A red card is illuminated by red light. Part A What color will the card appear? What color will the card appear? a. Red b. Black c. White d. Green

Answers

Answer 1
Red light reflects off the card into your eyes and you see the red card as red. The light will just make the card brighter. So A
Answer 2

The color that is reflected when a red card is illuminated by red light is white.

The color an object is perceived to have, depends on the frequency of light it reflects.

If white light incidents on a red filter, red is transmitted while blue and green are absorbed.

Consequently, when a red card is illuminated by red light, the red card will  reflect back almost all the incident light on it, causing it to appear brighter which creates an  illusion of white color to the eyes.

Thus, we can conclude the color that is reflected when a red card is illuminated by red light is white.

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Related Questions

A lamp in a child's Halloween costume flashes based on an RC discharge of a capacitor through its resistance. The effective duration of the flash is 0.220 s, during which it produces an average 0.520 W from an average 3.00 V.
A. How much charge moves through the lamp (C)?
B. Find the capacitance (F).
C. What is the resitance of the lamo?

Answers

Answer:

A. 0.0374C

B. 0.012F

C. 18 ohms

Explanation:

See attached file

Define the following, and give the letter which we will abbreviate them by:
Center of curvature:
Vertex:
Focal Point:
Radius of curvature:
Focal length:

Answers

Answer:

As in explanation.

Explanation:

A) Centre of Curvature: This is defined as the point in the center of the sphere from which the mirror was sliced. It is represented by the letter "C"

B) Vertex: It is defined as the point on the mirror's surface where the principal axis meets the mirror. It is represented by the letter A.

C) Focal Point: This is defined as the Midway point between the vertex and the center of curvature. It is represented by the letter "F"

D) Radius of Curvature: This is defined as the distance from the vertex to the center of curvature. It is represented by the letter "R"

E) Focal Length: This is defined as the distance from the mirror to the focal point. It's represented by the letter "f"

A Galilean telescope adjusted for a relaxed eye is 36.2 cm long. If the objective lens has a focal length of 39.5 cm , what is the magnification

Answers

Answer:

The magnification is  [tex]m = 12[/tex]

Explanation:

From the question  we are told that

   The object distance is [tex]u = 36.2 \ cm[/tex]

     The focal length is  [tex]v = 39.5 \ cm[/tex]

From the lens equation we have that

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

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

substituting values

       [tex]\frac{1}{v} = \frac{1}{39.5} - \frac{1}{36.2}[/tex]

       [tex]\frac{1}{v} = -0.0023[/tex]

=>   [tex]v = \frac{1}{0.0023}[/tex]

=>   [tex]v =-433.3 \ cm[/tex]

The magnification is mathematically represented as

         [tex]m =- \frac{v}{u}[/tex]

substituting values

        [tex]m =- \frac{-433.3}{36.2}[/tex]

         [tex]m = 12[/tex]

         

A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick is then sawed off to a length of 0.76 m, rebalanced at its center, and set into oscillation. With what period does it now oscillate?

Answers

Answer:

The new time period is  [tex]T_2 = 3.8 \ s[/tex]

Explanation:

From the question we are told that

  The period of oscillation is  [tex]T = 5 \ s[/tex]

   The  new  length is  [tex]l_2 = 0.76 \ m[/tex]

Let assume the original length was [tex]l_1 = 1 m[/tex]

Generally the time period is mathematically represented as

         [tex]T = 2 \pi \sqrt{ \frac{ I }{ mgh } }[/tex]

Now  I is the moment of inertia of the stick which is mathematically represented as

           [tex]I = \frac{m * l^2 }{12 }[/tex]

So

        [tex]T = 2 \pi \sqrt{ \frac{ m * l^2 }{12 * mgh } }[/tex]

Looking at the above equation we see that

        [tex]T \ \ \ \alpha \ \ \ l[/tex]

=>    [tex]\frac{ T_2 }{T_1} = \frac{l_2}{l_1}[/tex]

=>    [tex]\frac{ T_2}{5} = \frac{0.76}{1}[/tex]

=>     [tex]T_2 = 3.8 \ s[/tex]

13. A sinusoidal wave of frequency f is traveling along a stretched string. The string is brought to rest, and a second traveling wave of frequency 2f is established on the string. What is the wavelength of the second wave?

Answers

Answer:

It will be half that if the first wave

Explanation:

Because the wave speed remains the same, the result of doubling the frequency is that the wavelength is half as large as it

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answers

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

The magnitude of the magnetic field at the point P will be "9.6 × 10⁻⁴ T".

Magnetic field

The region of the environment close to something like a magnetic entity or a current-carrying body wherein this same magnetic forces caused by the body as well as a current might well be sensed.

According to the question,

Current, I = 2.4 A

Radius, R = 5 cm or,

                = 0.05 m

Constant, μ₀ = 4π × 10⁻⁷ H/m

We know the relation,

The magnetic field, B = [tex]\frac{\mu_0 I}{2 \pi R}[/tex]

By substituting the values in the above relation, we get

                                    = [tex]\frac{4 \pi\times 10^{-7}\times 2.4}{2 \pi\times 0.05}[/tex]

                                    = 9.6 × 10⁻⁴ T

Thus the above answer is appropriate.

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Design a voltage divider to provide the following approximate voltages with respect to ground using a 30 V source: 8.18 V, 14.7 V, and 24.6 V. The current drain on the source must be limited to no more than 1 mA. The number of resistors, their values, and their wattage ratings must be specified. A schematic showing the circuit arrangement and resistor placement must be provided

Answers

Answer:

R₁ = 14.7 10³ Ω , R₂ = 8.18 10³ Ω ,  R₃ = 1.72 10³ Ω ,  R₄ = 5.4 10³ Ω    1/8 W resistor

Explanation:

For this exercise we must use a series circuit since the sum of the voltage on each resin is equal to the source voltage (V = 30 V)

Therefore we build a circuit with 4 resistors in series, in such a way that

   V = i R

let the voltage

1st resistance

         V = i R

         R₁ = V / i

         R₁ = 14.7 / 1 10⁻³

         R₁ = 14.7 10³ Ω

power is

        P = V i

        P = 14.7 1 10⁻³

        P = 14.7 10⁻³ W = 0.0147 W

a resistance of ⅛ W is indicated

2nd resistance

          R₂ = 8.18 / 1 10⁻³

          R₂ = 8.18 10³ Ω

Power

          P = 8.18 1 10⁻³

          P = 0.00818W

a 1/8 W resistor

3rd resistance

this resistance is calculated in such a way that

          V₁ + V₂ + V₃ = 24.6

          V₃ = 24.6 - V₁ -V₂

          V₃ = 24.6 - 14.7 - 8.18

          V₃ = 1.72 V

          R₃ = 1.72 / 1 10⁻³

          R₃ = 1.72 10³ Ω

           

power

          P = Vi

          P = 1.72 10⁻³

          P = 0.00172 W

a resistance of ⅛ W

To obtain the voltage of 24.6 we use this three resistors together

4th resistance

The value of this resistance is calculated so that the sum of all the voltages reaches the source voltage

           30 = V₁ + V₂ + V₃ + V₄

           V₄ = 30 - V₁ -V₂ -V₃

           V₄ = 30 -14.7 - 8.18 - 1.72

           V₄ = 5.4 V

          R₄ = 5.4 / 1 10⁻³

          R₄ = 5.4 10³ Ω

Power

         P = V i

         P = 5.4 10⁻³

         P = 0.0054 W

⅛ W resistance

The values ​​of these resistance are commercially

Let's check the consumption of the circuit

  R_total = R₁ + R₂ + R₃ + R₄

  R_total = (14.7 + 8.18 + 1.72 + 5.4) 10³

   R_total = 30 10³

the current circulating in the circuit is

     i = V / R_total

     i = 30/30 10³

     i = 1 10⁻³ A

therefore it is within the order requirement.

for connections see attached diagram

A particle undergoes damped harmonic motion. The spring constant is 100 N/m, the damping constant is 8.0 x 10-3 kg.m/s, and the mass is 0.050 kg. If the particle starts at its maximum displacement, x = 1.5 m, at time t = 0. What is the amplitude of the motion at t = 5.0 s?

Answers

Answer:

The amplitude [tex]A(5) = 1 \ m[/tex]

Explanation:

From the question we are told that

     The  spring constant is  [tex]k = 100 \ N/m[/tex]

      The  damping constant is  [tex]b = 8.0 *10^{-3} \ kg \cdot m/s[/tex]

       The mass is  [tex]m = 0.050 \ kg[/tex]

       The  maximum displacement is [tex]A_o = 1.5 \ m \ at t = 0[/tex]

       The  time  considered is  t =  5.0 s

Generally the displacement(Amplitude) of damped harmonic motion is mathematically represented as

           [tex]A(t) = A_o * e ^{ - \frac{b * t}{2 * m} }[/tex]

substituting values

         [tex]A(5) = 1.5 * e ^{ - \frac{ 8.0 *10^{-3} * 5}{2 * 0.050} }[/tex]

         [tex]A(5) = 1 \ m[/tex]

       

a radio antenna emits electromagnetic waves at a frequency of 100 mhz and intensity of what is the photon density

Answers

Answer:

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

Explanation:

given data

frequency f = 100 mhz = 100 × [tex]10^{6}[/tex] Hz

we consider here intensity I = 0.2 W/m²

solution

we take here plank constant is h i.e = 6.626 × [tex]10^{-34}[/tex] s

and take energy density is E

so here

E × C = I  

E = [tex]\frac{I}{C}[/tex]   ................1

here C = 3 × [tex]10^{8}[/tex] m/s

so photon density is

photon density = [tex]\frac{I}{C} \times \frac{1}{f \times h}[/tex]     ...............2

photon density = [tex]\frac{0.2}{3 \times 10^8} \times \frac{1}{100 \times 10^6 \times 6.626 \times 10^{-34} }[/tex]

photon density = 1.0 × [tex]10^{16}[/tex] photon/m³

A positive point charge q is placed at the center of an uncharged metal sphere insulated from the ground. The outside of the sphere is then grounded as shown. Then the ground wire is removed. A is the inner surface and B is the outer surface. Which statement is correct

Answers

Explanation:

the missing figure in the Question has been put in the attachment.

Then from the figure we can observe that

the center of the sphere is positive, therefore, negative charge will be  induced at A.

As B is grounded there will not be any charge on B

Hence the answer is A is negative and B is charge less.

A rock weighing 20 N (mass = 2 kg) is swung in a horizontal circle of radius 2 m at a constant speed of 6 m/s. What is the tension in the cord?

Answers

Answer:

The tension in the cord provides by centripetal force

T = Fc

= mv^2/r

= 2kg ( 6)^2/2

=36 N

please i dont understand

Answers

Give me points I will help you

Are Quantum Physics, Quantum mechanics,Quantum Engagement same?
or, Do they branch of each others ​

Answers

Answer:

The topic of quantum entanglement is at the heart of the disparity between classical and quantum physics: entanglement is a primary feature of quantum mechanics lacking in classical mechanics. ... In the case of entangled particles, such a measurement will affect the entangled system as a whole

Explanation:

Answer:

quantum entanglement is thought to be one of the trickiest concepts in science, but the core issues are simple. And once understood, entanglement opens up a richer understanding of concepts such as the “many worlds” of quantum theory.

Explanation:

g Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?

Answers

Answer:

constructive interferencia  0, 1 , 2 m

destructive inteferencia   1/4, 3/4. 5/4, 7/4 m

Explanation:

This exercise is equivalent to the double slit experiment, the two sources are in phase and separated by a distance, therefore the waves observed in the line between them have an optical path difference and a phase difference, given by the expression

            Δr / λ = Φ / 2π

            Δr = Φ/2π   λ

let's apply this expression to our case

λ = 1 m

            Δr = Φ 1 / 2π

We have constructive interference for angle of  Φ = 0, 2π, ...

let's find the values ​​where they occur

  Φ         Δr

   0          0

  2π         1

  4π        2

Destructive interference occurs by    Φ = π /2, 3π / 2, ...

 Φ          Δr

 π/2       ¼ m

 3π /2    ¾ m

5π /2     5/4 m

7π /2      7/4 m

1 of 3 : please help got an extra day for a test and i don’t get this (must show work) points and brainliest!

Answers

Answer:

y = 1/2at^2

we could also write it as-

y = (at^2)/2

2y = at^2

2y/a = t^2

√2y/a = t

hope it helps

Calculate the answers to the appropriate number of significant
12.21 x 9.19 =

Answers

the answer for 12.22 times 9.29 is 112.2099

Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is 4.30×10^−5 N/m, and the wires repel each other. The current in one wire is 0.520 A.Required:a. What is the current in the second wire? b. Are the two currents in the same direction or in opposite directions?

Answers

Answer:

10.75 A

The current is in opposite direction since it causes a repulsion force between the wires

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

[tex]F/l[/tex] = [tex]\frac{u_{0}I_{1} I_{2} }{2\pi r }[/tex]

where [tex]F/l[/tex] is the force per unit length on the wires

[tex]u_{0}[/tex] = permeability of vacuum = 4π × 10^−7 T-m/A

[tex]I_{1}[/tex] = current on the first wire = 0.520 A

[tex]I_{2}[/tex] = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = [tex]\frac{4\pi *10^{-7}*0.520*I_{2} }{2\pi *0.026}[/tex] =  [tex]\frac{ 2*10^{-7}*0.520*I_{2} }{0.026}[/tex]

4.30×10^−5 = 4 x 10^-6 [tex]I_{2}[/tex]

[tex]I_{2}[/tex] = (4.30×10^-5)/(4 x 10^-6) = 10.75 A

The current is in opposite direction since it causes a repulsion force between the wires.

A force of 16.88 N is applied tangentially to a wheel of radius 0.340 m and gives rise to an angular acceleration of 1.20rad / (s ^ 2) . Calculate the rotational inertia of the wheel. A. 2.77 kg - m ^ 2 B. 0.73 kg - m ^ 2 C. 4.41 kg - m ^ 2 O. 4.78 kg - m ^ 2

Answers

Given.

force = 16.88 N is

radius = 0.340m

an angular acceleration = 1.20rad/s^2

the formula for torque is

F*r = I*a

where I is moment of inertia

16.88*.34 = I*1.2

I = 4.78Kg-m^2

so rotational inertia I = 4.78Kg-m^2

Light with an intensity of 1 kW/m2 falls normally on a surface and is completely absorbed. The radiation pressure is

Answers

Answer:

The radiation pressure of the light is 3.33 x 10⁻ Pa.

Explanation:

Given;

intensity of light, I = 1 kW/m²

The radiation pressure of light is given as;

[tex]Radiation \ Pressure = \frac{Flux \ density}{Speed \ of \ light}[/tex]

I kW = 1000 J/s

The energy flux density = 1000 J/m².s

The speed of light = 3 x 10⁸ m/s

Thus, the radiation pressure of the light is calculated as;

[tex]Radiation \ pressure = \frac{1000}{3*10^{8}} \\\\Radiation \ pressure =3.33*10^{-6} \ Pa[/tex]

Therefore, the radiation pressure of the light is 3.33 x 10⁻ Pa.

6. You push an object, initially at rest, across a frictionless floor with a constant force for a time interval t, resulting in a final speed of v for the object. You then repeat the experiment, but with a force that is twice as large. What time interval is now required to reach the same final speed v?

Answers

Answer:

   t = t₀ / 2

Explanation:

In this exercise we must use Newton's second law

          F = m a

          a = F / m

now we can use kinematics

  as in object part of rest (v₀ = 0)

        v =a t₀

        t₀ = v / a

these results are with the first experiment

now repeat the experiment, but F = 2F₀

           a = 2F₀ / m = 2 a₀

          v = 2 a₀ t

          t = v / 2a₀

          t = t₀ / 2

The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].

Given the following data:

Initial speed = 0 m/s (since the object is at rest)Final speed = VTime = [tex]\Delta t[/tex]Speed = V

To find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

[tex]F = \frac{M(V-U)}{t}[/tex]

Where:

F is the force.V is the final velocity.U is the initial velocity.t is the time.

Substituting the given parameters into the formula, we have;

[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]

When the experiment is repeated, the magnitude of the force is doubled:

[tex]F = 2F[/tex]

Now, we can find the time interval that is required to reach the same final speed (V):

[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]

Substituting the value of F, we have:

[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]

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Explain why water, with its high specific heat capacity, is utilized for heating systems such as hot-water radiators.

Answers

Answer:

Answer in explanation

Explanation:

Water is mainly used as coolant in heating systems like hot-water radiators. The main function of water in such systems, is to absorb as much heat as possible, in order to decrease the temperature of the system and as a result cool it.

The specific heat capacity is the measure of heat energy that is required to raise the temperature of unit mass of a substance through 1 °C. In other words, specific heat capacity quantifies the amount of heat that can be stored by a unit mass of a substance having a degree rise in temperature.

Thus, the more specific heat a substance has, the more heat it can absorb from the hot system. Hence, the specific heat capacity of a coolant must be high.

This is the reason why water, with its high specific heat capacity, is utilized for heating systems, such as radiators.

how many stars are in our solar system?

Answers

Answer:

there are over 100 billion stars in our galaxy.

Light of wavelength 519 nm passes through two slits. In the interference pattern on a screen 4.6 m away, adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern. What is the separation of the two slits?

Answers

Answer:

The separation of the two slits is 0.456 mm.

Explanation:

Given the wavelength of light = 519 nm

The indifference pattern = 4.6 m

Adjacent bright fringes = 5.2 mm

In the interference, the equation required is Y = mLR/d

Here, d sin theta = mL

L = wavelgnth

For bright bands, m is the  order = 1,2,3,4  

For dark bands,  m = 1.5, 2.5, 3.5, 4.5

R = Distance from slit to screen (The indifference pattern)

Y = Distance from central spot to the nth  order fringe or fringe width

Thus,  here d = mLR/Y

d = 1× 519nm × 4.6 / 5.2mm

d = 0.459 mm

What will be the nature of the image formed from both a convex lens and a concave
lens of 20 centimeter focus distance, when the object is placed at a distance of
10 centimeters?​

Answers

Answer:

Explanation:

Using the lens formula

1//f = 1/u+1/v

f is the focal length of the lens

u is the object distance

v is the image distance

For convex lens

The focal length of a convex lens is positive and the image distance can either be negative or positive.

Given f = 20cm and u = 10cm

1/v = 1/f - 1/u

1/v = 1/20-1/10

1/v = (1-2)/20

1/V = -1/20

v = -20/1

v = -20 cm

Since the image distance is negative, this shows that the nature of the image formed by the convex lens is a virtual image

For concave lens

The focal length of a concave lens is negative and the image distance is negative.

Given f = -20cm and u = 10cm

1/v = 1/f - 1/u

1/v = -1/20-1/10

1/v = (-1-2)/20

1/V = -3/20

v = -20/3

v = -6.67 cm

Since the image distance is negative, this shows that the nature of the image formed by the concave lens is a virtual image

One day, after pulling down your window shade, you notice that sunlight is passing through a pinhole in the shade and making a small patch of light on the far wall. Having recently studied optics in your physics class, you're not too surprised to see that the patch of light seems to be a circular diffraction pattern. It appears that the central maximum is about 1 cm across, and you estimate that the distance from the window shade to the wall is about 4 m.

Estimate:
a. The average wavelength of the sunlight (in nm)
b. The diameter of the pinhole (in mm).

Answers

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.

a. The molecules of a magnet are independent...​

Answers

Answer:

variable

Explanation:

As a skydiver falls, his potential energy ___ and his kinetic energy __​
increases,increases
increases,decreases
decreases,increases
decreases, decreases

Answers

Answer:

Hey there!

PE=mgh, so as height decreases, so does the potential energy.

KE=mv^2, so as velocity increases, kinetic energy increases.

Thus, the correct answer would be Decreases, Increases.

Let me know if this helps :)

Without actually calculating any logarithms, determine which of the following intervals the sound intensity level of a sound with intensity 3.66×10^−4W/m^2 falls within?

a. 30 and 40
b. 40 and 50
c. 50 and 60
d. 60 and 70
e. 70 and 80
f. 80 and 90
g. 90 and 100

Answers

Answer:

f. 80 and 90

Explanation:

1 x 10⁻¹² W/m² sound intensity falls within 0 sound level

1 x 10⁻¹¹ W/m² sound intensity falls within 10 sound level

1 x 10⁻¹⁰ W/m² sound intensity falls within 20 sound level

1 x 10⁻⁹ W/m² sound intensity falls within 30 sound level

1 x 10⁻⁸ W/m² sound intensity falls within 40 sound level

1 x 10⁻⁷ W/m² sound intensity falls within 50 sound level

1 x 10⁻⁶ W/m² sound intensity falls within 60 sound level

1 x 10⁻⁵ W/m² sound intensity falls within 70 sound level

1 x 10⁻⁴ W/m² sound intensity falls within 80 sound level

1 x 10⁻³ W/m² sound intensity falls within 90 sound level

Given sound intensity (3.66 x 10⁻⁴ W/m²) falls with 1 x 10⁻⁴ W/m² of intensity which is within 80 and 90 sound level.

f. 80 and 90

Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.

Answers

Answer:

137.69°

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Phase angle:

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

Learn more about RLC circuit:

https://brainly.com/question/372577?referrer=searchResults

An electric device delivers a current of 5.0 A to a circuit. How many electrons flow through this circuit in 5 s?

Answers

Answer:

1.6×10²⁰

Explanation:

An ampere is a Coulomb per second.

1 A = 1 C / s

The amount of charge after 5 seconds is:

5.0 A × 5 s = 25 C

The number of electrons is:

25 C × (1 electron / 1.6×10⁻¹⁹ C) = 1.6×10²⁰ electrons

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