The induced emf is mathematically represented as

    [tex]\epsilon =- N * \frac{d\phi }{dt }[/tex]

Where  [tex]d\phi[/tex] is the change in magnetic field which is mathematically represented as

        [tex]d\phi = dB * A * cos\theta[/tex]

=>   [tex]d\phi = [B_f - B_i ] * A * cos\theta[/tex]

Here  [tex]\theta = 0[/tex] given that the axis of the coil is parallel to the field

Also A is the cross-sectional area which is mathematically represented as

       [tex]A = \pi r^2[/tex]

substituting values

      [tex]A = 3.142 * [1.1*10^{-2}]^2[/tex]

       [tex]A = 3.8 *10^{-4] \ m^2[/tex]

So

    [tex]d\phi = [0 - 0.14 ] * 3.8*10^{-4}[/tex]

    [tex]d\phi = -5.32*10^{-5} \ weber[/tex]

So  

     [tex]\epsilon =- 1300 * \frac{-5.32*10^{-5} }{ 9.0*10^{-3} }[/tex]

    [tex]\epsilon =7.68 \ V[/tex]

Ohm’s Law
pls answer this photos​

Answers

Answer:

Trial 1: 2 Volts, 0 %

Trial 2: 2.8 Volts, 0%

Trial 3: 4 Volts, 0 %

Explanation:

Th experimental values are given in the table, while the theoretical value can be found by using Ohm/s Law:

V = IR

TRIAL 1:

V = IR

V = (0.1 A)(20 Ω)

V = 2 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2 - 2)/2| x 100%

% Difference = 0 %

TRIAL 2:

V = IR

V = (0.14 A)(20 Ω)

V = 2.8 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(2.8 - 2.8)/2.8| x 100%

% Difference = 0 %

TRIAL 3:

V = IR

V = (0.2 A)(20 Ω)

V = 4 volts

% Difference = [tex]|\frac{Theoretical Value - Exprimental Value}{Theoretical Value}|[/tex] x 100%

% Difference = |(4 - 4)/4| x 100%

% Difference = 0 %

A 0.210-kg metal rod carrying a current of 11.0 A glides on two horizontal rails 0.490 m apart. If the coefficient of kinetic friction between the rod and rails is 0.200, what vertical magnetic field is required to keep the rod moving at a constant speed?

Answers

Answer:

The  magnetic field is  [tex]B = 0.0764 \ T[/tex]

Explanation:

From the question we are told that  

    The mass of the metal is  [tex]m = 0.210 \ kg[/tex]

     The current is  [tex]I = 11.0 \ A[/tex]

      The distance between the rail(length of the rod ) is  [tex]d = 0.490 \ m[/tex]

      The coefficient of kinetic friction is  [tex]\mu_k = 0.200[/tex]

Generally the magnetic force is mathematically represented as

      [tex]F_b = B * I * d[/tex]

Given that the rod is moving at a constant velocity, it

=>    [tex]F_b = F_k[/tex]

Where [tex]F_k[/tex] is the kinetic frictional force which is mathematically represented as

       [tex]F_k = \mu_k * m * g[/tex]

So

    [tex]B * I * d = \mu_k * m * g[/tex]

=>   [tex]B = \frac{\mu_k * m * g}{I * d }[/tex]

substituting values

=>   [tex]B = \frac{0.200 * 0.210 * 9.8 }{ 11 * 0.490 }[/tex]

=>   [tex]B = 0.0764 \ T[/tex]

A large power plant generates electricity at 12.0 kV. Its old transformer once converted the voltage to 390 kV. The secondary of this transformer is being replaced so that its output can be 515 kV for more efficient cross-country transmission on upgraded transmission lines. (a) What is the ratio of turns in the new secondary compared with turns in the old secondary

Answers

Answer:

1.32 is the turns ratio

Explanation:

Note that The transformer steps up the voltage from 12000 V to 390000V

12000 V is the primary and in the secondary it is 390000 V in old transformer

If n₁ be no of turns in primary coil and n₂ be no of turns in secondary coils

the formula is

n₂ / n₁ = voltage in secondry / voltage in primary

n₂ / n₁ = 390000 / 12000

ratio of turns in old transformer is 32.5

ratio of turns in new transformer

n₃ / n₁ = 515 / 12 ( n₃ is no of turns in the secondary of new transformer )

= 42.9

T he ratio of turns in the new secondary compared with the old secondary

n₃ / n₂ = 42.9 / 32.5

= 1.32

two resistors of resistance 10 ohm's and 20 ohm's are connected in parallel to a batery of e.m.f 12V. Calculate the current passing through the 20hm's resister​

Answers

Current through 20 ohm resistor is 0.6 A

The specific heat of a certain type of cooking oil is 1.75 J/(g⋅°C). How much heat energy is needed to raise the temperature of 2.01 kg of this oil from 23 °C to 191 °C?

Answers

Answer:

Q = 590,940 J

Explanation:

Given:

Specific heat (c) = 1.75 J/(g⋅°C)

Mass(m) = 2.01 kg = 2,010

Change in temperature (ΔT) = 191 - 23 = 168°C

Find:

Heat required (Q)

Computation:

Q = mcΔT

Q = (2,010)(1.75)(168)

Q = 590,940 J

Q = 590.94 kJ

A small branch is wedged under a 200 kg rock and rests on a smaller object. The smaller object is 2.0 m from the large rock and the branch is 12.0 m long.
(a) If the mass of the branch is negligible, what force must be exerted on the free end to just barely lift the rock?
(b) What is the mechanical advantage of this lever system?

Answers

Answer:

a

  [tex]F =326.7 \ N[/tex]

b

  [tex]M = 6[/tex]

Explanation:

From the question we are told that

          The mass of the rock is  [tex]m_r = 200 \ kg[/tex]

          The  length of the small object from the rock is  [tex]d = 2 \ m[/tex]

          The  length of the small object from the branch [tex]l = 12 \ m[/tex]

An image representing this lever set-up is shown on the first uploaded image

Here the small object acts as a fulcrum

The  force exerted by the weight of the rock is mathematically evaluated as

      [tex]W = m_r * g[/tex]

substituting values

     [tex]W = 200 * 9.8[/tex]

     [tex]W = 1960 \ N[/tex]

 So  at  equilibrium the sum  of the moment about the fulcrum is mathematically represented as

         [tex]\sum M_f = F * cos \theta * l - W cos\theta * d = 0[/tex]

Here  [tex]\theta[/tex] is very small so  [tex]cos\theta * l = l[/tex]

                               and  [tex]cos\theta * d = d[/tex]

Hence

       [tex]F * l - W * d = 0[/tex]

=>    [tex]F = \frac{W * d}{l}[/tex]

substituting values

        [tex]F = \frac{1960 * 2}{12}[/tex]

       [tex]F =326.7 \ N[/tex]

The  mechanical advantage is mathematically evaluated as

          [tex]M = \frac{W}{F}[/tex]

substituting values

        [tex]M = \frac{1960}{326.7}[/tex]

       [tex]M = 6[/tex]

Light from an argon laser strikes a diffraction grating that has 4,917 lines per cm. The first-order principal maxima are separated by 0.4 m on a wall 1.62 m from the grating. What is the wavelength of the laser light in nm

Answers

Answer:

Wavelength is 4.8x10^-7m

Explanation:

See attached file

Need help understanding this. If anyone help, that would be greatly appreciated!

Answers

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

a) acceleration:

ā=v^2/r

ā=(15m/s)^2/27m

ā=225/27 m/s^2

ā=8.333 m/s^2

force:

F=mā. where the is equal to v^2/r

F=1000kg*8.3 m/s^2

F=8333.3 N

Answer:

8.33` m/s^2 and 8333.3 N

Explanation:

A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done by gravity during the roll, in Joules

Answers

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

[tex] W = F*d [/tex]

Where:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

[tex]F = W_{x} = mgsin(\theta)[/tex]

Where:

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

[tex]F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N[/tex]    

Now, we can find the work:

[tex]W = F*d = 24.53 N*20 m = 490.6 J[/tex]      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

An electron moves in a circular path perpendicular to a magnetic field of magnitude 0.285 T. If the kinetic energy of the electron is 2.10 10-19 J, find the speed of the electron and the radius of the circular path. (a) the speed of the electron

Answers

Answer:

The speed of the electron is 6.79 x 10⁵ m/s

The radius of the circular path is 1.357 x 10⁻⁵ m

Explanation:

Given;

magnetic field, B = 0.285 T

energy of electron, E = 2.10 x 10⁻¹⁹ J

The kinetic energy of the electron is calculated as;

[tex]K.E = \frac{1}{2} m_eV^2[/tex]

Where;

[tex]m_e[/tex] is the mass of electron = 9.11 x 10⁻³¹ kg

V is the speed of the electron

[tex]K.E = \frac{1}{2} m_eV^2\\\\V^2 = \frac{2.K.E}{m_e} \\\\V = \sqrt{\frac{2K.E}{m_e} } \\\\V = \sqrt{\frac{2*(2.1*10^{-19})}{9.11*10^{-31}} }\\\\V = 6.79 *10^{5} \ m/s[/tex]

The radius of the circular path is given by;

[tex]R = \frac{M_eV}{qB}[/tex]

where;

q is the charge of the electron = 1.6 x 10⁻¹⁹ C

[tex]R = \frac{M_eV}{qB} \\\\R = \frac{9.11 *10^{-31}*6.79 *10^{5}}{1.6*10^{-19}*0.285} \\\\R = 1.357 *10^{-5} \ m[/tex]

What is the distance in m between lines on a diffraction grating that produces a second-order maximum for 775-nm red light at an angle of 62.5°?

Answers

Answer:

The distance is [tex]d = 1.747 *10^{-6} \ m[/tex]  

Explanation:

From the question we are told that  

       The order of maximum diffraction is  m =  2

         The wavelength is   [tex]\lambda = 775 nm = 775 * 10^{-9} \ m[/tex]

         The angle is  [tex]\theta = 62.5^o[/tex]

Generally the   condition for  constructive  interference for diffraction grating  is mathematically represented as

          [tex]dsin \theta = m * \lambda[/tex]

where  d is  the distance between the lines on a  diffraction grating

     So  

            [tex]d = \frac{m * \lambda }{sin (\theta )}[/tex]

substituting values  

           [tex]d = \frac{2 * 775 *1^{-9} }{sin ( 62.5 )}[/tex]

          [tex]d = 1.747 *10^{-6} \ m[/tex]

   

When a current of 2.0 A flows in the 100-turn primary of an ideal transformer, this causes 14 A to flow in the secondary. How many turns are in the secondary

Answers

Answer:

14.29 turns.

Explanation:

From the question,

Ns/Np = Ip/Is........................ Equation 1

Where Ns = Secondary turn, Np = Primary turn, Is = current flowing in the secondary turn, Ip = current flowing in the primary turn.

Make Ns the subject of the equation

Ns = NpIp/Is.................... Equation 2

Given: Np = 100 turns, Ip = 2.0 A, Is = 14 A.

Substitute these values into equation 2

Ns = 100(2.0)/14

Ns = 14.29 turns.

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