Question 2.
In the US, lengths are often measured in inches, feet, yards and miles. Let's do
some conversions. The definition of the inch is: 1 inch = 25.4 mm, exactly. A foot is
12 inches and a mile is 5280 ft, exactly. A centimetre is exactly 0.01 m or 10 mm.
Sammy is 5 feet and 5.3 inches tall.
a). What is Sammy's height in Inches? (answer to 3 significant figures)
(3)
b). What is Sammy's height in Feet? (answer to 3 significant figures)
what is Sammy's hight in feet according to this statement ​

Answer:

Yardangs are formed on vertical strata while zeugen on horizontal strata. ... Yardangs are formed on vertical hard/soft layers of rock, while zeugen (this is its plural form) are formed on horizontal bands of hard/soft rocks giving it a more mushroom-like shape. The Great Sphinx of Giza has been sculpted in a yardang

Explanation:

In order of Increasing Wavelength of the Electromagnetic Spectrum :

B) X rays

D) Visible light

A) Microwave

C) Radio Waves

Electromagnetic waves in order of decreasing wavelength  is X-rays,visible light,microwaves and radio waves.

The electromagnetic radiation consists of waves made up of electromagnetic field which are capable of propogating through space and carry the radiant electromagnetic energy.

The radiation are composed of electromagnetic waves which are synchronized oscillations of electric and magnetic fields . They are created due to change which is periodic in electric as well as magnetic fields.

In vacuum ,all the electromagnetic waves travel at the same speed that is with the speed of air.The position of an electromagnetic wave in an electromagnetic spectrum is characterized by it's frequency or wavelength.They are emitted by electrically charged particles which undergo acceleration and subsequently interact with other charged particles.

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Answer:

The power of the eye is 51.64 diopters

Explanation:

The power of the eye is given by;

[tex]P = \frac{1}{f} = \frac{1}{d_o} +\frac{1}{d_i}[/tex]

where;

P is the power of the eye in diopter

f is the focal length of the eye

[tex]d_o[/tex] is the distance between the eye and the object

[tex]d_i[/tex] is the distance between the eye and the image

Given;

[tex]d_o[/tex] = 61.0 cm = 0.61 m

[tex]d_i[/tex] = 2.0 cm = 0.02 m

[tex]P = \frac{1}{d_o} +\frac{1}{d_i} \\\\P = \frac{1}{0.61} + \frac{1}{0.02} \\\\P = 51.64 \ D[/tex]

Therefore, the power of the eye is 51.64 diopters.

The power P of the eye when viewing an object 61.0 cm away is 51.639D

The power of a lens is a reciprocal of its focal length and it is expressed as:

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

According to the mirror formula

[tex]\frac{1}{f} =\frac{1}{d_i} +\frac{1}{d_0}[/tex]

where

[tex]d_i[/tex] is the distance from the lens to the image = 61.0cm = 0.61m

[tex]d_0[/tex] is the distance from the lens to the object = 2.00cm = 0.02m

[tex]P=\frac{1}{f} =\frac{1}{0.02} +\frac{1}{0.61}\\P=50+1.639\\P=51.639D[/tex]

Hence the power P of the eye when viewing an object 61.0 cm away is 51.639D

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Given that,

Central maximum = 1 cm

Distance from the window shade to the wall =4 m

We know that,

The visible range of the sun light is 400 nm to 700 nm.

(a). We need to calculate the average wavelength

Using formula of average wavelength

[tex]\lambda_{avg}=\dfrac{\lambda_{1}+\lambda_{2}}{2}[/tex]

Put the value into the formula

[tex]\lambda_{avg}=\dfrac{400+700}{2}[/tex]

[tex]\lambda_{avg}=550\ nm[/tex]

(b). We need to calculate the diameter of the pinhole

Using formula for diameter

[tex]w=\dfrac{2.44\lambda L}{D}[/tex]

[tex]D=\dfrac{2.44\lambda L}{w}[/tex]

Put the value into the formula

[tex]D=\dfrac{2.44\times550\times10^{-9}\times4}{1\times10^{-2}}[/tex]

[tex]D=0.537\ mm[/tex]

Hence, (a). The average wavelength 550 nm.

(b). The diameter of the pinhole is 0.537 mm.

Answer:

The answer is A

Explanation:

Answer:

implausible or untestable scientific claims

Answer:

n= speed of light in vacuum/ speed of light in the other medium.

Explanation:

If light is moving from medium 1 into medium 2 where medium 1 is vacuum (approximated to mean air) and we are required to find the velocity of light; then we can confidently write;

n= speed of light in vacuum/ speed of light in the other medium.

Hence;

n= c/v

Where;

n= refractive index of the material

c= speed of light in vacuum

v = speed of light in another medium.

Note that the refractive index is the amount by which a transparent medium decreases the speed of light.

Answer:

A.   number of neutrons of Magnesium Mg = 13

B.   The average mass of Mg = 22.29 amu

C.   the magnesium composition on Mars is not the same as that on Earth.

Explanation:

Isotopes are atoms with the same atomic number but different mass number. This is due to the difference in mass of the neutrons.

The atomic number of Magnesium Mg = 12

The atomic number of an element is the number of protons present in the atomic nucleus of the element

i.e Atomic number = number of protons = 12

The mass number of an element is the sum of the protons and neutrons in the atomic nucleus of the element.

Mass number = number of protons + number of neutrons

Given that the mass number of Mg = 25

Then;

25 = 12 + number of neutrons

25 - 12 = number of neutrons

13 = number of neutrons

number of neutrons of Magnesium Mg = 13

B. What is the average atomic mass of magnesium in these rocks?

The average atomic mass of an element which exhibit isotopy is the average mass of its various isotopes as they occur naturally in any quantity of the element.

Therefore the average atomic mass of magnesium can be calculated as:

= [tex]\mathtt{\dfrac{(23.9872 \times 79.70) + ( 24.9886 \times 10.13) + (25.9846 \times 10.17) }{79.7 + 10.13 +10.17}}[/tex]

= [tex]\mathtt{\dfrac{(1911.77984) + ( 53.134518) + (264.263382) }{100}}[/tex]

= [tex]\mathtt{\dfrac{2229.17774 }{100}}[/tex]

The average mass of Mg = 22.29 amu

C. Is the magnesium composition on Mars the same as that on Earth? Explain.

The average atomic weight of magnesium on Earth is said to be 24.305 amu while that of Mars is 22.29 amu.

There difference in the average atomic weight result into difference in their composition. Therefore,the magnesium composition on Mars is not the same as that on Earth.

Answer:

The critical angle phenomenon.

Explanation:

Critical angle in optics is the smallest angle of incidence of a wave, that will give total reflection of the wave. This phenomenon occurs at the boundary of two medium, where light will normally move from one medium to another.

To prevent light from entering the water, the angle of incidence of the light incident on the water must exceed the critical angle.

Answer:

the magnitude of the electric field is 1.25 N/C

Explanation:

The induced emf in the cube ε = LB.v where B = magnitude of electric field = 5 T , L = length of side of cube = 1 cm = 0.01 m and v = velocity of cube = 1 m/s

ε = LB.v = 0.01 m × 5 T × 1 m/s = 0.05 V

Also, induced emf in the cube ε = ∫E.ds around the loop of the cube where E = electric field in metal cube

ε = ∫E.ds

ε = Eds since E is always parallel to the side of the cube

= E∫ds  ∫ds = 4L since we have 4 sides

= E(4L)

= 4EL

So,4EL = 0.05 V

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

So, the magnitude of the electric field is 1.25 N/C

The magnitude of the electric field is 1.25 N/C

But before that the following calculations need to be done.

ε = LB.v = 0.01 m × 5 T × 1 m/s

= 0.05 V

Now

ε = ∫E.ds

here ε = Eds because E is always parallel to the side of the cube

So,

= E∫ds  ∫ds

= 4L so we have 4 sides

Now

= E(4L)

= 4EL

So,4EL = 0.05 V

Now

E = 0.05 V/4L

= 0.05 V/(4 × 0.01 m)

= 0.05 V/0.04 m

= 1.25 V/m

= 1.25 N/C

hence, The magnitude of the electric field is 1.25 N/C

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Answer:

[tex]f=1.13s^{-1}=1.13Hz[/tex]

Explanation:

Hello,

In this case, a frequency stands for a rate in which some action is done per unit of time. In this case, for the heartbeat, since 68 actions (heartbeats) occur in 1.0, the frequency turns out:

[tex]f=\frac{68}{1.0min}=68min^{-1}[/tex]

Or as most commonly used in Hz ([tex]s^{-1}[/tex]):

[tex]f=68\frac{1}{min} *\frac{1min}{60s}=1.13s^{-1}=1.13Hz[/tex]

Best regards.

Answer:

Increase in frequency of the laser

Explanation:

Because An increase in frequency will result in more lines per centimeter and a smaller distance between each consecutive line. And a decrease in distance between each gratin

Answer:

317.22

Explanation:

Given

Circular platform rotates ccw 93.1kg, radius 1.93 m, 0.945 rad/s

You 69.7kg, cw 1.01m/s, at r

Poodle 20.2 kg, cw 1.01/2 m/s, at r/2

Mutt 17.7 kg, 3r/4

You

Relative

ω = v/r

= 1.01/1.93

= 0.522

Actual

ω = 0.945 - 0.522

= 0.42

I = mr^2

= 69.7*1.93^2

= 259.6

L = Iω

= 259.6*0.42

= 109.4

Poodle

Relative

ω = (1.01/2)/(1.93/2)

= 0.5233

Actual

ω = 0.945- 0.5233

= 0.4217

I = m(r/2)^2

= 20.2*(1.93/2)^2

= 18.81

L = Iω

= 18.81*0.4217

= 7.93

Mutt

Actual

ω = 0.945

I = m(3r/4)^2

= 17.7(3*1.93/4)^2

= 37.08

L = Iω

= 37.08*0.945

= 35.04

Disk

I = mr^2/2

= 93.1(1.93)^2/2

= 173.39

L = Iω

= 173.39*0.945

= 163.85

Total

L = 109.4+ 7.93+ 36.04+ 163.85

= 317.22 kg m^2/s

Answer:

Explanation:

Given data

mass of canister= 3.4 kg

force acting on canister= 3 N

initial velocity u= 2.5 m/s

final velocity v= 4.8 m/s

The work done on the canister is the change in kinetic energy on the canister

change in [tex]KE= Kfinal- Kinitial[/tex]

K.E initial

[tex]Kintial= \frac{1}{2} mv^2\\\\Kintial= \frac{1}{2}*2*2.5^2\\\\KInitial= \frac{1}{2} *2*6.25\\\\Kinitial= 6.25J[/tex]

K.E final

[tex]Kfinal= \frac{1}{2} mv^2\\\\ Kfinal= \frac{1}{2}*2*4.8^2\\\\ Kfinal= \frac{1}{2} *2*23.04\\\\ Kfinal= 23.04J[/tex]

The net work done is [tex]KE= Kfinal- Kinitial[/tex]

[tex]W net= 23.04-6.25= 16.79J[/tex]

Explanation:

Answer:

Qsinθ/4πε₀R²θ

Explanation:

Let us have a small charge element dq which produces an electric field E. There is also a symmetric field at P due to a symmetric charge dq at P. Their vertical electric field components cancel out leaving the horizontal component dE' = dEcosθ = dqcosθ/4πε₀R² where r is the radius of the arc.

Now, let λ be the charge per unit length on the arc. then, the small charge element dq = λds where ds is the small arc length. Also ds = Rθ.

So dq = λRdθ.

Substituting dq into dE', we have

dE' = dqcosθ/4πε₀R²

= λRdθcosθ/4πε₀R²

= λdθcosθ/4πε₀R

E' = ∫dE' = ∫λRdθcosθ/4πε₀R² = (λ/4πε₀R)∫cosθdθ from -θ to θ

E' = (λ/4πε₀R)[sinθ] from -θ to θ

E' = (λ/4πε₀R)[sinθ]

= (λ/4πε₀R)[sinθ - sin(-θ)]

= (λ/4πε₀R)[sinθ + sinθ]

= 2(λ/4πε₀R)sinθ

= (λ/2πε₀R)sinθ

Now, the total charge Q = ∫dq = ∫λRdθ from -θ to +θ

Q = λR∫dθ = λR[θ - (-θ)] = λR[θ + θ] = 2λRθ

Q = 2λRθ

λ = Q/2Rθ

Substituting λ into E', we have

E' = (Q/2Rθ/2πε₀R)sinθ

E' = (Q/θ4πε₀R²)sinθ

E' = Qsinθ/4πε₀R²θ where θ is in radians

Answer:

[tex]q = -461532.5 \ C[/tex]

Explanation:

From the question we are told that

     The  electric filed is  [tex]E = 102 \ N/C[/tex]  

Generally according to Gauss law

=>   [tex]E A = \frac{q}{\epsilon_o }[/tex]

Given that  the electric field is pointing downward  , the equation become

    [tex]- E A = \frac{q}{\epsilon_o }[/tex]

Here   [tex]q[/tex] is the excess charge on the surface of the earth

          [tex]A[/tex] is the surface  area of the of the earth which is mathematically represented as

     [tex]A = 4\pi r^2[/tex]

Where r is the radius of the earth which has a value [tex]r = 6.3781*10^6 m[/tex]

 substituting values

    [tex]A = 4 * 3.142 * (6.3781*10^6 \ m)^2[/tex]

    [tex]A =5.1128 *10^{14} \ m^2[/tex]

So

   [tex]q = -E * A * \epsilon _o[/tex]

Here [tex]\epsilon_o[/tex] s the permitivity of free space with value

          [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

substituting values

     [tex]q = -102 * 5.1128 *10^{14} * 8.85 *10^{-12}[/tex]

     [tex]q = -461532.5 \ C[/tex]

Answer:

The correct option is  d

Explanation:

From the question we are told that

     The  electric potential is  [tex]V = 3000 \ V[/tex]

      The  power is  [tex]P = 60 \ W[/tex]

      The  charge delivered is  [tex]q = 3nC = 3.0 *10^{-9} \ C[/tex]

Generally the power generated is mathematically represented as

         [tex]P = I V[/tex]

=>      [tex]I = \frac{P}{V }[/tex]

=>       [tex]I = \frac{60 }{3000 }[/tex]

=>     [tex]I = 0.02 \ A[/tex]

This  current flow is mathematically represented as

           [tex]I = \frac{q -q_o}{\Delta t }[/tex]

Where [tex]q_o[/tex] is the charge delivered at t=0 s which is 0s

     So

             [tex]0.02 = \frac{ (3.0 *10^{-9}) -0 }{t - 0 }[/tex]

               [tex]t = 1.50 *10^{-7 } \ s[/tex]

               [tex]t = 150 *10^{-9 } \ s[/tex]

Answer:

Option (A) : 1m

Explanation:

According to Hooke's law:

F (spring elastic force) =

k( spring const.) * x(displacement)

Case-1

2 N = k * 0.4m

k = 5

Case- 2

5 N = 5 * x

x ( displacement) = 1 m

The displacement of the spring if a 5-N force was applied is equal to 1m. Therefore, option (1) is correct.

The strain and stress are proportional to each other, and this is called Hooke’s Law. Hooke’s law states that the strain is proportional to the stress applied within the elastic limit of the material.

When the materials are stretched, the atoms or molecules deform and when the stress is removed, they will return to their original state.

The mathematical equation for Hooke's law is as follows:

F = –kx

where F is the force, x is displacement, and k is the spring constant in N/m.

Given, F = 2N and x = 0.4m

F = -kx

2 N = - k (0.4m)

k = 5 N/m where the negative sign is omitted.

Now, the spring constant of the spring, k = 5 N/m and F = 5N

F = -kx

5 N = - (5 N/m)(x)

x = - 1m

Therefore, the displacement of the spring is 1 m.

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Answer:

Study Earth as a whole

Explanation:

ex. oxygen around Earth, layers, formations, temperature, mountains and how they form etc.

Answer:

Geologists study rocks and help to locate useful minerals. Earth scientists often work in the field—perhaps climbing mountains, exploring the seabed, crawling through caves, or wading in swamps. They measure and collect samples (such as rocks or river water), then they record their findings on charts and maps.

Answer:

p-aminoazobenzene is formed

Explanation:

The reaction of benzene diazonium chloride and aniline takes place in a basic medium and leads to the formation of an azo compound which is also a dye. The terminal diazonium nitrogen of the benzene diazonium ion is coupled to the aniline at the para-position. The product of the reaction, p-aminoazobenzene is a yellow dye.

Benzene diazonium chloride is prepared by diazotization of aniline in the presence of hydrochloric acid. The full reaction of aniline and benzene diazonium chloride​ is shown in the image attached to this answer.

Answer:

The box will cover a distance of 0.9199m before coming to rest

Explanation:

We are given;

Angle of tilt; θ = 12°

Speed of sliding down; u = 1.5 m/s

Coefficient of kinetic friction; μ = 0.34

We are told that the box is sliding down an incline tilted at a 12° angle above horizontal.

Thus,

The components of the weight of the block would be;

Fx = mg sinθ = mg sin 12

Fy = mg cosθ = mg cos 12

For, the normal force on the block, it will be counter balanced by the Y component of weight of block and so we have;

Normal force; Fn = mg cos 12

Now formula for the frictional force would be given by;

Ff = μmg cos 12

So, Ff = 0.34mg cos 12

So, the net force along the inclined plane is;

Fnet = Fx - Ff

Fnet = mg sin 12 - 0.34mg cos 12

Where Fnet = mass x acceleration.

Thus;

ma = mg sin 12 - 0.34mg cos 12

m will cancel out to give;

a = g sin 12 - 0.34g cos 12

a = 9.81(0.2079) - 0.34(9.81 × 0.9781)

a = -1.223 m/s²

According to Newton's equation of motion, we have;

(v² - u²) = 2as

s = (v² - u²)/2a

Final velocity is zero. Thus;

s = (0² - 1.5²)/(2 × -1.223)

s = -2.25/-2.446

s = 0.9199 m

Thus, the box will cover 0.9199m before coming to rest

Answer:

The rms current in the circuit is 3.513 A

Explanation:

Given;

angular frequency of the inductor, ω = 363 rad/s

maximum voltage of the inductive AC, V₀ = 169 V

Inductance of the inductor, L = 0.0937 H

Inductive reactance is given by;

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

[tex]X_L = 363 *0.0937\\\\X_L = 34.0131 \ ohms[/tex]

The rms voltage is given by;

[tex]V_{rms} = \frac{V_o}{\sqrt{2} } \\\\V_{rms} =\frac{169}{\sqrt{2} } \\\\V_{rms} = 119.5 \ V[/tex]

The rms current in the circuit is given by;

[tex]I_{rms} = \frac{V_{rms}}{X_L} \\\\I_{rms} = \frac{119.5}{34.0131} \\\\I_{rms} = 3.513 \ A[/tex]

Therefore, the rms current in the circuit is 3.513 A

Answer: The properties of magnets are used to make electricity. Moving magnetic fields pull and push electrons. Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current.

Answer:

a. The reactance of the inductor is XL = V₀/I₀

b. The inductance of the inductor is L = V₀/2πfI₀

Explanation:

PART A

Since the voltage across the inductor V₀ = I₀XL where V₀ = e.m.f of voltage source, I₀ = current amplitude and XL = reactance of the inductor,

XL = V₀/I₀

So, the reactance of the inductor is XL = V₀/I₀

PART B

The inductance of the inductor is gotten from XL = 2πfL where f = frequency of voltage source and L = inductance of inductor

Since XL = V₀/I₀ = 2πfL

V₀/I₀ = 2πfL

L = V₀/2πfI₀

So the inductance of the inductor is L = V₀/2πfI₀

A) The reactance XL of the inductor :  [tex]\frac{V_{0} }{I_{0} }[/tex]  

B) The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]  

A) Expressing the Reactance of the inductor

Voltage across the Inductor = V₀ = I₀XL   ---- ( 1 )

Where :  V₀ = emf voltage ,  I₀ = current

from equation ( 1 )

∴ XL ( reactance ) = [tex]\frac{V_{0} }{I_{0} }[/tex]  

B ) Expressing the Inductance of the Inductor

Inductance of an inductor is expressed as : XL = 2πfL

from part A

XL = [tex]\frac{V_{0} }{I_{0} }[/tex] = 2πfL

∴ The inductance L of the Inductor expressed in terms of V₀, F and I₀

L = [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]

Hence we can conclude that The reactance XL of the inductor :  [tex]\frac{V_{0} }{I_{0} }[/tex]  and The Inductance L of the inductor : [tex]\frac{V_{0} }{2\pi fl_{0} }[/tex]  .

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Answer:

0 J

Explanation:

Since work done W = ∫F.dr and F(x, y, z)= z²i + 4xyj + 5y²k and dr = dxi + dyj + dzk

F.dr = (z²i + 4xyj + 5y²k).(dxi + dyj + dzk) = z²dx + 4xydy + 5y²dz

W = ∫F.dr = ∫z²dx + 4xydy + 5y²dz = z²x + 2xy² + 5y²z

We now evaluate the work done for the different regions

W₁ = work done from (0,0,0) to (1,0,0)

W₁ = {z²x + 2xy² + 5y²z}₀₀₀¹⁰⁰ = 0²(1) + 2(1)(0)² + 5(0)²(0) - [(0)²(0) + 2(0)(0)² + 5(0)²(0)] = 0 - 0 = 0 J

W₂ = work done from (1,0,0) to (1,5,1)

W₂ = {z²x + 2xy² + 5y²z}₁₀₀¹⁵¹ =   (1)²(1) + 2(1)(5)² + 5(5)²(1) - [0²(1) + 2(1)(0)² + 5(0)²(0)] =  1 + 50 + 125 - 0 = 176 J

W₃ = work done from (1,5,1) to (0,5,1)

W₃ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁵¹ =   1²(0) + 2(0)(5)² + 5(5)²(1) - [(1)²(1) + 2(1)(5)² + 5(5)²(1)]  = 125 - (1 + 50 + 125) = 125 - 176 = -51 J

W₄ = work done from (0,5,1) to (0,0,0)

W₄ = {z²x + 2xy² + 5y²z}₁₅₁⁰⁰⁰ =   (0)²(0) + 2(0)(0)² + 5(0)²(0) - [1²(0) + 2(0)(5)² + 5(5)²(1)] = 0 - 125 = -125 J

The total work done W is thus

W = W₁ + W₂ + W₃ + W₄

W = 0 J + 176 J - 51 J - 125 J

W = 176 J - 176 J

W = 0 J

The total work done equals 0 J

The question is not complete so i have attached it.

Answer:

The light source is 2 cm from the bottom of the lamp

Explanation:

From the attached image, we can see that the parabola opens up with its vertex at the origin.

Now, the standard form of equation for a parabola is:

x² = 4ay

Looking at the parabola in the attachment, the top right edge of the lamp has a coordinate of (12,18)

Thus, we have;

12² = 4a(18)

144 = 72a

a = 144/72

a = 2

Looking at the parabola again, the line of symmetry is at x = 0

Thus, axis of symmetry is at x = 0.

Thus, focus is at (0, 2)

So, if the light source is placed at the focus, the distance of the light source from the bottom of the lamp is 2 cm

The distance of the light source from the bottom of the lamp is 2 cm.

The given parameters;

Apply standard parabola equation to determine the distance of the light source from the bottom of the lamp;

[tex]x^2 = 4ay\\\\12^2 = 4a(18)\\\\144 = 72 a\\\\a = \frac{144}{72} \\\\a = 2 \ cm[/tex]

Thus, the distance of the light source from the bottom of the lamp is 2 cm.

"Your question is not complete, it seems to be missing the following information";

the top right edge of the lamp has a coordinate of (12,18)

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Answer:

9.43*10^3 year

Explanation:

For this question, we ought to remember, or know that the half life of carbon 14 is 5730, and that would be vital in completing the calculation

To start with, we use the formula

t(half) = In 2/k,

if we make k the subject of formula, we have

k = in 2/t(half), now we substitute for the values

k = in 2 / 5730

k = 1.21*10^-4 yr^-1

In(A/A•) = -kt, on rearranging, we find out that

t = -1/k * In(A/A•)

The next step is to substitite the values for each into the equation, giving us

t = -1/1.21*10^-4 * In(5.4/15.3)

t = -1/1.21*10^-4 * -1.1041

t = 0.943*10^4 year

Answer:

The width is [tex]w_c = 0.00252 \ m[/tex]

Explanation:

From the question we are told that

  The  width of the single slit is  [tex]a = 1.5 \ mm = 1.5 *10^{-3} \ m[/tex]

   The  wavelength is  [tex]\lambda = 420 *10^{-9} \ m[/tex]

   The distance of the screen is  [tex]D = 4.5 \ m[/tex]

Generally the width of the central maximum is  

        [tex]w_c = 2 * y[/tex]

where y is the width of the first maxima which is mathematically represented as

       [tex]y = \frac{\lambda * D}{a}[/tex]

=>   [tex]y = \frac{ 420 *10^{-9} * 4.5}{ 1.5*10^{-3}}[/tex]

=>  [tex]y = 0.00126 \ m[/tex]

So

    [tex]w_c = 2 *0.00126[/tex]

    [tex]w_c = 0.00252 \ m[/tex]

Answer:

The blade undergoes 40 revolutions, so neither of the given options is correct!

Explanation:

The revolutions can be found using the following equation:

[tex]\theta_{f} = \theta_{i} + \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

Where:

α is the angular acceleration

t is the time = 2.5 s

[tex]\omega_{i}[/tex] is the initial angular velocity = 1500 rev/min                

First, we need to find the angular acceleration:

[tex] \alpha = \frac{\omega_{f} - \omega_{i}}{t} = \frac{400 rev/min*2\pi rad*1 min/60 s - 1500 rev/min *2\pi rad*1 min/60 s}{2.5 s} = -46.08 rad/s^{2} [/tex]

Now, the revolutions that the blade undergo are:

[tex]\theta_{f} - \theta_{i} = \omega_{i}*t + \frac{1}{2}\alpha*t^{2}[/tex]

[tex]\Delta \theta = 1500 rev/min *2\pi rad*1 min/60 s*2.5 s - \frac{1}{2}*(46.08 rad/s^{2})*(2.5)^{2} = 248.7 rad = 39.9 rev[/tex]        

Therefore, the blade undergoes 40 revolutions, so neither of the given options is correct!

I hope it helps you!