Helium-neon laser light (λ = 6.33 × 10−7 m) is sent through a 0.30 mm-wide single slit. What is the width of the central maximum on a screen 1.0 m from the slit?

Answers

Answer 1

Answer:

The width is [tex]w_c = 0.00422 \ m[/tex]

Explanation:

From the question we are told that

   The  wavelength is  [tex]\lambda = 6.33*10^{-7} \ m[/tex]

    The  width of the slit is  [tex]d = 0.3\ mm = 0.3 *10^{-3} \ m[/tex]

    The distance of the screen is  [tex]D = 1.0 \ m[/tex]

     

Generally the central maximum is mathematically represented as

      [tex]w_c = 2 * y[/tex]

Here  y is the width of the first order maxima which is mathematically represented as

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

substituting values

      [tex]y = \frac{6.33*10^{-7} * 1.0}{ 0.30}[/tex]

       [tex]y = 0.00211 \ m[/tex]

So  

    [tex]w_c = 2 *0.00211[/tex]

     [tex]w_c = 0.00422 \ m[/tex]


Related Questions

A planar electromagnetic wave is propagating in the +x direction. At a certain point P and at a given instant, the electric field of the wave is given by = (0.082 V/m) . What is the magnetic vector of the wave at the point P at that instant? (c = 3.0 × 108 m/s)

Answers

Answer:

[tex]B=2.74\times 10^{-10}\ T[/tex]

Explanation:

It is given that,

A planar electromagnetic wave is propagating in the +x direction.The electric field at a certain point is, E = 0.082 V/m

We need to find the magnetic vector of the wave at the point P at that instant.

The relation between electric field and magnetic field is given by :

[tex]c=\dfrac{E}{B}[/tex]

c is speed of light

B is magnetic field

[tex]B=\dfrac{E}{c}\\\\B=\dfrac{0.082}{3\times 10^8}\\\\B=2.74\times 10^{-10}\ T[/tex]

So, the magnetic vector at point P at that instant is [tex]2.74\times 10^{-10}\ T[/tex].

The magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]

The formula relating electric field and the magnetic field is given as;

[tex]c=\frac{E}{B}[/tex]

E is the electric field strengthB is the magnetic vector of the wavec is the speed of light

From the formula shown:

[tex]B=\frac{E}{c}\\B=\frac{0.082}{3.0\times 10^8}\\B=2.73 \times 10 ^{-10}T[/tex]

Hence the magnetic vector of the wave at the point P at that instant is [tex]2.73 \times 10^{-10}T[/tex]

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Rank the ultraviolet, infrared, and visible regions of theelectromagnetic spectrum in terms of lowest to highest energy,frequency, and wavelength.
Energy: < <
Frequency: < <
Wavelength: <

Answers

Answer:

1. Energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared >> visible > ultraviolet

Explanation:

Electromagnetic waves are a class of waves that do not require material medium for their propagation, and travel at the same speed. They are arranged with respect to either their decreasing wavelength or increasing frequency to form a spectrum called an electromagnetic spectrum.

Comparing the energy, frequency and wavelength of ultraviolet, infrared and visible regions, it can be deduced that:

1. Energy: ultraviolet has the highest energy, then followed by visible, and infrared has the lowest energy.

i.e energy: ultraviolet>> visible> infrared

2. Frequency: ultraviolet radiation has the highest frequency, visible region has a greater frequency than that of infrared.

i.e frequency: ultraviolet>> visible > infrared

3. Wavelength: infrared radiation has the highest wavelength, followed by visible region, and ultraviolet radiation has the lowest wavelength.

i.e  wavelength: infrared >> visible > ultraviolet

In terms of lowest to the highest energy,frequency, and wavelength is;

Energy: infrared > visible light > ultraviolet

Frequency: infrared > visible light > ultraviolet

Wavelength: ultraviolet > visible light > infrared

The electromagnetic spectrum:

The electromagnetic spectrum is made up of all the electromagnetic waves (ultraviolet, infrared, and visible) arranged according to their energy,frequency, and wavelength.

The ultraviolet: This wave is seen in the sunlight and is made up of wavelength of 10nm to 400nm. A frequency of [tex]10^{16}[/tex](Hz).Infrared wave: They are invisisble to the human eye but can be felt as heat. It has frequency of [tex]10^{12}[/tex]Hz and a wavelength of 780nm to 1mm.Visible light: This is part of the electromagnetic wave that the eye can view. It has frequency of [tex]10^{15}[/tex]Hz and a wavelength of 380 to 700nm.

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In France, the wall sockets provide an AC voltage with Vrms = 230 V. You want to use an appliance designed to operate in the United States (Vrms = 120 V) and decide to build a transformer to convert the power line voltage in France to the value required by your appliance.
(a) Should you use a "step-down" transformer (to make Vrms smaller) or a "step-up" transformer (which makes Vrms larger)?
a "step-up" transformer
a "step-down" transformer
(b) If the input coil of your transformer has 2760 turns, how many turns should the output coil have?
_____ turns

Answers

Answer:

a)step-down" transformer

b) 1440 turns

Explanation:

There are two types of transformers; step up transformers and step down transformers. A step down transformer converts a higher voltage to a lower voltage.

In a stepdown transformer, there are more turns in the primary coil than in the secondary coil, the turns ratio Ns/Np is less than 1 for a stepdown transformer.

If

Number of turns in primary coil Np= 2760

Number of turns in secondary coil Ns= unknown

Voltage in primary coil Vp= 230 V

Voltage in secondary coil Vs= 120 V

Ns/Np= Vs/VP

NsVp= NpVs

Ns= NpVs/VP = 2760 × 120/230

Ns= 1440 turns

Copper Pot A copper pot with a mass of 2 kg is sitting at room temperature (20°C). If 200 g of boiling water (100°C) are put in the pot, after a few minutes the water and the pot come to the same temperature. What temperature is this in °C?

Answers

Answer:

The final temperature is 61.65 °C

Explanation:

mass of copper pot [tex]m_{c}[/tex] = 2 kg

temperature of copper pot [tex]T_{c}[/tex] = 20 °C  (the pot will be in thermal equilibrium with the room)

specific heat capacity of copper [tex]C_{c}[/tex]= 385 J/kg-°C

The heat content of the copper pot = [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{c}[/tex] = 2 x 385 x 20 = 15400 J

mass of boiling water [tex]m_{w}[/tex] = 200 g = 0.2 kg

temperature of boiling water [tex]T_{w}[/tex] = 100 °C

specific heat capacity of water [tex]C_{w}[/tex] = 4182 J/kg-°C

The heat content of the water = [tex]m_{w}[/tex][tex]C_{w}[/tex][tex]T_{w}[/tex] = 0.2 x 4182 x 100 = 83640 J

The total heat content of the water and copper mix [tex]H_{T}[/tex] = 15400 + 83640 = 99040 J

This same heat is evenly distributed between the water and copper mass to achieve thermal equilibrium, therefore we use the equation

[tex]H_{T}[/tex] =   [tex]m_{c}[/tex][tex]C_{c}[/tex][tex]T_{f}[/tex] + [tex]m_{w}[/tex][tex]C_{w}[/tex]

where [tex]T_{f}[/tex] is the final temperature of the water and the copper

substituting values, we have

99040 = (2 x 385 x [tex]T_{f}[/tex]) + (0.2 x 4182 x

99040 = 770[tex]T_{f}[/tex] + 836.4

99040 = 1606.4[tex]T_{f}[/tex]

[tex]T_{f}[/tex] = 99040/1606.4 = 61.65 °C

An expensive vacuum system can achieve a pressure as low as 1.53 ✕ 10−7 N/m2 at 26°C. How many atoms are there in a cubic centimeter at this pressure and temperature?

Answers

Answer:

The  value is  [tex]N = 3.708*10^{7} \ \ atoms[/tex]

Explanation:

From the question we are told that

    The pressure is  [tex]P = 1.53 *10^{-7} \ N/m^2[/tex]

    The  temperature is  [tex]T = 26 + 273 = 299 \ K[/tex]

     The volume is  1 cubic cm = [tex]1 * 10^{-6} m^3[/tex]

Generally according to the ideal gas law we have that

      [tex]PV = NkT[/tex]

here  k is the Boltzmann constant with a value  [tex]k = 1.38 *10^{-23} \ J/K[/tex]

  =>  [tex]N = \frac{PV}{ k T}[/tex]

=>     [tex]N = \frac{ 1.53 *10^{-7} * (1* 10^{-6})}{ 1.38*10^{-23} * 299}[/tex]

=>    [tex]N = 3.708*10^{7} \ \ atoms[/tex]

     

A speeding car has a velocity of 80 mph; suddenly it passes a cop car but does not stop. When the speeding car passes the cop car, the cop immediately accelerates his vehicle from 0 to 90 mph in 4.5 seconds. The cop car has a maximum velocity of 90 mph. At what time does the cop car meet the speeding car and at what distance?

Answers

Answer:

Distance= 4 miles

Time = 36.3 seconds

Explanation:

80 mph = 178.95 m/s

90 mph = 201.32 m/s

V = u +at

201.32= 0+a(4.5)

201.32/4.5= a

44.738 m/s² = a

Acceleration of the cop car

= 44.738 m/s²

Distance traveled at 4.5seconds

For the cop car

S= ut + ½at²

S= 0(4.5) + ½*44.738*4.5

S= 100.66 meters

Distance traveled at 4.5seconds

For the speeding car

4.5*178.95=805.275

The cop car will still cover 704.675 +x distance while the speeding car covers for their distance to be equal

X/178.95= (704.675+x)/201.32

X-0.89x= 626.37

0.11x= 626.37

X= 5694.3 meters

The time = 5694.3/178.95

Time =31.8 seconds

So the distance they meet

= 5694.3+805.275

= 6499.575 meters

= 4.0 miles

The Time = 4.5+31.8

Time = 36.3 seconds

A chemist must dilute 55.6 ml of 1.48 M aqueous silver nitrate (AgNO3)solution until the concentration falls to 1.00 M. He'll do this by adding distilled water to the solution until it reaches a certain final volume. Calculate this final volume, in milliliters. Round your answer to 3 significant digits.

Answers

Answer:

82.2 mL

Explanation:

The process of adding water to a solution to make it more dilute is known as dilution. The formula for dilution is;

C1V1=C2V2

Where;

C1= concentration of stock solution

V1= volume of stock solution

C2= concentration of dilute solution

V2= volume of dilute solution

V2= C1V1/C2

V2= 1.48 × 55.6/ 1.0

V2= 82.2 mL

1.) When the acceleration is zero, what can you say about the velocity of an object?

Answers

Answer:

it is either constant or zero

Explanation:

A stereo speaker produces a pure "G" tone, with a frequency of 392 Hz. What is the period T of the sound wave produced by the speaker?

Answers

Answer:

The  period is [tex]T = 0.00255 \ s[/tex]

Explanation:

From the question we are told that

  The  frequency is  [tex]f = 392 \ Hz[/tex]

Generally the period is mathematically represented as  

           [tex]T = \frac{1}{f}[/tex]

=>       [tex]T = \frac{1}{ 392}[/tex]

=>       [tex]T = 0.00255 \ s[/tex]

Did the kinetic frictional coefficient (for the wood/aluminum and felt/aluminum cases) vary with area of contact

Answers

Answer:

Explanation:

Friction is defined as a force which acts at the surface of separation between two objects in contact and tends to oppose motion of one over the other.

While kinetic friction is the force that must be overcome so that a body can move with uniform speed over another.

Hence let consider one of the laws of friction which states that: '' Frictional force is independent of the area of the surfaces in contact.''

The value did not vary with area. This is because when calculating the kinetic fiction, the total contact area is not relevant and only the total weight of the system as well of as the block is put into consideration.

A circular loop of wire of area 25 cm2 lies in the plane of the paper. A decreasing magnetic field B is coming out of the paper. What is the direction of the induced current in the loop?

Answers

Answer:

counterclockwise

Explanation:

given data

area = 25 cm²

solution

We know that a changing magnetic field induces the current and induced emf is express as

[tex]\epsilon = -N \frac{d \phi }{dt}[/tex]     ..................................1

and we will get here direction of the induced current in the loop that is express by the Lens law that state that the direction of induces current is such that the magnetic flux due to the induced current opposes the change in magnetic flux due to the change in magnetic field

so when magnetic field decrease and point coming out of the paper.

so induced current in the loop will be counterclockwise

An atom in the ground state has a collision with an electron, then emits a photon with a wavelength of 1240 nm. What conclusion can you draw about the initial kinetic energy of the electron

Answers

Answer:

attached below is the free body diagram of the missing  illustration

Initial kinetic energy of the electron = 3 eV

Explanation:

The conclusion that can be drawn about the kinetic energy of the electron is

[tex]E_{e} = E_{3} - E_{1}[/tex]

E[tex]_{e}[/tex] = initial kinetic energy of the electron

E[tex]_{1}[/tex] = -4 eV

E[tex]_{3}[/tex] = -1 eV

insert the values into the equation above

[tex]E_{e}[/tex] = -1 -(-4)  eV

   = -1 + 4 = 3 eV

A square loop 16.0 cm on a side has a resistance of 6.35 Ω . It is initially in a 0.510 T magnetic field, with its plane perpendicular to B , but is removed from the field in 40.5 ms.

Required:
Calculate the electric energy dissipated in this process.

Answers

Answer:

Explanation:

change in magnetic flux = .16 x .16 x .510 - 0

= .013056 weber .

rate of change of flux = change in flux / time

= .013056 / 40.5 x 10⁻³

= .32237

voltage induced = .32237 V

electrical energy dissipated = v² / R where v is voltage , R is resistance

= .32237² / 6.35

= 16.36 x 10⁻³ J .

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, how many counts per second would be observed when the detector is 3 meters from the sample?

Answers

Answer:

6000 counts per second

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

2000 counts per second = 1 meter ... 1

In order to know the number of counts per second that would be observed when the detector is 3 meters from the sample, we will have;

x count per second = 3 meter ... 2

Solving the two expressions simultaneously for x we will have;

2000 counts per second = 1 meter

x counts per second = 3 meter

Cross multiply to get x

2000 * 3 = 1* x

6000 = x

This shows that 6000 counts per second would be observed when the detector is 3 meters from the sample

A mass M is attached to an ideal massless spring. When this system is set in motion with amplitude A, it has a period T. What is the period if the amplitude of the motion is doubled

Answers

Answer:

The period of the motion will still be equal to T.

Explanation:

for a system with mass = M

attached to a massless spring.

If the system is set in motion with an amplitude (distance from equilibrium position) A

and has period T

The equation for the period T is given as

[tex]T = 2\pi \sqrt{\frac{M}{k} }[/tex]

where k is the spring constant

If the amplitude is doubled, the distance from equilibrium position to the displacement is doubled.

Increasing the amplitude also increases the restoring force. An increase in the restoring force means the mass is now accelerated to cover more distance in the same period, so the restoring force cancels the effect of the increase in amplitude. Hence, increasing the amplitude has no effect on the period of the mass and spring system.

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Answers

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

16. If one body is positively charged and another body is negatively charged, free electrons tend to
O A. move from the negatively charged body to the positively charged body
O B. remain in the positively charged body
OC. move from the positively charged body to the negatively charged body
O D. remain in the negatively charged body

Answers

Answer:

Hey there!

The correct answer would be option A. If one body is positively charged and another body is negatively charged, free electrons tend to move from the negatively charged body to the positively charged body

Let me know if this helps :)

The resistor used in the procedures has a manufacturer's stated tolerance (percent error) of 5%. Did you results from Data Table agree with the manufacturer's statement? Explain.
Resistor Measured Resistance
100 99.1

Answers

Answer:

     e% = 0.99%   this value is within the 5% tolerance given by the manufacturer

Explanation:

Modern manufacturing methods establish a tolerance in order to guarantee homogeneous characteristics in their products, in the case of resistors the tolerance or error is given by

          e% = | R_nominal - R_measured | / R_nominal 100

where R_nominal is the one written in the resistance in your barcode, R_measured is the real value read with a multimeter and e% is the tolerance also written in the resistors

let's apply this formula to our case

R_nominal = 10 kΩ = 10000 Ω

R_measured = 100 99 Ω

        e% = | 10000 - 10099.1 | / 10000 100

        e% = 0.99%

this value is within the 5% tolerance given by the manufacturer

Two long straight wires carry currents perpendicular to the xy plane. One carries a current of 50 A and passes through the point x = 5.0 cm on the x axis. The second wire has a current of 80 A and passes through the point y = 4.0 cm on the y axis. What is the magnitude of the resulting magnetic field at the origin?

Answers

Answer:

450 x10^-6 T

Explanation:

We know that the magnetic of each wire is derived from

ByB= uoi/2pir

Thus B1= 4 x 3.14 x 10^-7 x50/( 2 x 3.142x 0.05)

= 0.2 x 10^ -3T

B2=

4 x 3.14 x 10^-7 x80/( 2 x 3.142x 0.04)

= 0.4 x 10^ -3T

So

(Bnet)² = (Bx)² + ( By)²

= (0.2² + 0.4²)mT

= 450 x10^-6T

The magnitude of magnetic field at the origin is required.

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

x = Location at x axis = 5 cm

y = Location at y axis = 4 cm

[tex]I_x[/tex] = Current at the x axis point = 50 A

[tex]I_y[/tex] = Current at the y axis point = 80 A

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi\times 10^{-7}\ \text{H/m}[/tex]

Magnitude of the magnetic field is given by

[tex]B=\dfrac{\mu_0I}{2\pi r}[/tex]

Finding the net magnetic field using the Pythagoras theorem

[tex]B^2=B_x^2+B_y^2\\\Rightarrow B^2=\left(\dfrac{\mu_0I_x}{2\pi x}\right)^2+\left(\dfrac{\mu_0I_y}{2\pi y}\right)^2\\\Rightarrow B=\dfrac{\mu_0}{2\pi}\sqrt{\left(\dfrac{I_x}{x}\right)^2+\left(\dfrac{I_y}{y}\right)^2}\\\Rightarrow B=\dfrac{4\pi\times 10^{-7}}{2\pi}\sqrt{\left(\dfrac{50}{0.05}\right)^2+\left(\dfrac{80}{0.04}\right)^2}\\\Rightarrow B=0.0004472=447.2\ \mu\text{T}[/tex]

The magnitude of resulting magnetic field at origin is [tex]447.2\ \mu\text{T}[/tex]

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A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

Convert 7,348 grams to kilograms

Answers

0.00735 kilograms is your answer. I’m sorry If I didn’t explain it right :(,

A car is travelling west at 22.2 m/s when it accelerated for 0.80 s to the west at 2.68 m/s2. Calculate the car's final velocity. Show all your work.

Answers

Answer:

24.34 m/s

Explanation:

recall that one of the equations of motions takes the form:

v = u + at

where,

v = final velocity

u = initial velocity (given as 22.2 m/s)

a = acceleration (given as 2.68m/s²)

t = time elapsed during acceleration (given as 0.80s)

since we are told that the the acceleration is in the direction of the intial velocity, we can simply substitute the known values into the equation above:

v = u + at

v = 22.2 + (2.68) (0.8)

v = 24.34 m/s

Approximating the eye as a single thin lens 2.70 cmcm from the retina, find the focal length of the eye when it is focused on an object at a distance of 265 cmcm

Answers

Answer:

0.37 cm

Explanation:

The image is formed on the retina which is at a constant distance of 2.70 cm to the lens. Therefore, image distance = 2.70 cm.

The object is at a distant of 265 cm to the lens of the eye.

From lens formula,

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

where: f is the focal length, u is the object distance and v is the image distance.

Thus, u = 265.00 cm and v = 2.70 cm.

[tex]\frac{1}{f}[/tex] = [tex]\frac{1}{265}[/tex] + [tex]\frac{27}{10}[/tex]

  = [tex]\frac{10+7155}{2650}[/tex]

[tex]\frac{1}{f}[/tex]  = [tex]\frac{7165}{2650}[/tex]

⇒ f = [tex]\frac{2650}{7165}[/tex]

      = 0.37

The focal length of the eye is 0.37 cm.

what is defect of vision​

Answers

Answer:

The vision becomes blurred due to the refractive defects of the eye. There are mainly three common refractive defects of vision. These are (i) myopia or near-sightedness, (ii) Hypermetropia or far – sightedness, and (iii) Presbyopia. These defects can be corrected by the use of suitable spherical lenses.

An object is made of glass and has the shape of a cube 0.13 m on a side, according to an observer at rest relative to it. However, an observer moving at high speed parallel to one of the object's edges and knowing that the object's mass is 2.0 kg determines its density to be 7300 kg/m3, which is much greater than the density of glass. What is the moving observer's speed (in units of c) relative to the cube

Answers

Answer:

The  velocity is  [tex]v = 2.6*10^{8} \ m/s[/tex]

Explanation:

From the  question we are told that

   The side of the cube is  [tex]l = 0.13 \ m[/tex]

   The  mass of the object is  [tex]m = 2.0 \ kg[/tex]

   The  density of the object is  [tex]\rho = 7300 \ kg / m^3[/tex]

Generally the volume of the object according to the moving observer is mathematically represented  as

        [tex]V =\frac{m}{\rho}[/tex]

        [tex]V =\frac{2}{7300}[/tex]

       [tex]V = 2.74*10^{-4} \ m^3[/tex]

Therefore the length of the side as observed by the observer on high speed is mathematically represented as

     [tex]L = \sqrt[3]{V}[/tex]        

     [tex]L = \sqrt[3]{2.74 *10^{-4}}[/tex]    

     [tex]L =0.065[/tex]

Now the original length of side is mathematically represented as

      [tex]L= l * \sqrt{ (1 - ( \frac{ v}{c})^2 )}[/tex]

Where c is the speed of light with value [tex]c = 3.0*10^{8} \ m/s[/tex]

So

     [tex]v = \sqrt{1 - [\frac{L}{l}]^2} * c[/tex]

=>  [tex]v = \sqrt{1 - [\frac{0.065}{0.13}]^2} * c[/tex]

=>   [tex]v = 2.6*10^{8} \ m/s[/tex]

y=k/x, x is halved.
what happens to the value of y

Answers

Answer:

y is doubled

Explanation:

If x is halved, that means the value is doubled. Here is an exmaple:

y=1/2. If the denominater is doubled, y would equal y=1/1. So, the value of y has doubled from 0.5 to 1. Therefore, if the denominator is halved, the solution will be doubled.

If a disk rolls on a rough surface without slipping, the acceleration of the center of gravity (G) will _ and the friction force will b

Answers

Answer:

Will be equal to alpha x r; less than UsN

A car accelerates uniformly from rest and reaches a speed of 22.7 m/s in 9.02 s. Assume the diameter of a tire is 58.5 cm. (a) Find the number of revolutions the tire makes during this motion, assuming that no slipping occurs. rev (b) What is the final angular speed of a tire in revolutions per second? rev/s

Answers

(a) The car is undergoing an acceleration of

[tex]a=\dfrac{22.7\frac{\rm m}{\rm s}-0}{9.02\,\mathrm s}\approx2.52\dfrac{\rm m}{\mathrm s^2}[/tex]

so that in 9.02 s, it will have covered a distance of

[tex]x=\dfrac a2(9.02\,\mathrm s)^2\approx102\,\mathrm m[/tex]

The car has tires with diameter d = 58.5 cm = 0.585 m, and hence circumference π d ≈ 1.84 m. Divide the distance traveled by the tire circumference to determine how many revolutions it makes:

[tex]\dfrac{102\,\mathrm m}{1.84\,\mathrm m}\approx55.7\,\mathrm{rev}[/tex]

(b) The wheels have average angular velocity

[tex]\omega=\dfrac{\omega_f+\omega_i}2=\dfrac{\theta_f-\theta_i}{\Delta t}[/tex]

where [tex]\omega[/tex] is the average angular velocity, [tex]\omega_i[/tex] and [tex]\omega_f[/tex] are the initial and final angular velocities (rev/s), [tex]\theta_i[/tex] and [tex]\theta_f[/tex] are the initial and final angular displacements (rev), respectively, and [tex]\Delta t[/tex] is the duration of the time between initial and final measurements. The second equality holds because acceleration is constant.

The wheels start at rest, so

[tex]\dfrac{\omega_f}2=\dfrac{55.7\,\rm rev}{9.02\,\rm s}\implies\omega_f\approx12.4\dfrac{\rm rev}{\rm s}[/tex]

A 70 kg human body typically contains 140 g of potassium. Potassium has a chemical atomic mass of 39.1 u and has three naturally occurring isotopes. One of those isotopes, 40K,is radioactive with a half-life of 1.3 billion years and a natural abundance of 0.012%. Each 40K decay deposits, on average, 1.0 MeV of energy into the body. What yearly dose in Gy does the typical person receive from the decay of 40K in the body?

Answers

Answer:

0.03143 Gy

Explanation:

Mass of the human body = 70 kg

Mass of potassium in the human body = 140 g

chemical atomic mass of potassium = 39.1

From avogadros number, we know that 1 atomic mass of an element contains 6.023 × 10^(23) atoms

Thus,

140g of potassium will contain;

(140 × 6.023 × 10^(23))/(39.1) = 2.1566 × 10^(24) atoms

We are told that the natural abundance of one of the 40K isotopes is 0.012%.

Thus;

Number of atoms of this isotope = 0.012% × 6.023 × 10^(23) = 7.2276 × 10^(19) K-40 atoms

Formula for activity of K-40 is given as;

Activity = (0.693 × number of K-40 atoms)/half life

Activity = (0.693 × 7.2276 × 10^(19))/1300000000

Activity = 3.85 × 10^(10)

We are told that each decay deposits 1.0 MeV of energy into the body.

Thus;

Total energy absorbed by the body in a year = 3.85 × 10^(10) × 1 × 365 = 1405.25 × 10^(10) MeV

Now, 1 MeV = 1.602 × 10^(-13) joules

Thus;

Total energy absorbed by the body in a year = 1405.25 × 10^(10) × 1.602 × 10^(-13) = 2.25 J

1 Gy = 1 J/kg

Thus;

Yearly dose = 2.25/70 = 0.03143 Gy

A piano string having a mass per unit length equal to 4.80 ✕ 10−3 kg/m is under a tension of 1,300 N. Find the speed with which a wave travels on this string.

Answers

Answer:

Velocity of wave (V) = 5.2 × 10² m/s

Explanation:

Given:

Per unit length mass (U) = 4.80 × 10⁻³ kg/m

Tension (T)= 1,300 N

Find:

Velocity of wave (V)

Computation:

Velocity of wave (V) = √T / U

Velocity of wave (V) = √1300 / 4.80 × 10⁻³

Velocity of wave (V) = √ 270.84 × 10³

Velocity of wave (V) = 5.2 × 10² m/s

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