A diffraction grating with 161 slits per centimeter is used to measure the wavelengths emitted by hydrogen gas. At what angles in the first-order spectrum would you expect to find the two violet lines of wavelength 434 nm and of wavelength 410 nm

Answers

Answer 1

Answer:

[tex]\theta_1 = 0.400^o[/tex]

[tex]\theta_2 =0.378^o[/tex]

Explanation:

From the question we are told that

    The  number of slits per cm is  k =  [tex]161\ slits\ per\ cm = 161 \ slits\ per\ 0.01 m[/tex]

    The order of the maxima is  n =  1

    The wavelength are  [tex]\lambda_1 = 434 nm = 434 *10^{-9} \ m \ \ \ , \lambda_2 = 410nm = 410 *10^{-9} \ m[/tex]

The  spacing between the slit is mathematically represented as

           [tex]d = \frac{ 0.01}{k}[/tex]

=>       [tex]d = \frac{ 0.01}{161}[/tex]

=>         [tex]d = 6.211 *10^{-5} \ m[/tex]

Generally the condition for constructive interference is  

        [tex]n\lambda = d \ sin \theta[/tex]

At  [tex]\lambda_1[/tex]

      [tex]\theta _1 = sin^{-1} [ \frac{1 * 434 *10^{-9}}{6.211 *10^{-5}} ][/tex]

      [tex]\theta_1 = 0.400^o[/tex]

At  [tex]\lambda_2[/tex]

       [tex]\theta _2 = sin^{-1} [ \frac{1 * 410 *10^{-9}}{6.211 *10^{-5}} ][/tex]

       [tex]\theta_2 =0.378^o[/tex]


Related Questions

You make a telephone call from New York to a friend in London. Estimate how long it will take the electrical signal generated by your voice to reach London, assuming the signal is (a) carried on a telephone cable under the Atlantic Ocean, and (b) sent via satellite 36,000 km above the ocean. Would this cause a noticeable delay in either case

Answers

Answer:

a)  t = 2 10⁻² s ,  t = 2.4 10-1 s

Explanation:

In this exercise they indicate the delay in two signals

a) A signal travels on an electrical cable, between New York and London.

The wave formed in this wire for the signal. This wave travels at the speed of light, c = 3 108 m / s, so the delay is very small

                t = d / c

               t = 6000 10³/3 10⁸

               t = 2 10⁻² s

b) The signal points to a satellite in geostationary orbit

                   distance traveled = √ (2 36000 10³)² + 6000²

                  distance = 72 10⁶ m

the waiting time is

                t = d / c

                t = 72 10⁶/3 10⁸

               t = 2.4 10-1 s

    we can see that the signal sent by the satellites has more delay because its distance is much greater

A 207-kg merry-go-round in the shape of a uniform, solid, horizontal disk of radius 1.50 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.750 rev/s in 2.00 s

Answers

Answer:

366 N

Explanation:

  τ = Iα

FR = ½mR²α

  F = ½mR(Δω/t)

  F = ½(207)(1.50)(0.75)(2π) /2.00

  F = 365.79919...

A resistor made of Nichrome wire is used in an application where its resistance cannot change more than 1.35% from its value at 20.0°C. Over what temperature range can it be used (in °C)?

Answers

Answer:

Pls seeattached file

Explanation:

A resistor made of Ni chrome wire is used in an application where its resistance cannot be more than 1.35 % so its temperature range will be from 33.75 to -33.75 °C.

What is Resistance?

Electrical resistance, or resistance to electricity, is a force that opposes the flow of current. Ohms are used to expressing resistance values.

When there is an electron difference between two terminals, electricity will flow from high to low. In opposition to that flow is resistance. As resistance rises, the current declines. On the other side, when the resistance falls, the current rises.

According to the question,

R = R₀ (1 + α ΔT)

(1 + 0.0135)R₀ = R₀(1 + α ΔT)

ΔT = (1 + 0.0135) / α

= 0.0135 / 0.0004

= 33.75 °C.

ΔT = [(1 - 0.0135) -1]/0.004

= -33.75 °C

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a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g

Answers

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

A brick weighs 50.0 N, and measures 30.0 cm × 10.0 cm × 4.00 cm. What is the maximum pressure it can exert on a horizontal surface due to its weight?

Answers

Answer:

Pressure, P = 1250 Pa

Explanation:

Given that,

Weight of a brick, F = 50 N

Dimension of the brick is 30.0 cm × 10.0 cm × 4.00 cm

We need to find the maximum pressure it can exert on a horizontal surface due to its weight. Pressure is equal to the force acting per unit area. Pressure exerted is inversely proportional to the area of cross section. So, we need to minimize area. Taking to smaller dimensions.

A = 40 cm × 10 cm = 400 cm² = 0.04 m²

So,

Pressure,

[tex]P=\dfrac{50\ N}{0.04\ m^2}\\\\P=1250\ Pa[/tex]

So, the maximum pressure of 1250 Pa it can exert on a horizontal surface.

The maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

What is pressure?

The force applied perpendicular to the surface of an item per unit area across which that force is spread is known as pressure. It is denoted by P.

The given data in the problem is;

W is the weight of a brick = 50 N

The dimension of the brick = 30.0 cm × 10.0 cm × 4.00 cm

A is the area,

The area is found as;

A=40 cm × 10 cm = 400 cm² = 0.04 m²

The pressure is the ratio of the force and area

[tex]\rm P = \frac{F}{A} \\\\ \rm P = \frac{50}{0.04} \\\\ \rm P =1250 \ Pascal[/tex]

Hence the maximum pressure it can exert on a horizontal surface due to its weight will be 1250 Pascal.

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You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb?

a. at the point, because all rays bouncing off the mirror will be parallel.
b. at the focal point of the mirror
c. at the radius of curvature of the mirror
d. none of the above, you cant make parallel rays wilth a concave mirror

Answers

Answer:

Explanation:

Concave mirrors is otherwise known as converging mirrors: These are mirrors that are caved inwards (reflecting surface is on the outside curved part). It is called a converging mirror due to the fact that light converges to a point when it strikes and reflects from the surface of the mirror. This type of mirror is used to focus light; parallel rays that are directed towards it will be concentrated to a point.

For a concave mirror to reflect light with properties that are the same as a spotlight (directed light rays parallel to each other), one has to consider its property to gather light to a point after reflecting. Meaning that, we can achieve the spotlight by locatng the point where the rays will be parallel, this point is called the focal point.

Therefore, the light bulb should be placed at the focal point of the mirror.

You need to make a spring scale to measure the mass of objects hung from it. You want each 1.0 cm length along the scale to correspond to a mass difference of 0.10 kg. What should be the value of the spring constant?

Answers

Answer:

The spring constant should be:

[tex]k= 98\, \frac{N}{m}[/tex]

Explanation:

Use Hooke's law for this problem, knowing that the magnitude of the force (F) on the spring equals the stretching it experiences [tex]\Delta x[/tex] times the spring constant "k":

[tex]F=k\,\Delta x[/tex]

in our case, since the mass hanging is given in kg, we need to multiply it by "g" to get the force exerted:

Then if we add to the spring in its relaxed state, a mass of 0.10 kg, and we want for that a displacement of 1 cm (0.01 m), then the value of the spring constant should be:

[tex]k=\frac{F}{\Delta x} \\k=\frac{9.8\,(0.1)}{0.01} \, \frac{N}{m} \\k= 98\, \frac{N}{m}[/tex]

A rod of length L is hinged at one end. The moment of inertia as the rod rotates around that hinge is ML2/3. Suppose a 2.50 m rod with a mass of 3.00 kg is hinged at one end and is held in a horizontal position. The rod is released as the free end is allowed to fall. What is the angular acceleration as it is released?

Answers

Answer:

6 rad/s²

Explanation:

Sum the torques about the hinge.

∑τ = Iα

mg(L/2) = mL²/3 α

g/2 = L/3 α

α = 3g/(2L)

α = 3 (10 m/s²) / (2 × 2.50 m)

α = 6 rad/s²

Which does not account for the fact that fish can survive the winter in ponds in temperate climate zones? 1. the density of ice versus that of water 2. the unique properties of water 3. the intermolecular bonding of water 4. the tendency for water to freeze from the bottom up

Answers

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

a 1010 W radiant heater is constructed to operate at 115 V. (a) What is the current in the heater when the unit is operating?

Answers

Answer:

8.78 Amps

Explanation:

Given data:

power rating of the heater P= 1010 W

voltage of the heater V= 115 volts

current taken by the heater I= ?

We can apply the power formula to solve for the current in the heater

i.e P= IV

Making I the current subject of formula we have

I= P/V

Substituting our given data into the expression for I we have

I=1010/115= 8.78 A

Hence the current when the unit/heater is operating is 8.78 Amp

A car starts from rest and accelerates with a constant acceleration of 5 m/s2 for 4 s. The car continues for 18 s at constant velocity. How far has the car traveled from its starting point

Answers

110m/s Or 36meters or miles, I think this is the answers

Hope this helped ♥︎

A metal blade of length L = 300 cm spins at a constant rate of 17 rad/s about an axis that is perpendicular to the blade and through its center. A uniform magnetic field B = 4.0 mT is perpendicular to the plane of rotation. What is the magnitude of the potential difference (in V) between the center of the blade and either of its ends?

Answers

We are being given that:

The length of a metal blade = 300 cmThe angular velocity at which the metal blade is rotating about its axis is ω = 17 rad/sThe magnetic field (B) = 4.0 mT

A pictorial view showing the diagrammatic representation of the information given in the question is being attached in the image below.

From the attached image below, the potential difference across the conducting element of the length (dx) moving with the velocity (v) appears to be perpendicular to the magnetic field (B).

The magnitude of the potential difference induced between the center of the blade in relation to either of its ends can be determined by using the derived formula from Faraday's law of induction which can be expressed as:

[tex]\mathsf{E = B\times l\times v}[/tex]

where;

B = magnetic fieldl = lengthv = relative speed

From the diagram, Let consider the length of the conducting element (dx) at a distance of length (x) from the center O.

Then, the velocity (v) = ωx

The potential difference across it can now be expressed as:

[tex]\mathsf{dE = B*(dx)*(\omega x)}[/tex]

For us to determine the potential difference, we need to carry out the integral form from center point O to L/2.

[tex]\mathsf{E = \int ^{L/2}_{0}* B (\omega x ) *(dx)}[/tex]

[tex]\mathsf{E = B (\omega ) \times \Big[ \dfrac{x^2}{2}\Big]^{L/2}_{0}}[/tex]

[tex]\mathsf{E = B (\omega ) * \Big[ \dfrac{L^2}{8}\Big]}[/tex]

Recall that,

magnetic field (B) = 4 mT = 4 × 10⁻³  TLength L = 300 cm = 3mangular velocity (ω) = 17 rad/s

[tex]\mathsf{E = (4\times 10^{-3}) * (17) \Big[ \dfrac{(1.5)^2}{8}\Big]}[/tex]

[tex]\mathsf{E = 19.13 mV}[/tex]

Thus, we can now conclude that the magnitude of the potential difference as a result of the rotation caused by the metal blade from the center to either of its ends is 19.13 mV.

Learn more about Faraday's law of induction here:

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An LR circuit consists of a 35-mH inductor, ac resistance of 12 ohms, an 18-V battery, and a switch. What is the current 5.0 ms after the switch is closed

Answers

Answer:

 I = 1.23 A

Explanation:

In an RL circuit current passing is described by

           I = E / R (1 - [tex]e^{-Rt/L}[/tex])

Let's reduce the magnitudes to the SI system

        L = 35 mH = 35 10⁻³ H

        t = 5.0 ms = 5.0 10⁻³ s

let's calculate

         I = 18/12 (1 - [tex]e^{-12 .. 5 {10}^{-3}/35 .. {10}^{-3} }[/tex]e (- 5 10-3 12/35 10-3))

         I = 1.5 (1- [tex]e^{-1.715}[/tex])

         I = 1.23 A

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 microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.

A)The power output of the microwave oven is?
B) The intensity of the microwave beam is?
C) The electric field amplitude is?
D) The force on the base due to the radiation is?

Answers

Answer:

power = 600000 W

intensity = 6666666.66 W/m²

Em = 70880.18 N/m

F = 2 × [tex]10^{-3}[/tex] N  

Explanation:

given data

frequency f = 2400 MHz

height oven cavity h = 25 cm = 0.25 m

base area measures A =  30 cm by 30 cm

total microwave energy content of cavity E = 0.50 mJ = 0.50 × [tex]10^{-3}[/tex]  

solution

first, we get here total time taken from top to bottom that is express as

Δt = [tex]\frac{h}{c}[/tex]  ...............1

Δt = [tex]\frac{0.25}{3\times 10^8}[/tex]  

Δt = 8.33 × [tex]10^{-10}[/tex] s

and

power output will be

power = [tex]\frac{E}{\Delta t}[/tex]   ..............2

power = [tex]\frac{0.50 \times 10^{-3}}{8.33 \times 10^{-10}}[/tex]

power = 600000 W

and

intensity of the microwave beam is  

intensity = power output ÷ base area    ..............2

intensity =  [tex]\frac{600000}{30 \times 30 \times 10^{-4}}[/tex]  

intensity = 6666666.66 W/m²

and

electric field amplitude is

as we know intensity  I = [tex]\frac{E^2}{c \mu o}[/tex]     ...............3

[tex]E(rms) = \sqrt{Ic\ \mu o} \\E(rms) = \sqrt{6666666.66 \times 3 \times 10^{8} \times 4 \pi \times 10^{-7} }[/tex]

E(rms) = 50119.87 N/m

and we know

[tex]E(rms) = \frac{Em}{\sqrt{2}}\\50119.87 = \frac{Em}{\sqrt{2}}[/tex]  

Em = 70880.18 N/m

and

force on the base due to the radiation is by the radiation pressure

[tex]Pr = \frac{l}{c}[/tex]    ..................4

[tex]\frac{F}{A} = \frac{l}{c}[/tex]  

so

F = [tex]\frac{6666666.66 \times 900 \times 10^{-4}}{3\times 10^8}[/tex]  

F = 2 × [tex]10^{-3}[/tex] N  

You have a lightweight spring whose unstretched length is 4.0 cm. First, you attach one end of the spring to the ceiling and hang a 1.8 g mass from it. This stretches the spring to a length of 5.1 cm . You then attach two small plastic beads to the opposite ends of the spring, lay the spring on a frictionless table, and give each plastic bead the same charge. This stretches the spring to a length of 4.3 cm .

Requried:
What is the magnitude of the charge (in nC) on each bead?

Answers

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

An intergalactic rock star bangs his drum every 1.30 s. A person on earth measures that the time between beats is 2.50 s. How fast is the rock star moving relative to the earth

Answers

Answer:

v = 0.89 c = 2.67 x 10⁸ m/s

Explanation:

The time dilation consequence of the special theory of relativity shall be used here, From time dilation formula we have:

t = t₀/√[1 - v²/c²]

where,

t = time measured by the person on earth = 2.50 s

t₀ = rest time of the intergalactic rock star = 1.30 s

v = relative speed of the rock star = ?

Therefore,

2.5 s = (1.3 s)/√[1 - v²/c²]

√[1 - v²/c²] = 1.3/2.5

√[1 - v²/c²] = 0.52

[1 - v²/c²] = 0.52²

[1 - v²/c²] = 0.2074

v²/c² = 1 - 0.2074

v²/c² = 0.7926

v/c = √0.7926

v = 0.89 c

where,

c = speed of light = 3 x 10⁸ m/s

v = (0.89)(3 x 10⁸ m/s)

v = 0.89 c = 2.67 x 10⁸ m/s

Large capacitors can hold a potentially dangerous charge long after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 μF capacitor from a camera flash unit retains a voltage of 140 V when an unwary student removes it from the camera. If the student accidentally touches the two terminals with his hands, and if the resistance of his body between his hands is 1.8 kΩ, for how long will the current across his chest exceed the danger level of 50 mA?

Answers

Answer:

93.3x10^-3s

Explanation:

If

Resistance = 1.8 kΩ

Current = 50 mA

Capacitor = 120 μF

Voltage = 140 V

to calculate the discharge current

Applying the formula of discharge current

io=vo/R

io= 140/ 1.8x 10³

= 0.078A

to calculate the time

Applying the formula of current

io= vo/R e-t/RC

50= 140/1800e-t/RC

0.649= e-t/RC

-t/RC= ln( 0.649)

t = 0.432x 120x10^-6x 1800

t=93.3 x 10^-3seconds

Does the moon light originate from the moon only

Answers

Answer:

No

Explanation:

Moon has no light of its own. It just shines because its surface reflects light from the sun and that's what we see.

:-)

a trombone can be modeled like an open closed air tube. the trombone plays a fifth harmonic of 159 hz. the speed of sound is 343 m/s. what is the wavelength of this sound

Answers

Answer:

The wavelength is 2.16 m.

Explanation:

Given the speed of the sound = 343 m/s

Trombone generate the frequency = 159 Hz

Now we have to find the wavelength of the sound. Here, we can find the wavelength by dividing the speed of the sound with frequency.

The wavelength of the sound = Speed of sound/frequency

Wavelength of the sound = 343 / 159 = 2.16 m

Suppose you exert a force of 185 N tangential to the outer edge of a 1.73-m radius 76-kg grindstone (which is a solid disk).

Required:
a. What torque is exerted?
b. What is the angular acceleration assuming negligible opposing friction?
c. What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

Answers

Answer:

a. 320.06 Nm b. 2.814 rad/s² c. 2.811 rad/s².

Explanation:

a. The torque exerted τ = Frsinθ where F = tangential force exerted = 185 N, r = radius of grindstone = 1.73 m and θ = 90° since the force is tangential to the grindstone.

τ = Frsinθ

= 185 N × 1.73 m × sin90°

= 320.05 Nm

So, the torque τ = 320.05 Nm

b. Since torque τ = Iα where I = moment of inertia of grindstone = 1/2MR² where M = mass of grindstone = 76 kg and R = radius of grindstone = 1.73 m

α = angular acceleration of grindstone

τ = Iα

α = τ/I = τ/(MR²/2) = 2τ/MR²

substituting the values of the variables, we have

α = 2τ/MR²

= 2 × 320.05 Nm/[76 kg × (1.73 m)²]

= 640.1 Nm/227.4604 kgm²

= 2.814 rad/s²

So, the angular acceleration α = 2.814 rad/s²

c. The opposing frictional force produces a torque τ' = F'r' where F' = frictional force = 20.0 N and r' = distance of frictional force from axis = 1.50 cm = 0.015 m.

So  τ' = F'r' = 20.0 N × 0.015 m = 0.3 Nm

The net torque on the grindstone is thus τ'' = τ - τ' = 320.05 Nm - 0.3 Nm = 319.75 Nm

Since τ'' = Iα

α' = τ''/I where α' = its new angular acceleration

α' = 2τ/MR²

= 2 × 319.75 Nm/[76 kg × (1.73 m)²]

= 639.5 Nm/227.4604 kgm²

= 2.811 rad/s²

So, the angular acceleration α' = 2.811 rad/s²

How much heat is required to convert 5.0 kg of ice from a temperature of - 20 0C to water at a temperature of 205 0F

Answers

Answer:

Explanation:

To convert from °C to °F , the formula is

( F-32 ) / 9 = C / 5

F is reading fahrenheit scale and C is in centigrade scale .

F = 205 , C = ?

(205 - 32) / 9 = C / 5

C = 96°C approx .

Let us calculate the heat required .

Total heat required = heat required to heat up the ice at - 20 °C  to 0°C  + heat required to melt the ice + heat required to heat up the water at  0°C to

96°C.

=  5 x 2.04 x (20-0) +  5 x 336 + 5 x ( 96-0 ) x 4.2  kJ .

= 204 + 1680 + 2016

= 3900 kJ .

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answers

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Which of the following describes wavelength?
A.
the height of a wave
B.
the distance between crests of adjacent waves
C.
the distance a wave travels in a given amount of time
D.
the number of waves that pass a point in a given amount of time

Answers

D. The number of wave that pass a point in a given amount of time

Sonar is used to determine the speed of an object. A 38.0-kHz signal is sent out, and a 40.0-kHz signal is returned. If the speed of sound is 341 m/s, how fast is the object moving?

Answers

Answer:

The velocity is  [tex]v = 8.743 \ m/s[/tex]

Explanation:

From the question we are told that

    The frequency of the signal sent out  is  [tex]f_s = 38.0 \ kHz = 38.0 *10^{3} \ Hz[/tex]

    The frequency of the signal received is  [tex]f_r = 40.0 \ kHz = 40.0 *10^{3} \ Hz[/tex]

     The  speed of sound is  [tex]v_s = 341 \ m/s[/tex]

Generally the frequency of the sound received is  mathematically represented as

         [tex]f_r = f_s [\frac{v_s + v}{v_s - v} ][/tex]

where v is the velocity of the object

       =>      [tex]40 *10^{3} = 38 *10^{3} * [\frac{341 + v}{341 - v} ][/tex]

       =>      [tex]1.05263 = \frac{341+v }{341-v}[/tex]

       =>   [tex]358.94 - 1.05263v = 341 + v[/tex]

      =>    [tex]17.947 = 2.05263 v[/tex]

      =>    [tex]v = 8.743 \ m/s[/tex]

The square armature coil of an alternating current generator has 200 turns and is 20.0 cm on side. When it rotates at 3600 rpm, its peak output voltage is 120 V.
A) What is the frequency of the output voltage?
B) What is the strength of the magnetic field in which the coil is turning?

Answers

Answer:

A) 60 Hz

B) 0.04 T

Explanation:

Given that.

Number of turns, N = 200

Length of the side, l = 20 cm = 0.2 m

Speed if rotation, w = 3600 rpm

Voltage, V = 120 V

First, we try to convert the speed from rpm to rad/s

3600 * (2π/60)

3600 * 0.10473

3600 rpm = 377 rad/s

Now, we use that as our w, speed of rotation

Frequency of output, f =

w/2π

f = 377 / 6.284

f = 59.99 Hz or approximately, 60 Hz.

B

Strength of the magnetic field in which the coil is turning

E• = NABw

Where, A = l² = 0.2² = 0.04, on substituting the values to the equation, we have

120 = 200 * 0.04 * 377 * B

120

Making B subject of formula,

B = 120/ 3016

B = 0.04 T..

The frequency of the output voltage is 60 Hz and the strength of the magnetic field is 0.04 T

How many turns of wire are needed in a circular coil 13 cmcm in diameter to produce an induced emf of 5.6 VV

Answers

Answer:

Number of turns of wire(N) = 3,036 turns (Approx)

Explanation:

Given:

Diameter = 13 Cm

emf = 5.6 v

Note:

The given question is incomplete, unknown information is as follow.

Magnetic field increases = 0.25 T in 1.8 (Second)

Find:

Number of turns of wire(N)

Computation:

radius (r) = 13 / 2 = 6.5 cm = 0.065 m

Area = πr²

Area = (22/7)(0.065)(0.065)

Area = 0.013278 m²

So,

emf = (N)(A)(dB / dt)

5.6 = (N)(0.013278)(0.25 / 1.8)

5.6 = (N)(0.013278)(0.1389)

N = 3,036.35899

Number of turns of wire(N) = 3,036 turns (Approx)

A fish in a flat-sided aquarium sees a can of fish food on the counter. To the fish's eye, the can looks to be 43 cmcm outside the aquarium.

Required:
What is the actual distance between the can and the aquarium?

Answers

Answer:

The actual distance is  [tex]d_a = 0.3233\ m[/tex]

Explanation:

From the question we are told that

     The  distance of the can is d =  43 cm  =  0.43 m

     

Generally the actual distance is mathematically represented as

         [tex]d_a = [\frac{ n_a }{n_w} ]* d[/tex]

Where  [tex]n_a , n_w[/tex] are the refractive index of air and water and their value is  

         [tex]n_a = 1 , \ \ \ n_w = 1.33[/tex]

So

       [tex]d_a = [\frac{ 1 }{1.33} ]* 0.43[/tex]

       [tex]d_a = 0.3233\ m[/tex]

Find the focal length of contact lenses that would allow a nearsighted person with a 130 cmcm far point to focus on the stars at night.

Answers

Answer:

130cm

Explanation:

The lens equation is expressed as;

1/f = 1/u+1/v where;

f is the focal length of the lens

u is the object distance

v is the image distance

Since the near sighted person wants focus the starts at nigt, the stars at night are the images located that infinity. Hence the image distance v = ∞.

The object distance u = 130cm

Substituting the given parameters in the formula to get the focal length f

[tex]\frac{1}{f} = \frac{1}{\infty} + \frac{1}{130} \\\\As \ x \ tends \ to \ \infty, \, \frac{a}{x} \ tends \ to \ 0 \ where\ 'a' \ is \ a\ constant \\\\} \\\\[/tex]

[tex]\frac{1}{f} = 0+ \frac{1}{130}\\\\[/tex]

[tex]\frac{1}{f} =\frac{1}{130}\\cross\ multiply\\\\f = 130*1\\\\f = 130cm[/tex]

Hence the focal length of contact lenses that would allow a nearsighted person with a 130 cm far point to focus on the stars at night is 130cm

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).

Answers

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

[tex]B = \frac{N\mu_o I}{2R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

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